QUANTITATIVE SECTION : What’s Covered • 37 Questions in 2 format o Problem Solving : 19 Questions o Data Sufficiency : 18 Questions • The Quantitative section of the GMAT Test measures basic mathematical skills. It tests a candidate's understanding of elementary concepts in mathematics learnt at school and the ability to reason quantitatively. The section involves solving quantitative problems, reasoning using quantitative techniques and interpreting graphic data. • Preparing for the GMAT Quant section would involve learning and mastering concepts in Arithmetic Elementary Algebra and Basic Geometry. General Advice… •
Always reread the last line of the problem to make sure you’ve actually answered the question. •
Use process of elimination To eliminate obviously wrong answers quickly To avoid overworking the problem To maximize opportunity to “guess” correctly. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 1
CONTENT Arithmetic: •
•
•
•
•
•
•
•
•
•
Number Theory & Fractions..…………………………………………………………………..5 Ratio, Proportion and Variation……………………………………………………………..15 Percentages……………….………………………………………………………………………….20 Profit and Loss……………………………………………………………………………………….23 Mean, Median, Mode, Standard Deviation…………………………………………….25 Mixtures and Alligations………………………………………………………………………..28 Speed, Time and Distance………………………………………………………………………31 Circular and Straight Races…………………………………………………………………….33 Pipes, Cisterns and Work, Time………………………………………………………………36 Simple and Compound Interest……………………………………………………………..38 Algebra and Geometry: •
•
•
•
•
•
•
•
•
•
•
Linear Equations……………………………………………………………………………………40 Quadratic Equations………………………………………………………………………………43 Sequence & Series…………………………………………………………………………………46 Probability…………………………………………………………………………………………….52 Permutation and Combination………………………………………………………………57 Set Theory……………………………………………………………………………………………..61 Exponential Equations…………………………………………………………………………..65 Inequalities……………………………………………………………………………………………69 Geometry………………………………………………………………………………………………73 Co‐ordinate Geometry…………………………………………………………………………..95 Mensuration………………………………………………………………………………………..102 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 2
Algebra •
o
o
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o
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Cosmic Problems: asked to write an equation that will answer a question regardless of what a “certain number x” is PLUG IN: pick numbers for the letters in the problem Plug in from the outside in (A,E,B,D,C) Avoid using 0,1 and numbers in the problem Why not solve algebraically? Makes difficult problems easy Test designed with many ways for you to screw up using algebra Solving Non‐Cosmic Problems 
o
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Non‐cosmic problems have specific answers Working BACKWARDS Start with choice C, plug number into problem to see if it works If choice C is too small, choose the next larger number If choice C is too big, choose the next smaller number Geometry  YOU MUST MEMORIZE ALL FORMULAS  Study problems that are drawn to scale to eliminate crazy answers o Not all diagrams are drawn to scale Geometry problems always involve more than 1 step  Topics Tested o Degree and angles, triangles, circles, four‐sided objects, solids and volume Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 3
Data Sufficiency  Consists of a question followed by 2 statements  You don’t decide what the answer is  You decide WHETHER the question can be answered based on the information in the 2 statements Data Sufficiency Answers 
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
The first statement ALONE answers the question The second statement ALONE answers the question You need both statements TOGETHER to answer the question Both statements SEPERATELY answer the question NEITHER statement together or separately answers the question How to Crack Data Sufficiency  Look at one statement at a time  Memorize AD or BCE o If statement (1) answers the question, write down AD  If statement (2) also answers the question, the final answer is D o If statement (1) DOES NOT answers the question, write down BCE  Evaluate statement (2) to narrow down the remaining choices A Hidden Trick  If a data sufficiency problem asks a yes/no question o A statement is sufficient if it ALWAYS gives the SAME answer, yes or no o If the answer is SOMETIMES yes and sometimes no, the statement is insufficient. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 4
Number Theory • The set of integers, I, is composed of all the counting numbers (i.e., 1, 2, 3, . . .), zero, and the negative of each counting number; that is, I = {…,‐3,‐2,‐1, 0, 1, 2, 3,…} • The remainder is r when p is divided by q” means p=qz+r; the integer z is called the quotient. Ex: The remainder is 1 when 7 is divided by 3 means 7=3 (2) +1 • The division of one integer by another yields either a zero remainder, sometimes called “dividing evenly,” or a positive‐integer remainder. • When we say that an integer N is divisible by an integer x, we mean that N divided by x yields a zero remainder. • The multiplication of two integers yields a third integer. The first two integers are called factors, and the third integer is called the product. The product is said to be a multiple of both factors, and it is also divisible by both factors (providing the factors are nonzero). Therefore, since (2) (7) = 14, we can say that 2 and 7 are factors and 14 is the product, 14 is a multiple of both 2 and 7, and 14 is divisible by both 2 and 7. • A number is even if the remainder is zero when n is divided by 2: n= 2z+0, or n=2z. • A number is odd if the remainder is one when n is divided by 2: n= 2z+1. • The following properties for odd and even numbers are very useful. • even x even= even • odd x odd = odd • even x odd = even • even + even = even • odd + odd = even • even + odd = odd • Some rules that are occasionally forgotten include: (i) Multiplication by 0 always results in 0; e.g., (0)(15) = 0. (ii) Division by 0 is not defined; e.g., 5 ÷ 0 has no meaning. • Consecutive integers are written as x, x+1, x+2,… Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 5
• Consecutive even or odd integers are written as x, x+2, x+4,… • The integer zero is neither positive nor negative, but it is even: 0=2(0). • A prime number is an integer that is divisible only by itself and 1. The prime numbers are 2,3,5,7,11,13,17,19,23,29,31,37,41. Prime Number has exactly two distinct positive divisors, 1 and N. • The integer 1 is not a prime number because it has only one positive divisor. • A number is divisible by 3 if the sum of its digits is divisible by 3. • The absolute value of a number N, denoted by |N| , is defined to be N if N is positive or zero and –N if N is negative. For example, |1/2|=1/2; |0|=0 and |‐2.6|= ‐(‐2.6)= 2.6. Note that the absolute value of a number cannot be negative. • Basic rules about divisors: 2 will be a divisor (or factor) of x if x is even 3 will be a divisor (or factor) of x if the sum of x's digits is divisible by 3 4 will be a divisor (or factor) of x if last 2 digits of x is divisible by 4. 5 will be a divisor (or factor) of x if x's last digit is 0 or 5 6 will be a divisor (or factor) of x if x is divisible by both 2 and 3 8 will be a divisor (or factor) of x if last 3 digits of x is divisible by 8. 9 will be a divisor (or factor) of x if the sum of x's digits is divisible by 9 10 will be a divisor(or factor) of x if x's last digit is 0 Dividing by 7 (2 Tests) Take the last digit in a number. Double and subtract the last digit in your number from the rest of the digits. Repeat the process for larger numbers. Example: 357 (Double the 7 to get 14. Subtract 14 from 35 to get 21 which is divisible by 7 and we can now say that 357 is divisible by 7. NEXT TEST Take the number and multiply each digit beginning on the right hand side (ones) by 1, 3, 2, 6, 4, 5. Repeat this sequence as necessary Add the products. If the sum is divisible by 7 ‐ so is your number. Example: Is 2016 divisible by 7? 6(1) + 1(3) + 0(2) + 2(6) = 21 21 is divisible by 7 and we can now say that 2016 is also divisible by 7. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 6
• All prime numbers except 2 and 5 end in 1, 3, 7 or 9, since numbers ending in 0, 2, 4, 6 or 8 are multiples of 2 and numbers ending in 0 or 5 are multiples of 5. Similarly, all prime numbers above 3 are of the form (6k+1) or (6k‐1) , because all other numbers are divisible by 2 or 3. • If p is a prime divisor, then the remainder obtained when is divided by p is 0. Number Theory‐ Questions Q1. If n is a positive integer, then the remainder when (n^7‐n) is divided by 7 is A. 3 B. 0 C. 5 D. 4 E. None Q2. If both 112 and 33 are factors of the number a * 43 * 62 * 1311, then what is the smallest possible value of a? A. 121 B. 3267 C. 363 D. 33 E. None of the above Q3. How many different positive integers exist between 106 and 107, the sum of whose digits is equal to 2? A. 6 B. 7 C. 5 D. 8 E. 18 Q4. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the same divisor, the remainder is 11. What is the value of the divisor? A. 13 B. 59 C. 35 D. 37 E. 12 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 7
Q5. How many keystrokes are needed to type numbers from 1 to 1000? A. 3001 B. 2893 C. 2704 D. 2890 E. None of these Q6. When 242 is divided by a certain divisor the remainder obtained is 8. When 698 is divided by the same divisor the remainder obtained is 9. However, when the sum of the two numbers 242 and 698 is divided by the divisor, the remainder obtained is 4. What is the value of the divisor? A. 11 B. 17 C. 13 D. 23 E. None of these Q7. How many integral divisors does the number 120 have? A. 14 B. 16 C. 12 D. 20 E. None of these Q8. How many trailing zeros will be there after the rightmost non‐zero digit in the value of 25!? A. 25 B. 8 C. 6 D. 5 E. 2 Q9. What is the remainder when 1044 * 1047 * 1050 * 1053 is divided by 33? A. 3 B. 27 C. 30 D. 21 E. 18 Q10. When the integer n is divided by 2, the quotient is u and the remainder is 1. When the integer n is divided by 5, the quotient is v and the remainder is 3. Which one of the following Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 8
must be true? A. 2u + 5v = 4 B. 2u ‐ 5v = 2 C. 4u + 5v = 2 D. 4u ‐ 5v = 2 E 3u ‐ 5v = 2c Fractions • To understand fractions you should be familiar with the terms part and whole. The whole is everything and the part is just a piece of it. Fractions are used to figure out just how much of the whole the part represents. Part = Fraction x Whole. Example: A whole pizza is ordered for a family of 4 and cut into 8 slices. If everyone agrees to eat the same amount, how much of the pizza will each person eat? Part = Fraction x Whole Part = 1/4 x 8 or Part = 2 slices • A fraction is a number of the form a/b, where a and b are integers and b≠0. The a is called the numerator of the fraction, and b is called the denominator. • If the numerator and denominator of the fraction a/b are both multiplied by the same integer, the resulting fraction will be equivalent to a/b. • To add two fractions with the same denominator, simply add the numerators and keep the denominator the same. • If the denominators are not the same, apply the technique mentioned above to make them the same before doing the addition. The same method applies for subtraction. • To multiply two fractions, multiply the two numerators and multiply the two denominators (the denominators need not be the same). • To divide one fraction by another, first invert the fraction you are dividing by, and then proceed as in multiplication. • Proper Fraction ‐ A fraction whose numerator is less than its denominator. • Improper Fraction ‐ A fraction whose numerator is greater than or equal to its denominator. • Mixed Fraction (aka Mixed Number) ‐ The combination of a whole number and a proper fraction. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 9
Ex: the t fraction
n 13/8 is a proper p
fraction since it combines a whole number (i.ee., 1) and aa 3
proper ffraction (i.e., /8). • Equivvalent Fracction ‐ A fraaction that has the saame numerrical value ((i.e., decimal value) ass another fraction (although thee numerato
or and deno
ominator off the equivaalent fractio
ons may nott be the eexact same number). Ex: thee following ffractions arre all equivalent fractions since their values are equal to 0.5: 1/2 == 2
/4 = 3/6 = 4/8 = 5/10 The Ideaa of a Fraction In orderr to concep
ptualize the concept off a fraction
n, it can be helpful to see the same fraction
n 3
represen
nted in diffferent wayss. The following show
ws some of the ways the t fraction
n /8 can bee conceptualized. ntation Graphicaal Represen
One wayy to undersstand a fracction is as aa part of a whole. Forr example, tthe fraction
n 3/8 can bee thought of as repre
esenting 3 p
pieces of pizzza from an
n 8 piece pizzza. Numberr Line Repre
esentation Since a ffraction is aa number, itt occupies aa place on tthe numberr line. Mostt fractions aare not integerss and as a re
esult, most fractions occupy a plaace on the n
number linee between integers. Decimall Representtation A fraction can be
e representted as a numerical value by dividing th
he numerator by thee 3
nator. For e
example, the fraction /8 represen
nts the num
merical valuee 0.375 sincce 3 divided
d denomin
by 8 is 0
0.375 Percent Representaation on can be represented
d as a perceent by divid
ding the numerator byy the denom
minator and
d A fractio
3
1
then mu
ultiplying byy 100. For eexample, /4 = 75% and
d /2 = 50%
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0
Comparing Fraction
ns ns, there aree two options: In order to compare the valuee of fraction
d 1: Convertt to Decimaal Method
Each fraaction can be conveerted into a decimal through dividing th
he numerator by thee denomin
nator. Once
e in decimall format, the value of eeach fractio
on can be co
ompared. 5
7
Which frraction is laarger: /7 or /9? 5
/7 = .714 and 7/9 = .777 equivalent o
of 7/9 is largger than thaat of 5/7, 7/9 > 5/7. Since the decimal e
d 2: Cross M
Multiply Method
Anotherr means to
o compare fractions is to crosss multiply such thatt there is a common
n denomin
nator. With
h a commo
on denomin
nator betw
ween the tw
wo fractions (i.e., with
h the samee number in the denominator)), the num
merators can be comp
pared and the fractio
on with thee numerator iis the largesst fraction.
largest n
Which frraction is laarger: 5/7 or 7/9? Adding &
& Subtractiing Fraction
ns Fractions can be ad
dded or sub
btracted. Th
here are fivee steps to th
his process:: • Cheeck to enssure the fractions haave a com
mmon deno
ominator. Adding A
fracctions with
h diffferent deno
ominators p
produces th
he wrong an
nswer. • Find a commoon denominator (if neecessary). The T smallest common
n denominaator can bee und through
h the least ccommon multiple. fou
• Alter each fraction so thaat each fracction sharees a commoon denominnator (i.e., eensure eachh he same number in thee denominaator). fracction has th
• Addd the numeerators whiile leaving tthe denominator unchhanged. Whhen you addd fractionss, thee denominaator does no
ot change. • Sim
mplify the ad
dded fractio
on. The above five step
ps are best u
understood
d through an example. Common Mistake: Adding Fraactions With Differentt Denominaators Apphelp GMAT coach
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In order to add two
o fractions, they must sshare a com
mmon deno
ominator. e Denomin
nators Common Mistake: Adding The
a
fracctions, do not add th
he denominators. Insstead, add the numerator whilee When adding keeping the commo
on denomin
nator the saame value.
he Denominator Common Mistake: Splitting Th
or of a fraction will reesult in a wrongg answerr. Splittingg the denominato
However, you can ssplit the num
merator. Fractio
ons‐ Quesstions Q1. A.
B. 14 C. 13 D. 12 E.
Q2. A. 13/15 B. 12/13 C. 11/12 D. 17/18 E. 10/13 Q3. Which of the frractions bellow is the laargest? A. 55/100 Apphelp GMAT coach
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2
B.
C.
D.
E.
2/150 8/20 12/25 5/10 Q4. For w
which of th
he followingg values of ’’n’ is A. 1 B. 2 C. 3 D. 4 E. 5 not an intteger? Q5. Whaat is the deccimal equivvalent of A. ‐0
0.333... B. 0 C. 0.166... D. 0.333... E. 0.5 Q6. Which of the fo
ollowing is cclosest to A. 1/25 B. 1/11 C. 1/10 D. 1/6 E. 1/5 ? ?
? Q7. A.
B.
C.
D.
E.
Q8. 3/8 ½
½ ¾
¾ 7/8 2 Apphelp GMAT coach
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3
A.
B.
C.
D.
E.
1/25 2/15 1/6 1/5 25/6 Q9. ½ A. ½
B. 2/3 C. 5/6 D.
E.
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Ratios • Wheen you discuuss the ratio between different things you aare relating the quantitty of one of those things t
to the otheer. Ratios are usually represented ass fractionss or with
h colons. Example: What is th
he ratio of b
books to m
magazines if there are 4
4 times as m
many bookss as there are magazines? As a ffraction: U
Using a colo
n: 4 : 1 or 4x : 1
1x o is 4 to 1, so for everry 4 books there is 1 m
magazine. IIf there aree 100 bookss, (or 25 x 4
4 The ratio
books), tthen there are 25 maggazines, (25
5 x 1 magazzines). Therre will alwaays be 4 tim
mes as manyy books ass there are magazines.. • The ratio of thee number 99 to the num
mber 21 can be expressed in sevveral ways; Ex: 9 to 21
1; 9:21; 9/2
21 Since a rratio is in faact an implieed division,, it can be reduced to llowest term
ms. Thereforre, the ratio
o above cou
uld also be w
written: 3 tto 7; 3:7; 3//7 • Wheen ratios aree used in coomparing unnits of measure, the units should be the sam
me. Inverse
e Ratio It is ofteen desirable to compare the nu
umbers of a a ratio in the t inverse order. To do this, wee simply in
nterchange
e the numerator and tthe de‐ nom
minator. Thus, the inveerse of 15:2
20 is 20:15
5. When th
he terms off a ratio are interchangged, the INV
VERSE RATIO
O results. Proporrtion • Closeely allied w
with the stu
udy of ratio is the subjject of prop
portion. When two raatios are set equal to
o each othe
er, a proporrtion is form
med. The p
proportion m
may be wriitten in threee different ways as in the following examples: The last two forms are the mo
ost common. All thesee forms are read, "15 is to 20 as 3
3 is to 4." In
n ords, 15 has the same ratio to 20 as 3 has to
o 4. other wo
• The nnumerator of the first ratio and thhe denominnator of thee second arre called thee extremes. • The second andd third term
ms (the insside terms) are called the MEAN
NS. The means are thee Apphelp GMAT coach
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denomin
nator of the
e first ratio and the numerator of f the second
d • In anny proportioon, the prodduct of the means equuals the prooduct of thee extremes
• The m
mean propo
ortional bettween two quantities is the square root of ttheir product. This rulee is stated
d algebraically as follow
ws: • The ffour selecteed numberss are in proportion by INVERSION
N in the form
m Variatiion • Whenn two quanttities are intterdependeent, changees in the value of one m
may have a predictablee effect on the value
e of the otther. Variattion is the name giveen to the study of thee effects of on which o
occur frequently in thee changess among rellated quanttities. The tthree typess of variatio
study off scientific p
phenomenaa are DIRECTT, INVERSE,, and JOINTT. • Dire
ect Variattion • If a quanntity can be expressed in terms off a second qquantity muultiplied byy a constantt, itt is said to VARY DIREC
CTLY as thee second qu
uantity. Forr example iff x and y are variabless and k is a co
onstant, x vaaries directly as y, if x = ky. Thus, as y increases x increaases, and ass n x caused
d by any ch
hange in yy. y decreasess, x decreases. There is a direcct effect on
H
However, it i
is usually w
written in the form x = ky. • If one quuantity varies directly as a seconnd quantity,, the ratio oof the first quantity too he second q
quantity is aa constant. Thus, whatever the value of x, w
where it is d
divided by yy, th
th
he result will w always be the sam
me value, k. A quantitty that varies directly as another quantity is aalso said to be DIRECTLLY PROPORTTIONAL to tthe second quantity. In
n x = ky, thee oefficient o
of x is 1. co
• The relatioonship x = kky can be w
written in prroportion fo
orm as O Or Apphelp GMAT coach
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• The constant, k, is the CONSTANT OF PROPORTIONALITY. • Inverse Variation • A quantity VARIES INVERSELY as another quantity if the product of the two quantities is a constant. For example, if x and y are variables and k is a constant, the fact that x varies inversely as y is expressed by x=k Or x= k/y • If values are substituted for x and y, we see that as one increases, the other must decrease, and vice versa. Otherwise, their product will not equal the same constant each time. • If a quantity varies inversely as a second quantity, it is INVERSELY PROPORTIONAL to the second quantity. In xy = k, the coefficient of k is 1. The equality xy = k can be written in the form or • Joint Variation • A quantity VARIES JOINTLY as two or more quantities, if it equals a constant times their product. For example, ff x, y, and z are variables and k is a constant, x varies jointly as y and z, if x = kyz. Note that this is similar to direct variation, except that there are two variable factors and the constant with which to contend in the one number; whereas in direct variation, we had only one variable and the constant. The equality, x = kyz, is equivalent to • If a quantity varies jointly as two or more other quantities, the ratio of the first quantity to the product of the other quantities is a constant. Ratio, Proportion and Variation‐ Questions Q1. Three friends Alice, Bond and Charlie divide $1105 amongst them in such a way that if $10, $20 and $15 are removed from the sums that Alice, Bond and Charlie received respectively, then the share of the sums that they got will be in the ratio of 11 : 18 : 24. How much did Charlie receive? A. $495 B. $510 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 17
C. $480 D. $375 E. $360 Q2. Mary and Mike enter into a partnership by investing $700 and $300 respectively. At the end of one year, they divided their profits such that a third of the profit is divided equally for the efforts they have put into the business and the remaining amount of profit is divided in the ratio of the investments they made in the business. If Mary received $800 more than Mike did, what was the profit made by their business in that year? A. $2000 B. $6000 C. $4000 D. $1333 E. $3000 Q3. A, B and C, each of them working alone can complete a job in 6, 8 and 12 days respectively. If all three of them work together to complete a job and earn $ 2340, what ill be C's share of the earnings? A. $1100 B. $520 C. $1080 D. $1170 E. $630 Q4. In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it? A. 1 : 2 B. 2 : 1 C. 1 : 3 D. 3 : 1 E. 2 : 3 Q5. A class of 42 students has 14 girls. The ratio of boys to girls is A. 3 : 1 B. 2 : 1 C. 1 : 2 D. 3 : 2 E. 1 : 3 Q6. In a partnership, two men invest $2000 and $3000. If the net profit of $13500 at the end of the year is divided in accordance with the amount each partner invested, then the man who invested more gets A. $2700 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 18
B.
C.
D.
E.
$4500 $5400 $8100 $9000 Q7. 24‐carat gold is pure gold. 18‐carat gold is 3/4 gold and 20‐carat gold is 5/6 gold. The ratio of pure gold in 20‐carat gold to pure gold in 18‐carat gold is A. 9 : 10 B. 8 : 5 C. 10 : 9 D. 5 : 8 E. 5 : 4 Q8. If there are 16 ounces in 1 pound, the ratio of 6 ounces to 3¾ pounds is A. 10 : 1 B. 1 : 10 C. 5 : 8 D. 8 : 5 E. 2 : 1 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 19
Percentage • The term percent means per hundred or divided by one hundred. Therefore, 43% = 43/100 = .43 300% = 300/100 = 3 • To find out what 30% of 350 is, you multiply 350 by either 0.30 or 30/100, 30% of 350 = (350) (0.30) = 105 or 30% of 350 = (350) 30/100 = 105 • To find out what percent of 80 is 5, you set up the following equation and solve for x: 5/80 = x/100 x = 6.25. So 5 is 6.25% of 80. The number 80 is called the base of the percent. Another way to view this problem is to simply divide 5 by the base, 80, and then multiply the result by 100 to get the percent. • If a quantity increases from 600 to 750, then the percent increase is found by dividing the amount of increase, 150, by the base, 600, which is the first (or the smaller) of the two given numbers, and then multiplying by 100: (150/600) x 100% = 25% • If a quantity decreases from 500 to 400, then the percent decrease is found by dividing the amount of decrease, 100, by the base, 500, which is the first (or the larger) of the two given numbers, and then multiplying by 100: (100/500) x 100% = 20% • Other ways to state these two results are “750 is 25 percent greater than 600” and “400 is 20 percent less than 500”. • In general, for any positive numbers x and y, where x < y, y is ((y‐x)/x) (100) percent greater than x. x is ((y‐x)/y) (100) percent greater than y. Note that in each of these statements, the base of the percent is in the denominator. Percentage‐ Questions Q1. If the price of gasoline increases by 25% and Ron intends to spend only 15% more on gasoline, by what % should he reduce the quantity of petrol that he buys? A. 10% B. 12.5% Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 20
C. 8% D. 12% E. 6.66% Q2. Robin earns 30% more than Erica. Charles earns 60% more than Erica. How much % is the wages earned by Charles more than that earned by Robin? A. 23% B. 18.75% C. 30% D. 50% E. 100% Q3. In an election contested by two parties, Party D secured 12% of the total votes more than Party R. If party R got 132,000 votes, by how many votes did it lose the election? A. 240,000 B. 300,000 C. 168,000 D. 36,000 E. 24,000 Q4. The difference between the value of a number increased by 12.5% and the value of the original number decreased by 25% is 30. What is the original number? A. 60 B. 80 C. 40 D. 120 E. 160 Q5. What is the % change in the area of a rectangle when its length increases by 10% and its width decreases by 10%? A. 0% B. 20% increase C. 20% decrease D. 1% decrease E. Insufficient data Q6. What percent of 60 is 25? A. 60% B. 50% C. 42% D. 25% E. 10% Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 21
Q7. A house sold for $500,000 in 1990 and sold ten years later for $400,000. By what percent did the value of the house change? A. 20% B. 0% C. ‐10% D. ‐20% E. ‐25% Q8. In a large forest, 300 deer were caught, tagged, and returned during 2001. During 2002, 500 deer were caught at random, of which only 20 had tags from the previous year. If the percent of deer in the forest that had tags during the second year and were caught in the 500 deer sample is representative of the percent of the total deer population in the forest with tags, what is the total deer population in the forest (assuming no change in population between 2001 and 2002)? A. 300 B. 500 C. 5000 D. 6000 E. 7500 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 22
Profit and Loss • Basic Terms • Cost Price (C.P): Price at which one buys anything • Selling Price (S.P): Price at which one sells anything • Profit/Loss: Difference between the SP and CP. If the difference is positive it is called the profit and if negative it is called as loss. • Profit/Loss %: Profit /Loss as a percentage of CP. • Margin: Normally is in % terms only. This is the profit as a percentage of SP. • Marked Price: Price of the product as displayed on the label. • Discount: Less given on the marked price before selling. • Markup: Increment on the cost price before selling. • Formulae • Gain= SP‐CP • Profit % = (Profit in Rs/CP) X 100 • SP = ((100 + P%)/100) X CP • CP = ((100‐D%)/(100+P%)) X MP • Loss= CP‐SP • Loss % = (Loss in Rs/CP) X 100 • SP = ((100 ‐ L%)/100) X CP • CP = ((100‐D%)/(100‐ L%)) X MP • Successive Discount: In case of successive discounts we can treat the problem as the problem of successive percentage change and can use the formula “ If a% and b% are the two successive discounts ,then Net Discount = ( a + b – (aXb)/100) % • Marked Price: It is also known as List price or Tag price which is written on the item. The markup price written is always greater than the actual CP of the item and the percentage rise in the markup price is on the CP of the item. Percentage increase in the Markup Price = ((MP‐CP)/CP) X 100 Profit and Loss‐ Questions Q1. If the cost price of 20 articles is equal to the selling price of 25 articles, what is the % profit or loss made by the merchant? A. 25% loss B. 25% profit Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 23
C. 20% loss D. 20% profit E. 5% profit Q2. Sam buys 10 apples for $1. At what price should he sell a dozen apples if he wishes to make a profit of 25%? A. $0.125 B. $1.25 C. $0.25 D. $1.5 E. $1.8 Q3. By selling an article at 80% of its marked price, a merchant makes a loss of 12%. What will be the percent profit made by the merchant if he sells the article at 95% of its marked price? A. 5% profit B. 1% loss C. 10% profit D. 5.5% profit E. 4.5% profit Q4. What is the maximum percentage discount that a merchant can offer on her Marked Price so that she ends up selling at no profit or loss, if she had initially marked her goods up by 50%? A. 50% B. 20% C. 25% D. 16.67% E. 33.33% Q5. merchant who marked his goods up by 50% subsequently offered a discount of 20%. What is the percentage profit that the merchant make after offering the discount? A.
B.
C.
D.
E.
30% 125% 25% 20% 16.66% Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 24
Mean, Median,, Mode and Standard Deviaation • The mean is the average aand is computed as thhe sum of aall the obseerved outcoomes from
m of events. the sample divided by the totaal number o
where n is the sam
mple size an
nd the x corrrespond to
o the observved valued.. • The median is the centrall point of a a data set. To find thee median, you y would list all dataa n ascendingg order and
d simply pick the entrry in the middle m
of th
hat list. If th
here are an
n points in
even nu
umber of ob
bservationss, the median is not u
unique, so o
one often ttakes the m
mean of thee two mid
ddle values. • For lists, the mode m
is the most common (frequ
uent) valuee. A list can
n have morre than onee A data set has no mode when all the nu
umbers app
pear in thee data with
h the samee mode. A frequency. A data set has mu
ultiple mod
des when two t
or mo
ore values appear a
with the samee bution with
h two modees is called bimodal. A
A distributio
on with threee modes iss frequency. A distrib
called trrimodal. • The sttandard deeviation is the most common c
m
measure of statistical dispersion, measuringg how wid
dely spread
d the valuess in a data set are. If tthe data po
oints are clo
ose to the mean, then
n the standard deviattion is smalll. Converseely, if many data pointss are far fro
om the meaan, then thee d deviation is large. If all the dataa values aree equal, theen the standard deviattion is zero
o. standard
It is defined as the square roott of the varriance. o be We define the vvariance to
deviation to
o be
and the standard d
Mean, Median,, Mode and Standard Deviaation‐ Qu
uestions Apphelp GMAT coach
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5
Q1. If the mean of numbers 28, x, 42, 78 and 104 is 62, then what is the mean of 128, 255, 511, 1023 and x? A. 395 B. 275 C. 355 D. 415 E. 365 Q2. The arithmetic mean of the 5 consecutive integers starting with 's' is 'a'. What is the arithmetic mean of 9 consecutive integers that start with s + 2? A. 2 + s + a B. 2 + a C. 2s D. 2a + 2 E. 4 + a Q3. The average weight of a group of 30 friends increases by 1 kg when the weight of their football coach was added. If average weight of the group after including the weight of the football coach is 31kgs, what is the weight of their football coach in kgs? A. 31 kgs B. 61 kgs C. 60 kgs D. 62 kgs E. 91 kgs Q4. The average wages of a worker during a fortnight comprising 15 consecutive working days was $ 90 per day. During the first 7 days, his average wages was $ 87/day and the average wages during the last 7 days was $ 92 /day. What was his wage on the 8th day? A. $ 83 B. $ 92 C. $ 90 D. $ 97 E. $ 104 Q5. The average of 5 quantities is 6. The average of 3 of them is 8. What is the average of the remaining two numbers? A. 4 B. 5 C. 3 D. 3.5 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 26
E. 0.5 Q6. The average age of a group of 10 students was 20. The average age increased by 2 years when two new students joined the group. What is the average age of the two new students who joined the group? A. 22 years B. 30 years C. 44 years D. 32 years E. None of these Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 27
Mixtures and Alligations Mixtures Mixing of two or more qualities of things produces a mixture. When two items of different qualities are thus mixed, the quality of the resultant mixture lies in between the qualities of the original constituent items, i.e., it will be higher than the lowest quality and lower than the highest quality of the items being mixed. In the above examples that we took, the “quality” that we looked at was the height of the students. We could also have taken their weights or the marks scored by them or any other “quality” or “parameter” and calculated the “weighted average” value of that particular “quality” for the entire group. Similarly, if two types of a product of different prices per unit are mixed, the unit price of the resultant mixture will lie between the prices of the two types that form the mixture. Here, the average quality is essentially the weighted average of the two constituent items. If Q1 is the quantity (or number of items) of one particular item of quality P1 and Q2 be the quantity (or number of items) of the second item of quality P2 are mixed together to give a new mixture, then the weighted average value (P) of the quality of the mixture is given by P=( P1*Q1 + P2 *Q2 )/ (Q1+Q2) Even if there are more than two groups of items mixed, the weighted average rule can be applied. A mixture can also be a solution‐ that is, a liquid mixed with another liquid which is normally water. The concentration of the solution is expressed as the proportion (or percentage) of the liquid in the total solution. Ex: if 10 litres of pure alcohol is mixed with 40 litres of water, then in a total solution of 50 litres, there is 10 litres of alcohol. Hence the concentration of this solution is 0.2 (=10/50) or 20%. Alligations We will take the weighted average rule and rewrite the formula such that the quantity terms come on one side and the price terms come on the other side. If we do this we get the rule Q1/Q2 = (P ‐ P2) / (P1 ‐ P) This is called the RULE OF ALLIGATION. This rule connects quantities and prices in mixtures. This Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 28
can also be written as Q1/Q2 = (P2 ‐ P) / (P – P1) In a descriptive manner, the Rule of Alligation can be written as Quantity of Cheaper / Quantity of Dearer = (Rate of Dearer – Average Rate) / ( Average Rate – Rate of Cheaper) This rule is a very powerful rule and is useful in problems on weighted averages and mixtures. This rule is also useful in a number of problems which can be treated as mixtures and applied to parameters other than price also. In actual practice, to apply allegation rule, we do not need to remember the above formula at all. It can be made very simple by representing the rule pictorially. The above formula can be represented as follows: The ratio of the two quantities in the bottommost line will give us the ratio of the quantities of Dearer and Cheaper varieties. Mixtures and Alligations‐ Questions Q1. Solution Y is 30 percent liquid X and 70 percent water. If 2 kilograms of water evaporate from 8 kilograms of solution Y and 2 kilograms of solution Y are added to the remaining 6 kilograms of liquid, what percent of this new solution is liquid X? A. 30% B. 100/3 % C. 75/2 % D. 40% E. 50% Q2. A sum of Rs 39 was divided among 45 boys and girls. Each girl gets 50 paise, whereas a boy gets one rupee. Find the number of boys and girls. A. Boy: 12, Girl: 33 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 29
B.
C.
D.
E.
Boy: 33, Girl: 12 Boy: 16, Girl: 29 Boy: 29, Girl: 16 Boy: 15, Girl: 30 Q3. There are two alloys A and B. Alloy A contains zinc, copper and silver, as 80% 15% and 5% respectively, whereas alloy B also contains the same metals with percentages 70%, 20%, 10% respectively. If these two alloys are mixed such that the resultant will contain 8% silver, what is the ratio of these three metals in the resultant alloy? A. 7:4 B. 4:9 C. 9:4 D. 4:7 E. 7:6 Q4. A 20 litre mixture of milk and water contains milk and water in the ratio 3 : 2. 10 litres of the mixture is removed and replaced with pure milk and the operation is repeated once more. At the end of the two removal and replacement, what is the ratio of milk and water in the resultant mixture? A. 17:3 B. 9:1 C. 3:17 D. 5:3 E. 6:19 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 30
Speed,, Distance
e and Tim
me Time an
nd Speed Distancee = Speed
d * Time Speed = Distan
nce / Time Distan
nce / Speed
d Time = 1 km/h == 5/18 m/sec 1 m/sec = 18/5 km/h me then aveerage speed
d = 2*x*y / (x + y) If distance traveled be the sam
MPORTANTT FORMULA
AE ‐ from Te
est's point o
of view SOME IM
• If an a object travels t
a distance d
D1
1 and D2 at a a speed
d of S1 and S2 Miless per Hourr, resspectively, in
n time T1 and T2 then the total tiime taken 'TT' is given b
by: The totaal distance ccovered 'D' is given byy: •
Wh
hile a certaiin distance 'D' is covered, if A maan changess his speed in the ratio
o X : Y, then
n thee ratio of tim
me taken beecomes Y : X. •
IF two t
objectts, say A and a B, startt at the saame time in i oppositee directionss from two
o diffferent pointts (P and Q
Q) and arrivee at the opp
posite points in 'a' and
d 'b' hours rrespectivelyy afteer having m
met. Then thhe ratio of ttheir speed is: Speed,, Distance
e and Tim
me‐ Questtions Q1. A trrain travelin
ng at 72 km
mph crossess a platform
m in 30 secconds and aa man stand
ding on thee platform
m in 18 seco
onds. What is the lengtth of the plaatform in m
meters? A. 240 meters B. 360 meters C. 420 meters D. 600 meters C
etermined E. Cannot be d
Q2. A trrain travelin
ng at 100 kkmph overtakes a mottorbike travveling at 64
4 kmph in 4
40 secondss. What is the length of the train
n in meters?? A. 1777 meterss Apphelp GMAT coach
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1
B.
C.
D.
E.
1822 meters 400 meters 1111 meters None of these Q3. Jim travels the first 3 hours of his journey at 60 mph speed and the remaining 5 hours at 24 mph speed. What is the average speed of Jim's travel in mph? A. 42 mph B. 36 mph C. 37.5 mph D. 42.5 mph E. 48 mph Q4. A runs 25% faster than B and is able to give him a start of 7 meters to end a race in dead heat. What is the length of the race? A. 10 meters B. 25 meters C. 45 meters D. 15 meters E. 35 meters Q5. Jane covered a distance of 340 miles between city A and city taking a total of 5 hours. If part of the distance was covered at 60 miles per hour speed and the balance at 80 miles per hour speed, how many hours did she travel at 60 miles per hour? A. 2 hours 30 minutes B. 3 hours C. 2 hours D. 1 hour 45 minutes E. None of these Q6. Steve traveled the first 2 hours of his journey at 40 mph and the remaining 3 hours of his journey at 80 mph. What is his average speed for the entire journey? A. 60 mph B. 56.67 mph C. 53.33 mph D. 64 mph E. 66.67 mph Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 32
Circular and Straight Races When two persons P and Q are running a race, they can start the race at the same time or one of them may start a little later than the other. In the second case, suppose P starts the race and after 5 seconds, Q starts. Then we say P has a “start” of 5 seconds. Alternatively, in a race between P and Q, P starts first and then when P has covered a distance of 10metres, Q starts. Then we say that P has a “start” of 10 metres. In a race between P and Q where Q is the winner, by the time Q reaches the winning post, if P still has another 15 metres to reach the winning post 10 seconds after Q reaches it, then we say that Q has won the race by 10 seconds. When two or more persons running around a circular track ( starting at the same point and at the same time), then we will be interested in two main issues: 1. When they will meet for the first time and 2. When they will meet for the first time at the starting point. To solve the problem on circular tracks, you should keep the following points in mind. When two persons are running around a circular track in OPPOSITE directions 1. The relative speed is equal to the sum of the speeds of the two individuals and 2. From one meeting point to the next meeting point, the two of them TOGETHER cover a distance equal to the length of the track When two persons are running around a circular track in SAME directions 1. The relative speed is equal to the difference of the speeds of the two individuals and 2. From one meeting point to the next meeting point, the faster person covers one COMPLETE ROUND more than the slower person. Circular and Straight Races When two people are running around a circular track Let the two people A and B with respective speeds of a and b (a>b) be running around a circular track (of length L) starting at the same point and at the same time. Then When the two persons are When the two persons are running in the SAME running in the OPPOSITE direction direction Time taken to meet for the L/ (a‐b)
FIRST TIME EVER L/ (a+b)
Time taken to meet for the LCM of { L/a, L/b}
first time at the STARTING POINT LCM of { L/a, L/b} Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 33
When three people are running around a circular track Let the three people A, B and C with respective speeds of a, b and c (a>b>c) be running around a circular track (of length L) starting at the same point and at the same time. In this case we consider the three persons running in the same direction as the general case. Time taken to meet for the LCM of { L/(a‐b) , L/(b‐c)}
FIRST TIME EVER Time taken to meet for the LCM of { L/a, L/b, L/c}
first time at the STARTING POINT The logic in obtaining the above is as follows: A and B will be together with a time gap of L/ (a‐b); B and C will be together with a time gap of L/ (b‐c); For A,B and C to be together, A and B should be together as well as B and C should be together. Hence LCM of the two timings L/(a‐b) and L/(b‐c) will give the time when A, B and C will all be together. Circular and Straight Races‐ Questions Q1. P and Q run around a circular track of radius 49 meters starting from the same point at the same time and in the same direction. If 'P' runs at a speed of 15m/min and 'Q' runs at a speed of 10m/min, when will they meet again for the first time? A. 1 hour, 1 minute and 36 seconds B. 1 hour, 1 minute C. 1 hour, 5 minute and 6 seconds D. 1 hour, 36 seconds E. 1 hour, 1 minute and 40 seconds Q2. In a 200m race, if A gives B a start of 25 metres, then A wins the race by 10 seconds. Alternatively, if A gives B a start of 45 metres the race ends in a dead heat. How long does A take to run 200m? A. 100 seconds B. 112.5 seconds C. 77.5 seconds Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 34
D. 87.5 seconds E. 82.44 Q3. In a kilometre race, A can give B a start of 100 m or 15 seconds. How long does A take to complete the race? A. 150 seconds B. 165 seconds C. 135 seconds D. 66.67 seconds E. 180 seconds Q4. A can give B a start of 50 metres or 10 seconds in a kilometer race. How long does A take to complete the race? A. 200 seconds B. 140 seconds C. 220 seconds D. 190 seconds E. 150 seconds Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 35
Pipes, Cistern, Time and Work 1. If A can finish a piece of work in n days then A’s 1 day’s work = 1/h 2. If A is twice as good a workman as B, then time taken to finish the work by A = ½ *(time taken by B) 3. If the number of persons engaged to do a piece of work be increased (or decreased) in a certain ratio ,the time required to do the same work will be decreased (or increased) in the same ratio. 4. Time and work are always in direct proportion. 5. If two taps or pipes P and Q take ‘m’ and ‘n’ hours respectively to fill a cistern or tank, then the two pipes together fill (1/m + 1/n) part of the tank in 1 hour and the entire tank is filled in 1/ (1/m + 1/n) = m*n/(m + n) hours. 6. If a leak in the bottom of a tank empties it in ‘n ’hours, then Leak's one hour's work= ‐1/n (negative work) 7. If a pipe fills a tank in ‘m’ hours and a leak in the bottom empties it in ‘n’ hours, Pipe and leak together one hour's work= 1/m – 1/n Pipes, Cistern, Time and Work‐ Questions Q1. Working together, Jose and Jane can complete an assigned task in 20 days. However, if Jose worked alone and complete half the work and then Jane takes over the task and completes the second half of the task, the task will be completed in 45 days. How long will Jose take to complete the task if he worked alone? Assume that Jane is more efficient than Jose. A. 25 days B. 30 days C. 60 days D. 65 days E. 36 days Q2. A can complete a project in 20 days and B can complete the same project in 30 days. If A and B start working on the project together and A quits 10 days before the project is completed, in how many days will the project be completed? A. 18 days B. 27 days Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 36
C. 26.67 days D. 16 days E. 12 days Q3. Ram, who is half as efficient as Krish, will take 24 days to complete a work if he worked alone. If Ram and Krish worked together, how long will they take to complete the work? A. 16 days B. 12 days C. 8 days D. 6 days E. 18 days Q4. Four men and three women can do a job in 6 days. When five men and six women work on the same job, the work gets completed in 4 days. How long will a woman take to do the job, if she works alone on it? A. 18 days B. 36 days C. 54 days D. 58 days E. None of these Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 37
Simple
e and Com
mpound IInterest
Interest is the cost of borrowiing money. An intere
est rate is th
he cost statted as a perrcent of thee amount borrowed per period of time, usually one yeear. Simple Interest Simple interest is ccalculated on the orriginal princcipal only. The intereest is calculaated on thee mount of m
money in each time period, and, ttherefore, the interestt t earned in
n each timee same am
period iss the same.. Simple Interest = P * R * n w
where, borrowed or loaned) P = principal (originaal amount b
nterest rate for one period R = in
n = nu
umber of periods und Interest Compou
Compou
und interesst is calculated each
h period on the origginal princiipal and all a interest accumullated durin
ng past periods. Altho
ough the in
nterest mayy be stated
d as a yearrly rate, thee compounding perio
ods can be yearly, sem
miannually, q
quarterly, o
or even con
ntinuously. TThe interestt principal of the previou
us period to
o become the principaal earned iin each period is added to the p
for the n
next period. nterest is co
ompounded
d, the total amount is ccalculated u
using the fo
ormula, When in
Compo
ound Intere
est = A‐P d the conversion periiod. At the end of thee The period for whiich interestt is calculatted is called
ncipal to geet the new principal. TThe formulaa conversiion period, the interesst is added to the prin
to calcullate the new
w rate of interest with respect to the converrsion period
d is Time period = nu
umber of yeears e and Com
mpound IInterest‐ Question
ns Simple
Q1. Brau
un invested a certain ssum of mon
ney at 8% p..a. simple in
nterest for 'n' years. Att the end of 'n' yearss, Braun gott back 4 tim
mes his original investm
ment. What is the valuee of n? A. 50 years B. 25 years Apphelp GMAT coach
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8
C. 12 years 6 months D. 37 years 6 months E. 40 years Q2. Shawn invested one half of his savings in a bond that paid simple interest for 2 years and received $ 550 as interest. He invested the remaining in a bond that paid compound interest, interest being compounded annually, for the same 2 years at the same rate of interest and received $605 as interest. What was the value of his total savings before investing in these two bonds? A. $ 5500 B. $ 11000 C. $ 22000 D. $ 2750 E. $ 44000 Q3. Ann invested a certain sum of money in a bank that paid simple interest. The amount grew to $240 at the end of 2 years. She waited for another 3 years and got a final amount of $300. What was the principal amount that she invested at the beginning? A. $ 200 B. $ 150 C. $ 210 D. $ 175 E. Insufficient data Q4. Peter invested a certain sum of money in a simple interest bond whose value grew to $300 at the end of 3 years and to $ 400 at the end of another 5 years. What was the rate of interest in which he invested his sum? A. 12% B. 12.5% C. 6.67% D. 6.25% E. 8.33% Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 39
Linear Equation
ns • Lineaar equation
ns are funcctions which
h have two variables. They havve an independent and
d dependeent variable
e. • Independ
dent Variab
bles ‐ variab
bles that yo
ou put into tthe equatio
ons • Dependeent Variablees‐ variablees that you solve for. y = 2x ‐ 4 he equation
n above, y h
has to be th
he Dependeent variablee. Because it is the variable that iss Using th
being so
olved for. bles, there are many p
possible com
mbinations of answerss. Becausee linear equations havee two variab
Those possible solu
utions are w
written as o
ordered paiirs. Usually an orderred pair is w
written with
h y = 3x ‐ 5 the indeependent vaariable first then the dependent vvariable. Example: Becausee we solved
d for y, then
n x is the in
ndependen
nt variable. The ordereed pair wou
uld then bee written aas (x, y). SLOPE pe of a line is the chan
nge in y – vvalues divid
ded by the change in xx – values ffor any two
o The slop
y2  y1
points on the line. m
x2  x1
where (x1 , y1) and (x2 , yy2) are any ttwo points on the linee. Parallel line : Their slopes will be EQUAL.. dicular liness : Their slo
opes will be the negativve reciprocaal of each o
other. Perpend
Y INTERCEPTS X AND Y
• The xx‐intercept is the x‐cooordinate of a point w
where the ggraph crossees the x‐axis i.e. when
n y=0. • The yy‐intercept is the y‐cooordinate of a point w
where the ggraph crossees the y‐axis i.e. when
n x=0. SLOPE‐IN
NTERCEPT FORM OF A
A LINE The slop
pe interceptt form of a line is y = m
mx + b, wheere “m” rep
presents the slope of tthe line and
d “b” reprresents the y‐interceptt. Apphelp GMAT coach
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0
Standard Form Ax + By = C Two‐point form where (x1,y1) &(x2,y2) are two points on the line with x2 ≠ x1 & slope is explicitly given as (y2−y1) / (x2−x1). Intercept form where a and b must be nonzero. The graph of the equation has x‐intercept a and y‐intercept b. FIND EQUATION OF A LINE GIVEN 2 POINTS 1. Find the slope between the two points. 2. Plug in the slope in the slope‐intercept form. 3. Pick one of the given points and plug in numbers for x and y. 4. Solve and find b. 5. Rewrite final form. Steps if given the slope and a point on the line. 1. Substitute the slope into the slope‐intercept form. 2. Use the point to plug in for x and y. 3. Find b. 4. Rewrite equation. Linear Equations‐ Questions Q1. Find the value of x that satisfies the following equation: A. X=1 B. X=5 C. X=1/4 D. X=1/5 E. X=3 Q2. Which of the following is NOT a solution to the equation 2x + 3y = 2y ‐ 5x + 10? A. x=1, y=3 B. x=7, y=‐38 C. x=0, y=10 D. x=5, y=‐25 E. x=2, y=‐4 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 41
Q3. A poultry farm has only chickens and pigs. When the manager of the poultry counted the heads of the stock in the farm, the number totaled up to 200. However, when the number of legs was counted, the number totaled up to 540. How many chickens were there in the farm? A. 70 B. 120 C. 60 D. 130 E. 80 Q4. Three years back, a father was 24 years older than his son. At present the father is 5 times as old as the son. How old will the son be three years from now? A. 12 years B. 6 years C. 3 years D. 9 years E. 27 years Q5. For what values of 'k' will the pair of equations 3x + 4y = 12 and kx + 12y = 30 not have a unique solution? A. 12 B. 9 C. 3 D. 7.5 E. 2.5 Q6. The basic one‐way air fare for a child aged between 3 and 10 years costs half the regular fare for an adult plus a reservation charge that is the same on the child's ticket as on the adult's ticket. One reserved ticket for an adult costs $216 and the cost of a reserved ticket for an adult and a child (aged between 3 and 10) costs $327. What is the basic fare for the journey for an adult? $111 $52.5 $210 $58.5 $6 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 42
Quadratic Equations A Quadratic Equation in x is an equation that can be written in the general form where a, b, and c are real numbers ax2 + bx + c = 0 We can solve by several methods: • By Factoring and setting each factor equal to 0 • Extracting Square Roots • Completing the Square • Using the Quadratic Formula Quadratic Equations ‐ Factoring Factor ax2 + bx + c = (Ax+B)(Cx+D) = 0 • Set each factor = 0. (Ax+B) = 0, (Cx+D) = 0 • Solve for x Quadratic Equations – Extracting Square Roots • If you can manipulate the equation so that: x2=d Then, x= + √d and x= ‐ √d or x= ± √d Quadratic Equations – Completing the Square Given : x2 + bx we can make it a "perfect square“ If ‘a’ the coefficient of is not 1, we divide both sides by a. Quadratic Equations – The Quadratic Formula Quadratic Equations – Types of Solutions Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 43
Quadratic Equations‐ Questions Q1. What is the highest integral value of 'k' for which the quadratic equation x2 ‐ 6x + k = 0 have two real and distinct roots? A. 9 B. 7 C. 3 D. 8 E. 12 Q2. If one of the roots of the quadratic equation x2 + mx + 24 = 0 is 1.5, then what is the value of m? A. ‐22.5 B. 16 C. ‐10.5 D. ‐17.5 E. Cannot be determined Q3. For what value of 'm' will the quadratic equation x2 ‐ mx + 4 = 0 have real and equal roots? A. 16 B. 8 C. 2 D. ‐4 E. Choice (B) and (C) Q4. x2 = 9x ‐ 20; Solve for x. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 44
A.
B.
C.
D.
E.
X=10 X
x = 3 or x = 6
6 5 x = 4 or x = 5
x = 5 N
No Solution Exists Q5. the expression x2 + 6x + 6 = 0, solve ffor x A.
B.
C.
D.
E.
0 = 0; x = ? Q6. 15x2 ‐ 60x ‐ 180
A. x==5 or x=20//3 B. x==‐3 or x=‐5//9 C. x==3 or x=5/9
9 D. x==2 or x=‐6 E. x==‐2 or x=6 Apphelp GMAT coach
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5
Sequence and Series Sequences • A sequence is a set of real numbers a1, a2,…an,… which is arranged (ordered) according to some definite rule. • Each number ak is a term of the sequence. • We called a1 ‐ First term. • The nth term an is called the general term of the sequence. Infinite Sequences An infinite sequence is often defined by stating a formula for the nth term, an by using {an}. Example:  The sequence {2} has nth term a=2 .  Using the sequence notation, we write this sequence as follows Recursively Defined Sequences A sequence is said to be defined recursively if the first term a1 is state together with a rule for obtaining any term ak +1 from the preceding term ak whenever k ≥ 1. Example: A sequence is defined recursively as follows Thus the sequence is 3, 6, 12, 24, … where Periodic Sequences A periodic sequence is a sequence with terms which are repeated after a certain fixed number of term. Example: 1 1
1. 1, 2 ,1, 2 , n
2. an  sin 2
The Summation Notation Of Sequences Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 46
The symbol ∑ (sigma) is called the summation sign. This symbol will represents the sum of the first m terms as follows: Series 
In general, given any infinite sequence,a1, a2,…an,… the expression 
a n  a 1  a 2  
a n   n 1
is called an infinite series or simply a series. Theorem of Sums Sequence Of Partial Sums • If n is positive integer, then the sum of the first n terms of an infinite sequence will be denoted by Sn. • The sequence S1, S2,…Sn,… is called a sequence of partial sums. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 47
Arithmetic Sequences • A sequence a1, a2,…an,… is an arithmetic sequence if there is a real number d such that for ak 1  ak  d
every positive integer k, • The number d  a k 1  a k is called the common difference of the sequence. The nth Term Of An Arithmetic Sequences • An arithmetic sequence with first term a1 and common different d, can be written as follows: a , a  d , a  2d , a  3d ,
1 1
1
1
• The nth term, an of this sequence is given by the following formula: an  a1   n  1 d
The nth Partial Sum Of An Arithmetic Sequences If a1, a2,…an,… is an arithmetic sequence with common difference d, then the nth partial n
sum Sn (that is the sum of the first nth terms) is given by either Sn   2a1   n  1 d 
2
n
Or Sn   a1  an 
2
The Arithmetic Mean Of An Arithmetic Sequences • The arithmetic mean of two number a and b (average of a and b) is defined by (a + b) / 2 . • Then the following sequence is true if d = (b – a ) / 2 . 1
a,  a  b  , b
2
• If c1, c2,…ck are real numbers such that a,c1, c2,…ck,b is a finite arithmetic sequence, then c1, c2,…ck are k arithmetic means between the numbers a and b. Geometric Sequences • A sequence a1, a2,…an,… is a geometric sequence if a1 ≠ 0 and if there is a real number r ≠ 0 such that for every positive integer k, ak 1  ak r
• a
•The number r  k 1 is called the common ratio of the sequence. ak
The nth Term Of An Arithmetic Sequences • A geometric sequence with first term a1 and common ratio r, can be written as follows: a1 , a1r , a1r 2 , a1r 3 ,
• The nth term, an of this sequence is given by the following formula: an  a1r n 1
The nth Partial Sum Of An Geometric Sequences Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 48
If a1, a2,…an,… is a geometric sequence with common ratio r ≠ 0 , then the nth partial sum Sn (that is the sum of the first nth terms) is given by  1 rn 
S n  a1 

 1 r 
The Geometric Mean Of An Geometric Sequences • The geometric mean of two number a and b (average of a and b) is defined by c . • If the common ratio is r, then a, c, b  r  c  b or c 2  ab
a c
• If c1, c2,…ck are real numbers such that a,c1, c2,…ck,b is a finite geometric sequence, then c1, c2,…ck are k geometric means between the numbers a and b. The Sum Of An Infinite Geometric Series If |r| < 1 , then the infinite geometric series a 1  a 1 r  a 1r 2  a1 r 3 ,
 a 1 r n  1 ,  has the sum S  a1
n
1 r
Harmonic Progression A sequence of numbers is said to form a harmonic progression if their reciprocals form an arithmetic progression. Note: 1. The series formed by the reciprocals of the terms of a geometric series is also a geometric series 2. There is no general method of finding the sum of a harmonic progression To find the nth term of an H.P To find the nth term of an H.P, find the nth term of the corresponding A.P. obtained by the reciprocals of the terms of the given H.P. Now the reciprocal of the nth term of an A.P. will be the nth term of the H.P. The Harmonic Mean (HM) of two numbers a and b HP= 2ab/(a+b) Relation between Arithmetic Mean (AM), Geometric Mean (GM) and Harmonic Mean (HM) AM * HM = GM *GM that is, AM, GM, HM are in Geometric Progression. For two positive numbers, AM ≥ GM ≥ HM equality holding for equal numbers. For n non‐zero positive numbers, AM ≥ GM ≥ HM Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 49
Binomiaal Expansion For Positiive Integerss • Any eexpression containing two terms is called a bbinomial; eg: (a + b), (xx – y). • If n is a positivve integer, then a genneral formuula for exppanding is given g
by th
he binomiaal m. theorem
• The ffollowing sppecial casess can be obttained by m
multiplicatio
on a  b2
 a  b   a 2  2ab
3
 a  b   a3  3a 2b  3ab2  b3
2
Binomiaal Theorem The bino
omial theorrem states tthat a binom
mial expanssion can be expanded as follows: a  b
n
 a n  n c1a n 1b1  n c2 a n  2b 2    n cr a n r b r   b n
n
  n cr a n  r b r
r 1
n
omial coeffiicient with k !  k  k  1 k  2   2 1 and 0!!  1
Where n C r     n ! is called a bino
r  r ! n  r  !

Propertiies Of Binomial Theorrem 1. There are n +
+ 1 terms in
n expansion
n, first beingg ‘an ‘ and the last ‘bn ‘ . of a decreasse by 1 and the powerr of b increaase by 1 along the expaansion. 2. The power o
powers of a
a and b in each term iss always equ
ual to n. 3. The sum of p
n
nr r
h term is c r a b . 4. The (n + 1)th
natural num
mbers, 1+2+3+ ... + n = Sum of tthe first n n
2
Sum of squaares of the ffirst n naturral numberss, 12+22+32+ ... + n
+
= Sum o
of the cubees of the firsst n natural numbers, 13+23+33+ ... + n3 =
Sequen
nce and SSeries‐ Qu
uestions
Q1. What is the W
sum of all 3
3 digit num
mbers that leeave a remaainder of '2
2' when diviided by 3? A. 897 B. 164,850 C. 164,749 D. 149,700 E. 156,720 Apphelp GMAT coach
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0
Q2. How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5? A. 128 B. 142 C. 143 D. 141 E. 129 Q3. The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m+2? A. m + 4 B. n + 6 C. n + 3 D. m + 5 E. n + 4 Q4. The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression? A. 300 B. 120 C. 150 D. 170 E. 270 Q5. If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P? A. 3 B. 1/3 C. 2 D. 9 E. 1/9 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 51
Probability Definitions Event ‐ An outcome to a random occurrence. For example, the following are events: drawing a black card from a 52‐card deck, a coin landing on heads, rolling an even number on a 6‐sided die. Probability, P(A) ‐ The likelihood of event A occurring. For example, the probability of a coin landing on heads is 0.5 since there is a 50% chance (i.e., probability) of the coin turning up heads when flipped. Intersection of Events, P(A∩B) ‐ When two events are fulfilled simultaneously. For example, the intersection of the events "rolling an even number" and "rolling a number less than three" is rolling a 2 since rolling a 2 fulfills both events (i.e., 2 is both even and less than 3). Union of Events, P(A B) ‐ When either of two events is fulfilled. For example, the union of the events "rolling an even number" and "rolling a number less than three" is rolling a 2, 4, 6 (an even number) or rolling a 1, 2 (a number less than three). Dependent Event ‐ An event whose probability of occurring is influenced by (i.e., dependent on) whether another event occurs. For example, the probability of drawing a red card from a normal 52‐card deck after you draw another card from the 52‐card deck without replacing this other card is a dependent event. The probability of the second card you draw being red depends on what card was drawn the first time. In particular, the probability of the second card being red depends on whether you drew a red card the first time (in which case one less red would be in the deck). A dependent event is the opposite of an independent event. Independent Events ‐ Two events are independent if the occurrence of one event does not affect the probability of the occurrence of the other. For example, the probability of flipping a coin twice and the coin landing on heads the second time is not affected by (i.e., is independent of) whether the first coin flip turned up heads or tails. An independent event is the opposite of a dependent event. Conditional Probability, P(A|B) ‐ The likelihood of event A occurring given that event B already occurred. For example, the probability of drawing a red card from a 52 card deck is not the same as the probability of drawing a red card from a 52 card deck given that you have already drawn three red cards and did not put them back into the deck. This later probability is an example of conditional probability. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 52
Mutually Exclusive
e Events ‐ Tw
wo events aare mutually exclusive if they cannot occur together. mple, the events e
"rolling an even number" and "rollin
ng an odd number" are mutuallyy For exam
exclusivee since by d
definition rolling an evven numbeer means yo
ou cannot rroll an odd number. Byy contrastt, the even
nts "rolling an odd nu
umber" an
nd "rolling a number two or lesss" are not mutuallyy exclusive since you ccould roll th
he number o
one, which is both odd
d and a num
mber two or less. ment of an
n Event, P(A
A') or P(AC) ‐ The even
nt that is co
omposed off all the outtcomes that Complem
are not iin another e
event. For exam
mple, the co
omplementt of flippingg a coin and
d it landing on heads iis flipping aa coin and it landing on tails. The complement of ro
olling an evven numbeer is rollingg an odd number. Thee ment of draawing a heaart is drawiing either aa spade, clu
ub, or diamond. The complement complem
of the teemperature
e being 30 d
degrees is th
he temperaature not beeing 30 deggrees. Graphicaal Represen
ntation The follo
owing table
e elucidates the relatio
onship betw
ween the diffferent typees of eventss. Mutuaally Exclusivve
Independ
dent
Yees
No
Yes
Im
mpossible
Possible
No
Po
ossible
Possible
The following chaart shows tthe relationship betw
ween the iintersectio
on of two eevents, thee of two events, and th
he compleement of aan event. In each casse, the pro
obability in
n union o
question is repressented by the area in gray. For example,, the gray in the mid
ddle of thee & B. far left ssection of the graph representss the interssection of events A &
Basic Probability obability of o an even
nt occurrin
ng is the likelihood of it hap
ppening exxpressed in
n The pro
mathem
matical term
ms. The likelihood of an
n event occcurring is th
he number o
of ways thaat particular event can occur divided d
byy the numb
ber of wayys any possible outccome can occur. Said
d Apphelp GMAT coach
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3
differenttly, the pro
obability off an event occurring is the num
mber of waays the speecific event outcomee can occurr divided byy the numbeer of ways aany possible event can
n occur. Said
d one other way, thee probabilityy of an event is the nu
umber of ou
utcomes that fulfill thaat event divvided by thee number of total possible outco
omes. 0 < P(A) < 1 he probabillity of certaain events can be comp
plex, there is one rule that alwayss While caalculating th
applies: 0 < P(A) < 1 obability off an event occurring ccan never b
be less than
n Regardleess of the ccircumstancces, the pro
zero or greater thaan one. The rationalee for why the probabiility of A occurring caan never bee would require the num
merator bein
ng larger than the denominator in
n greater tthan one iss that this w
the abo
ove formulaa for P(A). Since the denominattor includees the num
merator (i.ee., the totaal number of ways an
ny outcomee can occur includes th
he number of ways ou
utcome A caan occur), it he numerattor to be laarger than the denom
minator. Sim
milarly, a value for P(A
A) is imposssible for th
less than zero wou
uld require a negativee number of o ways an outcome could c
occurr and this iss e. logically impossible
Mutually Exclusive
e Events utually exclu
usive if theyy cannot occcur simultaaneously. In
n other worrds, if event Two eveents are mu
A occurss, then eve
ent B cannot occur. There T
is no
o overlap between events A and
d B if thesee events aare mutuallyy exclusive. P(A∩B) == 0 for all m
mutually excclusive even
nts A and B Examplees of mutuaally exclusive events: Drawingg a red card and drawin
ng a black ccard Rolling aan even num
mber and ro
olling a 3 Conditio
onal Probab
bility The concept of con
nditional prrobability p
pertains to the probab
bility of a certain even
nt occurringg given that another event alreaady occurreed. The con
ncept of conditional probability is important hange the probability o
of another eevent. since thee occurrencce of one evvent can ch
The conditional pro
obability of A given B, d
denoted P(A|B): ndent vs. De
ependent EEvents Indepen
Two eveents are ind
dependent iif the occurrrence of one event do
oes not chaange the prrobability of the occu
urrence of tthe other event. For exxample, if tthe probabiility of even
nt A = x beffore event B
B occurs aand the prob
bability of eevent A = x after eventt B occurs, tthe two aree independeent. Two eveents are dependent if the occurrrence of on
ne event do
oes change the probab
bility of thee occurren
nce of the other even
nt. For exam
mple, if thee probabilitty of eventt A = x befo
ore event B
B occurs b
but the prob
bability of eevent A = y after eventt B occurs, tthe two eveents are dep
pendent. Apphelp GMAT coach
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4
For Independent Events: P(A|B) = P(A) For Two Dependent Events: P(A|B) ≠ P(A) Union of Events: Or The union of events includes not only the probability of A and B, but also the probability of only A and the probability of only B. P(A B) = P(A) + P(B) ‐ P(A∩B) The last term is important since it removed items that appear in both A and B, thereby avoiding double counting. Intersection of Events: And The intersection of events includes only the probability that both events A and B occur simultaneously. If event A occurs alone or if event B occurs alone, this does not fall within the intersection of events A & B. P(A∩B) = probability of the intersection of events A and B P(A∩B) = P(A)P(B|A) P(A∩B) = P(B)P(A|B) If events A and B are mutually exclusive (i.e., if one event occurs then the other cannot occur), the formula can be simplified: P(A∩B) = probability of the intersection of events A and B P(A∩B) = P(A)P(B) Complement of an Event: Not The formula for the complement of an event, denoted P(AC): P(A') or P(AC) = 1 ‐ P(A) Stated Differently: P(AC) + P(A) = 1 One common misconception that some students make is to assume that the complement of an event is simply its opposite. For example, it would be wrong to assume that the complement of a positive number is a negative number. Instead, the complement of a positive number is everything that is not a positive number (i.e., negative numbers and zero). Probability‐ Questions Q1. Event E is defined to be rolling an even number on a 6‐sided die and Event F is defined to be rolling a 1, 2 or 3. Calculate the probability of rolling a die such that events E and F occur simultaneously on a single roll of the die. A. 1/2 B. 1/6 C. 0 D. 5/6 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 55
E. 1 Q2. If E is defined to be drawing a black card out of a normal 52 card deck and F is defined to be drawing a heart, what is the probability of either E or F coming true? A. 1/2 B. 2/3 C. 1 D. 5/8 E. ¾ Q3. A man can hit a target once in 4 shots. If he fires 4 shots in succession, what is the probability that he will hit his target? A. 1 B. 1/256 C. 81/256 D. 175/256 E. 144/256 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 56
Permu
utations aand Comb
binations Definitio
ons Combinaatorics ‐ Th
he branch o
of mathematics that d
deals with ccollections of objects that satisfyy specified
d criteria (e
e.g., countin
ng arrangem
ments, perm
mutations, and combinations). Fo
or examplee, combinaatorics would answer tthe questio
on "how maany different ways can you arrangge a 10‐songg playlist if you have 45 songs to
o choose fro
om?" e branch off combinato
orics wheree changing tthe order o
of the objeccts does not Combinaations ‐ The
create a new scenaario. For eexample, th
he question "how man
ny teams of 9 baseball players can
n ble from a rroster of 18
8 players?"" is a combinations qu
uestion sincce changingg a managger assemb
the ordeer in which the player iis chosen does not creeate a new aarrangement e branch off combinato
orics wheree changing tthe order o
of the objeccts creates aa Permutaations ‐ The
new scenario. mple, the question q
"h
how many different ways w
can a a baseball pitcher who throws 6
6 For exam
unique pitches (e.g., curveb
ball, fastballl, changeu
up, etc.) th
hrow the next n
3 pitcches?" is aa wing order creates a new arrangeement (e.g.., permutaations quesstion because changingg the throw
throwingg fastball, sinker, changeup is diffferent than throwing sinker, changgeup, fastb
ball) Factoriaals Solving p
permutatio
ons and com
mbinations p
problems u
using a form
mula requirees the use o
of factorialss. x factorial, denoted
d x!, is the p
product of aall positive integers lesss than or eequal to x.
3)(2)(1) x factorial = x! = x(xx‐1)(x‐2)...(3
Simplifyying Factorials Complicated expressions using multiple factorials o
often contaain overlapp
ping terms and can bee simplifieed considerably. Permu
utations aand Comb
binations Permutaations Permutaations involve problem
ms in which the arrangement of th
he items in question d
does matterr. Some exxamples of common problems p
a arrangin
are ng books on a booksh
helf, creatin
ng a seatingg chart or making a sschedule. In
n all of these situation
ns, moving things arou
und createss a different ment, charrt, or sched
dule. (If ch
hanging thee order of the items in question creates aa arrangem
uniquelyy different aarrangemen
nt, you are dealing witth a permuttation). Apphelp GMAT coach
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There are two gen
neral ways to solve th
hese probleems. The first method
d involves drawing d
thee m. The secon
nd involvess the use off a formula.. For simpleer problemss it is often possible to
o problem
draw thee arrangem
ment. However, as the problems b
become mo
ore difficult, drawing the problem
m can beco
ome more p
prone to errrors and ussing the form
mula may b
be best. n = the n
number of o
objects to cchoose from
m k k = the number of objectss selected
d Combinaations Unlike permutatio
p
ns, combin
nations do not consid
der order. The numbeer of possible euchree hands or the numb
ber of team
ms that can be created
d in a gym class are ju
ust a few eexamples. In
n der does no
ot matter. A team consisting of Jim
mmy, Dan, aand Bob is tthe same ass both casses, the ord
a team cconsisting o
of Dan, Jimm
my, and Bob
b. Unlike permutatio
p
ns, combin
nations can
nnot be eaasily solved
d by diagraamming the problem
m. Instead, use the equation belo
ow to solve combinatio
on problem
ms. n = the number of k = th
o objects to t choose from he numberr of objectss selected
d uation can b
be read as having n objects and choosing to
o arrange kk of them (ii.e., you aree This equ
selectingg k objects from a pile of n objeects). The k! k on the bottom b
reprresents how
w switchingg two objeects around
d no longer creates a n
new arrangeement. Permu
utations aand Comb
binations‐ Questio
ons Q1. How
w many diffe
erent uniqu
ue combinations of lettters can be created byy rearrangin
ng the lettters in math
hematics? A. 2! B. 10! C. 8!/3 D. 11!/6 E. 11!/8 ow many w
ways can thee letters of the word A
ABACUS be rrearranged such that tthe vowels Q2. In ho
alw
ways appearr together? A. 6! / 2! B. 3! * 3! C. 4! / 2! Apphelp GMAT coach
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8
D. 4! * 3! / 2! E. 3! * 3! / 2! Q3. If a class of 10 students has five men, how many ways can the men and women be arranged in a circle so that no two men sit next to each other? A. 5!4! B. 5!5! C. 4!4! D. 10! E. 10!/5! Q4. How many different four letter words can be formed (the words need not be meaningful) using the letters of the word "MEDITERRANEAN" such that the first letter is E and the last letter is R? A. 59 B. C. 56 D. 23 E. Q5. What is the probability that the position in which the consonants appear remain unchanged when the letters of the word Math are re‐arranged? A. 1/4 B. 1/6 C. 1/3 D. 1/24 E. 1/12 Q6. There are 6 boxes numbered 1, 2,....6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is A. 5 B. 21 C. 33 D. 60 E. 6 Q7. In how many ways can 5 letters be posted in 3 post boxes, if any number of letters can be Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 59
posted in all of the three post boxes? A. 5 C 3 B. 5 P 3 C. 53 D. 35 E.
25 Q8. Ten coins are tossed simultaneously. In how many of the outcomes will the third coin turn up a head? A. 210 B. 29 C. 3 * 28 D. 3 * 29 E. None of these Q9. In how many ways can the letters of the word "PROBLEM" be rearranged to make 7 letter words such that none of the letters repeat? A. 7! B. 7C7 C. 77 D. 49 E. None of these Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 60
Set Theory Set A set is a well‐defined collection of elements. Ex.1: A = {Ganga, Yamuna, Saraswathi, Kaveri} Well‐defined  rivers Ex. 2: B = {2, 4, 5, 6, 8….} Well‐defined  even integers Remark: In Ex.1, we have a finite set and in example Ex.2, we have an infinite set. Subset: A set A is called a subset of B, if all the elements of A are also the elements of B. we writs A B Ex. Let A = {4, 6, 8} and B = {2, 4, 6, 8} then A B Equal sets: Two sets A and B are said to be equal if all the elements in A are elements in B and vice‐versa. Notation: A = B Ex. If A = {2, 4, 8}, B = {4, 2, 8} Then A = B Note that order is not important. Universal set: In any discussion on sets, there exists a very large set which is such that all the sets under discussion are subsets of this large set. Such a large set is called Universal set and is denoted by U. Singleton set: A set which contains only one element is called a singleton set. Ex. A = {a} B = {2} Null set or Empty set: A set with no elements is called a null set or empty set. A null set is denoted by {} (phi) Union of two sets: The union of two sets A and B is the set consisting of all the elements which belong to A or B or both A and B. It is denoted by A  B Ex: A = {1, 2, 3} B ={3, 4, 5, 6} A  B = {1, 2, 3, 4, 5, 6} Intersection of two sets: The intersection of two sets A and B is the set consisting of all the elements which belong to both A and B. It is denoted by A  B A  B = {3} Ex: A = {1, 2, 3} B = {3, 4, 5, 6} Note: Union  or Intersection  and Difference of two sets: Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 61
The diffeerence of ttwo sets A and B is th
he set of alll elements which are in A but no
ot in B. It iss denoted
d by A – B Ex: A = {1, 3, 5, 7} B = {0, 1, 2, 3, 8}
A – B = {5, 7} Observee that B – A = {0, 2, 8} Disjoint sets: } Two setss A and B arre said to be disjoint, it A  B = {
Ex: A = {1, 2, 3} A  B = {
} B = {4, 5, 6} ment of a sset: Complem
The com
mplement of o a set A is i the set of o all thosee elements of universaal set U exxcluding thee elementts of A. Ex: U = {{0, 1, 2, 3, 4
4, 5, 6, 7, 8, 9} A = {1, 3, 5, 7}
A’ = {0, 2, 4, 6, 8, 9} Represen
ntation: Diagrammatic R
1. U U  Rectangle  Un
niversal set A  circlle  set 2. A  B (shadeed portion A  B) 3. A  B (shaded portion A 
 B) Apphelp GMAT coach
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2
4. A’ (shaded portion A’) Laws of set theory C
Law 1. Cumulative A  B = B 
 A and A  B = B  A A
2. Associative Law A  (B  C
C) = (A  B))  C  (B  C) == (A  B)  C and A 
D
e Law 3. Distributive
A  (B  A
C) = (A  B)  (A 
 C) and A  (B
B  C) = (A 
 B)  (A  C) et Identitie
es Other Se
1.Inttersection aand Union with universal set: A 
 U = A  U = U and A 
ouble Comp
plement Law
w: 2.Do
c c
(A ) = A empotent LLaws: 3.Ide
A 
 A = A A
A = A and A  e Morgan’s Laws: 4.De
(A 
B)c = Ac 
 Bc and (A  B)c = Ac  Bc 5.Ab
bsorption Laaws: A 
 (A  B) = A A  B) = A and A  (A
6.Altternate Rep
presentatio
on for Diffe
erence: A –– B = A  Bc 7.Inttersection aand Union with a subsset: iff A  B, then A  B = A
A Apphelp GMAT coach
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A  B = B Set Theory‐ Questions Q1. In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects? A. 19 B. 41 C. 21 D. 57 E. 26 Q2. Of the 200 candidates who were interviewed for a position at a call center, 100 had a two‐wheeler, 70 had a credit card and 140 had a mobile phone. 40 of them had both, a two‐wheeler and a credit card, 30 had both, a credit card and a mobile phone and 60 had both, a two wheeler and mobile phone and 10 had all three. How many candidates had none of the three A. 0 B. 20 C. 10 D. 18 E. 25 Q3. In a class of 40 students, 12 enrolled for both English and German. 22 enrolled for German. If the students of the class enrolled for at least one of the two subjects, then how many students enrolled for only English and not German? A. 30 B. 10 C. 18 D. 28 E. 32 Q4. In a class 40% of the students enrolled for Math and 70% enrolled for Economics. If 15% of the students enrolled for both Math and Economics, what % of the students of the class did not enroll for either of the two subjects? A. 5% B. 15% C. 0% D. 25% E. None of these Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 64
Exponential Equations Definitions Base ‐ the number that is multiplied by itself a certain quantity of times. Ex: in the expression 23, the 2 is the base. Exponent ‐ the number of times a quantity is multiplied by itself. Ex: in the expression 23, the 3 is the exponent. Exponents signify repeated self‐multiplication. E.g.,: 23 = 2*2*2 Radical ‐ the sign used to denote the square or nth root of a number. Ex: the value of "radical 4" is 2 and the value of "radical 9" is 3. Exponential Expression ‐ an expression or term with a power or exponent that is not one. Ex: x2 is an exponential expression while x is not an exponential expression. Similarly, x1/2 (called the square root of x) is an exponential expression while 2x is not an exponential expression. Exponential Equation ‐ an equation with a term that has an exponent greater than one. Ex: x3/2 + 2x + 1 is an exponential expression while 2x + 3 is not an exponential expression. Similarly, x3 = 27 is an exponential equation while x + 2 = 29 is not an exponential equation. Exponents are a shorthand way of representing repeated multiplication. Consider the following examples, which are all exponential equations because a term is multiplied by itself multiple times: x3 = x*x*x 8 = 2*2*2 = 23 (‐2)3 = ‐8 = (‐2)(‐2)(‐2) Laws of Exponents There are many laws of exponents that should be memorized and practiced in order to be thoroughly understood. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 65
Solving Exponential Equations Exponential Equations & the Number of Solutions One property of exponential equations that is initially confusing to some students is determining how many solutions an equation will have. 1. Exponential equations with one term and an even power will have up to 2 solutions. x4=‐2 has no solutions and x4=16 has solutions x=‐2 and x=2 2. Exponential equations with one term and an odd power will have exactly one solution. x3=27 has one solution x=3. Notice x=‐3 is not a solution as (‐3)*(‐3)*(‐3) = ‐27 3. Exponential equations with multiple terms and both even and odd exponents can have many solutions. x3 ‐ 7x + 6 = 0 has three solutions: x=1, 2, ‐3 An even exponent hides the sign of its roots (e.g., x2=4; x = 2 and x = ‐2). Techniques for Solving Exponential Equations As noted above, an exponential equation has one or more terms with a base that is raised to a power that is not 1. While there is no formula for solving an exponential equation, the following examples provide some insight into common techniques used in finding the unknown value in an exponential equation. Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 66
Techniqu
ue 1: Isolatte and Raise
e to the Invverse Expon
nent Arrange the term with w an exp
ponent on one side of o the equation and th
he other teerms on thee de of the eq
quation. Raise both sid
des of the eequation to the inversee exponent. 3x4 + 6
6 other sid
= 54 e x4 term byy subtracting 6 from bo
oth sides an
nd then diviiding both ssides by 3.
Work to isolate the
4
4
3x = 48 and x = 16 o the 4/1 po
ower, raise both sidess to the invverse power In orderr to isolate x, since x iis raised to
(i.e., 1/4
4). Techniqu
ue 2: Solve Through Factoring Isolatingg an expone
ent often m
makes solvin
ng an equation easier.
Arrange all similar termss on one side of o the equal e
sign
n and th
hen factorr. by 2, which
h is a comm
mon factor, and then ssubtract thee number o
on the right Divide eeach term b
side of the equation. Using factoring rule
es, simplify and solve tthe exponential equation. ents Multiplyying Expone
The form
mulas abovve can be used to multiply togeth
her exponeents in ordeer to solve eexponentiaal equation
ns. The mosst importan
nt formulas to use are the following: Dividingg Exponentss The aforrementione
ed formulass are helpfu
ul in dividing exponentts, especially these tw
wo exponent formulass. Apphelp GMAT coach
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Exponential Equations‐ Questions Q1. Solve for x, x5‐32 = 0 A. 3 B. 2 C. 7 D. 5 E. 4 Q2. Solve for x, (82x)2 = 64x‐2 A. ‐3 B. 5 C. ‐1 D. 4 E. ‐2 Q3. Solve for x: A. x=6 or x=‐6 B. x=6 C. x=5 or x=‐5 D. x=4 E. x=4 or x=‐4 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 68
Inequaalities Definitio
ons An inequ
uality comp
pares two sttatements w
with differeent values.
Inequaliity ‐ A com
mparison off two values or expresssions. For example, x x > 10 is an
n inequalityy whereass x = 10 is an equation. n ‐ A statem
ment declarring the equ
uality of two expressio
ons. For exaample, 5x + 5 = 30 is an
n Equation
equation
n whereas 5
5x + 5 < 30 is an inequality. pes of Inequ
ualities Five Typ
Greater Than (>)‐ W
With a greaater than in
nequality, alll numbers larger than
n x (but nott equal to xx) nequality. Fo
or examplee, any numb
ber larger th
han 7 fits th
he above in
nequality (e.g., 7.1, 7.2
2, fit the in
8, 9, 10, ...). Less Thaan (<)‐ With
h a less than inequalityy, all numbeers smaller than x (butt not equal to x) fit thee inequality. For exam
mple, any n
number smaaller than 7
7 fits the above inequaality (e.g., 6.9, 6.8, 6, 5
5, 00, ...) 0, ‐1, ‐10
Greater Than or Eq
qual to (>)‐ With a grreater than or equal tto inequalitty, all numb
bers greater than x (aand equal tto x) fit thee inequalityy. For examp
ple, any number that is 7 or is laarger than 7
7 fits the aabove inequ
uality (e.g., 7, 7.1, 7.2,, 8, ...). Less Thaan or Equal to (<)‐ W
With a less than or equ
ual to inequ
uality, all nu
umbers smaller than xx (and equ
ual to x) fitt the inequality. Exam
mple: any nu
umber thatt is 7 or is ssmaller thaan 7 fits thee above in
nequality (e
e.g., 7, 6.9, 6
6, 0, ‐1, ‐10
00, ...). Not Equ
ual to (≠) Apphelp GMAT coach
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With a n
not equal to
o inequalityy, any numb
ber that is n
not equal to
o the number in questiion (7 in this casee) fits the in
nequality. For examplee, any numb
ber that is n
not 7 fits the above ineequality (e.g., 7.1
1, 6.9, 6, 8, 10, 0, ‐100,, ...). Operatio
ons With In
nequalities When one action (e.g., addition, multiplication, sub
btraction, o
or division) is done to one side of quality, the same action must be done to the other side of thee inequalityy. Howeverr, the ineq
there is one major catch: if bo
oth sides off an inequaality are mu
ultiplied or divided byy a negativee r, then the inequality ssign must b
be flipped.
number,
If x+2 < y+2 is multtiplied by ‐1
1 on both ssides, the result is nott ‐2‐x < ‐y‐2
2. Instead, tthe result iss ‐2‐x > ‐y‐2. by a negativve number fflips the ineequality sign, you cann
not multiplyy Since divviding or multiplying b
or dividee by an unknown (i.e.., a variablee), as it cou
uld be negaative. This iis a commo
on trap. For examplee: xz < 10z cannot be solved by d
dividing botth sides by zz to get x < 10. uality woulld end up as x > 10. Unless a problem sttates that aa If z werre negative, the inequ
variable is positive or negativve, both sid
des cannot be divided or multiplied by an u
unknown ass not be certaain whetheer to flip thee inequalityy sign. you cann
e Inequalitiies Multiple
Just as it is possible to solvve two sim
multaneous equations, so it is possible p
to
o solve two
o ur, etc.). In solving mu
ultiple simu
ultaneous inequalitiess, the most inequalities (or thrree, or fou
o solve each
h inequalityy separatelyy and then ccombine th
hem. important part is to
ng With Ine
equalities: A
Absolute V
Value Operatin
Just as ttraditional e
equations w
with absolu
ute value teerm within tthem usually have two solutionss, so inequ
ualities with
h an absolu
ute value teerm within them typiccally have ttwo solutions. In order to solvee an inequ
uality with an absolu
ute value term, t
isolaate the absolute valu
ue, set thee expression within tthe absolute value braackets equal to a positive and neggative valuee, and solvee unknown vaariable. for the u
ng With Ine
equalities: EExponents
Operatin
In dealing with in
nequalities that invollve exponeents, thesee inequalities behave much likee nal equatio
ons. Inequaalities with an even exponent e
u
usually havee two solu
utions whilee tradition
inequalities with an
n odd expon
nent have o
one solution
n. ponents Odd Exp
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0
An inequality with an odd exponent behaves exactly like an inequality without an exponent or a traditional equation with an odd exponent. The one word of caution‐‐which also applies to inequalities without exponents‐‐is that you must know the sign of a variable before you can divide or multiply by it Even Exponents As stated above, an inequality with an even exponent typically has two solutions. The reason for this is that x can be either positive or negative. Consequently, when evaluating an even exponent within an inequality, we deal with two cases: x is positive, x is negative. Inequalities‐ Questions Q1. Solve for x: 4 ‐ 2x<10 A. x<3 B. x<‐5 C. x>‐3 D. x>5 E. x<6 Q2. What is the range of possible x values given: 2x+8<20 5x>15 A. 2<x<7 B. x>6 and x<0 C. x>10 and x<‐2 D. 3<x<6 E. x>3 and x<1 Q3. x2 + 9 < 34 A. x<6 and x>‐6 B. x<5 and x>‐5 C. x>5 and x<‐5 D. x>6 and x<‐6 E. x<3 and x>‐3 Q4. Which of the following inequalities have a finite range of values of "x" satisfying them? A. x2 + 5x + 6 > 0 B. |x + 2| > 4 C. 9x ‐ 7 < 3x + 14 D. x2 ‐ 4x + 3 < 0 E. (B) and (D) Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 71
Q5. For what range of values of 'x' will the inequality 15x – 2/x > 1? A. x > 0.4 B. x < 1/3 C. ‐ 1/3 < x < 0.4, x >15/2 D. ‐ 1/3< x < 0,x >2/5 E. x < ‐ 1/3 and x >2/5 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 72
Geome
etry‐ Anggles Definations Angle ‐ A
A figure forrmed by two
o lines sharring a comm
mon endpoiint. Ray ‐ A sstraight line
e extendingg from a point. Acute Angle ‐ An an
ngle less than 90° and greater thaan 0° Angle ‐ An angle betw
ween 90° and 180° Obtuse A
Right An
ngle ‐ An an
ngle that is eexactly 90° Complem
mentary An
ngles ‐ Two
o angles who
ose sum is 90° Supplem
mentary Angles ‐ Two aangles who
ose sum is 1
180° Adjacen
nt Angles ‐ TTwo angles that share a side (i.e., are adjacent). Angle Biisector ‐ A rray that divvides an anggle into two
o equal partts. Vertical Angles ‐ Th
he two non‐‐adjacent angles that sshare a verttex when tw
wo lines inttersect. Angles ‐ W
When two paarallel liness are cut byy a transverrsal, the two angles on
n Alternatte Interior A
oppositee sides of th
he transverssal and betw
ween the p
parallel liness . Alternatte Exterior Angles ‐ W
When two parallel liness are cut byy a transverrsal, the tw
wo angles on
n oppositee sides of th
he transverssal and outside the parallel lines . Types off Angles Acute Angle one whose measuremeent is less tthan 90°. A
Angle A, sho
own below, is an acutee An acutee angle is o
angle. Obtuse A
Angle An obtu
use angle iss one whose measurement is greeater than 90° and less than 180
0°. Angle B
B, shown b
below, is an obtuse anggle. Right An
ngle A right aangle is one
e whose meeasurementt is exactly 9
90°. Angle C
C, shown beelow, is a rigght angle.
Note: Th
he lines thaat surround
d the letter C are the fformal wayy of denotin
ng a right aangle. If you
u see this symbol, yo
ou know thaat you are d
dealing with a right an
ngle. You caannot assum
me an anglee is right ssimply becaause it appeears to be riight. Apphelp GMAT coach
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Interseccting Lines When tw
wo lines intersect, theyy form fourr angles. An
ngles that arre not adjaccent (i.e., d
do not sharee a comm
mon side) are called vertical an
ngles and these anggles are co
ongruent (o
or equal in
n measureement). Due to tthe properties of interssecting linees, vertical aangles are ccongruent. Consequen
ntly, angle D
D = angle FF and angle
e G = angle E. by a Transve
ersal Parallel Lines Cut b
wo parallell lines are cut by a trransversal (i.e., (
a third
d line interrsects the two t
paralleel When tw
lines), a number of relationshiips exist bettween the rresulting an
ngles. Alternatte Interior A
Angles Are EEqual: K = LL; O = J Alternatte Exterior A
Angles Are Equal: H = M
M; N = I Correspo
onding Anggles Are Equ
ual: K = N; J = M; H = O
O; I = L Non‐Alteernate Interior Angles Are Suppleementary: LL + J = 180; K + O = 180
0 It is imp
portant to n
note that if f two lines ccut by a traansversal have any of the above propertiess, then thee two lines must be paarallel. For eexample, if alternate in
nterior angles are equal, then thee two lines cut by a transversal m
must be parallel. d plane figu
ure formed by three orr more line segments, called the sides of thee Polygon: is a closed
polygon. Angles Sum of A
The sum
m of the intterior anglees of a polyygon is 180
0*(n‐2) wheere n is thee number of o sides. For Apphelp GMAT coach
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4
examplee, the sum of the interior angles of a trianggle is 180 == (180)(3‐2)) while the sum of thee interior angles of a square is 3
360 = (180)((4‐2). Geome
etry‐ Anggles ‐ Qu
uestions
Q1. A.
B.
C.
D.
E.
20° 55° 35° 125° 45° ollowing deepicts line l parallel to lline m? Q2. Which of the fo
A.
B.
C.
D.
E.
A A
B B C C A
A and C B
B and C Geome
etry‐ Areaa Apphelp GMAT coach
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Area e refers to tthe space co
ontained in
nside of the lines creating the shape. The areaa of a shape
e Triangle
The areaa, A, of a trriangle can be found u
using the eequation A = .5*B*H w
where B staands for thee base of tthe triangle
e and H stands for the height. Anyy side can b
be chosen tto be the baase, but thee height iss the line th
hat is perpendicular to the base and goes thrrough the o
opposing veertex. 1
Area = /2(Base)(He
eight) 1
Area = /2(B)(H) Circle point (i.e., th
he center). The area, A
A, of a circlee In a circlle, each poiint is equidistant from a central p
2
can be found f
using the equaation A = π*R
π wheree R is the radius r
of th
he circle an
nd π is pi == 3.1415.... π*(radius)2 A
Area = π*R2 2
Area = π
Square ery side is eequal (2) th
he opposingg sides are parallel (3) all the anggles are 90°°. In a square: (1) eve
quare can b
be found ussing the equ
uation A = aa2, where aa is the lenggth of a sidee The areaa, A, of a sq
of the square. s
Nottice that th
he equation
n for the area of a sq
quare is a specific s
verrsion of thee rectangle equation below. Area = (sside)2 Area = a2 Rectanggle In a rectangle: (1) opposing sides of a rectangle are a equal and a parallel (2) every side is not n length (3)) all of the aangles are 9
90°. The areea, A, of a rrectangle caan be found
d necessarily equal in
n A = l*w. In
n this case,, l stands fo
or the lengtth and w sttands for th
he width. It using the equation
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does nott matter wh
hich sides aare labeled tthe length o
or the widtth. Area = (llength)(wid
dth) Area = lw
w Parallelo
ogram In a paraallelogram, opposing sides are p
parallel and equal in leength. Oppo
osing anglees are equaal but not necessarilyy 90°. The area, a
A, of a parallelo
ogram can be found using u
the eq
quation A == b stands forr the base of the paraallelogram and h stands for the height. Anyy b*h. In tthis case, b
side can
n be chosen
n to be the base, but tthe height is the line tthat is perp
pendicular to the basee and opp
posing line. Area = (b
base)(heigh
ht) Area = b
bh Trapezo
oid In a trap
pezoid, one set of opposing sidess is parallel,, but not neecessarily eequal. The aarea, A, of aa trapezoid can be fo
ound using tthe equatio
on A = .5*(aa+c)*h. In th
his case, h sstands for tthe line that des a and c. is perpendicular to parallel sid
5*(Base1+Base2)*h Area = .5
Area = .5
5*(a+c)*h Rhombu
us In a rhom
mbus: (1) aall sides are equal in leength (2) op
pposite sidees are parallel (3) diago
onals bisect each oth
her (4) the iintersection
n of diagonaals forms a 90° angle.
Area = .5
5*(Diagonal1*Diagonal2) Apphelp GMAT coach
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Geome
etry‐ Areaa ‐ Qu
uestion Q1. A circle with a radius of 6 has a squ
uare inscrib
bed inside o
of it. What iis the area o
of the blackk region? A.
B.
C.
D.
E.
23π – 31 36π – 72 63π – 42 24π – 16 30π ‐ 62 Geome
etry‐ Circles Definations Circle ‐ A
An infinite sset of pointts that are aall equidistaant from a p
point called
d the centerr. erence ‐ Th
he edge (or boundary) of a circle. Circumfe
m the center of a circlee to the edgge of the cirrcle. Radius ‐‐ A line from
er ‐ A line that passes through the center off the circle aand has its endpoints at the edgee Diamete
of the circle. dge of the ccircle. Chord ‐ A line that lies within tthe circle and connectts two pointts on the ed
Angle ‐ An angle whosse vertex is the center of the circle. Central A
Arc ‐ A p
portion of th
he circumfeerence of a circle (i.e., a segment of the edgee of a circlee). of a Circle ‐ A slice (or sliver) of th
he circle en
nclosed on o
one side byy the edge o
of the circlee Sector o
and on tthe other sides by radii. Apphelp GMAT coach
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o a Circle ‐ The poin
nt in a circle from which all poin
nts on the edge of the circle aree Center of equidistant. nds from the outside) a polygon
n Circumscribed Circcle ‐ A circlle that enccompasses (or surroun
he polygon.. such thaat the circle's circumference intersects the veertices of th
An inscribed
d circle is aa circle thatt lies insidee a figure su
uch that po
oints on thee Inscribed Circle ‐ A
ure. edge of the circle are tangent to the sides of the figu
w the same center point but not necesssarily the same s
radiuss Concenttric Circles ‐ Circles with length. Propertiies of a Circcle A circle is formed b
by an infinitte number o
of points th
hat are equidistant from a center.. The length
h quidistant p
points from
m the centerr is the radius, r. of the eq
es Formulaas for Circle
2 Area = π
πr
Diameteer, d = 2r Circumfeerence = 2π
πr = πd Central A
Angle = 2(In
nscribed An
ngle) x
2 Area of SSector = ( /360)πr
Length o
of an Arc = ((x/360)Circum
mference = (x/360)2πr Arcs & SSectors Arcs ween A and
d An arc iss a portion of the edgee of a circlee. For example, the portion of thee circle betw
B is calleed arc AB. TThe length of the arc is proportional to thee central an
ngle that forms the arcc (i.e., anggle ACB in this examplee). Sectors Apphelp GMAT coach
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A sectorr is a portion of the areea of a circle. For exam
mple, the po
ortion of the circle bettween center C
C and arc AB
B forms a seector (this ssector is shaaded in black below). TThe area off the sectorr is propo
ortional to the central aangle that fforms it. Inscribed Angle & C
Central Anggle A centraal angle is an angle whose vertex is the centeer of the cirrcle and wh
hose endpo
oints are thee edge of the circle. A
Angle ACB is a central angle. An iinscribed angle is an aangle whosee vertex liess e
of the
e circle and
d whose en
ndpoints liee on anotheer part of the t edge of the circlee. on the edge Angle AD
DB and anggle AEB are both examp
ples of inscribed anglees. or a central angle aand an inscrribed angle with the saame endpoints: oints are equal All inscriibed angless with the same endpo
ADB = AEB d Angle = (1/2)(Central Angle) Inscribed
ADB = (1/2)ACB Inscribed Triangle e equality mentioned above bettween an inscribed angle a
and half of thee As a ressult of the
measureement of a central anggle, the follo
owing prop
perty holds true: if a trriangle is inscribed in aa circle su
uch that on
ne side of that trianggle is a diameter of the t circle, then t
the angle of thee triangle that is opposite the diiameter is aa right anglee. Apphelp GMAT coach
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The reasson that an
ngle ADB is a right anggle is becau
use central angle ACB is 180° and
d ADB is an
n inscribed
d angle who
ose endpoints are the same as AC
CB. Consequ
uently: 1
ADB = /2(ACB) ADB = 1/2(180) = 90
0 For the above to hold true: (1
1) C must be the centeer of the cirrcle (2) AB must be a d
diameter of the centter Inscribed Circles & Circumscriibed Circless Inscribed Circle of the circlee An inscrribed circle is a circle tthat lies insside a figurre such thatt points on the edge o
are tanggent to the ssides of thee figure. Circumscribed Circles mscribed cirrcle is a circle that enco
ompasses aa polygon such that the circle tou
uches all thee A circum
vertices of the polyygon. Geome
etry‐ Circles ‐ Qu
uestion Q1. Find the are
F
ea of the bllack region. Assume th
hat the basse of the triangle is a d
diameter of the circle and the radius of thee circle is 12
2.5 A.
B.
C.
D.
E.
121π‐160 0 156.25π‐150
5 120.25π‐145
2 152.75π‐132
133.5π‐125 Geome
etry‐ Line
es Definitio
ons Line ‐ A series of co
onnected po
oints that eextends out infinitely. Apphelp GMAT coach
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Line Seggment ‐ A portion of a line. Parallel Lines ‐ Two
o or more lines that neever intersect. dicular Line
es ‐ Two linees whose intersection fforms a 90°° angle. Perpend
Line Bassics In the world w
of ge
eometry, a line refers to a straigght line thaat extends out infinittely in both
h direction
ns. For anyy two poin
nts, there iss one and only one line that connects c
th
hem. A linee segmentt is a portio
on of a line ffrom one point to ano
other. For exxample: m, sometimes denoted
d nm. The p
portion of th
he line from
m n to m is aa Line l can be called the line nm
ment. nm can also den
note the line segment as well as tthe length o
of the line segment. line segm
Parallel Lines me plane are said to b
be parallel iff they neveer cross or in
ntersect. In
n coordinatee Two linees in the sam
geometrry, parallel llines have tthe same slo
ope but diff
fferent x‐inttercepts. Perpend
dicular Line
es Two linees are perpe
endicular w
when they in
ntersect and
d form 90° angles. Geome
etry‐ Periimeter The Con
ncept of Perrimeter The perimeter of an a object refers to the total lenggth of the lines formiing the outtside of thee Apphelp GMAT coach
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shape. In order to calculate the perimetter of an ob
bject, add u
up the lenggth of each side of thee object. Circle nd using the equation C = 2πr, wh
here r is thee The periimeter or ciircumferencce, C, of a ccircle is foun
radius. mference = 2
2πr where rr is the radius [CE or CII or CD] Perimeteer = Circum
Perimeteer = Circum
mference = π
πd where d is the diam
meter [DI] e Triangle
The periimeter of a triangle is the sum off its three sides. It should be noteed that the sum of anyy two sidees of a trian
ngle must bee larger thaan the lengtth of the third side. Quadrilaateral The periimeter, P, o
of a quadrilaateral is thee sum of th
he four sidees. There are shortcutts to findingg the perimeter for ssome quadrrilaterals. Fo
or squares,, P = 4*e where e is th
he length off a side. For p
ams, P = 2xx + 2y wherre x and y are the len
ngths of thee rectangles, P = 2w + 2l. For parallelogra
parallel sides. Square *e where e is the lengtth of a side PerimeteerSquare = 4*
ogram Parallelo
PerimeteerParallelogram = 2x + 2y w
where x and
d y are the llengths of tthe parallel sides Note: In a parallelo
ogram, oppo
osite sides h
have the same length. Apphelp GMAT coach
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hapes Other Sh
In some cases, the figure doess not fit into
o any specific geometric shape. In these casses, one can
n nized geom
metric figurees. find the perimeter by splittingg the shape into recogn
eter of figurre ABCDE, shown abovve, if AC = 4,, BC = 3, ED
D = 2.5, AE == 3, DC = 3
What is the perime
er of the traapezoid: ED
D + AE + DC
The outeer perimete
2.5 + 3 ++ 3 = 8.5 er of the triangle: The outeer perimete
In orderr to find th
his, one must first fin
nd the lenggth of AB, which is 5 5 by the Pythagorean
P
n theorem
m. Outside perimeter of a triangle = AB + BC
C (do not co
ount AC sincce AC is on the interiorr). AB + BC = 5 + 3 = 8 he triangle
Total Perimeter = outer perimeter of the trapezoid ++ outer periimeter of th
Total Perimeter = 8.5 + 8 = 16.5 Geome
etry‐ Periimeter ‐ Question Q1. If square ABC
CD is inscribed insidee the circle and DB an
nd AC are both b
diameeters of thee circle, w
what is the ccircumferen
nce of the circle if the p
perimeter o
of the squarre is 4(200)(1/2)? A.
B.
C.
D.
E.
25π 20π 5π 10π 12π Geome
etry‐ Pyth
hagorean
n Theorem
m The Pyth
hagorean TTheorem Apphelp GMAT coach
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A right ttriangle hass one angle that is 90° (i.e., a righ
ht angle). Th
he side opp
posite the riight angle iss called th
he hypotenu
use, often labeled c, w
while the otther two sid
des of the triangle (leggs) are often
n labeled a and b. For a rigght triangle with legs a and b and hypotenuse c: a2 + b2 = c2 Note: Th
he Pythagorrean Theoreem only wo
orks for righ
ht triangles.. Note: When W
takingg the squarre root, normally one must takee the positive and neggative valuee (i.e., ±5)). However,
r, since a neegative length has no
o meaning when w
solving for the length of aa side, thee negative vvalue is igno
ored. Conversse If a2 + b2 = c2, then the trianglee is right In otherr words, if aa2 + b2 = c2, a triangle ,
exists with
h sides a, b and c such
h that a righ
ht angle liess between
n the sides of length a and b orean Triple
es Pythago
There are three co
ommon relations betw
ween the sides s
of a right trianggle. Recogn
nizing thesee n side len
ngths can save conssiderable calculation c
time. Ratther than using thee common
Pythago
orean theorrem to calculate the missing siide length, the length of the side s
can bee determined by notiicing the paattern. + 42 = 52 3‐4‐5 triangles: 32 +
9‐12‐15, an
nd 12‐16‐20
0 triangles aare simply m
multiples off the 3‐4‐5 rrule. 6‐8‐10, 9
orean Triple
es Pythago
5‐12‐13 triangles: 5
5 2 + 12 2 = 1
13 2 6 is another common w
way for thiss ratio to ap
ppear. 10‐24‐26
72 8‐15‐17 triangles:82 + 15 2 = 17
wing example: Consider the follow
If a rightt triangle haas legs 10 and 24, whaat is the length of the tthird side? ming the caalculation to
o find the third side, recognize that this triangle t
is aa Instead of perform
13 pattern
n. multiplee of 5‐12‐13, wheree each side of the triangle iss double the 5‐12‐1
Consequ
uently, the tthird side iss 2(13) = 26
6. Special Right Trriangles Apphelp GMAT coach
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Isosceles Right Triaangle: 45°‐4
45°‐90° As the above diagraam indicatees, a triangle with anglles 45°‐45°‐‐90° will havve sides in tthe ratio off 1‐1‐21/2 Bisected
d Equilateraal Triangle: 30°‐60°‐90
0° As the above diagraam indicatees, a triangle with anglles 30°‐60°‐‐90° will havve sides in tthe ratio off 1‐31/2‐2 etry‐ Pyth
hagorean
n Theorem
m ‐Question Geome
Q1. What is the a
W
area of the ttriangle BCD? Apphelp GMAT coach
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6
A.
B.
C.
D.
E.
40 42 44 46 48 Geome
etry‐ Quaadrilateraals A quadrrilateral is aa four‐sided
d polygon. TThe five main quadrilateral types (listed in order from
m most to least symm
metric) are:: squares, rrectangles, parallelograams, rhombuses, and trapezoidss. pe of quadrrilateral has its own d
defining feaatures and equations tto determin
ne the areaa Each typ
and the perimeter. The interio
or angles for all quadrillaterals must sum to 3
360°. Square All sidess are equal iin length & Opposite ssides are paarallel All anglees are right (i.e., 90°) Perimeteer = 4(side)) Area = (sside)2 Rectanggle Oppositee sides are equal in len
ngth & Opp
posite sides are paralleel All anglees are right (i.e., 90°) Perimeteer = 2(width) + 2(lengtth) Area = (llength)(wid
dth) Parallelo
ogram Apphelp GMAT coach
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Oppositee sides are equal in len
ngth & Op
pposite sidees are paralllel Oppositee angles are
e equal (i.e., bottom‐right = uppeer‐left) Adjacent angles are
e supplemeentary Perimeteer = 2(width) + 2(lengtth) Area = (w
width)(heigght) Rhombu
us All sidess are equal iin length Adjacent angles are
e supplemeentary e equal Oppositee angles are
Diagonals form righ
ht angles Diagonals bisect eaach other nd angles Diagonals bisect en
qual, all anggles are nott necessarilyy equal. Note: In a rhombuss, while all ssides are eq
Perimeteer = 4(side)) Area = 1/2(diagonal1 * diagonaal2) Trapezoid One pairr of sides is parallel Perimeteer = A + B +
+ C + D 1
Area = /2(height)(b
base1 + basee2) Apphelp GMAT coach
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Geome
etry‐ Quaadrilateraals ‐ Q
Question
Q1. In the diagram below, if f ABCD is a ssquare who
ose area is 100, CDEF iis a parallellogram, and
d the perimeter of AB
BCFED is 80
0, what is th
he length off CF? A.
B.
C.
D.
E.
10 20 15 22 30 Geome
etry‐ Triangles Propertiies of All Trriangles A triangle is a three
e‐sided shaape whose tthree innerr angles mu
ust sum to 1
180°. The laargest anglee m the longeest side while the smaallest angle will be acrross from tthe shortest will be aacross from
side of the triangle. If and onlyy if two sidees of a trian
ngle are equ
ual, the anggles oppositte them willl be equal as well. Angles is 18
80 Sum of A
The sum
m of the inte
erior angless in any triangle is 180°° e Inequalityy Theorem Triangle
The sum
m of any two
o sides of a triangle mu
ust be greater than thee third sidee of a trianggle. osite Largestt Angle Longest Side Oppo
onversely, tthe smallest The longgest side off a triangle iis opposite the largestt angle of a triangle. Co
side of aa triangle is opposite th
he smallestt angle of a triangle. Exteriorr Angle Theorem Apphelp GMAT coach
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The exteerior angle ttheorem states: Angle EFFG + Angle EGF = Anglee DEG Types Trriangles Right Triiangle One anggle is 90° Scalene Each side has a diffferent lengtth Each anggle has a different meaasurement Obtuse One anggle of the triangle is greeater than 90° Acute ngle of the ttriangle is leess than 90
0° Every an
eral Equilate
Every an
ngle of the ttriangle is eequal (i.e., 6
60°) Every sid
de of the triangle is eq
qual in lengtth Isosceles Two anggles of the ttriangle are equal Two sidees of the triiangle are eequal in lenggth es Similarr Triangle
When A
Are Triangles Similar? Apphelp GMAT coach
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Two triangles are siimilar if anyy one of thee following three possiible scenariios is met:
• AAA
A [Angle Angle A
Anglee] ‐ The co
orrespondin
ng angles of o each triiangle havee the samee meeasurementt. In otherr words, thee above triaangles are similar if: Angle Q; An
ngle M = An
ngle P Angle L = Angle O; Angle N = A
An angle in one trianggle is the saame measu
urement as an angle in
n 2. SASS [Side Anggle Side] ‐ A
thee other trian
ngle and the two sidess containingg these angles have the same ratio. In otherr words, thee above triaangles are similar if: Angle L = Angle O; Side LM/Side OPP = Side LN/Sidee OQ ombination of side, anggle, side alsso proves similarity. Note: Any other co
orrespondin
ng sides havve the samee ratio. 3. SSSS [Side Side Side] ‐ Each pair of co
In otherr words, thee above triaangles are similar if: Side LM
/Sidee OP = Side LN/Side OQ = Side MN/Side PQ Propertiies of Similar Triangless 1. Correspondingg angles are the same m
measuremeent. 2. The perimeter of each triaangle is in tthe same raatio as the ssides. me proporttion. 3. Correspondingg sides are aall in the sam
ent Triangle
es Congrue
Are Triangles Congruen
nt? When A
Two triangles are co
ongruent iff any one off the follow
wing three p
possible scenarios is met: • SASS [Side Anggle Side] ‐ TTwo pairs off corresponnding sides are equal aand the corrrespondingg anggle between
n the sides is equal. In otherr words, thee above triaangles are ccongruent iff: W = Side UV; Angle W = Angle V; Siide WR = Side VT Side SW
ombination of side, anggle, side alsso proves co
ongruence. Note: Any other co
A [Angle Side S
Anglee] ‐ Two pairs of correspond
c
ing angless are equaal and thee 2. ASA
corrrespondingg side betw
ween them is equal. In otherr words, thee above triaangles are ccongruent iff: Angle R = Angle T; SSide RW = SSide TV; Angle W = Angle V ombination of angle, siide, angle aalso proves congruencee. Note: Any other co
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1
onding sides are equal. 3. SSSS [Side Side Side] ‐ All tthree pairs of correspo
In otherr words, thee above triaangles are ccongruent iff: Side RS = Side TU; == Side RW == Side TV; SSide SW = Siide UV ngles Propertiies of Conggruent Trian
1. Correspondi
C
ing angles h
have the same measurrement. 2. Correspondi
C
ing sides haave the sam
me measurement. Area of a Triangle owing formula: The areaa of a trianggle is given by the follo
1
Area = /2(base)(he
eight) The heigght of the ttriangle is the length o
of the line w
which is perpendicularr to the basse and goess through the opposiite vertex (i.e., line KH in the trian
ngle below)). To reiterrate, the arrea of a triaangle can b
be found ussing the eq
quation: A == 1/2bh. In this case, b
b stands fo
or the base
e of the triangle and h stands for tthe height. Any side caan be choseen to be thee base, bu
ut the heigh
ht is the linee that is perrpendicularr to the basse and goess through th
he opposingg vertex. TThe perimeter of a triaangle is the sum of thee three sides. Consider the follow
wing example: ove, what iss the area o
of triangle K
KIJ if KJ = 13
3, KH = 5, IH
H Referring to the triaangle immeediately abo
= 3, and JI = 9? Area = 1/2(JI)(HK) Area = 1/2(9)(5) = 22
2.5 Geome
etry‐ Triangles ‐ Questiions Q1. If a cube hass a volume of 1000, w
what is the length of aa diagonal that goes tthrough thee of the cube?? center o
A. 15*(7)(1/2) Apphelp GMAT coach
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B.
C.
D.
E.
16*(6)(1/2) 20*(2)(1/2) 12*(5)(1/2) 10*(3)(1/2) W
value of x? Q2. What is the v
A.
B.
C.
D.
E.
6 7 8 9 e Determineed Itt Cannot Be
Geome
etry‐ Volu
ume Rectanggular Solids A rectan
ngular solid is a three‐d
dimensionaal object composed of rectangulaar sides. Eacch of the sixx rectangu
ular sides iss a face. Thee solid and dotted linees are the eedges. Oppo
osing faces are paralleel and have the same
e length and
d width. If all of the six sides aree of equal length, the rectangular be. solid is ccalled a cub
Volume = (length)(w
width)(heigght) = lw
wh Surface Area h) + 2(heigh
ht)(width) ++ 2(length)(height) Surface Area = 2(length)(width
SA = 2lw
w + 2hw + 2lh Apphelp GMAT coach
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Right Cyylinders A right circular cylinder c
has two circcular basess of equal size who
ose centerss are both
h dicularly inttersected by b the samee line. In th
he diagram
m below, thee surface with w point P
P perpend
and thee surface with w
point Q are both the same size. Surrface P and
d surface Q Q are both
h perpend
dicularly intersected byy line PQ, m
making thesse surfaces parallel. Volume = (area of b
base) * (heiight) 2
ove diagram
m V = πr h where r is the radius, which is QR in the abo
e volume is equal to th
he area of th
he base mu
ultiplied by tthe height.
In other words, the
urface Areaa Su
Surface Area = 2(pi)(radius)2 ++ 2(pi)(radiu
us)(height)
2
πr ) + 2πrh SA = 2(π
In other words, the
e surface areea is equal to the summation of tthe area of each of thee faces. Geome
etry‐ Volu
ume ‐ Question
n Q1. If a gas tank iin the shape of a right circular cylinder is 10
0 meters higgh and is beeing filled at 10 minutes, what is thee radius of the base of a rate off 4π gallonss a minute until it is filled after 1
the cylin
nder? A. 2 B. 3 C. 4 D. 5 E. 6 Apphelp GMAT coach
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Coordiinate Geo
ometry Cartesiaan Coordinaate System The following terms are
e used when interactin
ng with coo
ordinate planes: 1. X
X‐Axis ‐ The horizontal line runningg through tthe center o
of the graph
h from left tto right. 2. Y‐Axis ‐ The vertical linee running th
hrough the center of the graph frrom bottom
m to top. O
Paiir ‐ The meeans of ideentifying a point thro
ough its coo
ordinates. The proper 3. Ordered notation is: ((X, Y) wheree (0, 0) is th
he intersecttion of the xx and y‐axiss. e, For example
4) Point A: (2, 4
0) Point E: (0, 0
Point B: (‐7 ,, 5) Point D: (2, ‐‐1) O
he center off the coordinate planee where thee x and y axxis intersect point in th
4. Origin ‐ The (0
0, 0). The origin is poin
nt E in this ggraph. nts
Quadran
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Each coo
ordinate plaane is divided up into four quadraants, labeleed below. (N
Note: Somee graphs only sho
ow one quadrant. In th
his case, thee other quadrants still exist, but they are meerely not shown). In the firrst quadran
nt, both x an
nd y are positive whilee in the seco
ond quadraant x is negaative and y is positivve. The chart below deepicts the sign of x and
d y [denoted (X, Y)]. Quadran
nt I: (+, +) Quadran
nt II: (‐, +) Quadran
nt III: (‐, ‐) Quadran
nt IV: (+, ‐) Slope bove graph,, point A is at (‐8, 4) an
nd point B is at (8, ‐4).
In the ab
Slope Caategories Apphelp GMAT coach
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There arre four type
es of slope: Positive Slope Negativee Slope Slope off Zero Undefined Slope Lines b y = mx+b
In the ab
bove coord
dinate planees, the liness appeared without an
ny explanattion as to w
why the linee pointed in a certaain directio
on at a certain steep
pness. The location and a
slant of o a line iss equation. Itt is the line that graphss out all thee points thaat satisfy this equation
n. determined by an e
his is impo
ortant, it bears b
repeeating: a line on a coordinate c
plane is a graphicaal Since th
represen
ntation of aa series of p
points that ffulfill a mathematical eequation. ndard form in which liinear equattions which
h are graphed appear on a coord
dinate planee The stan
is: y = mx ++ b paces verticcally above or below the x‐axis) y is the yy‐coordinatte (or the number of sp
m is the slope, or th
he degree o
of steepnesss of the linee, as defineed above umber of sp
paces horizontally righ
ht or left fro
om the y‐axxis) x is the xx‐coordinatte (or the nu
b is the y‐interceptt, which is tthe numberr of units ab
bove or bellow the horrizontal axiss where thee line crossses the verrtical axis Horizontal and Verrtical Lines 1. A
A horizontal line can bee written ass y = b sincee for each vvalue on thee line, the y‐coordinate
e will be thee same (reggardless of tthe x‐coord
dinate). Sincce the line d
does not riise when it runs, the slope, m, is 0
0. In the coordinate pllane above,, the light b
blue line can
n be written as: y = 3 Apphelp GMAT coach
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2. A vertical line can be written as x = n since for each value on the line, the x‐coordinate will be the same (regardless of the y‐coordinate). However, since the line does not run when it rises, the slope is ∞/0, which is undefined since you cannot divide by zero. In the coordinate plane above, the red line can be written as: x = 3 Writing the Equation of a Line It is important to know how to take a pair of points (whether from a graph or from a word problem) and write an equation for a line that satisfies the two points. This process involves solving for m and b in the equation y = mx + b. Consider the following example: Find the equation of a line that passes through the points (3, 6) and (‐2, ‐4). The goal is to take these two points and write an equation in the form y = mx + b that passes through the points. • Find m, the slope Slope = rise/run = (6 ‐ [‐4])/(3 ‐ [‐2]) = 10/5 = 2 2. Plug in a point (it does not matter which one) and solve for b y = 2x + b ; 6 = 2(3) + b ; 6 = 6 + b b = 0 3. Plug in m and b to write an equation: : y = 2x Axis Intercepts The x and y‐intercept are important properties of a line and it is often necessary to find the exact location where a line intersects the x‐axis and y‐axis. The best means to find an intercept is algebraically. X‐Axis Intercept When a line crosses the x‐axis, its y‐value will be zero. Consequently, by setting y = 0 and solving for x, the x‐coordinate at which the line crosses the x‐axis can be found. Y‐Axis Intercept When a line crosses the y‐axis, its x‐value will be zero. As a result, by setting x = 0 and solving for y, the y‐coordinate at which the line crosses the y‐axis can be found. If an equation is in y = mx+b format, recall that since setting x = 0 yields y = b, the value of b is the y intercept. Parallel Lines Parallel lines are lines that never intersect. In order to never intersect, two lines must have the same angle (technically called slope). If two lines do not have the same slope, they will eventually intersect. However, if two distinct lines have the same slope, they will never intersect. What is the slope of a line parallel to the line that connects points (1, 4) and (5, 16)? Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 98
The linee that is paarallel will have the same s
slopee as the lin
ne that con
nnects the two pointss mention
ned above. Slope off line conne
ecting two p
points: rise/ruun = (16‐4)/(5‐11) = 12/4 = 3
Consequ
uently, any line with a slope of 3 w
will be paraallel with th
he line that connects (1
1, 4) and (5
5, 16). dicular Line
es Perpend
Line A iss perpendiccular to Line B if Line A
A intersects Line B at a 90° anglee. The most important propertyy of perpen
ndicular linees is as follo
ows: The slop
pe of perpendicular lines forms neegative reciiprocals Slope off Line A
1
/ 2 3
‐5
4
‐ / 7 Slop
pe of Line P
Perpendicular to Line A
A
‐2
1
‐ /3
1
7
/5
/4
e Between Points Distance
In order to find the
e distance between two
o points, either: •
usee the distan
nce formula [to be deriived] or •
raw
w a triangle and use the Pythagorrean theoreem. e Formula Distance
The proccess above can be simplified using the follow
wing formula: Coordiinate Geo
ometry ‐ Questio
ons Q1. Whiich of the fo
ollowing is the equatio
on of a linee that goes through th
he point (10
0,5) and hass an x‐inteercept of 5. A. y = x – 4 B. y = x – 5 C. y = 2x – 4 D. y = 2x – 5 E. y = 3x ‐ 4 Apphelp GMAT coach
hing © Copyright
C
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Q2. What quadrants do the line y = 2x ‐ 4 go through? A. 1 and 2 B. 1 and 3 C. 1 and 4 D. 1, 2 and 3 E. 1, 3 and 4 Q3. What is the shortest distance between points (‐2, 1) and (2,‐2)? A. 1 B. 2 C. 3 D. 4 E. 5 Geometry ‐ Questions (Mixed) Q1. Vertices of a quadrilateral ABCD are A(0, 0), B(4, 5), C(9, 9) and D(5, 4). What is the shape of the quadrilateral? A. Square B. Rectangle but not a square C. Rhombus D. Parallelogram but not a rhombus E. Kite Q2. What is the area of an obtuse angled triangle whose two sides are 8 and 12 and the angle included between two sides is 1500? A. 24 sq units B. 48 sq units C.
D.
E. Such a triangle does not exist Q3. What is the measure of the radius of the circle that circumscribes a triangle whose sides measure 9, 40 and 41? A. 6 B. 4 C. 24.5 D. 20.5 E. 12.5 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 100
Q4. If the sum of the interior angles of a regular polygon measures up to 1440 degrees, how many sides does the polygon have? A. 10 sides B. 8 sides C. 12 sides D. 9 sides E. None of these Q5. What is the radius of the incircle of the triangle whose sides measure 5, 12 and 13 units? A. 2 units B. 12 units C. 6.5 units D. 6 units 7.5 units Q6. How many diagonals does a 63‐sided convex polygon have? A. 3780 B. 1890 C. 3843 D. 3906 E. 1953 Q7. If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? A. 7 B. 12 C. 9 D. 13 E. 11 Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 101
Mensuration That branch of applied geometry which gives rules for finding the length of lines, the areas of surfaces, or the volumes of solids, from certain simple data of lines and angles. Formulas 1.Area of a circle of radius (r) is ∏ r2. 2.Perimeter (or circumference) of a circle of radius (r) is 2∏ r. 3.Area of Sector = (x/360)πr2 where x is an angle at the centre of circle 4.Length of a Sector = (x/360) 2πr 5.Volume of a sphere of radius (r) is 4/3 ∏ r3 6.Surface area of a sphere of radius (r) is 4∏ r2 7.Volume of right circular cone is 1/3 ∏ r2h where (r) is the radius of the base and (h) is the height of the cone. 8.Curved surface area of a cone is ∏ rl where (l) is the slant height of the cone. 9.Total surface area of a cone is ∏ rl + ∏ r2 10.Volume of a right circular cylinder is ∏ r2 h. 11.Curved surface area of a right circular cylinder is 2 ∏ rh. 12.Total surface area of a right circular cylinder is 2 ∏ rh + 2 ∏ r2. 13.Area of a triangle=1/2 (base) (height). 14.Area of a rectangle = (length) (breadth). 15.Perimeter of a rectangle= 2 (length) +2 (breadth). 16.Area of a square = (Side) 2 17.Perimeter of a square = 4 (Side). Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 102
18.Volume of a cube= x3 where (x) is the length of the side. 19.Surface area of a cube= 6x2 20.Area of a trapezium=1/2 (sum of parallel side) (distance between the parallel sides). 21.Area of an equilateral triangle = where (a) is the length of a side and (p) is the length of an altitude. Mensuration – Questions Q1. A cube of side 5cm is painted on all its side. If it is sliced into 1 cubic centimer cubes, how many 1 cubic centimeter cubes will have exactly one of their sides painted? A. 9 B. 61 C. 98 D. 54 E. 64 Q2. A wheel of a car of radius 21 cms is rotating at 600 RPM. What is the speed of the car in km/hr? A. 79.2 km/hr B. 47.52 km/hr C. 7.92 km/hr D. 39.6 km/hr E. km/hr Q3 The area of a square field is 24200 sq m. How long will a lady take to cross the field diagonally at the rate of 6.6 km/hr? A. 3 minutes B. 0.04 hours C. 2 minutes D. minutes E. 2 minutes 40 seconds Q4. A lady grows cabbages in her garden that is in the shape of a square. Each cabbagetakes 1 square feet of area in her garden. This year, she has increased her output by 211 cabbages Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 103
as compared to last year. The shape of the area used for growing the cabbages has remained a square in both these years. How many cabbages did she produce this year? A. 11236 B. 11025 C. 14400 D. 12696 E. Cannot be determined Q5. The length of a rope, to which a cow is tied, is increased from 19m to 30m. How much additional ground will it be able to graze? Assume that the cow is able to move on all sides with equal ease. A. 1696 sq m B. 1694 sq m C. 1594 sq m D. 1756 sq.m E. 1896 sq.m Apphelp GMAT coaching © Copyright 2006. All rights reserved Page 104