3
RMAT—Success Master
1
Number System
Introduction
In the decimal number system, numbers are expressed by means of symbols 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, called
digits. Here, 0 is called an insignificant digit whereas 1, 2, 3, 4, 5, 6, 7, 8, 9 are called significant digits. We can
express a number in two ways :
Notation Representing a number in figures is known as notation as 350.
Numeration Representing a number in words is known as numeration as ‘Five hundred and forty five’.
Indian System of Numeration
Ten Arb Arb Ten Crore Crore Ten Lac
1010
109
108
107
106
Lac
Ten Thousand Thousand
Place
Place
105
104
103
Hundred
Place
102
Ten Place Unit Place
101
100
Face Value and Place Value of a Digit
Face Value It is the value of the digit itself eg, in 3452, face value of 4 is ‘four’, face value of 2 is ‘two’.
Place Value It is the face value of the digit multiplied by the place value at which it is situated eg, in
2586, place value of 5 is 5 × 10 2 = 500.
Categories of Numbers
Natural Numbers (N ) Numbers that are used
in counting are called natural numbers.
If N is the set of natural numbers, then we write
N = {1, 2, 3, 4, 5, 6, ......... }
The smallest natural number is 1.
Whole Numbers (W ) When zero is included in
the set of natural numbers, the numbers are known
as whole numbers.
If W is the set of whole numbers, then we write
W = { 0, 1, 2, 3, 4, 5, .......... }
The smallest whole number is 0.
Integers (I ) Integers are whole numbers,
negative of whole numbers including zero.
If I is the set of integers, then we write I =
{ ........... – 3, – 2, – 1, 0, 1, 2, 3, .......... }
Rational Numbers Numbers which can be
p
expressed in the form of , where p and q are both
q
integers and q ≠ 0 are called rational numbers.
Basic Rules on Natural Numbers
1. One digit numbers are from 1 to 9. There are 9 one digit
numbers. ie, 9 × 10 0 .
2. Two digit numbers are from 10 to 99. There are 90 two digit
numbers. ie, 9 × 10 1.
3. Three digit numbers are from100 to 999. There are 900 three
digit numbers ie, 9 × 10 2.
In general the number of n digit numbers are 9 × 10 ( n − 1).
4. Sum
of
the
first
n
natural
numbers
ie,
n (n + 1)
.
1+ 2 + 3 + 4 + K + n =
2
5. Sum of the squares of the first n natural numbers ie,
n (n + 1) (2n + 1)
.
12 + 2 2 + 3 2 + 4 2 + K + n 2 =
6
6. Sum of the cubes of the first n natural numbers ie,
2
n (n + 1) 
13 + 23 + 33 + K + n3 =
 2 
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Mathematical Ability : Number System
3 −7
,
, 5, – 2, ........... .
2 9
There exists infinite number of rational numbers between any two rational numbers.
Irrational Numbers Non-recurring and non-terminating decimals are irrational numbers. These numbers
p
cannot be expressed in the form of .
q
eg,
eg,
Real Numbers
3, 5, 29, ............ .
Real number includes both rational and irrational numbers.
Different Types of Numbers
Even Numbers Numbers which are exactly divisible by 2
Extra Punch !
are called even numbers. eg, 2, 4, 6, ..... . Sum of first n even
l 2 is the only even prime number.
numbers = n ( n + 1)
l 1 is not a prime number because it has two equal
factors.
Odd Numbers Numbers which are not exactly divisible
l Every prime number greater than 3 can be written in
by 2 are called odd numbers. eg, 1, 3, 5, ..... . Sum of first n odd
the form of (6K + 1) or (6K − 1) where K is an integer.
numbers = n 2
l There are 15 prime numbers between 1 and 50 and
Prime Numbers Prime numbers are divisible by one
10 prime numbers between 50 and 100.
and ‘itself’ only. eg, 2, 3, 5, 7, 11, ...... .
Relative Prime Numbers Two numbers are said to be
relatively prime if they do not have any common factor other
than 1. eg, (3, 5), (4, 7), (11, 15), (15, 4), ......
Twin Primes Two prime numbers which differ by 2 are
called twin primes.
eg,
(3, 5), (5, 7), (11, 13), .... .
Note 1 is neither prime nor composite.
Composite Numbers Numbers which are not prime are called
composite numbers. eg, 4, 6, 9, 15, ...... .
Perfect Numbers A number is said to be a perfect number, if
the sum of all its factors excluding itself is equal to the number itself.
eg,
Factors of 6 are 1, 2, 3 and 6.
Sum of factors excluding 6 = 1 + 2 + 3 = 6.
6 is a perfect number.
∴
Other examples of perfect numbers are 28, 496, 8128 etc.
Tests for Divisibility
Divisibility By 2 A number is divisible by 2 when the digit at ones place is 0, 2, 4, 6 or 8. eg, 3582, 460,
28, 352, .... .
Divisibility By 3 A number is divisible by 3 when sum of all digits of a number is a multiple of 3. eg,
453 = 4 + 5 + 3 = 12. 12 is divisible by 3 so, 453 is also divisible by 3.
Divisibility By 4 A number is divisible by 4, if the number formed with its last two digits is divisible by 4.
eg, If we take the number 45024, the last two digits form the number 24. Since, the number 24 is divisible
by 4, the number 45024 is also divisible by 4.
Divisibility By 5 A number is divisible by 5, if its last digit is 0 or 5. eg, 10, 25, 60, ..... .
Divisibility By 6 A number is divisible by 6, if it is divisible by both 2 and 3. eg, 48, 24, 108, ..... .
Divisibility By 7 A number is divisible by 7 when the difference between twice the digit at ones place and
the number formed by other digits is either zero or a multiple of 7.
eg,
658
65 − 2 × 8 = 65 − 16 = 49
As 49 is divisible by 7 the number 658 is also divisible by 7.
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RMAT—Success Master
Divisibility By 8 A number is divisible by 8, if the number formed by the last 3 digits of the number is
divisible by 8. eg, If we take the number 57832, the last three digits form 832. Since, the number 832 is
divisible by 8, the number 57832 is also divisible by 8.
Divisibility By 9 A number is divisible by 9, if the sum of all the digits of a number is a multiple of 9. eg,
684 = 6 + 8 + 4 = 18. 18 is divisible by 9 so, 684 is also divisible by 9.
Divisibility By 10 A number is divisible by 10, if its last digit is 0. eg, 20, 180, 350, ............. .
Divisibility By 11 When the difference between the sum of its digits in odd places and in even places is
either 0 or a multiple of 11.
Extra Punch !
eg,
30426
l If p and q are coprimes and both are factors of a
3 + 4 + 6 = 13
number r, then their product p × q will also be a
0+ 2 = 2
factor of r. eg, Factors of 24 are 1, 2, 3, 4, 6, 8, 12
13 − 2 = 11
and 24 prime factors of 24 are 2 and 3, which are
As the difference is a multiple of 11 the number 30426 is
also coprime. Product of 2 × 3 = 6, 6 is also a factor
also divisible by 11.
of 24.
l If ‘p’ divides ‘q’ and ‘r’, then ‘p’ also divides their
In a sum of division, we have four quantities.
sum or difference. eg, 4 divides 12 and 20.
They are (i) Dividend (ii) Divisor (iii) Quotient and (iv)
Sum of 12 and 20 is 32 which is divisible by 4.
Remainder.
Difference of 20 and 12 is 8 which is divisible by 4.
These quantities are connected by a relation.
l If a number is divisible by another number, then it
(a) Dividend = Divisor × Quotient + Remainder.
must be divisible by each of the factors of that
(b) Divisor = (Dividend – Remainder) ÷ Quotient.
number. 48 is divisible by 12. Factors of 12 are 1, 2,
(c) Quotient = (Dividend – Remainder) – Divisor.
3, 4, 6, 12. So, 48 is divisible by 2, 3, 4 and 6 also.
Illustration 1 In a sum of division the dividend is 258, the quotient is 17 and the
remainder is 3. Find the divisor.
Solution. Divisor = (Dividend – Remainder) ÷ Quotient = (258 − 3) ÷ 17 = 15
Hence, the divisor is 15.
Illustration 2 In a sum of division, the quotient is 110, the remainder is 250, the
divisor is equal to the sum of the quotient and remainder. What is the dividend ?
Solution. Divisor = (110 + 250) = 360
Dividend = (360 × 110) + 250 = 39865
Hence, the dividend is 39865.
Illustration 3 Find the total number of numbers upto 600 which are divisible by 14.
Solution. Divide 600 by 13, the quotient obtained is 46. Thus, there are 46 numbers less than
600 which are divisible by 14.
Factors and Multiples
Factor
A number which exactly divides a given number is called a
factor of the given number.
eg,
24 = 1 × 24, 2 × 12, 3 × 8, 4 × 6
Thus, 1, 2, 3, 4, 6, 8, 12 and 24 are factors of 24.
Number of factors of a number If N is a composite
number such that N = a m b n c o K where a, b, c, ... are prime
factors of N and m, n, o, ... are positive integers, then the
number of factors of N is given by the expression
( m + 1) ( n + 1) ( o + 1) ........ .
Extra Punch !
l
l
l
l
l
l
1 is a factor of every number.
A number is a factor of itself.
The smallest factor of a given number is 1 and the
greatest factor is the number itself.
If a number is divided by any of its factors, the
remainder is always zero.
Every factor of a number is either less than or at the
most equal to the given number.
Number of factors of a number are finite.
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Mathematical Ability : Number System
Illustration 4 Find the number of factors that 224 has
Solution. 224 = 25 × 71
Hence, 224 has (5 + 1) (1 + 1) = 6 × 2 = 12 factors.
Multiple
Extra Punch !
A multiple of a number is a number obtained by
multiplying it by a natural number eg, Multiples of 5 are 5, 10,
15, 20, .......
Multiples of 12 are 12, 24, 36, 48, .....
l
l
l
l
Every number is a multiple of 1.
The smallest multiple of a number is the number
itself.
We cannot find the greatest multiple of a number.
Number of multiples of a number are infinite.
LCM & HCF of Numbers
Lowest Common Multiple (LCM)
The lowest common multiple of two or more given numbers is the least of their common multiples.
Multiples of 25 are 25, 50, 75, 100, 125, 150, ....
Multiples of 30 are 30, 60, 90, 120, 150, 180, ....
Lowest common multiple is 150.
Highest Common Factor (HCF)
The highest common factor of two or more given numbers is the largest of their common factors.
Factors of 20 are 1, 2, 4, 5, 10, 20.
Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
Common factors are 1, 2 and 4.
Highest common factor is 4.
LCM and HCF by Division Method
Illustration 5 Find the LCM and HCF of 12 and 52.
Solution. To find LCM
2
2
3
13
12 52
6
26
3
13
1
13
1
1
Thus, LCM of 12 and 52 is 2 × 2 × 3 × 13 = 156
Thus, LCM of 12 and 52 is 2 × 2 × 3 × 13 = 156
To find HCF
12)52(4
48
4) 12 (3
12
×
So, 4 is the HCF of 12 and 52.
Illustration 6 Find the LCM and HCF of 15, 20 and 30.
Solution. LCM
5
2
3
2
LCM = 2 × 2 × 3 × 5 = 60.
15
3
3
1
1
20
4
2
2
1
30
6
3
1
1
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RMAT—Success Master
HCF Let us first find HCF of any two numbers (15 and 20).
15)20(1
15
5)15(3
15
×
So, 5 is the HCF of 15 and 20.
Now, we find HCF of 5 and 30.
5)30(6
30
×
Thus, HCF of 15, 20 and 30 is 5.
Illustration 7 Find the HCF of 321 and 428.
Solution.
321)428(1
321
107)321(3
321
×
HCF of 321 and 428 is 107.
LCM of Numerators
HCF of Denominators
HCF of Numerators
To find HCF of fractions =
LCM of Denominators
To find LCM of fractions =
Illustration 8 Find the LCM and HCF of
3 4 9
,
, .
5 15 8
LCM of 3, 4, 9 36 − 36
=
HCF of 5, 15, 8
1
HCF of 3, 4, 9
1
HCF of fractions =
=
LCM of 5, 15, 8 120
Solution. LCM of fractions =
Illustration 9 Arrange the following in ascending order
3 4 7
, , .
5 7 10
Solution. LCM of 5, 7 and 10 = 70
3 3 × 14 42 4 4 × 10 40 7
7 × 7 49
; =
;
=
=
=
=
=
5 5 × 14 70 7 7 × 10 70 10 10 × 7 70
4 3 7
∴ Ascending order is , ,
.
7 5 10
Extra Punch !
l
l
l
l
l
The least number which is exactly divisible by a, b and c is the LCM of a, b, c.
The greatest number that will divide a, b, c is the HCF of a, b, c.
If x is a factor of a and b, then x is also a factor of a + b, a − b and ab.
HCF of given numbers must be a factor of their LCM.
The product of the LCM and HCF of two numbers is equal to the product of the two numbers.
Illustration 10 LCM of two numbers is 56 times of their HCF. Sum of LCM and HCF is
456. If one of them is 56. Find the other.
Solution. Let the HCF be x; then LCM = 56x
x + 56x = 456
Now,
⇒
x=8
∴ HCF = 8 and LCM = 448
8 × 448
So, other number =
= 64
8 × 448 = 56 × other number
56
Practice Exercises
A. Model Questions
1. Find the sum of the smallest four-digit number and the
largest five-digit number.
(a) 199900
(b) 100999
(c) 211000
(d) 211999
2. A positive integer which when added to 999 gives a sum
which is greater than when it is multiplied by 999. Which
of the following could be the value of the positive integer ?
(a) 3
(b) 4
(c) 1
(d) 2
3. What is the total number of prime numbers less than
100 ?
(a) 25
(b) 21
(c) 29
(d) 27
4. If 43p is divisible by 3, then what is the largest value p
can take ?
(a) 8
(b) 2
(c) 5
(d) 0
5. If the number 6pq 5 is divisible by both 3 and 5, which of
the following digits can replace p and q?
(a) 5,9
(b) 9,5
(c) 5,7
(d) 9,7
6. There are four prime numbers written in ascending
order. The product of the first three is 1001 and that of
the last three is 2431. Find the first prime number.
(a) 7
(b) 11
(c) 3
(d) 5
7. What is the remainder when 152 × 698 is divided by 5.
(a) 1
(b) 2
(c) 4
(d) 3
8. What is the remainder when 75 3 is divided by 4 ?
(a) 1
(b) 1
(c) 3
(d) 2
9. What is the remainder when 412 × 363 × 312 is divided by
15 ?
(a) 2
(b) 6
(c) 4
(d) 3
10. Find the largest four digit number exactly divisible by
55.
(a) 9900
(b) 9955
(c) 9945
(d) 9960
11. What must be added to 581173 to make it exactly
divisible by 11 ?
(a) 1
(b) 2
(c) 0
(d) 3
12. An even number, total of whose digits is 27, is always
divisible by
(a) 18
(b) 9
(c) 2
(d) 6
13. By what least number 250 must be multiplied to get a
multiple of 15.
(a) 3
(b) 5
(c) 2
(d) 15
14. What is the sum of all numbers between 250 and 550,
which are divisible by 19.
(a) 6635
(b) 6640
(c) 6000
(d) 6695
15. A man is typing numbers 5 to 700 on his computer. How
many times does he press the key of his computer ?
(a) 988
(b) 1988
(c) 1888
(d) 1990
16. Find the smallest number to be added to the least three
digit number so that it is exactly divisible by 45.
(a) 35
(b) 25
(c) 45
(d) 15
17. What least number must be added to 15763 so that it is
exactly divisible by 18 ?
(a) 5
(b) 18
(c) 12
(d) 13
18. What least number must be subtracted from 178669 so
that it is exactly divisible by 36 ?
(a) 18
(b) 1
(c) 36
(d) 5
19. How many numbers between 1000 and 3050 are
divisible by 24 ?
(a) 86
(b) 82
(c) 78
(d) 92
20. Find the numbers of prime factors contained in the
product of 2311 × 74 × 35 .
(a) 25
(b) 17
(c) 30
(d) 20
21. A number when divided by 225 gives a remainder of 32 .
What will be the remainder when the same number is
divided by 15 ?
(a) 4
(b) 2
(c) 3
(d) 1
22. Two numbers when divided by a certain divisor give
remainders 35 and 20 respectively and when their sum
is divided by the same divisor, the remainders is 15.
Find the divisor.
(a) 30
(b) 40
(c) 35
(d) 45
23. What is the smallest number by which 1250 be divided
in order to make it a perfect square ?
(a) 5
(b) 25
(c) 2
(d) 4
24. What is the smallest number by which 3000 be divided
in order to make it a perfect cube ?
(a) 27
(b) 9
(c) 3
(d) 6
25. A certain number when successively divided by 5 and 7
leaves the remainders 3 and 4 respectively. What is the
remainder, if the same number is divided by 35.
(a) 21
(b) 23
(c) 24
(d) 29
26. What is the difference between the greatest and the
smallest five digit numbers formed using the digits 7, 0,
3, 4 and 2 without repeating the digits ?
(a) 65672
(b) 53978
(c) 54672
(d) 65978
27. A boy had to do a multiplication. Instead of taking 35 as
one of the multipliers, he took 53. As a result the product
went up by 540. What is the new product?
(a) 1590
(b) 1450
(c) 1550
(d) 1420
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RMAT—Success Master
28. If both 52 and 73 are factors of the number p × 43 × 72
× 135 , then what is the least possible value of p ?
(a) 175
(b) 145
(c) 190
(d) 156
29. A worker earns ` 40 the first day and spends ` 25 on the
second day. He earns ` 40 on the third day and spends
` 25 on the fourth day and so on. On which day would he
have ` 100.
(a) 10
(b) 9
(c) 8
(d) 11
30. The sum of two numbers is 85 and their difference is 9.
What is the difference of their squares ?
(a) 765
(b) 845
(c) 565
(d) 645
31. Find the least number, which is exactly divisible by 20,
36 and 128.
(a) 5766
(b) 5760
(c) 5780
(d) 5790
32. Find the greatest four digit number which when divided
by 8, 9 and 11 gives a remainder 7 in each case.
(a) 9504
(b) 9541
(c) 9511
(d) 9527
33. Find the least four digit number which when divided by
5, 6 and 7 gives a remainder 3 in each case.
(a) 1053
(b) 1063
(c) 1073
(d) 1083
34. A shepherd has 324 sheeps, 396 goats and 428 cows. He
wants to arrange them into groups of equal size without
mixing. Find the least number of groups required to
arrange them.
(a) 285
(b) 287
(c) 284
(d) 289
35. The LCM of 69, 115 and a third number is 2760. Their
HCF is 23. Find the third number.
(a) 170
(b) 186
(c) 172
(d) 184
36. A number 475 is divided into two parts in such a way
that the LCM and HCF of the two parts is 2250 and 25
respectively. Find the two numbers.
(a) 250, 225 (b) 200, 275 (c) 175, 300 (d) 240, 235
37. The LCM of two numbers 45 times their HCF one
number is 250 and the sum of their HCF and LCM is
2300. Find the other number.
(a) 400
(b) 450
(c) 460
(d) 420
38. The sum of the first 100 numbers, 1 to 100 is always
divisible by
(a) 2
(b) 2 and 4
(c) 2, 4 and 8
(d) None of these
39. Find the HCF of 36a 2b3 c2,18a3 b2c4 and 12ab4c3 .
(a) 6ab2c2
(b) 12a 2bc
(c) 6a 2bc2
(d) 12ab2c2
40. The length, breadth and height of a room are 6 m 30 cm,
5 m 85 cm and 3 m 60 cm respectively. What will be the
greatest length of a tape which can measure the
dimensions of room exact number of times ?
(a) 45 cm
(b) 50 cm
(c) 60 cm
(d) 40 cm
41. The traffic lights at three different road crossings
change after every 50 s, 75 s and 100 s respectively.If
they start changing simultaneously at 10 am, after how
much time will they change again simultaneously ?
(a) 6 min
(b) 5 min
(c) 3 min
(d) 8 min
42. The number of students in three sections of a medical
college are 40, 45 and 60. Find the minimum number of
books required for the class library for equal
distribution in the three sections.
(a) 340
(b) 360
(c) 320
(d) 380
43. Three boys steps off together from the same spot. Their
steps measures 30 cm, 35 cm and 40 cm respectively.
What is the minimum distance each should cover so that
all can cover the distance in complete steps ?
(a) 860 cm
(b) 840 cm
(c) 820 cm
(d) 875 cm
44. Find the unit digit in the expansion of 4223 .
(a) 4
(b) 8
(c) 6
(d) 2
45. Find the unit digit in the expansion of 32545 is
(a) 0
(b) 5
(c) 1
(d) 2
46. Find the unit digit in the expansion of 31448.
(a) 6
(b)4
(c) 8
(d) 2
47. A worker was engaged for a certain number of days and
was promised to be paid ` 1755. He remained absent for
some days and was paid ` 1365 only. What were his daily
wages ?
(a) ` 182
(b) ` 195
(c) ` 185
(d) ` 192
48. A heap of stones can be made up into groups of 7. When
made up into groups of 16, 18, 20 and 25 there are
4 stones left in each case.How many stones at least can
there be in the heap ?
(a) 18000
(b) 18004
(c) 18002
(d) 18006
49. If 2/7 of a number increased by 25 gives 45. Find the
number
(a) 81
(b) 63
(c) 72
(d) 70
50. Out of four consecutive prime numbers, the product of
first three is 385 and the product of the last three is
1001. Find the last number.
(a) 7
(b) 11
(c) 13
(d) 17
51. Three numbers are in the ratio 3 : 5 : 6. Sum of the
greatest and the smallest is equal to the sum of the
middle and 16. Find the smallest number.
(a) 12
(b) 20
(c) 24
(d) 16
52. Sum of squares of two numbers is 60 and difference of
the squares is 12. Find the sum of two numbers.
(a) 4
(b) 10
(c) 6
(d) 8
53. The product of two consecutive numbers is 600. Then,
the sum of their squares is
(a) 1203
(b) 1206
(c) 1201
(d) 1204
54. Thrice a number diminished by 2 is 19. Find the
number.
(a) 5
(b) 7
(c) 4
(d) 6
55. A sum of ` 640 is distributed among three persons P, Q
and R. Q gets ` 70 more than P. R gets `. 80 more than
Q. Find the share of Q.
(a) ` 210
(b) ` 215
(c) ` 200
(d) ` 220
10
Mathematical Ability : Number System
56. If the numerator of a fraction is increased by 2 and the
2
denominator is decreased by 1, then it becomes . If the
3
numerator is increased by 1 and the denominator is
1
increased by 2, then it becomes . Find the fraction.
3
2
2
1
1
(b)
(c)
(d)
(a)
9
7
6
5
57. A and B had some candies each. If A gave 5 candies to B,
then B will have twice as many as A. Instead of that, if B
were to give 5 candies to A, then they will both have the
same number of candies. How many did A have ?
(a) 28
(b) 21
(c) 26
(d) 25
58. There were 240 multiple choice questions in an entrance
examination of a medical college. A candidate was given
1
2 marks for every correct answer and penalised mark
2
for every wrong answer. The candidate scored 160
marks in the paper having answered all the questions.
Find the number of correct answers given by the
candidate.
(a) 108
(b) 112
(c) 116
(d) 104
59. 126 is divided into two parts such that 8 times the first
part added to 5 times the second part is 810. Find the
first part.
(a) 60
(b) 66
(c) 70
(d) 56
60. A worker was engaged for a certain number of days and
was promised to be paid ` 480. He was absent for
six days and was paid ` 432 only. What was his daily
wages ?
(a) 432
(b) 60
(c) 65
(d) 70
61. Find the LCM of 8, 10, 24.
(a) 100
(b) 120
(c) 240
(d) 48
62. Find the LCM of 3, 9, 12, 15, 6
(a) 240
(b) 90
(c) 180
(d) 120
63. Find the LCM of 23 × 34 × 52, 22 × 33 × 7, 52 × 72
(a) 23 × 34 × 52 × 72
(b) 23 × 52 × 72
4
2
2
(c) 3 × 5 × 7
(d) 23 × 72 × 5
4 8 3
64. Find the LCM of , ,
3 9 5
(a) 20
(b) 24
(c) 1/24
(d) 1/20
65. Find the HCF of 40 and 64.
(a) 4
(b) 8
(c) 2
(d) 10
66. Find the HCF of 120, 150, 180.
(a) 30
(b) 60
(c) 50
(d) 10
67. Find the HCF of 2378 and 3280.
(a) 86
(b) 84
(c) 80
(d) 82
68. Find the HCF of 22 × 33 × 52,23 × 32 × 5, 52 × 7
(a) 22 × 32 × 5 × 7
(b) 23 × 33 × 5
3
2
(c) 2 × 3 × 7
(d) 32 × 52 × 7
2 12 1
69. Find the HCF of ,
,
5 11 3
1
5
2
12
(a)
(b)
(c)
(d)
165
165
165
165
70. Find the LCM and HCF of 0.25, 0.5, 0.75
(a) 7.5, 0.05 (b) 0.75, 0.5 (c) 0.75, 0.05 (d) 7.5, 0.5
71. The product of 2 numbers is 3024 and their LCM is 36.
Find their HCF.
(a) 88
(b) 82
(c) 84
(d) 86
72. LCM of two numbers is 12 times of their HCF. Sum of
LCM and HCF is 195. If one of them is 60. Find the
other.
(a) 48
(b) 45
(c) 52
(d) 36
73. The sum of two numbers is 135 and their HCF is 9. How
many such pairs of numbers can be formed ?
(a) 6
(b) 2
(c) 5
(d) 4
74. HCF of two numbers is 12 and their product is 3600.
How many such pairs of numbers can be formed ?
(a) 0
(b) 1
(c) 2
(d) 4
75. The LCM of two numbers is 36 times that of their HCF.
The product of numbers is 3600. Find their HCF.
(a) 10
(b) 15
(c) 8
(d) 20
76. Two numbers are in the ratio 3 : 5. Their, LCM is 600.
Find the sum of the numbers.
(a) 320
(b) 160
(c) 340
(d) 280
77. Find the least number which is exactly divisible by
12, 15, 20 and 27.
(a) 650
(b) 520
(c) 600
(d) 540
78. Find the least number which when increased by 3 is
exactly by 10, 12, 14 and 16.
(a) 1680
(b) 1677
(c) 1697
(d) 1670
79. Find the smallest number which when decreased by 4 is
exactly divisible by 9, 12, 15 and 18.
(a) 188
(b) 182
(c) 186
(d) 184
80. Find the least number which when divided by 12, 15, 20
and 24 leaves the same remainder 2 in each case.
(a) 122
(b) 120
(c) 124
(d) 126
81. Find the least number of girls so that they can be
arranged in the groups of 15 or 20 or 30.
(a) 60
(b) 90
(c) 45
(d) 80
82. Find the least perfect square number which is exactly
divisible by 3, 4, 5, 6 and 8.
(a) 3600
(b) 3900
(c) 3500
(d) 3200
83. Find the greatest number less than 800, which is
divisible by 8, 12, 28.
(a) 624
(b) 672
(c) 636
(d) 684
84. Find the least number which when divided by 12, 15 and
20 leaves remainders 4, 7 and 12 respectively.
(a) 48
(b) 60
(c) 52
(d) 64
85. Find the least number which when divided by 5, 6, 7 and
8 leaves a remainder 3 but when divided by 9 leaves no
remainder.
(a) 1683
(b) 1725
(c) 1685
(d) 1723
86. Find the least number which when divided by 15, 20 and
24 leaves 4 as remainder but when divided by 13 leaves
no remainder.
(a) 364
(b) 370
(c) 484
(d) 244
87. Five bells begin to toll together and toll at intervals of
24, 40, 64, 72 and 120 s. After what interval of time will
they toll again together ?
(a) 42 min
(b) 36 min
(c) 48 min
(d) 54 min
11
RMAT—Success Master
88. Four wheels moving 12, 25, 20 and 30 revolutions in a
minute starting at a certain point on the circumference
downwards. After what interval of time will they come
together again in the same position ?
(a) 30 s
(b) 20 s
(c) 15 s
(d) 45 s
93. Find the greatest number which will divide 321,428 and
535 exactly.
(a) 105
(b) 107
(c) 109
(d) 102
89. Three athletes start together to travel the same way
around on a circular path of 12 km with speeds of
3 km/ph, 4 km/ph and 6 km/ph respectively. They meet
at the same point again after what time ?
(a) 12 h
(b) 8 h
(c) 16 h
(d) 10 h
95. Find the least number of square tiles required to pave the
floor of a room of 4 m 50 cm long and 5 m 25 cm broad.
(a) 40
(b) 42
(c) 36
(d) 38
94. Find the greatest possible length which can be used to
measure 4 m 3 cm, 4 m 34 cm and 4 m 65 cm exactly.
(a) 31 cm
(b) 29 cm
(c) 28 cm
(d) 32 cm
96. Find the greatest number which will divide 676, 481 and
767 exactly.
(a) 11
(b) 13
(c) 17
(d) 19
90. The traffic lights at three different road crossings
change after every 48,72 and 108 s respectively .If they
change simultaneously at 9 am,at what time will they
change again simultaneously ?
(a) 10 min 10 s
(b) 7 min 12 s
(c) 7 min 10 s
(d) 12 min 10 s
97. Find the greatest number which divides 48, 63 and 78
leaving 3 as the remainder in each case.
(a) 11
(b) 18
(c) 12
(d) 15
91. Find the smallest 4 digit number that is exactly divisible
by 8,10 and 12.
(a) 1080
(b) 1100
(c) 1050
(d) 1120
99. Find the greatest number that will divide 1196, 209 and
1449 to leave remainders 6,5 and 4 respectively.
(a) 15
(b) 17
(c) 21
(d) 14
92. Find the greatest five digit number exactly divisible by
6, 8, and 12.
(a) 99924
(b) 99988
(c) 99984
(d) 99972
100. Find the greatest number that will divide 228, 275 and
321 to leave remainder 3, 5, and 6 respectively.
(a) 45
(b) 36
(c) 42
(d) 39
98. Find the greatest number which divides 121, 134 and
147 leaving 4 as the remainder in each case.
(a) 11
(b) 19
(c) 15
(d) 13
B. MBA Entrances’ Gallery
1. The number of boys in a class is three times the number
of girls. Which one of the following numbers cannot
represent the total number of children in the class ?
(a) 48
(b) 44
[MAT Feb 2010]
(c) 42
(d) 40
2. If 8 + 12 = 2, 7 + 14 = 3, then 10 + 18 = ?
(a) 10
(b) 4
(c) 6
[IP 2009]
(d) 18
3. Which among the following is the least number which
when divided by 10 leaves a remainder of 9, when
divided by 9 leaves a remainder of 8, when divided by 8
leaves a remainder of 7, when divided by 7 leaves a
remainder of 6, when divided by 6 leaves a remainder of
5, when divided by 5 leaves a remainder of 4, when
divided by 4 leaves a remainder of 3, when divided by 3
leaves a remainder of 2, when divided by 2 leaves a
remainder of 1 ?
[IP 2009]
(a) 7559
(b) 839
(c) 5039
(d) None of these
4. What is the smallest number which when increased by 5
is completely divisible by 8, 11 and 24 ?
[IP 2009]
(a) 264
(b) 259
(c) 269
(d) None of these
5. What will be the value of the remainder when,
[IP 2009]
1693 × 797 × 3395 is divided by 17 ?
(a) 16
(b) 5
(c) 15
(d) 1
6. 25 times a number is 775. The number is
[IP 2008]
(a) 31
(b) 30
(c) 32
(d) None of these
7. Which of the following cannot be a digit in the unit’s
place of a perfect square ?
[IP 2008]
(a) 7
(b) 1
(c) 5
(d) 0
8. The smallest number, which when added to the sum of
square of 9 to 10 gives a perfect square is
[IP 2008]
(a) 0
(b) 3
(c) 8
(d) 15
9. (9992 − 9982) is equal to
(a) 1
(c) 1997
[IP 2008]
(b) 999
(d) 998
10. A number when divided by 119 leaves the remainder 19.
If the same number is divided by 17, the remainder will
be
[IP 2008]
(a) 19
(b) 10
(c) 7
(d) 2
11. The LCM of two numbers is 63 and their HCF is 9. If one
of the numbers is 27, the other number will be [IP 2008]
(a) 9
(b) 21
(c) 17
(d) 189
12. If n is any positive integer, then 34n − 43 n is always
divisible by
[IP 2008]
(a) 7
(b) 17
(c) 112
(d) 145
13. The number zero (0) is surrounded by the same 2 digit
number on both (left and right) sides: for example
25025, 67067 etc. The largest number that always
divides such a number is
[IP 2008]
(a) 7
(b) 11
(c) 13
(d) 1001
14. A number when divided by 45 leaves the remainder 30.
If the number is divided by 5, the remainder must be
(a) 3
(b) 2
[IP 2008]
(c) 1
(d) 0
12
Mathematical Ability : Number System
15. A boy was asked to write 25 ⋅ 92, but he wrote 2592. The
numerical difference between the values of 25 ⋅ 92 and
2592 is
[IP 2008]
(a) 0
(b) 1
(c) 2
(d) 3
16. The HCF of 168, 189 and 231 reduced by 8, gives
(a) 13
(b) 14
[IP 2008]
(c) 17
(d) 21
17. Which of the following fractions is less than 7 8 and
[RMAT 2008]
greater than 1 3 ?
(a) 1 4
(b) 23 24
(d) 17 24
(c) 11 12
18. Ashok had to do a multiplication. Instead of taking 35 as
one of the multipliers, he took 53. As a result, the
product went up by 540. What is the new product ?
(a) 1050
(b) 1590
[RMAT 2008]
(c) 1440
(d) None of these
19. If a, b, c are real numbers such that a − b + c < 1,
a + b + c > −1 and 9a + 3b + c < −4. What is the sign of a ?
(a) ‘+’ only
(b) ‘−’ only [MPMET 2008]
(c) ‘+’ or ‘−’
(d) cannot say
20. Which of the following is a multiple of 88 ?
(a) 1392578
(b) 138204 [MPMET 2008]
(c) 1436280
(d) 143616
21. How many of the natural numbers from 1 to 1000 have
none of their digits repeated ?
[MPMET 2008]
(a) 589
(b) 585
(c) 738
(d) 739
22. PQ ⋅ RS × D ⋅ AB
[MPMET 2008]
In the multiplication problem shown above, if each of
these letters represents a single digit number from 1 to
9, both inclusive and ‘.’ denotes decimal point, then the
product of the two numbers must be
(a) between 10 and 1000
(b) between 1 to 1000
(c) at most 100
(d) greater than 100
23. Find the values of x and y, if the number 5372 xy is
divisible by 45.
[MPMET 2008]
(a) {(1, 5);(5, 5)}
(b) {(1, 0);(5, 5)}
(c) {(5, 5):(0, 5)}
(d) {(3, 0);(5, 0)}
a3 + b3 + c3
.
abc
[MPMET 2008]
(b) 0
(d) 2
24. If a + b + c = 0, find the value of
(a) 3
(c) 1
25. If a and b are distinct primes, the LCM of a x and by were
x and y are positive integers is
[MPMET 2007]
(a) (ab) LCM of x and y
(b) (ab)xy
(c) a x / by HCF of (x, y)
(d) a x by
26. How many numbers lying between 3000 and 4000 can be
formed with the digits 1, 2, 3, 4, 5, and 6 ? The digits are
[MPMET 2007]
not repeated in any one number.
(a) 40
(b) 104
(c) 60
(d) 82
27. The LCM of two numbers is 280 and the ratio of the
numbers of 7 : 8. Find the numbers.
[MPMET 2006]
(a) 70 and 80
(b) 42 and 48
(c) 35 and 40
(d) 28 and 32
28. When a number is divided by 7, the remainder is 4.
When the same number is divided by 6, the remainder
[OJEE 2009]
is 3. What is the number ?
(a) 39
(c) 49
(b) 37
(d) 52
29. What is the least number to which when 5 is added, the
sum is divisible by 12, 15 and 18 respectively ?
(a) 165
(b) 175
[OJEE 2008]
(c) 185
(d) 195
30. The largest four digit number which is a multiple of
[OJEE 2008]
8, 10, 12 and 15 is
(a) 0120
(b) 9600
(c) 9840
(d) 9960
31. Which of the following pairs is equal in value, given that
‘a’ is a real number ?
[OJEE 2008]
(a) 812 a and 243a 10
(b) 125a 12 and 15625a 16
(c) 4a 4 and 8a 6
(d) 494a 3 and 3433 a 4
32. What is the number which when successively divided by
4, 5 and 6 leaves respective remainders of 3, 4 and 5 with
last quotient as 2 ?
[OJEE 2008]
(a) 247
(b) 231
(c) 351
(d) 359
Answers with Solutions
A. Model Questions
1. (b) The smallest four-digit number = 1000
The largest five-digit number = 99999
Required sum = 100999
∴
2. (c) Let the positive integer = x
Then, 999 + x > 999x
By Trial and error when x = 1
1000 > 999
So, the required positive integer is 1.
3. (a) The prime numbers less than 100 are 2, 3, 5, 7, 11, 13,
17, 19, 23, 29, 31, 37, 41, 43, 53, 59, 61, 67, 71, 73, 79, 83,
89, 97. These add upto 25.
4. (a) A number is divisible by 3 when sum of the digits is
divisible by 3.
4+3+ P = 7+ P
So, P can be 2, 5, 8. So the largest value is 8.
5. (d) If p is 9 and q is 7. The number is 6975. 6975 is
divisible by both 3 and 5.
6. (a) Let the four prime numbers be a, b, c, d.
abc = 1001 and bcd = 2431
abc 1001
7
=
=
bcd 2431 17
a
7
=
⇒a = 7
∴
d 17
7. (a) The remainder when 152 is divided by 5 = 2
The remainder when 698 is divided by 5 = 3
Their product = 2 × 3 = 6
The remainder when 6 is divided by 5 = 1.
8. (c) The remainder when 75 is divided by 4 is 3.
∴The remainder when 3 × 3 × 3 is divided by 4 is 3.
So, the required remainder is 3.
9. (b) 41, 36 and 31 when divided by 15, leave remainders
11, 6 and 1 respectively.
∴412 × 363 × 312 when divided by 15, leaves remainder
112 × 63 × 12 = 26136.
As remainder cannot exceed the divisor.
∴ Remainder is 6 ie, remainder obtained on dividing
26136 by 15.
10. (b) The largest four digit number = 9999.
55)9999(181
55
449
440
99
55
44
The required number = 9999 − 44 = 9955.
11. (a) Sum of digits at add places (from left side)
= 5 + 1 + 7 = 13
Sum of digits at even places (from left side)
= 8 + 1 + 3 = 12
Whatever we add to the given number will increase the
sum of digits which includes the units digit ie, to 12. If
we add 1 to given number difference between the sum of
digits becomes 13 − 13 = 0
∴1 is to be added to the given number.
12. (a) An even number is always divisible by 2.
Further any number sum of whose digits is 27 ie;
(2 + 7 = 9) is always divisible by 9.
∴ The number is divisible by both 2 and 9.
∴ The number is divisble by 2 × 9 = 18.
13. (d) 15 = 3 × 5
∴Any multiple of 15 must be divisible by 3 and 5.
250 = 2 × 5 × 5 × 5.
∴250 is divisible by 5, but not by 3.
So, 250 must be multiplied by 3 to make it a multiple of
15.
14. (c) Since, 250 = 13 × 19 + 3
Terms divisible by 19 upto 250 = 13.
And 550 = 28 × 19 + 18
∴Term divisible by 19 upto 550 = 28.
∴Total terms divisible by 19 between 250 and
550 = 28 − 13 = 15.
Midterm (Average) between 250 to 550 = 400.
∴Sum of 15 terms = 400 × 15 = 6000
15. (b) One digit numbers from 5 to 9 = 5
Two digit numbers from 10 to 99 = 90
Three digit numbers from 100 to 700 = 601
Total number of digits that are to be used
= (5 × 1) + (90 × 2) + (601 × 3)
= 5 + 180 + 1803 = 1988
16. (a) The least three digit numbers = 100
45)100(2
90
10
So, the smallest number to be added
= 45 − 10 = 35.
17. (a)
18)15763(875
144
136
126
103
90
13
∴The least number that should be added = (18 − 13) = 5.
14
Mathematical Ability : Number System
18. (b)
36)178669(4963
144
346
324
226
216
109
108
1
∴The least number that should be subtracted = 1.
19. (a) Quotient when 1000 and 3050 are divided by 24 are
= 41 and 127.
∴Between 1000 and 3050 there are
127 − 41 = 86 numbers.
20. (d) Number of prime factors = 11 + 4 + 5 = 20.
21. (b) Let the number be N. N = 225Q + 32 where Q the
quotient can have the values 1, 2, 3, etc.
N = 15 × 15Q + (15 × 2) + 2
Divide N by 15, we see that the remainder is 2.
22. (b) In such questions the divisor is r1 + r2 − r3 .
Divisor = 35 + 20 − 15 = 40
∴
23. (c) 1250 = 5 × 5 × 5 × 5 × 2
So, when 1250 is divided by 2, we get a perfect square.
24. (c) 3000 = 2 × 2 × 2 × 5 × 5 × 5 × 3
So, when 3000 is divided by 3, we get a perfect cube.
25. (b)
5
N
7
X −3
should include 52. , 73 is a factor of the given number. The
number has 72. Hence, P should include 71 to make it 73 .
∴P should be atleast 52 × 7
= 175 to have 52 and 73 as its factors.
29. (b) His savings for every 2 days = ` 40 − ` 25 = ` 15
His savings in 8 days = ` 60
On the 9th day he earns ` 40.
On the 9th day he has ` 60 + ` 40 = `100.
30. (a) Let the two numbers be x and y.
x + y = 85
x− y =9
Solving x = 47 and y = 38.
The difference of their squares is 472 − 382 = 765.
31. (b) Required number = LCM of 20, 36, 128 = 5760.
32. (c) The LCM of 8, 9 and 11 = 792.
The greatest four-digit number = 9999.
792)9999(122
792
2079
1584
495
The greatest four-digit number divisible by 792 is
9999 − 495 = 9504.
Now, the greatest four-digit number divisible by 792 and
leaving a remainder 7 is 9504 + 7 = 9511.
33. (a) The LCM of 5, 6 and 7 = 210.
Y −4
X = 7Y + 4
N = 5 (7Y + 4) + 3
= 35Y + 23
If
Y = 1, N = 58
∴When 58 is divided by 35 the remainder is 23.
26. (b) The greatest five-digit number that can be formed
using the given digits = 74320
The smallest five-digit number that can be formed using
the given digits = 20342.
∴ The required difference = 74320 − 20342
= 53978
27. (a) Let the number that the boy wanted to multiply be x.
He was expected to find the value of 35 x. Instead, he
found the value of 53x.
The difference between the value that he got and what
he was expected to get is 540.
ie,
53x − 35x = 450
x = 30
∴The new product = 30 × 53 = 1590.
28. (a) If 52 is a factor of the given number. The number does
not have a power or multiple of 11 as its factor. Hence, P
The least four- digit number = 1000.
210)1000(4
840
160
The least four-digit number divisible by 210 is
1000 + (210 − 160) = 1050.
Now, the least four-digit number divisible by 210 and
leaving a remainder 3 is 1050 + 3 = 1053.
34. (b) HCF of (324, 396, 428) = 4
So, the number of groups required
324 396 428
=
+
+
4
4
4
= 81 + 99 + 107
= 287
35. (d) Let the third number be 23x
LCM of 69, 115 and 23x is
23
69, 115, 23x
3, 5, x
LCM = 23 × 3 × 5 × x = 345x
Now,
345x = 2760
x=8
∴The third number = 23x = 23 × 8 = 184.
15
RMAT—Success Master
36. (a) Let the two numbers be 25x and 25 y.
25x + 25 y = 475
x + y = 19
Also, the LCM of 25x and 25 y = 25xy.
25xy = 2250
xy = 90
(x − y)2 = (x + y)2 − 4xy
192 = (x + y)2 − 4 (90)
Solving
x− y=1
Using Eqs. (i) and (iii)
x = 10
and
y = 9.
∴The two numbers are 250 and 225.
...(i)
...(ii)
…(iii)
37. (b) Let the HCF of the two numbers be x.
So, the LCM = 45x
Now,
x + 45x = 2300
x = 50
HCF = 50, LCM = 45x = 45 × 50 = 2250.
Product of the two numbers = Product of the LCM and
HCF of the two numbers.
Let the other number be N.
N × 250 = 50 × 2250
N = 450.
38. (a) The sum of the first n natural numbers
n (n + 1)
=
2
= The sum of the first 100 natural numbers.
100 × 101
=
= 50(101)
2
As 101 is an odd number and 50 is divisible by 2. So, the
sum is always divisible by 2.
39. (a) HCF of 36, 18 and 12 is 6.
The common factors are a, b, c and their highest powers
common to all are 1, 2, 2 respectively
∴ Required HCF = 6ab2c2
40. (a) Greatest length of tape = HCF of 630, 585 and 360
= 45 cm
41. (b) LCM of 50, 75 and 100 = 300 s.
∴Traffic lights will change simultaneously after 300 s.
= 5 min.
42. (b) LCM of 40, 45 and 60 = 360.
∴ 360 is the minimum number of books required for
equal distribution in each section.
43. (b) LCM of 30, 35 and 40 = 840 cm. 840 cm is the
minimum distance they should cover in complete steps.
44. (b) The unit digit comes from the unit place. So, we have
to find the unit digit is the expansion of 223.
Now,
21 = 2, 22 = 4, 23 = 8, 24 = 16,
25 = 32, 26 = 64 ........
We see that the unit digit is 21 is the same as 25 .
∴The unit digit in 223 is the same as 23 = 8.
45. (b) The unit digit in 32545 is the same as that of 545 .
Now, 51 = 5, 52 = 25, 53 = 125, 54 = 625, ........
We see that the unit digit is always 5.
∴The unit digit in 32545 is 5.
46. (a) The unit digits in the expansion of 31448 is the same
as that of 448.
Now,
41 = 4, 42 = 16 , 43 = 64,44 = 256.
We see that the unit digit in 41 is the same as that in 43 .
So, the unit digit is 448 is the same as is 448 = 6.
47. (b) The daily wages of the worker is the HCF of 1755 and
1365.
HCF of 1755 and 1365 is 195.
∴The daily wages of the worker is Rs 195.
48. (b) LCM of 16, 18, 20 and 25 = 3600.
1st number = 3600 × 1 + 4 = 3604 which is not divisible
by 7.
2nd number = 3600 × 2 + 4 = 7204 which is not divisible
by 7.
3rd number = 3600 × 3 + 4 = 10804 which is not divisible
by 7.
4th number = 3600 × 4 + 4 = 14404 which is not divisible
by 7.
5th number = 3600 × 5 + 4 = 18004 which is divisible
by 7.
∴There can be 18004 stones.
49. (d) Let the number be x.
2
x + 25 = 45. On solving x = 70
7
50. (c) Let the four numbers be p, q, r and s.
pqr = 385, qrs = 1001
qrs 1001
=
pqr 385
13
=
4
s 13
=
p 5
So,
p = 5 and s = 13.
∴The required number is 13.
51. (a) Let the numbers be 3x, 5x and 6x.
Now, 3x + 6x = 5x + 16
x=4
So, the smallest number is 3x = 12.
52. (b) Let the number be x and y.
x2 + y2 = 60, x2 − y2 = 12
Adding
2x2 = 72
x = 6 and y = 4
∴Required sum is 10.
53. (c) Let the two numbers be x and x + 1
x (x + 1) = 600
On solving x = 24, so the numbers are 24 and 25.
The sum of their squares is = 242 + 252 = 1201
16
Mathematical Ability : Number System
54. (b) Let the number be x.
62. (c)
2
3, 9, 12, 15, 6
2
3, 9, 6, 15, 3
On solving x = 17
3
3, 9, 3, 15, 3
∴The required number is 7.
3
1, 3, 1, 5, 1
3x − 2 = 19
55. (a)
P + Q + R = 640
...(i)
Q = 70 + P
5
R = 80 + (70 + P )
On solving, we get Q = Rs 210.
x
56. (b) Let the fraction be
y
x+2 2
=
y −1 3
x+1 1
=
y+2 3
x 2
On solving, we get = .
y 7
...(iii)
1, 1, 1, 1, 1
LCM = 2 × 2 × 3 × 3 × 5 = 180
63. (a) LCM of given numbers = 23 × 34 × 52 × 72
...(i)
...(ii)
(Take the greatest power of each term)
LCM of numerators
64. (b) LCM of fraction =
HCF of denomenators
LCM of 4, 8, 3 24
=
=
= 24
HCF of 3, 9, 5
1
65. (b)
b + 5 = 2 (a − 5)
b −5 = a + 5
On solving, we get a = 25.
Then,
...(i)
...(ii)
58. (b) Let the number of wrong answers marked = w
Let the number of right answers marked = r
...(i)
r + w = 240
w
...(ii)
2r − = 160
2
On solving, we get r = 112.
59. (a) Let the two parts be x and 126 − x.
8x + 5 (126 − x) = 810
8x + 630 − 5x = 810
3x = 810 − 630
= 180
x = 60
So, the first part is 60.
60. (b) Let the number of days for which the worker works = x
For x days he is paid Rs 480
For (x − 6) days he is paid Rs 432.
480
His daily wages =
x
432
=
x −6
On solving, x = Rs 60
2
2
2
3
5
8, 10, 24
4, 5, 12
2, 5, 6
1, 5, 3
1, 5, 1
1, 1, 1
LCM = 2 × 2 × 2 × 3 × 5 = 120
40)64(1
40
24)40(1
24
16)24(1
16
8)16(2
16
0
57. (d) Let A have a candies and B have b candies
61. (b)
1, 1, 1, 5, 1
...(ii)
∴ HCF = 8
66. (a) Step 1 120)150(1
120
30)120(4
120
0
Step 2 30) 180 (6
180
0
HCF of 30 and 180 is 30.
HCF of 120 and 150 is 30.
∴The required HCF is 30.
67. (d) 2378)3280(1
2378
902)2378(2
1804
574)902(1
574
328)574(1
328
246)328(1
246
82) 246(3
246
0
∴The required HCF is 82.
17
RMAT—Success Master
68. (a) HCF = 22 × 32 × 5 × 7 (Take the least powers of
common terms).
HCF of numerator
69. (a) HCF of fraction =
HCF of denominator
HCF of 2, 12, 1
1
=
=
LCM of 5, 11, 3 165
25 5 75
70. (a) LCM = LCM of
,
,
100 10 100
LCM of 25, 5, 75
75
=
=
= 7.5
HCF of 100, 10, 100 10
25 5 75
HCF = HCF of
,
,
100 10 100
HCF of 25, 5, 75
5
=
=
= 0.05
LCM of 100, 10, 100 100
71. (c) Product of (LCM × HCF) = Product of 2 numbers
36 × HCF = 3024
3024
HCF =
= 84
36
72. (b) Let the HCF = x
77. (d) Required number = LCM of 12, 15, 20 and 27
2
12, 15, 20, 27
2
6, 15, 10, 27
3
3, 15, 5, 27
3
1, 5, 5, 9
3
1, 5, 5, 3
5
1, 5, 5, 1
1, 1, 1, 1
∴ LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540.
78. (b) Required number = (LCM of 10, 12, 14, 16) −3
2
10, 12, 14, 16
2
5, 6, 7, 8
2
5, 3, 7, 4
2
5, 3, 7, 2
3
5, 3, 7, 1
5
5, 1, 7, 1
7
1, 1, 7, 1
LCM = 12x
x + 12x = 195
13x = 195
x = 15
∴HCF = 15 and LCM = (12) (15) = 180.
180 × 15
Other number =
= 45.
60
73. (d) The two numbers are always the multiples of the
HCF
∴Let the two numbers are 9x and 9y.
9x + 9 y = 135
⇒
x + y = 15
Now, the possible values of x and y are (1, 14), (2, 13),
(3, 12), (4, 11), (5, 10), (6, 9), (7, 8). Now, consider only the
coprime pairs. These are (1, 14), (2, 13), (4, 11), (7, 8).
∴4 pairs of numbers can be formed whose sum is 135 and
HCF is 9.
74. (b) Let the two numbers be 12x and 12 y.
12x × 12 y = 3600
xy = 25
Possible values of x and y are (1, 25) (5, 5).
But (5, 5) are not coprimes.
∴Only 1 pair of number can be formed.
75. (a) Let HCF = x
LCM = 36x
36x × x = 3600
x2 = 100 ⇒ x = 10
∴ HCF is 10.
76. (a) Let the numbers be 3x and 5x.
LCM is 15x = 600
x = 40
The two numbers are 120 and 200. Sum = 320.
1, 1, 1, 1
LCM = 2 × 2 × 2 × 2 × 3 × 5 × 7 = 168
∴Required number = 1680 − 3 = 1677.
79. (d) Required number = (LCM of 9, 12, 15, 18) + 4
2
9, 12, 15, 18
2
9, 6, 15, 9
3
9, 3, 15, 9
3
3, 1, 5, 3
5
1, 1, 5, 1
1, 1, 1, 1
LCM = 2 × 2 × 3 × 3 × 5 = 180
∴Required number = 180 + 4 = 184
80. (a) Required number = (LCM of 12, 15, 20, 24) + 2
2
12, 15, 20, 24
2
6, 15, 10, 12
3
3, 15, 5, 6
2
1, 5, 5, 2
5
1, 5, 5, 1
1, 1, 1, 1
LCM = 2 × 2 × 2 × 3 × 5 = 120
Required number = 120 + 2 = 122
81. (a) Required number = (LCM of 15, 20, 30)
2
15, 20, 30
2
15, 10, 15
5
15, 5, 15
3
3, 1, 3
1, 1, 1
LCM = 2 × 2 × 3 × 5 = 60
18
Mathematical Ability : Number System
82. (a) LCM of 3, 4, 5, 6 and 8 is 120
2
2
2
3
5
3, 4, 5, 6, 8
3, 2, 5, 3, 4
3, 1, 5, 3, 2
3, 1, 5, 3, 1
1, 1, 5, 1, 1
1, 1, 1, 1, 1
LCM = 2 × 2 × 2 × 3 × 5 = 120
120 = 23 × 3 × 5
∴The required perfect square = (23 × 3 × 5) × 2 × 3 × 5
= 3600
83. (b) LCM of 8, 12 and 28 = 168.
2
2
2
3
7
8, 12, 28
4, 6, 14
2, 3, 7
1, 3, 7
1, 1, 7
1, 1, 1
LCM = 2 × 2 × 2 × 3 × 7 = 168
168)800(4
672
128
∴The required number = 800 − 128 = 672
84. (c) LCM of 12, 15 and 20 = 60
2
2
3
5
12, 15, 20
6, 15, 10
3, 15, 5
1, 5, 5
1, 1, 1
LCM = 2 × 2 × 3 × 5 = 60
Now, 12 − 4 = 8, 15 − 7 = 8
and
20 − 12 = 8
So, required number = 60 − 8 = 52
85. (a) LCM of 5, 6, 7 and 8 = 840
2
2
2
3
5
7
5, 6, 7, 8
5, 3, 7, 4
5, 3, 7, 2
5, 3, 7, 1
5, 1, 7, 1
1, 1, 7, 1
1, 1, 1, 1
LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840
Multiples of 840 = 840, 1680, ...........
∴Required number = 1680 + 3 = 1683
Which is exactly divisible by 9.
86. (a) LCM of 15, 20 and 24 = 120
2
2
2
3
5
15, 20, 24
15, 10, 12
15, 5, 6
15, 5, 3
5, 5, 1
1, 1, 1
LCM = 2 × 2 × 2 × 3 × 5 = 120
Multiples of 120 = 120, 240, 360, 480, .....
∴Required number = 360 + 4 = 364
Which is exactly divisible by 13.
87. (c) LCM of 24, 40, 64, 72 are 124 = 2880 s
2
2
2
2
2
2
3
3
5
24, 40, 64, 72, 120
12, 20, 32, 36, 60
6, 10, 16, 18, 30
3, 5, 8, 9, 15
3, 5, 4, 9, 15
3, 5, 2, 9, 15
3, 5, 1, 9, 15
1, 5, 1, 3, 5
1, 5, 1, 1, 5
1, 1, 1, 1, 1
LCM = 26 × 32 × 5 = 2880
2880
∴Required time =
= 48 min
60
88. (c) Time taken for 1 revolution
1 1 1 1
min
=
,
,
,
12 25 20 30
LCM of Numerators 1
∴ Required time =
= min
HCF of Denominators 4
1
= × 60 = 15 s
4
89. (a) Time taken to complete 1 revolution
12 12 12
h = 4, 3, 2 h
=
,
,
3 4 6
∴Required time = LCM of 4, 3 and 2 = 12 h.
90. (b) LCM of 48, 72 and 180 = 432 s.
2
2
2
3
3
2
3
48, 72, 108
24, 36, 54
12, 18, 27
6, 9, 27
2, 3, 9
2, 1, 3
1, 1, 3
1, 1, 1
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432
Traffic lights will change simultaneously after
432 s = 7 min 12 s
91. (a) The smallest four–digit number exactly divisible by
8, 10 and 12 should also be divisible by the LCM of 8, 10
and 12.
LCM of 8, 10 and 12 = 120.
Smallest four-digit number = 1000
120)1000(8
960
40
So, the required number = 1000 + (120 − 40)
= 1000 + 80 = 1080
92. (c) The greatest five-digit number divisible by 6, 8 and
12 should also be divisible by the LCM of 6, 8 and 12.
LCM of 6, 8 and 12 = 24.
19
RMAT—Success Master
The greatest five-digit number = 99999
24)99999(416
96
39
24
159
144
15
So, the required number = 99999 − 15 = 99984
93. (b) Required number is the HCF of 321, 428 and 535.
Step 1 Find HCF of 321 and 428
321)428(1
321
107)321(3
321
0
HCF of 321 and 428 = 107
Step 2 Find HCF of 107 and 535.
107)535(5
535
0
So, HCF of 321, 428 and 535 = 107.
94. (a) Required measure = HCF of 403, 434 and 465.
Step 1 Find HCF of 403 and 434.
403)434(1
403
31)403(13
31
93
93
0
HCF of 403 and 434 = 31.
Step 2 Find HCF of 31 and 465
31)465(15
31
155
155
0
HCF of 31 and 465 = 31.
∴Required measure = 31 cm.
95. (b) Side of square tile = HCF of 450 and 525.
= 75 cm
450)525(1
450
75)450(6
450
0
∴Required number of tiles =
450 × 525
= 42.
75 × 75
96. (b) Required number is the HCF of 676, 481
and 767 = 13.
97. (d) Required number = HCF of (48 − 3), (63 − 3)
and (78 − 3)
= HCF of 45, 60, 75 = 15
98. (d) Required number = HCF of (121 − 4), (134 − 4)
and (147 − 4).
= HCF of 117, 130, 143 = 13
99. (b) Required number = HCF of (1196 − 6) , (209 − 5)
and (1449 − 4)
= HCF of 1190, 204, 1445 = 17
100. (a) Required number = HCF of (228 − 3), (275 − 5)
and (321 − 6)
= HCF of 225, 270, 315 = 45
B. MBA Entrances’ Gallery
1. (c) The ratio of Boys : Girls = 3 : 1
So, 42 is not possible as it is not divisible by (3 + 1) = 4
6. (a) Let the number be n ⇒ n × 25 = 775
⇒
n = 31
2. (a) 8 + 12 = 20 ⇒ 2 + 0 = 2
7 + 14 = 21 ⇒ 2 + 1 = 3
∴ 10 + 18 = 28 ⇒ 2 + 8 = 10
7. (a) Perfect square are 1, 4, 9, 16, 25, 36, 49, 64, 81, 100,
121, .....
Clearly 7 cannot be at unit’s digit.
3. (d) (10 − 9) = (9 − 8) = (8 − 7) = (7 − 6) = (6 − 5)
= (5 − 4) = (4 − 3) = (3 − 2) = 1
LCM of 10, 9, 8, 7, 6, 5, 4, 3, 2, = 2520
∴Required number = 2520 − 1 = 2519
4. (b) LCM of 8, 11, and 24 = 264
Required number = 264 − 5 = 259
∴
5. (c) Remainder when 1693 divided by 17 = 10
Remainder when 797 divided by 17 = 15
Remainder when 3395 divided by 17 = 12
∴Required number = Remainder when 10 × 15 × 12
divided by 17
= 1800 divided by 17 = 15
8. (d) 92 + 102 = 81 + 100 = 181, 196 is the nearest perfect
square. Hence, 15 should be added.
9. (c) 9992 − 9982 = (999 + 998) (999 − 998)
= (1997) (1)
= 1997
10. (d) The number is = 119n + 19 = 138, 257
When these numbers are divided by 17, then remainder
is 2.
11. (b) Product of numbers = HCF × LCM
⇒
27 × n = 63 × 9
63 × 9
⇒
n=
= 21
27
20
Mathematical Ability : Number System
12. (b) 34(1) − 43(1) = 81 − 64 = 17, 34( 2) − 43( 2)
= 812 − 642 = 2465
It is always divisible by 17.
13. (d) The largest such number is 1001.
14. (d) In this case the remainder must be 0.
15. (a) The difference between 25 ⋅ 92 and 2592 ⇒ 32 × 81
⇒
2592 − 2592 = 0
16. (a) HCF of 168, 189, 231 is 21.
Then, difference between 21 − 8 = 13.
7
1
17. (d)
= 0. 875, = 0. 333,
8
3
1
23
= 0. 25,
= 0. 958
4
24
11
17
= 0. 916,
= 0. 708
12
24
Clearly, 0.708 lies between 0.333 and 0.875.
17
.
∴ Required fraction is
24
18. (b) Suppose one of the multiplier is x then,
53x − 35x = 540
18x = 540
x = 30
∴ New multiplication = 30 × 53 = 1590 .
19. (b) Let us rearrange the equations :
a −b+ c=1−k
a + b + c = −1 + l
9 a + 3 b + c = −4 − m
Where k, l, m are positive.
Suppose λ and µ are two numbers such that in the
expression, 9a + 3b + c + λ (a − b + c) + µ (a + b + c), the
terms b and c vanish.
∴
− λ + µ = −3
λ + µ = −1
−4
µ=
= −2 and λ = 1
∴
2
Hence, 9a + 3b + c + 1 (a − b + c) − 2 (a + b + c)
= 9a + 3b + c + a − b + c − 2a − 2b − 2c
= 8 a = −4 − m + 1 − k + 2 − 2 l = −1 − k − m − 2 l
which is less than 0, since k, m and l are all + ve.
∴
a <0
It’s sign is ‘−’ ve only.
20. (d) Only 143616 is a multiple of both 11 and 8 so it is a
multiple of 88.
21. (c) We have to consider all the one-digit, two-digit and
three-digit numbers and see how many of these have
distinct digits.
(i) One-digit numbers, 1 to 9 are all distinct.
(ii) Two-digit numbers, ten’s place cannot be filled by 0.
Hence, 9 possibilities. Having filled that, 8
possibilities for the unit’s place and also zero.
Hence, 9 × 9 = 81 numbers are possible.
(iii) In a similar way, the three-digit numbers can be
determined
Hundred’s place = 9
Ten’s place = 9
Unit’s place = 8
∴ Number of three-digit numbers = 9 × 9 × 8 = 648
∴ Total numbers = 9 + 81 + 648 = 738.
22. (a) PQ ⋅ RS is less than 100 D ⋅ AB is less than 10
∴ PQ ⋅ RS × D ⋅ AB must be less than 1000 and definitely
greater than 10.
23. (b) If 5372 xy is to be divisible by 45, it must be divisible
by 9 and also by 5.
y must be 0 or 5 for divisibility by 5. For divisibility by 9.
5 + 3 + 7 + 2 + x + y must be a multiple of 9.
ie, 17 + x + y must be a multiple of 9.
∴ If y = 5, x must be 5
If y = 0, x must be 1
ie, x = 1 or 5 and y = 0 or 5
∴ {(1, 0);(5, 5)} is the correct answer.
24. (a) Since, a + b + c = 0, a + b = − c
(a + b)3 = (− c)3
3
3
a + b + 3ab(a + b) = − c3
a3 + b3 + 3ab(− c) = − c3
a3 + b3 − 3abc = − c3
a3 + b3 + c3 = 3abc
a3 + b3 + c3
∴
=3.
abc
25. (d) LCM of a x and by
Since, a and b are distinct primes there is no common
factor.
Hence,
LCM = a x by .
26. (c) The thousands place can be filled by 3 only. Then, the
hundreds place can be filled in 5 ways , the tens place in
4 ways and the units place in 3 ways.
Total number of ways is 1 × 5 × 4 × 3 or 60.
27. (c) Let the two numbers are 7x and 8x and LCM is 56x.
It is given that LCM = 280
ie,
56x = 280 and x = 5
ie,
Numbers are 35 and 40 .
28. (a) From the given options
Required number = 39
29. (b) LCM of 12, 15, 18 = 180
So, required number = 180 − 5 = 175
30. (d) LCM of 8, 10, 12 and 15 = 20
We know that 83 × 120 = 9960
and
84 × 120 = 10080
So, 9960 is the required number.
31. (c)
4a 4 = 2a 2
8a 6 = 2a 2
32. (d) 359 is the required number.
As
359 ÷ 4 (remainder is 3)
89 ÷ 5 (remainder is 4)
17 ÷ 6 (remainder is 5)