In base 7 number is written only using the digits 0,1..6.the number 135 in base 7 is
1*(7^2)+3*(7^1)+5*(7^0)=75 in bse 10.what is the sum of base 7 numbers 1234 and 6543 in base 7?
a)11101 b)11110 c)10111 d)11011
343+98+21+4=466 (1234)
2058+245+28+3=2334 (6543)
thus addition in decimal=2800
now convert 2800 to base 7= 11110
7!2800 ! 0
-------7!400 ! 1
-------7! 57 ! 1
-------7! 8 ! 1
-------1!1
2800 (in decimal)==11110 (in base 7)
The IT gaint Tirnop has recently crossed a head count of 150000 and earnings of $7 billions.As one of the
forerunners in the technology front,Tirnop continues to leasd the way in product and services in India.At Tirnop
all programmers are equal in every respect.They receive identical salaries & also write code at the same
rate.Suppose 11 lines of code in total.How many lines of code can be written by 88 programmers in 88 minutes.
a.)8 b.)704 c.)11 d.)88
formula=(programmer*time)/loc;
88*88/loc=11*11/11
loc=704an
How many 3-digit numbers have even number of factors?(a) 22 (b) 879 (c) 21 (d) 878
Every factor of a number has a pair, so there ought to be an even number of factors for every number. However, if a pair of the
factors are the same number, then there will be an odd number of factors, that is if the number is a perfect square.
Assuming three digit numbers are 100 to 999:
From 100 to 999, the perfect squares are 10^2 = 100 to 31^2 = 961, a total of 31-10+1 = 22 numbers.
So of the 999-100+1 = 900 three digit numbers, 22 have an odd number of factors, so 900-22 = 878 have an even number of
factors.
in which of the system,decimal number 194 is equal to 1234?
to find the solution in a easiest way...
we know while dividing the first remainder holds the unit value...so,one with remainder 4 builds the system...
we need to divide 2,3...9 with 194...
well it sounds like a lot but,,
2 gives remainder 0..
3 gives 2.. 4 gives 2.. 5 gives 4... 6 gives 2..
7 gives 5.. 8 gives 2.. 9 gives 5..
so the answer will be 5....
if the desimal equivalent of 5434 to base x is 1246,what is the value of x?
a) 8
b) 6 c)4 d) 2
2 and 4 is not possiable .
Now 5a^3+4a^2+3a+4=1246
put the a=6 equation will be satisfy.
The base address of a 2-D integer array NUMS of dimension 5*7 is 3000. What will be the address of element
NUMS[3][6], if array indices start from 0 and each integer takes two bytes. a) 3056 b) 3038 c) 3054 d) 3036
as indices starts from 0 so nums[3][6] means you have to access 7th element of 4th row (as i mentioned index starts from 0 ). so
before access this element we have to compute 3 rows and 6 elements of 4th row hence
(3 rows * elements in row * size)+( 6 elements * size) so that
(3*7*2)+(6*2)=54 so
address of NUMS[3][6]=base+54 = 3054
answer is c)3054
If P(x) = ax^4 + bx^3 + cx^2 + dx + e has roots at x = 1; 2; 3 and 4,
and P(0) = 48/ what is P(5)? a) 48 b) 24 c) 0 d) 50 e) none of these
if 1 is a root then x-1 is a factor of p(x)
similarly x-2 is a factor,x-3 ,x-4 are factors
but p(x) is 4th degree polynomial therefore it can be in the form
p(x) = k(x-1)(x-2)(x-3)(x-4) , where k is a constant
but given p(0) =48
therefore 24 k =48
k=2
p(x) =2(x-1)(x-2)(x-3)(x-4)
p(5)= 2*4*3*2*1 = 48
In a sequence of integers , A(n)= A(n-1)-A(n-2), where A(n) is the nth term in the sequence, n is the integer
and n>=3 , A(1)=1, A(2)=1,. calculate S(1000), where S(1000) is the sum of the first 1000 terms
ANS = 0
s(1000)= A(1)
+ A(2)
+ A(2)-A(1)
+ A(3)-A(2)
........
+ A(998)-A(997)
+ A(999)-A(998)
cross terms will cancel each other
=>s(1000)= A(2)+ A(999)
again by definition
A(n)= A(n-1)-A(n-2)= (A(n-2)-A(n-3)) - A(n-2)= -A(n-3)=A(n-5)
A(999)= A(994)=...=A(4)= -A(1)=-1
hence s(1000)=1+(-1)=0
7,5,9,12,18,36,47,72,?
5, 9, 12, 18, 26, 36, 47, 72, _?
it is 5 ...... 12....... 26......... 47
12= 5+7
26= 12+2×7
47=26+3×7
Next is 47+4×7=75
and 9, 18, 36, 72 hence next 144
Therefore 5, 9, 12, 18, 26, 36, 47, 72, 75, 144
Alok is attending a workshop 'How to do more with less1 and today's theme is working with fewer digits. The
speakers discuss how a lot of miraculous mathematics can be achieved if mankind (as well as womankind)
had only worked with fewer digits. The problem posed at the end of the workshop is 'How many 8 digit
numbers can be formed using the digits 1, 2, 3, 4, 5 (but with repetition) that are divisible by 4?' Can you help
Alok find the answer?
ans 5^7
no divisible by 4 .last 2 digit should be divided by 4..
possibility is 12,24,32,44,52..so one possibility.last two got placed...
remaining 6 digits.5^6 and 5(12,24,44,52,32)..
then 5^7
if g(0)=1, G(1)=-1 and G(N)=G(N-1)+G(N-2) then g(5)=?
g(0)=1
g(1)=-1
g(2)=g(1)+g(0) = -1 + 1 = 0
g(3) = g(2) + g(1) = 0 -1 = -1
g(4) = g(3) + g(2) = -1 + 0 = -1
g(5)= g(4) + g(3) = -1 + -1 = -2
The prime factorization of a number N is AxAxBxC, where A,B,C are all distinct prime integers. How many
factors does N have?
A has power of 2,B has power of 1,C has power 1
prime factorization can be calculated
if A^P+B^Q+C^R,den
prime factorization z (P+1)*(Q+1)*(R+1)
in dis ques P=2,Q=1,R=1
So (2+1)(1+1)(1+1)=12
The citizens of planet negate are 8 fingered and have thus developed their decimal system in base 8.
A certain street in negate contains 1000 (in base 8) buildings numbered 1 to 1000.
How many 3’s are used in numbering these buildings?
rule is 3n^2,where n is the base no=3*8^2=192
Find the no of zeros in the product of 1^1*2^2*3^3*.....*49^49??
consider the case of multiples of 5
5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x (5^2)^25 x 30^30 x 35^35 x 40^40 x 45^45
or
5^5 x 10^10 x 15^15 x 20^20 x 5^50 x 30^30 x 35^35 x 40^40 x 45^45
total zeros at the end of product 5^5 x 10^10 x 15^15 x 20^20 x 25^25 x 30^30 x 35^35 x 40^40 x 45^45 will be
5+10+15+20+50+30+35+40+45 = 250 zeros
The sequence {A(n)} is defined by A(1)=2 and A(n+1)=A(n)+2n. What is the value of A(100).
quickly find the values of n=2,3,4,5,6
we get for n=2 4
n=3 8 ( 1 less than its square i.e.9)
n=4 14 (2 less than its square i.e 16)
n=5 22 (3 less than its square i.e 25)
continuing ... for n=100 it will be 98 less than its square
so (100^2)-98=9902 ans
cost of 3 mango,4 pears cost 55 and cost of 1 apple, 5 pears cost 70 ,and cost of 2 mango, 5 apple cost
60...what the total cost of 1 app,1 mango and 1 pea?
answer is 35
3 mango,4 pear 3x+4y=55
5 pear ,1 apple 5y+z=70
5 apple, 2 mango 2x+5z=60
add all 5x+9y+6z=185 equation 1
equation 2 is 3x+4y = 55
by using z=70-5y put in to equation 2 and we get 5x-26y=-235 equation 3
subtract eq.2 - eq 3
we get y=10 , z= 20, x= 5
so x+y+z=35
like a number e.g(54321) was there then what should be added to that number so the remainder will be 35
when divided by 424.
if v divide 54321 wid 424 we get 49 as remainder
so we must add 424-49=375 to get 0 as remainder...but we want remainder 35 so 375+35=410 should b added to 54321 to get
remainder 35....
if p(x)=ax^4+bx^3+cx^2+dx+e has roots at x=1,2,3,4 & p(0)=48.what is p(5
if 1 is the root then (x-1) is the factor of px,similarly
(x-2),(x-3),(x-4) are factors.
but p(x) is the 4th degree polynomial therefor it can be in the form
p(x)=k(x-1)(x-2)(x-3)(x-4)
where k is a constant
but p(0)=48 given
therefore 24k=48
so k=2
p(x)=2(x-1)(x-2)(x-3)(x-4)
p(5)=2*4*3*2*1=48
given the following information who is youngest?
c is younger than a; a is taller than b;
c is younger than d;
c is older than b;
b is taller than c;
a is older than d
(1)d (2)b (3)a (4)c
(2)b because c is younger than both a and d, and c is older than b also so finally b is younger
sqrt (6+ sqrt(6+ sqrt (6+...))) is equal to a) 2 b) 5 c) 4 d) 3
let y=sqrt(6+sqrt(6+sqrt(+...)))
y=sqrt(y+6)
squaring both side
y2=y+6
y=3
Two numbers are such their diff ,their sum and their product are in the ratio 1;7:24.the product of the numbers
is a) 24 b) 36 c) 48 d) 60
x+y/x-y=7/1
7x+7y=x-y
6x=8y
3x=4y
also, x-y/xy=1/24
24x-24y=xy
8(3x)-24y=xy
replace 3x by 4y
32y-24y=xy
8y=xy
x=8
y=(3/4)*8=6
xy=8*6=48 ans
five different digits are given out of these digits 4 digit numbers are formed fine the no of 4 digit numbers
formed which has at lest one digit repeted
Answer: 5^4 - 5p4
Reason: number of 4 digit numbers formed using 5 numbers with repetition are 5^4
and number of 4 digit numbers formed from given 5 digit numbers without repetition are 5p4.Number of numbers formed with
atleast 1 digit repeated are 5^4 -5p4
1!+2!+3!...+50! divided by 5! remainder will be ?
1!+2!+3!...+50! divided by 5!
remainder will be whatever is obtained by dividing 1!+2!+3!+4! with 5!.
so remainder is obtained by dividing (1+2+6+24)= 33 with 5! ( 120).
so remainder is 33.
There are six multiple choice questions in the examination. How many sequences of anwers are possible, if the
first two question have 3 choices each, the next two have 4 choice each and last two have 5 choices each?
3c1*3c1*4c1*4c1*5c1*5c1=3600
for any one of first three option of Q.1 we have three options in Q.2 4 option in Q.3.....this goes on...
Suppose 12 such programmers take 12 minutes to write 12 lines of code in total.
How long will it take 72 programmers to write 72 lines of code in total?
18
72
6
12
12 programmers take 12 minute to write 12 lines
so 1 programmer will take 12 minutes to write 1 line
so 1 programmer will take 1 minute to write 1/12th of a line
so 72 programmers will take 1 minute to write (1/12)*72 = 6 lines of code
if 6 lines need 1 minute then 72 lines need (72/6) minutes = 12 minutes
Ans : 12
Alok one such child finds only bicycles and wheeled wagons in this schoolyard. He counts the total number of
wheels to be 36. What could be the possible number of bicycles? a)14 b)13 c)15 d)20
2x+4y=36
so from options we have to check the answer
as if x=14 the only we can get valid output 0f y = 2
so answer is 14
The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and
the 15 as remainder. What is the smaller number?
(a) 240 (b) 270 (c) 295 (d) 360
ans b)270
x-y=1365
divide larger number by the smaller, we get 6 as quotient and the 15 as remainder
x/y=6 quotient and 15 reminder convert this as
x=6y+15
apply this x in above eqn
therefore,
6y+15-y=1365
5y+15=1365
y+3=273
y=270
so ans is b)270.
23. A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg
packets are there?
(a) 100 (b) 114 (c) 200 (d) 208
answer : (b) 114
Let us say there are x number of 1kg packets and y number of 2kg packets.
Totally there are 150 packets so , x+y = 150
Next, total weight of box is given to be 264kg , therefore x(1kg)+y(2kg)= 264
Solving the two equations (eliminate x) we get y = 114
A man divides Rs.8600 among 5 sons, 4 daughters and 2 nephews. If each daughter receives four times as
much as each nephew, and each son receives five times as much as each nephew, how much does each
daughter receive?
(a) Rs.800 (b) Rs.600 (c) Rs.200 (d) Rs.700
Answer : (a) 800
Let us say each nephew received rupees x
So each daughter will get 4x and each son will get 5x
There are 2 nephews, 4 daughters and 5 sons
So 2(x)+4(2x)+5(5x) = 8600
43x = 8600
x = 200
sp each daughter gets 4x or 4*200 = 800
How many numbers can be formed using all of 1,2,3,4,5(without repetition),when the digit at the units place
must be greater than that in the tenth place? (a)54(b)40(c)60(d)48(e)36
total methods without any restriction = 5*4*3*2*1 = 120
now for units place to be greater than tens place , only one method out of the set of two will be correct. ex: 12 & 21 only one will
be valid so answer will be 120/2 = 60
Shreya writes 1st 100 whole num. let A and B be the num of times she writes 0 and 9 respectively,find the
value of A+B
first 100 whole nos=0 to 99.
thus A=no.of 0's=(0,10,20,30,40,50,60,70,80,90)=10
B=no. of 9's=(9,19,29,39,49,59,69,79,89,90,91,92,93,94,95,96,97,98,99)=20
A+B=30
The ratio of sum of squares of first n natural numbers to square of sum of first n natural numbers is 17:325. the
value of n is (a)15(b)25(c)35(d)30 (e)none of these
sum of square of n terms =n(n+1)(2n+1)/6 -------------- eq no.1
square of sum of n terms =n^2(n+1)^2/4 ---------------eq no. 2
eq 1/2
(2n+1)2 17
------- = ---n(n+1)3 325
only 25 satisfies this
The difference of two numbers is 1365. On dividing the larger number by the smaller, we get 6 as quotient and
the 15 as remainder. What is the smaller number? (a) 240 (b) 270 (c) 295 (d) 360
smaller number=x larger number=6x+15
6x+15-x=1365 or,5x=1350 or,x=270
A three digit number xyz is squared,which of the following could be the last digit of the square so obtained?
a)8 b)3 c)2 d)1
last digit could be 0,1,4,5,6,9
ans should be 1
find out last two digits of (2957^3661)+(3081^3643)
Options o 42 o 38 98
o 22
2957)^1=...57 as last two digts
(2957)^2=...49
(2957)^3=...93
(2957)^4=...01
(2957)^5=...57
as we observed dat last digit repeats for every power of five...
now consider (2957)^3661=(2957)^{3660) * 2957
since 3660 is multiple of 4 it gives 01 as last two digits and again multiplying with 2957 we get 57 as last digits..
similarly do for (3081)^3643... we get 41 as last two dits
now finally adding both 57+41= 98 therefore 98 is last two digits...
if N is a natural no and N^3 has 16 factors then how many maximum factors can N^4 have??a)21 b)24 c)25
d)26
lets say a term 6
Now 6^3=(2^3 )* (3^3)
so, no of factors=(3+1)*(3+1)=16(using prime factorization)
Now similarly
6^4=(2^4)*(3^4)
so factors, (4+1)*(4+1)= 25
So the ans is c)25
If 137+276=435; how much is 731+672?
A : 534
B: 1403
C: 1623
D: 1573
137+276=435
which is true for base 8 numbers where 8 is written as 10
9 as 11 and so on..
731
672
....
1623
If any doubt, check it by changing each number to base 8. in the same way solving 731+672 we get 1623!!!!!!!!!!!!!!!!!
a,b,c, chosen randomly and with replacement from the set {1,2,3,4,5},the probablity that a*b+c is even
conditions:
1. All the values of a,b,c are ODD
2. C is odd.. "atleast" one of a and b should be odd
i.e, if a is odd, b can be anything
if b is odd, a can be anything
Probability(a*b*c):(3/5 * 3/5 * 3/5) + ( 2/5 * 5/5 * 2/5 ) + (5/5 * 2/5 * 2/5)
solving it.. = 67/125
When numbers are written in base b, we have 12 x 25 = 333. The value of b is
a. None of the choices b. 7 c. 8
d. 6
for this problem we have to go fro options......
first take option b.7
LHS......12=7^0*2+7^1*1=2+7=9
LHS.......25=7^0*5=5+7^1*2=5+14=19
LHS....we obtain 9*19=171
for RHS....333=7^0*3+7^1*3+7^2*3=171
so LHS=RHS
note(all these calc mus b done from right 2 left)
ANS is ....7
12 divides, ab313ab (in decimal notation, where a,b are digits>0, the smallest value of a+b is
Option
o7
6
o2
o4
If a number is divisible by 12 then it should be divisible by 4&3
for divisible by 4 last [2 digit]no's should be divisible by 4
so last no's are 12,15,18,,,,,,and soon..
now least is 12 it,. 3 which is not the option,,
no other least is 16 it's sum 1+6=7..
so 7 is option :)
How many 6 digit even numbers can be formed from digits 1 2 3 4 5 6 7 so that the digit should not repeat and
the second last digit is even?
a)6480 b)320 c)2160 d)720
there are six digits..
a b c d e f(say)
f=possible digits are 3 -(2,4,6)
e=possible 2 digits - (2,4,6)
and then rest...
d=possible 5 digits
c=possible 4 digits
b=possible 3 digits
a=possible 2 digits
the digits goes on decreasing since digits are not supposed 2 b repeated...
hence ans=>3*2*5*4*3*2=720...:)
x^y+y^x=46
find x,y
take x = 45
y=1
45^1 + 1^45 = 46
How many positive int eger so lutions does the equation 2x+3y = 100 have?
(a) 50 (b) 33 (c) 16 (d) 17
divide the number by the L.C.M of the cofficient of the variable i.e.
100/L.C.M(2,3)=100/6=16.
Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?
337
5678%460=158
(460-158)+35=337
How would the decimal number 520 be represented in a base -7 number System?
(a) 1564 (b) 1234 (c) 1342 (d) 1562 (e) 1672
520/7 = 74 with remainder 2
74/7 = 10 with reminder 4
10/7 = 1 with remainder 3
1/7 = 0 with remainder 1
so the base 7 form of decimal number 520 is 1342
sum of 2 no is 50 and sum of there reciprocal is 1/12. So find these number
x+y=50--->(1)
(1/x)+(1/y)=(1/12)---(2)
on rearranging (2) (x+y)/(xy)=(1/12)
from(1)x+y=50
therefore 60/(xy)=(1/12)
on simplifying the above eqn (y^2)-50y+600=0
on solving the above eqn the factors will be 20 and 30
if a+b+c+d+e=fg, such that a,b,c, d, e are disinct nnumbers, for fg to be the maximum possible value. what is the value g be??
for making max of fg(ie 2 digit number)and a,b,c,d,e to be distinct... we take a,b,c,d,e = 9,8,7,6,5 respectively. On adding this we
get 5 as the unit digit of the sum... Hence 5 is the answer.