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Old May 31st, 2016, 09:40 AM
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Default Quantitative Aptitude For Elitmus

Hii sir, I am preparing for the Elitmus PH test will you please provide me some Sample questions of the Quantitative Section of the Elitmus PH test ?
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Old May 31st, 2016, 10:13 AM
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Default Re: Quantitative Aptitude For Elitmus

eLitmus Evaluation Private Limited was founded in the year 2005 by former employees of Infosys

About the eLitmus PH test

pH Test or Hiring Potential Test is an assessment exam conducted by eLitmus. Companies use pH Test as a criterion for hiring employees.

The test consists of three parts- the
Quantitative section,
Verbal section
Reasoning section


The Sample questions of the Quantitative Section eLitmus PH test is given below

. If v,w,x,y,z are non negative integers each less than 11, then how many distinct combinations are possible of(v,w,x,y,z) which satisfy v(11^4) +w(11^3)+ x(11^2)+ y(11) +z = 151001 ?

Ans :

Changing 151001 to base 11 number we get,
it will be
A34A4 i.e.
10 3 4 10 4 v(11^4) +w(11^3)+ x(11^2)+ y(11) +z=151001 where v=10, w=3,x=4, y=10, z=4

12. In a certain examination paper there are n questions. For j=1,2,3,.....n, there are 2^(n-1) students whoanswered j or more question wrongly. If the total number of wrong answers is 4096 then the value of n is
a)13
b)11
c)10
d)9

Ans

Given that,
2^(n-1) = 4096 = 2^12
i.e)n-1=12
==> n=13

How many six digit number can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at it's unit's place?

Ans

As each number will contain all the six digits and the sum of digits is = 1+2+3+4+5+6 = 21 which is divisible by 3. So each number is divisible by 3.
The numbers ending with digit 1 will be divisible 1.
The numbers ending with digit 2 will be divisible 2.
The numbers ending with digit 3 are divisible 3.
The numbers ending with digit 5 will be divisible 5.
The numbers ending with digit 6 will be divisible 6.
Except the numbers ending with last two digits as 14, 34 and 54 all other numbers ending with 4 are divisible by 4.

The no. of numbers ending with last two digits 14,34 and 54 are = 3*4*3*2 = 72. (As for a number to be divisible by 4 last two digits must be divisible by 4)
so six digit number that can be formed using the digits 1 to 6, without repetition, such that the number is divisble by the digit at unit's place = 720-72= 648

. A natural number has exactly 10 divisors including 1 and itself. how many distinct prime factors can this natural number can have?

Ans

Consider the no is 512, the divisors of this no. are
1,2,4,8,16,32,64,128,256,512;
so only one prime factors are there i.e. 2;
if the no is 48 the divisors are
1,2,3,4,6,8,12,16,24,48;
so the prime factors are 2,3;

15. If m and n are two positive integers, then what is the value of mn? Given:

(1)7m + 5n= 29
(2) m + n= 5

Ans

7m + 5n = 29
m + n = 5(multiply both sides by 7)
subtract equation second from first,we get
7m + 5n =29
7m + 7n = 35
n = 3
substitute value of n in any equation.
m = 2
mn = 6.


16. A natural number has exactly 10 divisors including 1 and itself. How many distinct prime factors can this natural number have?
A. Either 1 or 2
B. Either 1 or 3
C. Either 2 or 3
D. Either 1,2 or 3
Solution:
Ans) Either 1 or 2
Check on 29 i.e 512 , 39, 59 which have only 1 prime factor and 80 , 48 which are having 2
prime factors and total of 10 divisors.
Let us also consider the case of 3 prime factors. Let x, y ,z be the three prime factors of a
number. Therefore 1 , x ,y ,z ,xy ,yx, zx ,xyz must be the factors of that nos . We have
minimum 8 such factors with xyz as the nos or the factor of the nos.
When xyz is the nos then we will have exactly 8 divisors but if the nos is greater thn xyz that
is a multiple of xyz , either the nos is multiplied by any of these prime factors x , y , z then
we will get at least 12 divisors. So we donít get 3 prime factors with 10 divisors.
17. What is the remainder when 128^1000 is divided by 153?


128^1000 = (153-25)^1000 = (25^1000)mod153 = (625^500)mod153 = [(4*153+13)^500]mod153 = (13^500)mod153 = (169^250)mod153 = [(153+16)^250]mod153 = (16^250)mod153 = (256^125)mod153 = [(153+103)^125]mod153 = (103^125)mod153 = [(2*3*17+1)^125]mod153
At this point,observe that 153=17*(3^2);
Now,therefore clearly (2*3*17+1)^125 = [(125C124)*{(2*3*17)^1}*(1^124)+1]mod153.
Actually,the above line can be written since only except the last two
terms,every term of the expansion of (2*3*17+1)^125 has [(2*3*17)^2],i.e,
[153*68] as one of its factors.
Now, [(125C124)*{(2*3*17)^1}*(1^124)+1]= 125*2*3*17 + 1;
[125*2*3*17 + 1]/(153) = [125*2*3*17 + 1]/(3^2*17) = (125*2*3*17)/(3^2*17) + 1/(3^2*17) = 250/3 + 1/(3^2*17) =83 + (1/3)+ 1/(3^2*17) = 83 + (52/153);
which means [125*2*3*17 + 1] = 83*153 + 52;
which again implies 128^1000 = [125*2*3*17 + 1]mod153 = {83*153 + 52}mod153 = 52mod153 ;
So, remainder is 52.

18. Given a,b,c are in GP and a < b < c. How many different different values of a, b, c satisfy (log(a) + log(b) + log(c) ) = 6?


log(abc)=6
abc=10^6 because of b^2=ac=100,
b^3=10^6,
b=10^2=100
and the combinations are
(2,100,5000), (4,100,2500), (5,100,2000), (10,100,1000), (50,100,200),(25,100,400),(20,100,500).

19. What will be the remainder when expression 2^2+22^2+222^2+2222^2+....+22222...48times^2 is divided by 9?


First, let us consider a general case :
(222222222222.....{2 is repeated n times})^2
=[2(111111111111........{1 is repeated n times})]^2
=4(111111111111........{1 is repeated n times})^2
=4{(10)^(n-1) + (10)^(n-2) + (10)^(n-3) + (10)^(n-4) + .... +1}^2.......[(eqn1)]
Now, we know 9 divides (10^k -1) , since (10^k -1)=(999999....k times);
Where k is ofcourse a +ve integer. So,(10^k -1) is of the form (9A + 1);[A being a +ve integer].
Now, looking at the [(eqn1)] ,
we can find that here k =(n-1), (n-2), (n-3),....,1.
Now,for 10^(n-1) = 9*(a1) + 1;
10^(n-2) = 9*(a2) + 1;
10^(n-3) = 9*(a3) + 1;
10^(n-4) = 9*(a4) + 1;
.
.
.
.
10^1 =9 +1;
1 = 1
So,now in [(eqn1)] , we can write
(222222222222.....{2 is repeated n times})^2
=4(9*A +{1 + 1 + 1 + 1 + .........[1 is added n times]})^2
=4*(9*A + n)^2; here, A =(a1+a2+a3+a4+....+1),of course an integer.
=4*81*(A^2) + 4*2*(9*A)*n + 4*(n^2);...............[(eqn 2)]
Now this a general expression for (222222222222.....{2 is repeated n times})^2.
For the given problem,we can find that n=1,2,3,4,5,6,....48.
In the [(eqn 2)],since 1st two terms are already divisible by 9,so we only have
to consider the last term,i.e, 4*(n^2);
Summing up [4*(n^2)] for n=1,2,...48. we get
4{(1^2)+(2^2)+(3^2)+(4^2)+(5^2)+...(48^2)}
=4*[48*(48+1)*(2*48 +1)/6] ; using the formula[1^2 +2^2 +..+n^2=n(n+1)(2n+1)/6]
=4*8*49*97
=4(9- 1)(9*5 +4)(9*11 -2)
=9*B + 4(-1)*(4)*(-2); B an integer, to know its value is not important
=9*B + 32
=9*B + 9*3 + 5;
So, clearly the remainder will be 5.

20. Given that a number Q < 200, calculate sum of all Q such that when Q divided by 5 or 7 givesremainder 2?


for 5
7 12 17.........197....39 terms sum=3978
for 7
9 16 23 30....... 198...28 terms sum =2848
but some terms are counted more than 1 ..ie which are common in 5 and 7
eg 37 72 107 142 177..sum=535
so 6876-535=6341
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