Previous Years e-litmus Paper latest set 2

ELITMUS PREVIOUS QUESTIONS

1. How many five digit numbers can be formed by using the digits 0,1,2,3,4,5 such that the number is divisible by 4?

Explanation:
If a number has to be divisible by 4, the last two digit of that number should be divisible by 4.
So _ _ _ x y.  Here xy should be a multiple of 4.
There are two cases:
Case 1: xy can be 04, 20 or 40
In this case the remaining 3 places can be filled in 4×3×2 = 24.  So total 24×3 = 72 ways.
Case 2: xy can be 12, 24, 32, 52.
In this case, left most place cannot be 0.  So left most place can be filled in 3 ways.  Number of ways are 3×3×2 = 18.  Total ways = 18×4 = 72.
Total ways = 144

2. Data sufficiency question:
There are six people. Each cast one vote in favour of other five. Who won the elections?
i)  4 older cast their vote in favour of the oldest candidate
ii) 2 younger cast their vote to the second oldest

Explanation:
Total possible votes are 6.  Of which 4 votes went to the oldest person.  So he must have won the election. Statement 1 is sufficient.

3.

(Image taken while taking eLitmus Test, as you see eLitmus Test has New Layout from 2016)

4. Decipher the following multiplication table: (See and learn how to solve cryptographical elitmus problem here:)
M A D
B E
————-
M A D
R A E
————-
A M I D
————-

Explanation:
It is clear that E = 1 as MAD×E=MAD
From the hundred’s line, M + A = 10 + M or 1 + M + A = 10 + M
As A = 10 not possible, A = 9
So I = 0.
and From the thousand’s line R + 1 = A. So R = 8.
     M 9 D
         B 1
————-
     M 9 D
  8 9  1
————-
  9 M 0 D
————-
As B×D = 1, B and D takes 3, 7 in some order.
If B = 7 and D = 3, then M93×7 =   _51 is not satisfying. So B = 3 and D = 7.
     2 9 7
        3 1
————-
     2 9 7
8  9  1
————-
9 2 0 7
————-

5. If

log3N+log9Nlog3N+log9N is whole number, then how many numbers possible for N between 100 to 100?
Explanation:
log3N+log9Nlog3N+log9N = log3N+log32Nlog3N+log32N = log3N+12log3Nlog3N+12log3N =32log3N32log3N
Now this value should be whole number.
Let 32log3N32log3N = w
⇒log3N=23w⇒log3N=23w
N=3(23w)N=3(23w)
As N is a positive integer, So for w = 0, 3, 6 we get N = 1, 9, 81.
Three values are possible.

Ques. What will be obtained if 8 is subtracted from the HCF of 168, 189, and 231?
Op 1: 15
Op 2: 10
Op 3: 21
Op 4: None of these
Op 5:

Correct Op : 4

7. If a4 +(1/a4)=119 then a power 3-(1/a3) =

a. 32
b. 39
c.  Data insufficient
d.  36
Explanation:
Given that a4+1a4=119a4+1a4=119 , adding 2 on both sides, we get : (a2+1a2)2=121(a2+1a2)2=121
⇒a2+1a2=11⇒a2+1a2=11
Again, by subtracting 2 on both sides, we have, ⇒(a−1a)2=9⇒(a−1a)2=9
⇒a−1a=3⇒a−1a=3
Now, ⇒a3−1a3⇒a3−1a3 = (a−1a)(a2+1a2+1)(a−1a)(a2+1a2+1) = 12×3 = 36