# 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