# Elitmus previous questions

1. If a number x is in octal form and having zero as unit place then if we change that number x in decimal form then what is the probability that number x having zero as unit place?
a) 1/40
b) 3/560
c) 11/64
d) 1/8
e) data insufficient
Explanation:

Ans: 11/40

2. log x = 1; log y = 2 then then value of log y /log x
a) 1/4
b) 2/50
c) 1/6
d) 1
e) data insufficient

Ans: 1

3. Einstein walks on an escalator at a rate of 5steps per second and reaches the other end in 10 sec. while coming back, walking at the same speed he reaches the starting point in 40secs. What is the number of steps on the escalator?
a) 40
b) 60
c) 120
d) 80
e) data insufficient

Explanation:
Escalator problems are similar to boats and streams problems.  If we assume man's speed as 'a m/s' and escalator speed as 'b m/sec' then while going up man's speed becomes 'a -b' and while coming down 'a + b'.
In this question, Let the speed of escalator be b steps per sec.  And length of escalator be L.   Einstein's speed = 5 steps/ sec
While going down,  L5+x=10L5+x=10 ⇒⇒ L = 50 + 10x
While coming up, L5−x=40L5−x=40 ⇒⇒ L = 200 - 40x
Multiply the first equation by 4, and add to the second, we get L = 80

4.

(Screenshot taken while taking eLitmus Test)

5. Data sufficiency question:
Is x has only '3' factors?
a) x^2 has only '5' factors.
b) one of its factor must be prime number

Explanation:
Only squares of primes has 3 factors.  For example, 4 has three factors. 1, 2, 4.
Now when we square 4, we get 16, which has 5 factors. 1, 2, 4, 8, 16.  So Statement 1 is sufficient.
Statement two says that one of the factor must be prime number.  10 has 4 factors.  i.e., 1, 2, 5, 10 and one of which is prime. but we cannot say whether numbers of this format always have three factors or not. So statement 2 is insufficient.

6.  Data sufficiency question:
If solution (x) with alcohol (14%) and solution (y) with alcohol (20%) mixed together to form 24 liters of mixture.  Then the amount of solution(x) is?
a) The alcohol in solution(mixture) is 15%
b) If we inter changed the amounts of solutions 'x' and 'y' we get mixture of alcohol %,44%

Explanation:
Let a liters of first solution, and 'b' liters of second solution has been taken.
So using weighted average formula = n1×a1+n2×a2n1+n2n1×a1+n2×a2n1+n2 = n1×14+n2×20n1+n2=15n1×14+n2×20n1+n2=15
⇒n1=5n2⇒n1n2=51⇒n1=5n2⇒n1n2=51
As total volume was given as 24 liters, first solution = 16×24=416×24=4
So statement 1 is sufficient.

Let us take volume of the first solution is 'a' liters. Then volume of the second = 24 - a.
Again by using the question statement, and statement 2, (24−a)×14+a×2024=44(24−a)×14+a×2024=44
From this also, value of 'a' can be calculated.  So Statement 2 also sufficient.

7. How many solution  does the expression x+3‾‾‾‾‾√=xx+3‾‾‾‾‾√x+3=xx+3 has?

Explanation:
Squaring on the both sides, x+3=x2(x+3)x+3=x2(x+3)
Bring x+3 to one side and take common, (x2−1)(x+3)(x2−1)(x+3)
So roots are, 1, -1, 3. But -1 is not a root. So 1, -3 are the roots.

(∵∵ when you square the equation additional roots may be generated. So finally you have to check manually, whether the given roots satisfy the original equation or not)

8. In 2 bags, there are to be put together 5 red and 12 white balls, neither bag being empty. How must the balls be divided so as to give a person who draws 1 ball from either bag-
(i) the least chance of drawing a red ball ?
(ii) the greatest chance of drawing a red ball ?

Explanation:
(1) Put 1 white in first bag, and all the remaining ball in second bag.  Probability = 12×0+12×51612×0+12×516 = 532532
(2) To maximize, we put 1 red in the first bag, and all the remaining in the second. Probability = 12×1+12×43212×1+12×432 = 916916

9.

(Screenshot taken while taking eLitmus Test)

10. How many two digit numbers are there such that the product of their digits after reducing it to the smallest form is a prime number? for example if we take 98 then 9*8=72, 72=7*2=14, 14=1*4=4. Consider only 4 prime numbers (2,3,5,7)

Explanation:
2 = 12 or 21 So 1×2, 2×1, 3×4, 4×3, 2×6, 6×2, 3×7, 7×3
3 = 13 or 31 So 1×3, 3×1
5 = 15, 51 So 1×5, 5×1, 3×5, 5×3, 7×5, 5×7
7 = 17 or 71 So 1×7, 7×1
15 = 3×5 = 5×3
So total 18 numbers = 12,13,15,17,21,26,31,34,35,37,43,51,53,57,62,71,73,75

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