# Elitmus previous questions - 2

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 favor of other five. Who won the elections?
i)  4 older cast their vote in favor 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. Ram draws a card randomly among card 1-23 and keep it back.  Then Sam draws a card among those.  What is the probability that Sam has drawn a card greater than ram.
Explanation:
If Ram draws 1, then Sam can draw anything from 2 to 23 = 22 ways
If Ram draws 2, then Sam can draw anything from 3 to 23 = 21 ways
. . . .
. . . .
If Ram draws 23, Sam has no option = 0 ways.
Total required ways = 22 + 21 + 20 + . . . . + 0 =  $\frac{{22 \times 23}}{2}$ = 253
Total ways of drawing two cards = 23×23
Required probability = $\dfrac{{253}}{{529}} = \dfrac{{11}}{{23}}$

4. Decipher the following multiplication table:
M A D
B E
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M A D
R A E
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A M I D
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Explanation:
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
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M 9 D
8 9  1
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9 M 0 D
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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
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2 9 7
8  9  1
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9 2 0 7
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5. If ${\log _3}N + {\log _9}N$ is whole number, then how many numbers possible for N between 100 to 100?
Explanation:
${\log _3}N + {\log _9}N$ = ${\log _3}N + {\log _{{3^2}}}N$ = ${\log _3}N + \dfrac{1}{2}{\log _3}N$ =$\dfrac{3}{2}{\log _3}N$
Now this value should be whole number.
Let $\dfrac{3}{2}{\log _3}N$ = w
$\Rightarrow {\log _3}N = \dfrac{2}{3}w$
$N = {3^{\left( {\frac{2}{3}w} \right)}}$
As N is a positive integer, So for w = 0, 3, 6 we get N = 1, 9, 81.
Three values are possible.

6. If an ant moves a distance x,then turn left 120 degree and travel x distance again and turn right 120 dgree and travel, he is 9 inches from the starting point,  what is the value of x?
a.  3root(3)
b.  9root(3)
c.  3root(3)/2
d.  3root(3)/4
Explanation:

See the above diagram.  (A bit confusing though!!)
The ant starts from A travelled to B, turned 120 degrees left and travelled to C. and turned 120 degrees right and travelled to D.
From the diagram, $\angle ACB = {60^0}$.  As BC and AB are equal, $\angle BCA = \angle BAC = {60^0}$.
So $\Delta ABC$ is equilateral.
Also $\angle BCD = {60^0}$.
As, $\angle ACB = \angle BCD$ equal, CB divides AD into two equal parts. Aso DE = 9/2.
In the triangle CED, $Sin{\rm{ }}{60^0} = \dfrac{{\sqrt 3 }}{2} = \dfrac{{{\textstyle{9 \over 2}}}}{x}$
So $x = \dfrac{9}{{\sqrt 3 }}$

7. If a4 +(1/a4)=119 then a power 3-(1/a3) =
a. 32
b. 39
c.  Data insufficient
d.  36
Explanation:
Given that ${a^4} + \dfrac{1}{{{a^4}}} = 119$ , adding 2 on both sides, we get : ${\left( {{a^2} + \dfrac{1}{{{a^2}}}} \right)^2} = 121$
$\Rightarrow {a^2} + \dfrac{1}{{{a^2}}} = 11$
Again, by subtracting 2 on both sides, we have, $\Rightarrow {\left( {a - \dfrac{1}{a}} \right)^2} = 9$
$\Rightarrow a - \dfrac{1}{a} = 3$
Now, $\Rightarrow {a^3} - \dfrac{1}{{{a^3}}}$ = $\left( {a - \dfrac{1}{a}} \right)\left( {{a^2} + \dfrac{1}{{{a^2}}} + 1} \right)$ = 12×3 = 36