Elitmus previous questions - 2

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 =  22×232$\frac{22\times 23}{2}$ = 253
Total ways of drawing two cards = 23×23
Required probability = 253529=1123${\displaystyle \frac{253}{529}}={\displaystyle \frac{11}{23}}$

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:
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
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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 log3N+log9N${\log }_{3}N+{\log }_{9}N$ is whole number, then how many numbers possible for N between 100 to 100?
Explanation:
log3N+log9N${\log }_{3}N+{\log }_{9}N$ = log3N+log32N${\log }_{3}N+{\log }_{3^2}N$ = log3N+12log3N${\log }_{3}N+{\displaystyle \frac{1}{2}}{\log }_{3}N$ =32log3N${\displaystyle \frac{3}{2}}{\log }_{3}N$
Now this value should be whole number.
Let 32log3N${\displaystyle \frac{3}{2}}{\log }_{3}N$ = w
⇒log3N=23w$\Rightarrow {\log }_{3}N={\displaystyle \frac{2}{3}}w$
N=3(23w)$N=3^{\left(\frac{2}{3}w\right)}$
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, ∠ACB=600$\mathrm{\angle }ACB={60}^0$.  As BC and AB are equal, ∠BCA=∠BAC=600$\mathrm{\angle }BCA=\mathrm{\angle }BAC={60}^0$.
So ΔABC$\mathrm{\Delta }ABC$ is equilateral.
Also ∠BCD=600$\mathrm{\angle }BCD={60}^0$.
As, ∠ACB=∠BCD$\mathrm{\angle }ACB=\mathrm{\angle }BCD$ equal, CB divides AD into two equal parts. Aso DE = 9/2.
In the triangle CED, Sin600=3√2=92x$Sin{60}^0={\displaystyle \frac{\sqrt{3}}{2}}={\displaystyle \frac{\frac{9}{2}}{x}}$
So x=93√$x={\displaystyle \frac{9}{\sqrt{3}}}$

7. If a⁴ +(1/a⁴)=119 then a power 3-(1/a³) = 
a. 32 
b. 39
c.  Data insufficient
d.  36
Explanation:
Given that a4+1a4=119$a^4+{\displaystyle \frac{1}{a^4}}=119$ , adding 2 on both sides, we get : (a2+1a2)2=121${\left(a^2+{\displaystyle \frac{1}{a^2}}\right)}^2=121$
⇒a2+1a2=11$\Rightarrow a^2+{\displaystyle \frac{1}{a^2}}=11$
Again, by subtracting 2 on both sides, we have, ⇒(a−1a)2=9$\Rightarrow {\left(a-{\displaystyle \frac{1}{a}}\right)}^2=9$
⇒a−1a=3$\Rightarrow a-{\displaystyle \frac{1}{a}}=3$
Now, ⇒a3−1a3$\Rightarrow a^3-{\displaystyle \frac{1}{a^3}}$ = (a−1a)(a2+1a2+1)$\left(a-{\displaystyle \frac{1}{a}}\right)\left(a^2+{\displaystyle \frac{1}{a^2}}+1\right)$ = 12×3 = 36