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

-------------

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

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2 9 7

8 9 1

-------------

9 2 0 7

-------------

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 a

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

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

-------------

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 ${\log _3}N + {\log _9}N$ 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 a

^{4}+(1/a^{4})=119 then a power 3-(1/a^{3}) =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

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