XAT Miscellaneous Questions

Let the no. chosen by Amitabh be x

In the first step, the result got by Amitabh is 2x and the result got by Shashi is 2x+50.

In the second step, the result got by Amitabh is 4x+100 and the result got by Shashi is 4x+150.

In the third step, the result got by Amitabh is 8x+300 and the result got by Shashi is 8x+350.

In the fourth step, the result got by Amitabh is 16x+700 and the result got by Shashi is 16x+750.

In the fifth step, the result got by Amitabh is 32x+1500 and the result got by Shashi is 32x+1550.

In step 5, since both of their results are greater than 1000, so step 4 will be the last step.

Since Amitabh wins the game the result of Shashi must be greater than 1000.

Now, $$16x+750\ge\ 1000$$

or $$x\ge\ 16$$

Hence the minimum value of x should be 16 and the its sum of digits is 7.

VP (HR) visits the plant after a gap of 2 days i.e every third day. Similarly, VP (Operations) visits the plant every fourth day and VP (Sales) visits the plant every sixth day. To find the number of days after which all the three will come together is the lowest common multiple of 3, 4 and 6 which is 12. So, every 12th day, all three VPs would come together. If their meeting was held on 3rd January, then their subsequent meetings would be held on 15th January, 27th January, 8th February, 20th February and so on. However, the CEO is not available till 28th January. So, their next meeting would be held on 8th February.
hence, option C is the correct answer.

If the group had two members, the one in the first position would win.

If the group had three members, the one in the third position would win.

If the group had four members, the one in the first position would again win.

Similarly, for five members, the one in the third position will win

Similarly, for six members, the one in the fifth position will win.

So, for f(2n), the one at 2f(n)-1 will win

And for f(2n+1) members, the one at 2f(n)+1 will win.

When 2n = 2, the winner will be 1.

For 2n = 4, winner will be 2f(2) - 1. And since f(2)=1, for f(4), the winner will be 1.

Now, f(545) = 2f(242) + 1

f(272) = 2f(136) - 1

f(136) = 2f(68) - 1

f(68) = 2f(34) - 1

f(34) = 2f(17) - 1

f(17) = 2f(8) + 1

f(8) = 2f(4) - 1

f(4) = 2f(2) - 1

Since f(2) = 1, f(4) = 1,f(8) = 1,.......f(545) = 67

Thus, the correct option is B.

For f(2n), the one at 2f(n)-1 will win.

And for f(2n+1) members, the one at 2f(n)+1 will win.

When 2n = 2, the winner will be 1.

For 2n = 4, winner will be 2f(2) - 1. And since f(2)=1, for f(4), the winner will be 1.

Now, f(545) = 2f(242) + 1

f(272) = 2f(136) - 1

f(136) = 2f(68) - 1

f(68) = 2f(34) - 1

f(34) = 2f(17) - 1

f(17) = 2f(8) + 1

f(8) = 2f(4) - 1

f(4) = 2f(2) - 1

Since f(2) = 1, f(4) = 1,f(8) = 1, f(17)= 3,f(34)= 5, f(68) = 9, f(136) = 17, f(272) = 33, f(545) = 67

Now, f(544) = 65, f(543) = 63....

So when every second member is removed, for f(n) = A, f(n - m) = A - 2m.

When every x member is removed and f(n) = A, f(n - m) = A - x*m.

If A > f(n), then f(n) = A % f(n)...(i)

Now, f(542) = 437, and x = 300

f(543) = 437 + 300 = 637 % 543 = 194

f(544) = 196 + 300 = 494

f(545) = 496 + 300 = 794 % 545 = 249

Thus, the correct option is C.

Let $$P=\left(x_1,\ y_1\right)$$ and $$Q=\left(x_2,\ y_2\right)$$

$$P^2=P\times P=(x_1\times x_1, y_1+y_1-y_1\times y_1)=(x_1^2, y_1(2-y_1))$$

Since $$x_1$$ and $$y_1$$ are both less than 1, $$x_1^2<x_1$$ and $$y_1(2-y_1)>y_1$$

As P is raised to higher powers, the x-coordinate will keep on decreasing and the y-coordinate will keep on increasing.

As the value of n tends to infinity, $$x_1$$ and $$y_1$$ will tend towards 0 and 1, resptectively.

Thus, $$P^n=(0, 1)$$.

Similarly, $$Q^n=(0, 1)$$

Thus, $$P^n+Q^n=(0\times 0, 1+1-1\times 1)=(0, 1)$$

Hence, the answer is option C.

Let $$P=\left(x_1,\ y_1\right)$$ and $$Q=\left(x_2,\ y_2\right)$$

$$2P=P+P=(x_1+x_1-x_1^2, y_1^2)=(x_1(2-x_1), y_1^2)$$

Since $$x_1$$ and $$y_1$$ are less than 1, $$x_1(2-x_1)>x_1$$ and $$y_1^2<y_1$$.

As the value of n increases the value of x-coordinate tends towards 1 and the value of y-coordinate tends towards 0.

Thus, $$nP=(1, 0)$$

Similarly, $$nQ=(1, 0)$$

Thus, $$nP+nQ=(1+1-1\times 1,  0\times 0)=(1, 0)$$

Hence, the answer is option B.

Let the coupon payments of A, E, B, D and C be a-2d, a-d, a, a+d, and a+2d, respectively.

Since, the coupon payment of A is twice the common difference,

a - 2d = 2d

a = 4d

Thus, the coupon payments of A, E, B, D and C are 2d, 3d, 4d, 5d, and 6d, respectively.

Since the coupon payment on bond B is half the price of bond A, the price of bond A = 2 x 4d = 8d.

Let the price of bond E be $$x$$. Thus, the price of bond B = $$1.75x$$.

Let the face value of bond E be $$e$$. Thus, the face value of bond B = $$2e$$.

Since, the face value of bond C is equal to the price of bond A, the face value of C is 8d.

We know that three of the bonds mature in two years, one in three years and one in five years. We are also given that A matures in 2 and D matures in 5 years.

Tabulating the given information.

Applying the price formula for A, we get

8d = $$\frac{2d}{1.25}+\frac{2d}{1.25^2}+\frac{1000}{1.25^2}$$

Solving this, we get $$d=125$$

Thus, the coupon payments of A, B, C, D, and E are 125, 500, 750, 625 and 375, respectively.

Price of A = 8d = 1000 = Face value of C

We are given that the price of bond C is more than 1800. The time period for C could be 2 or 3 years.

Case 1: Time period is 2 years.

Applying the price formula for C.

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{1000}{1.25^2}$$

Price = 1720 (<1800)

Thus, the maturity period of bond C is 3 years and for bonds B and E will be 2 years.

Applying the price formula again, we get

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{750}{1.25^3}+\frac{1000}{1.25^3}$$

Price of C = 1976

Using the price formula on bond B, we get

1.75$$x$$ = $$\frac{500}{1.25}+\frac{500}{1.25^2}+\frac{2e}{1.25^2}$$ ...(1)

Using the price formula on bond E, we get

$$x$$ = $$\frac{375}{1.25}+\frac{375}{1.25^2}+\frac{2}{1.25^2}$$ ...(2)

Solving equations (1) and (2) simultaneously, we get

$$x=1440$$ and $$e=1406.25$$

Thus, the table becomes

Thus, the face value of bond E is 1406.25.

Hence, the answer is option A.

Let the coupon payments of A, E, B, D and C be a-2d, a-d, a, a+d, and a+2d, respectively.

Since, the coupon payment of A is twice the common difference,

a - 2d = 2d

a = 4d

Thus, the coupon payments of A, E, B, D and C are 2d, 3d, 4d, 5d, and 6d, respectively.

Since the coupon payment on bond B is half the price of bond A, the price of bond A = 2 x 4d = 8d.

Let the price of bond E be $$x$$. Thus, the price of bond B = $$1.75x$$.

Let the face value of bond E be $$e$$. Thus, the face value of bond B = $$2e$$.

Since the face value of bond C is equal to the price of bond A, the face value of C is 8d.

We know that three of the bonds mature in two years, one in three years and one in five years. We are also given that A matures in 2 and D matures in 5 years.

Tabulating the given information.

Applying the price formula for A, we get

8d = $$\frac{2d}{1.25}+\frac{2d}{1.25^2}+\frac{1000}{1.25^2}$$

Solving this, we get $$d=125$$

Thus, the coupon payments of A, B, C, D, and E are 125, 500, 750, 625 and 375, respectively.

Price of A = 8d = 1000 = Face value of C

We are given that the price of bond C is more than 1800. The time period for C could be 2 or 3 years.

Case 1: Time period is 2 years.

Applying the price formula for C.

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{1000}{1.25^2}$$

Price = 1720 (<1800)

Thus, the maturity period of bond C is 3 years and for bonds B and E will be 2 years.

Applying the price formula again, we get

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{750}{1.25^3}+\frac{1000}{1.25^3}$$

Price of C = 1976

Using the price formula on bond B, we get

1.75$$x$$ = $$\frac{500}{1.25}+\frac{500}{1.25^2}+\frac{2e}{1.25^2}$$ ...(1)

Using the price formula on bond E, we get

$$x$$ = $$\frac{375}{1.25}+\frac{375}{1.25^2}+\frac{2}{1.25^2}$$ ...(2)

Solving equations (1) and (2) simultaneously, we get

$$x=1440$$ and $$e=1406.25$$

Thus, the table becomes

Since Madhubala has Rs. 2500, she can either buy one bond each of A and E, or a single bond C.

Return from bonds A and E = 1000 + 2(250) + 1406.25 + 2(375) = 3656.25

Return from bond C = 1000 + 3(750) = 3250

Thus, the maximum return that Madhubala can get is Rs. 3656.25.

Hence, the answer is option C.

Let the coupon payments of A, E, B, D and C be a-2d, a-d, a, a+d, and a+2d, respectively.

Since, the coupon payment of A is twice the common difference,

a - 2d = 2d

a = 4d

Thus, the coupon payments of A, E, B, D and C are 2d, 3d, 4d, 5d, and 6d, respectively.

Since the coupon payment on bond B is half the price of bond A, the price of bond A = 2 x 4d = 8d.

Let the price of bond E be $$x$$. Thus, the price of bond B = $$1.75x$$.

Let the face value of bond E be $$e$$. Thus, the face value of bond B = $$2e$$.

Since, the face value of bond C is equal to the price of bond A, the face value of C is 8d.

We know that three of the bonds mature in two years, two in three years and one in five years. We are also given that A matures in 2 and D matures in 5 years.

Tabulating the given information.

Applying the price formula for A, we get

8d = $$\frac{2d}{1.25}+\frac{2d}{1.25^2}+\frac{1000}{1.25^2}$$

Solving this, we get $$d=125$$

Thus, the coupon payments of A, B, C, D, and E are 125, 500, 750, 625 and 375, respectively.

Price of A = 8d = 1000 = Face value of C

We are given that the price of bond C is more than 1800. The time period for C could be 2 or 3 years.

Case 1: Time period is 2 years.

Applying the price formula for C.

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{1000}{1.25^2}$$

Price = 1720 (<1800)

Thus, the maturity period of bond C is 3 years and for bonds B and E will be 2 years.

Applying the price formula again, we get

Price = $$\frac{750}{1.25}+\frac{750}{1.25^2}+\frac{750}{1.25^3}+\frac{1000}{1.25^3}$$

Price of C = 1976

Using the price formula on bond B, we get

1.75$$x$$ = $$\frac{500}{1.25}+\frac{500}{1.25^2}+\frac{2e}{1.25^2}$$ ...(1)

Using the price formula on bond E, we get

$$x$$ = $$\frac{375}{1.25}+\frac{375}{1.25^2}+\frac{2}{1.25^2}$$ ...(2)

Solving equations (1) and (2) simultaneously, we get

$$x=1440$$ and $$e=1406.25$$

Thus, the table becomes

The price of bond C = 1976.

Hence, the answer is option D.