XAT 2010

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

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, 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

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.

The alphabetical order = CCHJL 

Number of words starting with C = $$4! = 24$$

Number of words starting with H = $$\frac{4 !}{2} = 12$$

Number of words starting with J = $$\frac{4 !}{2} = 12$$

Total words till now = 24 + 12 + 12 = 48

First word starting with L (49th in dictionary) = LCCHJ

Therefore, the 50th word =  LCCJH

Let the number of units of A and B produced be $$x$$ and $$y$$ respectively.

For product A, time taken for grinding = $$2x$$ and polishing = $$3x$$

For product B, time taken for grinding = $$3y$$ and polishing = $$2y$$

Total number of hours of grinding done per day = $$10 \times 12 = 120$$ hrs

Total number of hours of polishing done per day = $$15 \times 10 = 150$$ hrs

=> $$2x + 3y = 120$$ ---------Eqn(I)

and $$3x+ 2y = 150$$ ----------Eqn(II)

Applying the operation : 3* Eqn(I) - 2* Eqn(II), we get :

=> $$(6x - 6x) + (9y - 4y) = 360 - 300$$

=> $$y = \frac{60}{5} = 12$$

=> $$x = 42$$

$$\therefore$$ Profit made by the manufacturer = $$(42 \times 5) + (12 \times 7)$$

= $$210 + 84 = Rs. 294$$

Volume of tank = $$150 \times 120 \times 100 = 18,00,000 cm^3$$

Volume of water in the tank = $$12,81,600 cm^3$$

Volume to be filled in the tank = $$18,00,000 - 12,81,600 = 5,18,400 cm^3$$

Let the number of bricks to be placed in the tank = $$x$$

Volume of $$x$$ bricks = $$x \times 20 \times 6 \times 4 = 480x cm^3$$

Each brick absorbs $$(\frac{1}{10})^{th}$$ of its volume in water

=> $$x$$ bricks will absorb = $$\frac{480 x}{10} = 48x cm^3$$

$$\therefore$$ $$5,18,400 + 48x = 480x$$

=> $$480x - 48x = 432x = 5,18,400$$

=> $$x = \frac{518400}{432} = 1200$$

Let the number of matches that India needs to play = $$x$$

Now, if $$1 - (\frac{5}{6})^x \geq \frac{1}{2}$$ , we can consider that India has a fair chance of winning the match.

=> $$(\frac{5}{6})^x \leq \frac{1}{2}$$

If, $$x = 3$$, we get $$\frac{125}{216}$$, which is greater than $$\frac{1}{2}$$

If, $$x = 4$$, we get $$\frac{625}{1296}$$, which is less than $$\frac{1}{2}$$

$$\therefore$$ The number of matches that India needs to play must be atleast 4

Since each carton should contain boxes of the same brand of chocolates and all boxes being of equal size, to get the minimum number of cartons, we should have the maximum number of boxes in each carton.

Thus, the number of boxes in each carton = H.C.F (96,240,336)

= 48

So, we will get minimum number of cartons if there are 48 boxes in each carton.

$$\therefore$$ Number of cartons = $$\frac{96}{48} + \frac{240}{48} + \frac{336}{48}$$

= $$2 + 5 + 7 = 14$$

Total population = 70000

Average consumer expenditure on Roses and carnations = 47.4

Total expenditure on both carnations and roses = 47.4*70000

                                                                       = 3318000

Given total expenditure on roses = 1136916

So the total expenditure on carnations = 3318000 - 1136916

                                                           = 2181084

Similarly in the month of March 

Average consumer expenditure on Roses and carnations = 49.5

Total expenditure on both carnations and roses = 49.5 *70000

                                                                       = 3465000

Given total expenditure on roses = 1137915

So the total expenditure on carnations = 3465000 - 1137915

                                                           = 2327085

Percentage increase in Expenditure  = (2327085-2181084)/2181084

=6.69%

Hence C is the correct answer

Expenditure on roses in January = 1136916

Price of roses per dozen = 99

Roses sold = 1136916/99

                =11484 dozens

Expenditure on roses in July = 1188432

Price of roses per dozen = 144

Roses sold = 1188432/144

                =8253 dozens 

Sales of roses in July as compared to January = [(8253-11484)/11484]*100

                                                                    =  - 28.13%

Since Change is negative, there is a decrease of 28.13%.
Hence B is the correct answer.