XAT 2010 Question Paper - QUANT

XADY = XA + AD + DY = 2/2 + (2 * 3.14 * 2)/4 + (30/360) * (2 * 3.14 * 2) = 5.19
XOBY = XO + OB + BY = 2/2 + 2 + (60/360) * (2 * 3.14 * 2) = 5.09
XODY = XO + OD + DY = 2/2 + 2 + (30/360) * (2 * 3.14 * 2) =  4.04
Hence, option D is the correct answer.

Expression : $$x^{2}+ 4xy+ 6y^{2}-4y+ 4$$

= $$(x^2 + 4xy + 4y^2) + (2y^2 - 4y + 4)$$

= $$(x + 2y)^2 + \frac{1}{2} (4y^2 - 8y + 8)$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y)^2 - 2 (2y) (2) + (2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} [(2y - 2)^2 + 4]$$

= $$(x + 2y)^2 + \frac{1}{2} (2y - 2)^2 + 2$$

Since, $$x$$ and $$y$$ are real, => Min value of $$(x + 2y)^2 = 0$$

Minimum value of $$(2y - 2)^2 = 0$$

$$\therefore$$ Minimum value of expression = $$0 + 0 + 2 = 2$$

Given that unit digit of $$X$$ and product of all 4 digits of $$X$$ are prime.

The product of two numbers to be prime is possible only when one of the numbers is prime and the other is ‘1’.

The possibilities for the prime unit digits are - 2,3,5,7     (Note that 1 is not a prime number)

Hence the possibility of remaining 3 digits, considering the product of all 4 digits to be prime is ‘111’ only.

Hence all the possible numbers are 1112,1113,1115,1117

$$\therefore$$ Total 4 integers are possible.

Let total employees in Sun Metals = $$100x$$

Number of employees who are general graduates = $$\frac{40}{100} \times 100x = 40x$$

=> Number of employees who are engineers = $$100x - 40x = 60x$$

Now, number of engineers who earn more than Rs. 5 lakh/year = $$\frac{75}{100} \times 60x = 45x$$

Number of employees (both general graduates and engineers) who earn more than Rs. 5 lakhs/year = $$\frac{50}{100} \times 100x = 50x$$

=> Number of general graduates who earn more than Rs. 5 lakhs/year = $$50x - 45x = 5x$$

Thus, number of general graduates who earn less than Rs. 5 lakhs/year = $$40x - 5x = 35x$$

$$\therefore$$ Proportion of the general graduates employed by the organisation earn Rs. 5 lakh or less

= $$\frac{35 x}{40 x} = \frac{7}{8}$$

Given : AB = AC = BC = 3  cm and BD = $$\frac{1}{2}$$ CD

AE is median.

To find : $$AD = ?$$

Solution : BD + CD = 3

=> $$BD + 2BD = 3BD = 3$$

=> $$BD = \frac{3}{3} = 1$$ cm

Also, since AE is media => $$BE = CE = \frac{3}{2}$$ cm

=> $$DE = BE - DE = \frac{3}{2} - 1 = \frac{1}{2}$$ cm

Also, AE = $$\frac{\sqrt{3}}{2} a = \frac{3 \sqrt{3}}{2}$$ cm

In $$\triangle$$ ADE

=> $$(AD)^2 = (AE)^2 + (DE)^2$$

=> $$(AD)^2 = (\frac{3 \sqrt{3}}{2})^2 + (\frac{1}{2})^2$$

=> $$(AD)^2 = \frac{27}{4} + \frac{1}{4} = \frac{28}{4}$$

=> $$AD = \sqrt{7}$$ cm

To find : $$EF = x = ?$$

Solution : In $$\triangle$$ ABC and $$\triangle$$ EFC

$$\angle ACB = \angle ECF$$  (common)

$$\angle ABC = \angle EFC = 90$$

=> $$\triangle ABC \sim \triangle EFC$$

=> $$\frac{x}{2} = \frac{5 - d}{5}$$ ----------Eqn(I)

Similarly, $$\triangle BCD \sim \triangle BFE$$

=> $$\frac{x}{3} = \frac{d}{5}$$ --------Eqn(II)

Adding Eqns (I) & (II), we get :

=> $$\frac{x}{2} + \frac{x}{3} = \frac{5 - d}{5} + \frac{d}{5}$$

=> $$\frac{3x + 2x}{6} = \frac{5}{5}$$

=> $$5x = 6 \times 1 = 6$$ 

=> $$x = \frac{6}{5} = 1.2 m$$

Let the volume of the solution with 30 % acid content lie between $$v_1$$ and $$v_2$$, where we get a 20% acid solution for $$v_1$$

For $$v_2$$, we get a 25 % acid solution as the resultant mixture.

=> $$15 \% (200) + 30 \% (v_1) = 20 \% (200 + v_1)$$

=> $$30 + 0.3 v_1 = 40 + 0.2 v_1$$

=> $$0.1 v_1 = 10$$ => $$v_1 = 10 \times 10 = 100$$ litres

Similarly, $$15 \% (200) + 30 \% (v_2) = 25 \% (200 + v_2)$$

=> $$30 + 0.3 v_2 = 50 + 0.25 v_2$$

=> $$0.05 v_2 = 20$$ => $$v_2 = 20 \times 20 = 400$$ litres

$$\therefore$$ For the acid content in the resultant mixture to lie between 20 % and 25 %, the volume of the 30 % concentration acid solution must lie between 100 litres and 400 litres.

Expected profit when brand ambassador is used = (3.5 - 3.45)[100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3] lakh = 0.05lakh*77000= 3850 lakh

Expected profit when brand ambassador is not used =  (3.5 - 3.45)[80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3] lakh =0.05lakh*53000= 2650 lakh

Difference = (3850 - 2650) = 1200 lakh = 12 crore.

This is the maximum amount of money that the company can afford to pay its brand ambassador.

Hence, option D is the correct answer.

Expected profit when brand ambassador is used = (3.5 - 3.45)[100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3] lakh = 0.05lakh*77000= 3850 lakh

Expected profit when brand ambassador is not used = (3.5 - 3.45)[80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3] lakh =0.05lakh*53000= 2650 lakh

Difference = (3850 - 2650) = 1200 lakh = 12 crore.

When profit is Rs.12.0 crores when Mr. Khan Advertises for the company.

Then profit is (12.0 - 9.0) = Rs.3 crore 

No. of cars sold when Mr. khan advertises = [100000 × 0.3 + 80000 × 0.4 + 50000 × 0.3]= 77000

No. of cars sold when no popular actor is for advertisement= [80000 × 0.3 + 50000 × 0.4 + 30000 × 0.3]=53000

No of extra vehicles sold due to Mr. Khan= 24000

Suppose p is the amount escalated in cost of putting a car on the road,

24000p = 3,00,00,000

p = Rs1250

Had the cost not been increased, the company would have a extra profit of Rs. 1250 due to brand ambassador.

So, this the maximum escalation for which the company can afford signing contract with Mr. Khan.

Hence, option C is the correct answer.

As calculated in the last question, the company would have an additional profit or Rs. 1250/unit.

But,  the cost is going up by Rs. 1000/unit

Therefore, profit/unit = Rs. 1250 - Rs. 1000 = Rs. 250

∴    Addition/increase in profit

Rs. $$3 * \dfrac{250}{1250}$$ = Rs. 60 lakh

Hence, option B is the correct answer.

Let us draw the diagram first,

Let L$$_{1}$$: 2x+ 3y -1 = 0

L$$_{2}$$: x+ 2y -3 = 0

L$$_{3}$$: 5x-6y- 1 = 0

With respect to L$$_{1}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are opposite side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ < 0

$$\Rightarrow$$ $$(2a+ 3a^2-1)(-1)<0$$

$$\Rightarrow$$ $$(3a^2+2a-1)>0$$

$$\Rightarrow$$ a < -1 or a > $$\frac{1}{3}$$   ... (1)

With respect to L$$_{2}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(a+ 2a^2-3)(-3)>0$$

$$\Rightarrow$$ $$(2a^2+a-3)<0$$

$$\Rightarrow$$ $$\frac{-3}{2}$$ < a < 1  ... (2)

With respect to L$$_{3}$$, we can see that the point $$(a, a^{2})$$ lies within the triangle and (0, 0) are on the same side. Therefore,

 L$$_{(a, a^2)}$$* L$$_{(0, 0)}$$ > 0

$$\Rightarrow$$ $$(5a- 6a^2-1)(-1)>0$$

$$\Rightarrow$$ $$(6a^2-5a+1)>0$$

$$\Rightarrow$$ a < $$\frac{1}{3}$$ or a > $$\frac{1}{2}$$   ... (3)

From equation (1), (2) and (3) we can say that 

a $$\epsilon$$ (-3/2,-1) ⋃ (1/2, 1). Hence, option C is the correct answer.

Let $$p \times \frac{0.34 mt}{mt}$$m packets of milk be prepared in unit time at the normal speed.

Now, at normal speed in $$t$$ time, the number of packets of milk that would be produced = $$mt$$

=> Number of packets of milk produced at fast speed = $$(\frac{110}{100} \times m) + (\frac{60}{100} \times t) = 0.66 mt$$

The target for the supervisor = $$mt$$ packets

Number of packets produced at normal speed = $$mt - 0.66 mt = 0.34 mt$$

Let the probability of a packet being damaged when produced at normal speed = $$p$$

=> Probability that a packet is damaged when produced at fast speed = $$2p$$

The probability that a packet selected at random will be damaged = 0.112

=> $$(p \times \frac{0.34 mt}{mt}) + (2p \times \frac{0.66 mt}{mt}) = 0.112$$

=> $$0.34p + 1.32p = 1.66 p = 0.112$$

=> $$p = \frac{0.112}{1.66} = 0.067$$

$$\therefore$$ Probability that a packet will not be damaged at normal speed = $$1 - 0.067 = 0.93$$

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> $$x_8 = a + 7d = -1 + 7(3) = 20$$

To find y coordinate, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

$$\therefore y_8 = p x_8 + q = \frac{1}{2} (20) + (-3) = 7$$ 

$$x_1 , x_2 , x_3,......., x_n$$ are in A.P. Let the first term be $$a$$ and common difference be $$d$$

Also, $$x_2 = a + d = 2$$

and $$x_{24} = a + 23d = 68$$

Solving above equations, we get : $$a = -1$$ and $$d = 3$$

=> x coordinates = {$$x_1 , x_2 , x_3, x_4 , x_5 ,.......$$} = {$${-1 , 2 , 5 , 8 , 11 , 14 ,......}$$}        -------------(i)

To find y coordinates, we will use $$y = px + q$$

$$\because A_2 (2, -2)$$ and $$A_{24} (68, 31)$$

Substituting in above equation, => $$-2 = 2p + q$$

and $$31 = 68p + q$$

Solving above equations, we get : $$p = \frac{1}{2}$$ and $$q = -3$$

=> $$y_1 = px_1 + q = \frac{1}{2}(-1) + (-3) = -3.5$$

Similarly, y coordinates = {$$y_1 , y_2 , y_3, y_4 , y_5 ,.......$$} = {$${-3.5 , -2 , \frac{-1}{2} , 1 ,......}$$}         -------(ii)

From (i) and (ii),

=> $$A_1 = (-1 , -3.5)$$

$$A_2 = (2 , -2)$$

$$A_3 = (5 , \frac{-1}{2})$$

$$A_4 = (8 , 1)$$

For the points to lie in the first quadrant, the coordinates of both $$(x_k , y_k)$$ must be positive.

Since, $$d$$ is positive and $$y$$ is a linear relation in $$x$$, the corresponding coordinates of $$A_k$$ i.e. $$A_5$$ onwards will be increasing.

Thus, only for $$A_1 , A_2$$ and $$A_3$$ we do not have both coordinates positive.

$$\therefore$$ Only 3 points do not lie in the first quadrant.

Expression : $$f(x + y) = f(x).f(y)$$

and $$f(1) = 2$$

Putting, $$x=y=1$$, => $$f(1 + 1) = f(1).f(1)$$

=> $$f(2) = 2 \times 2 = 4$$

If $$x = 2 , y = 1$$ => $$f(3) = f(2).f(1)$$

=> $$f(3) = 4 \times 2 = 8$$

Similarly, $$f(4) = f(3).f(1) = 8 \times 2 = 16$$

The pattern followed : $$f(n) = 2^n$$

Now,  $$\sum_{x=1}^nf(x) = 1022$$

= $$f(1) + f(2) + f(3) + ............+ f(n) = 1022$$

=> $$2^1 + 2^2 + 2^3 + ......... + 2^n = 1022$$

The series is a G.P. with first term, $$a = 2$$ and common ratio, $$r = 2$$

=> Sum of G.P. = $$\frac{a (r^n - 1)}{r - 1}$$

=> $$\frac{2 (2^n - 1)}{2 - 1} = 1022$$

=> $$2^n - 1 = \frac{1022}{2} = 511$$

=> $$2^n = 511 + 1 = 512 = 2^9$$

Comparing both sides, we get : $$n = 9$$

Amarendra took 25 minutes to cover 15 km. In the same time Dharmendra travel 2500m less, i,e, 12.5 km

=> Speed of Amarendra = $$\frac{15}{25} = 0.6$$ km/min

Speed of Dharmendra = $$\frac{12.5}{25} = 0.5$$ km/min

=> Time taken by Dharmendra to reach the station = $$\frac{15}{0.5} = 30$$ minutes

Next day, Dharmendra started 7 minutes early, so he will reach the station = 30 - 25 - 7 = -2 minutes

=> 2 minutes or 120 seconds before Amarendra.

Length of side of $$S_{n + 1}$$ = Length of diagonal of $$S_n$$

=> Length of side of $$S_{n + 1}$$ = $$\sqrt{2}$$ (Length of side of $$S_{n}$$)

=> $$\frac{\textrm{Length of side of }S_{n + 1}}{\textrm{Length of side of }S_n} = \sqrt{2}$$

=> Sides of $$S_1 , S_2 , S_3 , S_4,........, S_n$$ form a G.P. with common ratio, $$r = \sqrt{2}$$

It is given that, $$S_3 = ar^2 = 4$$

=> $$a (\sqrt{2})^2 = 4$$

=> $$a = \frac{4}{2} = 2$$

$$\therefore$$ $$n^{th}$$ term of G.P. = $$a (r^{n - 1})$$

= $$2 (\sqrt{2})^{n - 1}$$

=$$2^[{\frac{n+1}{2}}]$$

Angle covered by the hour hand in 12 hours = 360°

In 1 hour = $$\frac{360}{12} = 30^{\circ}$$

and in 1 minute = $$\frac{30}{60} = \frac{1}{2}^{\circ}$$

Similarly, angle covered by minute hand in 1 hour = 360°

In 1 minute = $$\frac{360}{60} = 6^{\circ}$$

=> Every minute, the angle between the two hands changes by = $$6 - \frac{1}{2} = \frac{11}{2}^{\circ}$$

$$\therefore$$ From 7:45 A.M. to 7:47 A.M.,i.e. in 2 minutes the angle between the two hands will change by

= $$2 \times \frac{11}{2} = 11^{\circ}$$

In the figure, $$\triangle AQR \sim \triangle APS$$

=> $$\frac{AQ}{AP} = \frac{QR}{PS} = \frac{AR}{AS} = k$$ --------Eqn(I)

Statement I : PQ = 3 cm , RS = 4 cm , $$\angle$$ QPS = 60°

In right $$\triangle$$ PQM

=> $$sin 60^{\circ} = \frac{QM}{QP}$$

=> $$\frac{\sqrt{3}}{2} = \frac{QM}{3}$$

=> $$QM = \frac{3 \sqrt{3}}{2} = RN$$

Similarly, $$sin (\angle RSN) = \frac{3 \sqrt{3}}{8}$$

=> $$\angle RSN = sin^{-1} (\frac{3 \sqrt{3}}{8})$$

$$\therefore$$ In $$\triangle$$ APS

=> $$\angle PAS = 180^{\circ} - \angle APS - \angle PSA$$

=> $$\angle PAS = 120^{\circ} - sin^{-1} (\frac{3 \sqrt{3}}{8})$$

Thus, statement I alone is sufficient. 

Statement II : PS = 10, QR = 5

From eqn(I), $$k = \frac{1}{2}$$

But, we do not know anything regarding the measures of the remaining sides or any of the angles.

So, statement II is not sufficient.

Expression : $$A_{n+1}=A_{n}^{2}+1$$ ---------Eqn(I)

Statement I : $$A_{0} = 1$$

Putting n = 0 in Eqn (I), => $$A_1 = A_0^2 + 1 = 1 + 1 = 2$$

Similarly, $$A_2 = A_1^2 + 1 = 4 + 1 = 5$$ and so on

We can find the values of $$A_{900}$$ and $$A_{1000}$$ and also their greatest common divisor.

Thus, statement I alone is sufficient.

Statement II : We have $$A_{1} = 2$$

In the above manner, we can determine $$A_{900}$$ and $$A_{1000}$$ and also their greatest common divisor.

Thus, statement II alone is sufficient.

$$\therefore$$ Either statement alone is sufficient.

Given that the numbers are in G.P.

Let the common ratio be ‘r’, hence the series a,b,c,d,e can also be expressed as:

$$a , ar , ar^2 , ar^3 , ar^4$$

lcm(a,b) = lcm$$(a,ar) = ar$$

lcm(b,c) = lcm$$(ar,ar^2) = ar^2$$

lcm(c,d) = lcm$$(ar^2,ar^3) = ar^3$$

lcm(d,e) = lcm$$(ar^3,ar^4) = ar^4$$

$$\therefore \frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$$

= $$\frac{1}{ar} + \frac{1}{ar^2} + \frac{1}{ar^3} + \frac{1}{ar^4}$$

= $$\frac{1}{a} (\frac{1}{r} + \frac{1}{r^2} + \frac{1}{r^3} + \frac{1}{r^4})$$

To get max value of this, ‘a’ and ‘r’ should be minimum.

It is given that $$1 \leq a$$ => Minimum value of ‘a’ = 1

For the values in the series to be integers, the minimum common ratio, r = 2   ($$r \leq 1$$ won’t work here as it is an increasing GP)

Substituting values of 'a' and 'r' in the expression, we get :

Max value = $$\frac{1}{1} (\frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \frac{1}{2^4})$$

= $$\frac{8 + 4 + 2 + 1}{16} = \frac{15}{16}$$

Let total cost price of Books and More = $$RS. 100x$$

It is given that Books and More have earned 20% profit overall

=> Total S.P. = $$\frac{120}{100} \times 100x = 120x$$

Thus, S.P. of music CDs = $$\frac{35}{100} \times 120x = 42x$$

S.P. of books = $$42x + \frac{50}{100} \times 42x = 63x$$

=> S.P. of DVDs = $$120x - 42x - 63x = 15x$$

40% profit is earned in music CDs and 25% profit in books.

=> C.P. of music CDs = $$\frac{100}{140} \times 42x = 30x$$

and C.P. of books = $$\frac{100}{125} \times 63x = 50.4x$$

=> C.P. of DVDs = $$100x - 30x - 50.4x = 19.6x$$

=> Loss made on DVDs = $$19.6x - 15x = 4.6x$$

$$\therefore$$ Loss % on DVDs = $$\frac{4.6 x}{19.6 x} \times 100 = 23.46 \%$$

Let total selling price of books and more = $$Rs. 100x$$

Thus, S.P. of music CDs = $$\frac{35}{100} \times 100x = 35x$$

S.P. of books = $$35x + \frac{50}{100} \times 35x = 52.5x$$

=> S.P. of DVDs = $$100x - 35x - 52.5x = 12.5x$$

40% profit is earned in music CDs and 25% profit in books.

=> C.P. of music CDs = $$\frac{100}{140} \times 35x = 25x$$

and C.P. of books = $$\frac{100}{125} \times 52.5x = 42x$$

It is given that Books and More made 50% loss in film DVDs

=> C.P. of DVDs = $$\frac{100}{50} \times 12.5x = 25x$$

Thus, total C.P. of books and more = $$25x + 42x + 25x = 92x$$

$$\therefore$$ Profit made by books and more = $$\frac{100x - 92x}{92x} \times 100$$

= $$\frac{8}{92} \times 100 = 8.69 \% \approx 8.7 \%$$

Area of parallelogram = $$ab sin 60 = 15\frac{\sqrt{3}}{2}$$

=> $$\frac{\sqrt{3}}{2} ab = 15 \frac{\sqrt{3}}{2}$$

=> $$ab = 15$$

Using cosine rule in $$\triangle$$ ABD

=> $$cos 120 = \frac{a^2 + b^2 - 7^2}{2 ab}$$

=> $$\frac{-1}{2} = \frac{a^2 + b^2 - 49}{30}$$

=> $$a^2 + b^2 = 49 - 15 = 34$$

Also, $$(a + b)^2 = a^2 + b^2 + 2ab$$

=> $$(a + b)^2 = 34 + 2(15) = 64$$

=> $$(a + b) = \sqrt{64} = 8$$

$$\therefore$$ Perimeter of parallelogram = $$2 (a + b) = 2 \times 8 = 16$$ cm

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 (Costs / Sales Revenue) * 100 = x%

(Profit / Sales Revenue) * 100 = (100-x) % 

(Profit / Sales Revenue) = (100-x) /100 % 

(Profit / Price * Sales Quantity) =(100-x) /100 % 

Profit / Sales Quantity =[(100-x) /100]* Price  

On tabulating the Profit rate of both the factories , we get 

Clearly C is the correct answer

For Jan- Mar 2009 Paharpur cement 

Sales Quantity = 543278 

Cost as percentage of revenue = 87.56 % 

Profit as a percentage of revenue = 100-87.56

                                                   = 12.44 % 

Profit = (12.44/100)* 543278 *304

         = 20545470

Profit rate = 37.82

For Jan- Mar 2009 Bahsin Cement 

Sales Quantity = 526532 

Cost as percentage of revenue = 91.34 % 

Profit as a percentage of revenue = 100-91.34 

                                                   = 8.66 % 

Profit = (8.66/100)* 526532*440

         = 20062975.3

Profit rate = 30.5

For Apr- Jun 2009 Paharpur cement 

Sales Quantity = 698236

Cost as percentage of revenue = 91.03% 

Profit as a percentage of revenue = 100-91.03

                                                   = 8.97 % 

Profit = (8.97/100)* 698236*340

         = 21294801.53

Profit rate = 30.5

For Apr- Jun 2009 Bahsin Cement 

Sales Quantity = 499874

Cost as percentage of revenue = 89.96% 

Profit as a percentage of revenue = 100-89.96 

                                                   = 10.04 % 

Profit = (10.04/100)* 499874*430

         = 21580560.33

Profit rate = 43.17

Hence E is the correct answer

Let the Sales of Paharpur Cement during Apr - Jun 2009 be x 

Therefore  profit = [(100-91.03)*340*x]/100

                          = 30.498x

the Sales of Paharpur Cement during Jul- Sept 2009 increased by 2.25%

Therefore Sales of Paharpur Cement during Jul- Sept 2009 is 31.54x 

Percentage increase in profit = (31.54x-30.498x)/30.498x

                                            = 3.42%

Hence E is the correct answer

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.

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

Sales of carnations in January = 13848 dozens

Price of carnation  per dozen  in January =2181084/13848

                                                              = 157.5 

Similarly In December 

Average consumer expenditure on Roses and carnations = 56.4

Total expenditure on both carnations and roses = 56.4*70000

                                                                       = 3948000

Given total expenditure on roses = 977688

So the total expenditure on carnations = 3948000 - 977688

                                                           = 2970312

Sales of carnations  = 18859 dozens

Price of carnation  per dozen  in January =2970312/18859

                                                              = 157.5

Therefore the price did not change

Hence E is the correct answer