Instructions

Questions are followed by two statements labelled as I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below:
A. Statement I alone is sufficient to answer the question.
B. Statement II alone is sufficient to answer the question.
C. Statement I and Statement II together are sufficient, but neither of the two alone is sufficient to answer the question.
D. Either Statement I or Statement II alone is sufficient to answer the question.
E. Both Statement I and Statement II are insufficient to answer the question

XAT 2010 - Question 81

A sequence of positive integer is defined as $$A_{n+1}=A_{n}^{2}+1$$ for each n ≥ 0. What is the value of Greatest Common Divisor of $$A_{900}$$ and $$A_{1000}$$ ?
I. $$A_{0} = 1$$
II. $$A_{1} = 2$$

Solution

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.

Instructions

For the following questions answer them individually

XAT 2010 - Question 82

a, b, c, d and e are integers such that 1 ≤ a < b < c < d < e. If a, b, c, d and e are geometric progression and lcm (m , n) is the least common multiple of m and n, then the maximum value of $$\frac{1}{lcm(a,b)}+\frac{1}{lcm(b,c)}+\frac{1}{lcm(c,d)}+\frac{1}{lcm(d,e)}$$ is

Solution

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

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

Instructions

Books and More sells books, music CDs and film DVDs. In December 2009, they earned 40% profit in music CDs and 25% profit in books. Music CDs contributed 35% towards their total sales in rupees. At the same time total sales in rupees from books is 50% more than that of music CDs.

XAT 2010 - Question 83

If Books and More have earned 20% profit overall, then in film DVDs they made

Solution

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 = $$\frac{4.6 x}{19.6 x} \times 100 = 23.46 \%$$

XAT 2010 - Question 84

If Books and More made 50% loss in film DVDs, then overall they made

Solution

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 = $$\frac{100x - 92x}{92x} \times 100$$

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

Instructions

For the following questions answer them individually

XAT 2010 - Question 85

ABCD is a parallelogram with $$\angle$$ ABC = 60°. If the longer diagonal is of length 7 cm and the area of the parallelogram ABCD is $$15\frac{\sqrt{3}}{2}$$ sq.cm, then the perimeter of the parallelogram (in cm) is

Solution

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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

Instructions

OABC is a square where O is the origin and AB = 1. Consider the set of points $$s = {(x_{i},y_{i})}$$ in the square such that $$x_{i}+y_{i}$$≤1. Let $$P (x_{1}, y_{1})$$ and $$Q (x_{2}, y_{2})$$ be two such points. Two operations addition (+) and multiplication (.) on S are defined as
$$P + Q = (x_{1}+x_{2} - x_{1}x_{2},y_{1}y_{2})$$
$$P.Q = (x_{1}x_{2},y_{1}+y_{2} - y_{1}y_{2})$$

XAT 2010 - Question 86

For a very large number n, $$P^{n}+ Q^{n}$$ is

Solution

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.

XAT 2010 - Question 87

For a very large number n, nP + nQ is

Solution

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.

Instructions

Answer thequestions based on the data given in table 2

Cost and price data for Portland cement manufactured by Paharpur Cement and Bahsin Cement, for four consecutive quarters, are given in table 2:

Sales Revenue = Price × Sales Quantity
Profit = Sales Revenue – Cost
Profit Rate = Profit / Sales Quantity

XAT 2010 - Question 88

Profit rate of Paharpur Cement is more than the profit rate of Bahsin Cement in:

Solution

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

XAT 2010 - Question 89

If between Jan - Mar 2009 and Apr - Jun 2009 sales of Paharpur Cement increased from 543278 to 698236 and that of Bahsin Cement decreased from 526532 to 499874, then which of the following is true?

Solution

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

XAT 2010 - Question 90

If between Apr - Jun 2009 and Jul - Sept 2009 sales of Paharpur Cement increased by 2.25%, its profit increased by

Solution

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

CAT Content

XAT 2010 Question Paper

XAT 2010 - Question 81

XAT 2010 - Question 82

XAT 2010 - Question 83

XAT 2010 - Question 84

XAT 2010 - Question 85

XAT 2010 - Question 86

XAT 2010 - Question 87

XAT 2010 - Question 88

XAT 2010 - Question 89

XAT 2010 - Question 90

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