XAT 2013 Question Paper

Rs. 15 lakhs is to be divided equally.

In the case of the younger son,
Principal = Rs. 750000, time = 9 years and Interest = Rs. 1350000
Rate of interest = $$\dfrac{1350000 * 100}{750000 * 9}$$ = 20%

In the case of the elder son, 
Principal = Rs. 750000, time = 6 years and Interest = Rs. 1350000
Rate of interest = $$\dfrac{1350000 * 100}{750000 * 6}$$ = 30%

Hence, option E is the correct answer.

Let the total time required be 12T (LCM of 2, 3 and 4 for the ease of calculation)
No. of pages read by Albela in 12T minutes at 2 minute per page = 6T
No. of pages read by Bob in 12T minutes at 3 minute per page = 4T
No. of pages read by Chulbul in 12T minutes at 4 minute per page = 3T
Total no. of pages read by all three of them = (6T + 4T + 3T) = 13T
According to the question, total pages = 78
So, 13T = 78
Or, T = 6
Therefore, no. of pages read by Bob = 4T = 24
Hence, option  A is the correct answer.

For the first 2 km, fare = Rs. 20
For the next 8 km, fare = Rs. 5 * 8 = Rs. 40
For the next 4 km, fare = Rs. 8 * 4 = Rs. 32
Total fare up to 14 km = Rs. 20 + Rs. 40 + Rs. 32 = Rs. 92
Total fare paid by Sardar Khan = Rs. 102
Money spent on bullock cart = Rs. 102 - Rs. 92 = Rs. 10
Charge for bullock cart = Rs. 2/km
Distance travelled by bullock cart = 5 km
Total distance travelled by Sardar Khan = (14 + 5) km = 19 km
Hence, option C is the correct answer.

The given expression can be written as $$\frac{a^2+a+1}{a}*\frac{b^2+b+1}{b}*\frac{c^2+c+1}{c}*\frac{d^2+d+1}{d}*\frac{e^2+e+1}{e}$$.
$$\frac{a^2+a+1}{a}=a+\frac{1}{a}+1$$
We know that for positive values, AM $$\geq$$ GM. 
$$\frac{a+\frac{1}{a}}{2} \geq \sqrt{a*\frac{1}{a}}$$
$$a+\frac{1}{a} \geq 2$$
The least value that $$a+\frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+\frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.

PQ = QR = RS = SP = 6 cm
PA = 2AS and PA + AS = 6cm
=> PA = 4 cm and AS = 2 cm
Similarly, RB = 2BS and RB + BS = 6 cm
=> RB = 4 cm and BS = 2cm

Area of △ ABQ = area of PQRS - area of △APQ - area of △RBQ - area of △ASB
 = (36 - 12 - 12 - 2) sq. cm
= 10 sq. cm

Hence, option C is the correct answer.

Let us analyse all numbers from 100 to 200 (both excluded).

Total numbers when unit digit is 2 = 10 {102,112,122, ..., 192}

Total numbers when tens digit is 2 = 10 {120,121,121, ..., 129}

We can see that the there is exactly one number{122} which have both tens and unit digit as 1. So total unique numbers from 100 to 200 (both excluded) which contain at least one 2 digit = 10 + 10  - 1 = 19

From 200 to 299 (both included), all numbers have at least one digit as 2. Total such number = 100.

From 300 to 400, 400 to 500, 500 to 600, 600 to 700 and 700 to 800 we will have 19 number for each pair with at least one 2 digit.

Therefore, total whole numbers between 100 and 800 contain the digit 2 = 100 + 19*6 = 214.

The product of 2 numbers A and B is maximum when A = B.
If we cannot equate the numbers, then we have to try to minimize the difference between the numbers as much as possible. 
pq will be maximum when p=q.
qr will be maximum when q=r.
qr will be maximum when r=p.
Therefore, p, q, and r should be as close to each other as possible. 
We know that p,q,and r are integers and p+q+r=10.
=> p,q, and r should be 3,3, and 4 in any order. 
Substituting the values in the expression, we get, 
pq+qr+pr+pqr = 3*3 + 3*4 + 3*4 + 3*3*4
= 9 + 12 + 12 + 36
= 69
Therefore, option C is the right answer.

As the number is between 10 and 100 and 100 cannot be the number we are looking for, we can assume the number to be of two-digits.
Let the number be xy.
According to the question, for the number to be interesting
x + y + xy = 10x + y
On solving, we get 
xy = 9x
or, x (9 - y) = 0
x cannot be 0, because we need a number greater than or equal to 10.
So, 9 - y = 0
=> y = 9
For all the numbers whose unit digit is 9 will be an interesting number.
So, the numbers are 19, 29, 39, 49, .....99
There are 9 such numbers out of 91 total numbers between 10 and 100 including both.
Required fraction = $$\dfrac{9}{91}$$ = 0.0989
As this is not given in any of the options, the answer will be "none of the above".
Hence, option E is the correct answer.

Let '100x' be the number of students who joined XCRI last year.

Let 'a', 'b', 'c' and d be the number of students who play 1 game, 2 games, 3 games and 4 games respectively. 

Therefore, 

a+b+c+d = 100x ... (1)

a+2b+3c+4d = 70x+75x+80x+85x

a+2b+3c+4d = 310x ... (2)

By equation (2) - (1)

b+2c+3d = 210x

We have to minimize 'd' for that we have to maximize c. But c $$\leq$$ 100x

At c$$_{max}$$ = 90x, d$$_{min}$$ = 10x

Therefore, we can say that the minimum percentage of students who play all four games = 10%.

$$p^q = q^p$$.
It has been given that $$q = 9p$$.

Substituting, we get, 
$$p^{9p}=(9p)^p$$
$$(p^p)^9 = 9^p*p^p$$
=> $$(p^p)^8 = 9^p$$
$$p^{8p}=9^p$$
Raising the power to $$\frac{1}{p}$$ on both sides, we get, 
$$p^8=9$$
$$p=\sqrt[8]{9}$$.
Therefore, option D is the right answer.