MAT 2012 Question Paper

If together they had 60 more balls, each of them would have had 100 balls on an average,

$$\frac{40 + X + Y + 60}{3} = 100$$

$$X + Y + 100 = 300 \Rightarrow X + Y = 200$$.........(1)

If B gives 20 balls to A, he is left with half as many balls as C,

$$X - 20 = \frac{1}{2}Y$$

$$2X - 40 = Y$$ (or) $$2X - Y = 40$$..................(2)

Add equations (1) and (2) 

$$3X = 240$$ (or) $$X = 80$$

Substitute $$X$$ value in equation (1)

$$80 + Y = 200$$ (or) $$Y = 120$$

$$X : Y = 80 : 120$$ (or) $$2:3$$

Hence, option B is the correct answer.

Given,

Total number of students in 4 schools of a city = 43,800 and

$$\frac{1}{2}(A) = \frac{2}{3}(B) = \frac{3}{4}(C) = \frac{4}{5}(C)$$

$$A : B = 4 : 3$$ and

$$B : D = 6 : 5$$

Now, $$A : B : D = 8 : 6 : 5$$

$$A : D = 8 : 5$$

Hence, option A is the correct answer.

Let the four numbers be 'a', 'b', 'c' and 'd'

Given ratio of first two numbers is 1:3 and four numbers are in proportion,

$$\therefore$$ a : b = c : d = 1 : 3 

$$\Rightarrow$$ b = 3a and d = 3c

Sum of the squares of the four numbers = 50

$$\Rightarrow$$ a$$^{2}$$ + b$$^{2}$$ + c$$^{2}$$ + d$$^{2}$$ = 50

$$\Rightarrow$$ a$$^{2}$$ + 9a$$^{2}$$ + c$$^{2}$$ + 9c$$^{2}$$ = 0

$$\Rightarrow$$ 10a$$^{2}$$ + 10c$$^{2}$$ = 0

$$\Rightarrow$$ a$$^{2}$$ + c$$^{2}$$ = 0........(1)

Sum of the means = 5 

b + c = 5 

3a + c = 5 (as b = 3a)

a = $$\frac{5 - c}{3}$$........................(2)

From equations (1) and (2)

$$(\frac{5 - c}{3})^{2} + c^{2} = 0$$ 

$$\frac{25 + c^{2} - 10c}{9} + c^{2} = 5$$

$$25 + c^{2} - 10c + 9c^{2} = 5$$

$$10c^{2} - 10c + 20 = 0$$

$$c^{2} - c + 10 = 0$$

After solving the equation we get, c = 2.

Substitute the value of 'c' in equation (2)

a = $$\frac{5 - 2}{3}$$ (or) a = 1

We know b = 3a (or) b = 3

We know d = 3c (or) d = 6

Average of the four numbers,

= $$\frac{1 + 3 + 2 + 6}{4}$$

= 3

Hence, option B is the correct answer.

Quantity of Petrol in the mixture,

= $$\frac{1}{2}$$ x 2 + $$\frac{3}{5}$$ x 3 + $$\frac{4}{5}$$ x 1

= $$1 + \frac{9 + 4}{5}$$

= $$\frac{18}{5}$$

Quantity of kerosene in the mixture,

= $$\frac{1}{2}$$ x 2 + $$\frac{2}{5}$$ x 3 + $$\frac{1}{5}$$ x 1

= $$1 + \frac{6 + 1}{5}$$

= $$\frac{12}{5}$$

Ratio of petrol and kerosene in the mixture,

= $$\frac{18}{5} : \frac{12}{5}$$

= $$3 : 2$$

Hence, option B is the correct answer.

Number of persons in the meeting = 9 + 1(host) = 10 persons

Except X, Y and the host the number of ways the other 7 persons are arranged  = 7!

Now, X and Y can be seated on either side of the host i.e in 2 ways.

Total number of ways = 7! x 2 = 10080

Hence, option A is the correct answer.

Three particular flowers are taken as '1' as these flowers are taken together.

Now total number of flowers is '8' and these flowers can be arranged in 8! ways. And

Three particular flowers are arranged in 3! ways.

Total number of ways = 8! x 3! = 241920

Hence, option D is the correct answer.

Number of ways the committee be selected with atleast one engineer and one manager,

= $$3_{C_{1}}3_{C_{1}}3_{C_{1}} + 3_{C_{1}}3_{C_{2}} + 3_{C_{2}}3_{C_{1}} + 3_{C_{2}}1_{C_{1}} + 3_{C_{2}}1_{C_{1}}$$

= 9 + 9 + 9 + 3 + 3 = 33

Hence, option A is the correct answer.

Number of different ways so that the team has at least 3 bowlers and wicket keeper,

= $$4_{C_{3}}2_{C_{1}}10_{C_{7}} + 4_{C_{4}}2_{C_{1}}10_{C_{6}} + 4_{C_{3}}2_{C_{2}}10_{C_{6}} + 4_{C_{4}}2_{C_{2}}10_{C_{5}}$$

= 4 x 2 x 120 + 1 x 2 x 210 + 4 x 1 x 210 + 1 x 1 x 252

= 960 + 420 + 840 + 252

= 2472

Hence, option A is the correct answer.

By using allegation method,

Ratio of first mixture to second mixture mixed in third mixture, 

$$\Rightarrow \frac{1}{99} : \frac{1}{63}$$

$$\Rightarrow \frac{1}{11} : \frac{1}{7}$$

$$\Rightarrow$$ 7 : 11

Required quantity,

$$\Rightarrow \frac{7}{18}$$ x 18 = 7

Hence, option B is the correct answer.

Let $$a,b,c,d$$ be the number of students in A, B, C and D respectively

Given, average marks of the students in section A, B, C and D of the school = 60

Then $$\frac{45a + 50b + 72c + 80d}{a + b + c + d} = 60$$%

= $$45a + 50b + 72c + 80d = 60a + 60b + 60c + 60d$$

= $$12c + 20d = 15a + 10b$$..........(1)

Average marks of the students of sections A and B together is 48%,

= $$\frac{45a + 50b}{a + b} = 48$$%

= $$45a + 50b = 48a + 48b$$

= $$3a =  2b$$ (or) $$15a = 10b$$...............(2)

Average marks of the students of sections B and C together is 60%

= $$\frac{72c + 80d}{c + d} = 60$$%

= $$72c + 80d = 60c + 60d

= $$12c = 20d$$............(3)

Substitute equations (2) and (3) in equation (1)

$$\Rightarrow 20d + 20d = 15a + 15a$$

$$\Rightarrow 40d = 30a$$

$$\Rightarrow a : d = 4 : 3$$

Hence, option B is the correct answer.