MAT 2012 Question Paper

Let $$E_{1}$$, $$E_{2}$$ and $$E_{3}$$ be the events that the toys are produced by the machines A, B and C respectively.

Probability of toys manufactured by A i.e P($$E_{1}$$) = $$\frac{25}{100}$$

Probability of toys manufactured by B i.e P($$E_{2}$$) = $$\frac{35}{100}$$

Probability of toys manufactured by C i.e P($$E_{3}$$) = $$\frac{40}{100}$$

Let D be the event of the toy being defective.

Now, P$$(\frac{D}{E_{1}})$$ = $$\frac{5}{100}$$; P$$(\frac{D}{E_{2}})$$ = $$\frac{4}{100}$$; P$$(\frac{D}{E_{3}})$$ = $$\frac{2}{100}$$

Probability that the toy drawn at random is defective and is manufactured on machine B,

P(B|D) = $$\frac{P(E_{2})P(D/E_{2})}{P(E_{1})P(D/E_{1}) + P(E_{2})P(D/E_{2}) + P(E_{3})P(D/E_{3})}$$

P(B|D) = $$\frac{(35/100)(4/100)}{(25/100)(5/100) + (35/100)(4/100) + (40/100)(2/100)}$$

P(B|D) = $$\frac{(35 \times 4)}{(25 \times 5) + (35 \times 4) + (40 \times 2)}$$

P(B|D) = $$\frac{140}{125 + 140 + 80}$$

P(B|D) = $$\frac{140}{345}$$

P(B|D) = $$\frac{28}{69}$$

Hence, option B is the correct answer.

The probability that A throws '6' = $$\frac{5}{36}$$ and 

The probability of A not getting '6' = $$1 - \frac{5}{36} = \frac{31}{36}$$ 

The probability that B throws '7' = $$\frac{6}{36}$$

The probability of A winning the game is given by,

= (In the first throw if A wins) or (In the 1st throw A looses, B also looses and in the 2nd throw A wins) or ( A and B looses in the 1st and 2nd throw's and A wins in the 3rd throw .....it goes on like that)

= $$\frac{5}{36} + \frac{31}{36}.\frac{30}{36}.\frac{5}{36} + \frac{31}{36}.\frac{30}{36}.\frac{31}{36}.\frac{30}{36}\frac{5}{36} + $$.......

It is in arithmetic progression,

= $$(\frac{5}{36})(\frac{1}{1 - \frac{31}{36}.\frac{30}{36}})$$

= $$(\frac{5}{36})(\frac{1296}{1296 - 930})$$

= $$(\frac{5}{36})(\frac{1296}{366})$$

= $$\frac{5 \times 6}{61} = \frac{30}{61}$$

Probability of B winning the game is given by,

= $$1 - \frac{30}{61}$$

= $$\frac{31}{61}$$

Hence, option B is the correct answer.

 Ratio of simple interest earned by a certain amount at the same rate of interest for 6 years and 9 years is

$$\frac{P1 \times R1 \times t1}{100} : \frac{P2 \times R2 \times t2}{100}$$

Where, P1 and P2 = P ; R1 and R2 = R (as the principle and rate of interest for both the cases are same)

t1 = 6 ; t2 = 9

$$\frac{P \times R \times 6}{100} : \frac{P \times R \times 9}{100}$$

$$ 6 : 9 $$ (or) $$ 2 : 3$$

Hence, option C is the correct answer.

Let the principle amount be 100

Simple interest for first 6 months = $$\frac{100 \times 1 \times 10}{100 \times 2}$$ = 5 

Simple interest for next 6 months = $$\frac{105 \times 1 \times 10}{100 \times 2}$$ = 5.25

Total principle amount after 1 year = 100 + 5 + 5.25 = 110.25

Effective rate of interest = 10.25%

Hence, option B is the correct answer.

Let the amount payable be P

Compound interest when compounded half-yearly = $$P(1 + \frac{R}{2 \times 100})^{2t} - P$$

$$\Rightarrow P(1 + \frac{10}{100})^{4} - P$$

$$\Rightarrow P[(\frac{11}{10})^{4} - 1] \Rightarrow P[\frac{14641 -10000}{10000}] = \frac{4641}{10000} P$$

Compound interest when compounded annually = $$P(1 + \frac{R}{100})^{t} - P$$

$$\Rightarrow P(1 + \frac{20}{100})^{2} - P$$

$$\Rightarrow P(1 + \frac{1}{5})^{2} - P \Rightarrow P(\frac{6}{5})^{2} - P = P[(\frac{36}{25}) - 1]$$

$$\Rightarrow P(\frac{11}{25})$$

According to the given question,

$$P(\frac{4641}{10000}) - P(\frac{11}{25}) = 482$$

$$P(\frac{4641- 4400}{10000}) = 482$$

$$P = \frac{4820000}{241}$$ (or) $$P = 20,000$$.

Hence, option B is the correct answer.

Let the amount of annual installment be 'P'

The amount for 1 year,

$$\Rightarrow P + \frac{P \times 1 \times 8}{100} = \frac{27P}{25}$$

The amount for 2 years,

$$\Rightarrow P + \frac{P \times 2 \times 8}{100} = \frac{29P}{25}$$

The amount for 3 years,

$$\Rightarrow P + \frac{P \times 3 \times 8}{100} = \frac{31P}{25}$$

From the given question,

$$\therefore \frac{27P}{25} + \frac{29P}{25} + \frac{31P}{25} = 2240$$

$$\Rightarrow \frac{27P + 29P + 31P}{25} = 2240$$

$$\Rightarrow \frac{112P}{25} = 2240 \Rightarrow P = \frac{2240 \times 25}{112} = 500$$

Hence, option A is the correct answer.

Surface area of the tent = 2$$\pi$$rh + $$\pi$$rl

= $$\pi$$r(2h + l)

= $$\frac{22}{7} \times 70(2 \times 3 + 80)$$

= $$18920$$

Width of the canvas = 2m

$$\therefore$$ Length of the Canvas required to make the tent = $$\frac{18920}{2} = 9460$$ m

Hence, option C is the correct answer.

Let length and breadth of the rectangular park be 3x and 2x

Distance covered by the man cycling around the park (in meters) is given by,

$$12 \times \frac{5}{18} \times 8 \times 60 = 1600$$

Perimeter of the rectangular park is equal to 1600

$$\therefore$$ 2(3x + 2x) = 1600 

5x = 800 (or) x = 160

Area of the park = $$\times$$ 6x$$^{2}$$

= 6(160 x 160)

= 153600 sq.m

Hence, option B is the correct answer.

Let the length of the edge of the be 'a'

Then according to the question,

a$$^{3}$$ = 6a$$^{2}$$

a = 6

Hence, option D is the correct answer.

Surface area of the metal sheet = 2(lb + bh + hl)

= 2(27 x 8 + 8 x 1 + 1 x 27)

= 2(216 + 8 + 27)

= 2(251) = 502 cm$$^{2}$$

Edge of the cube which is formed after melting = l x b x h

i.e a$$^{3}$$ = 27 x 8 x 1 (or) a = 6

Surface area of the cube = 6a$$^{2}$$ = 216

Required difference = 502 - 216 = 286 cm$$^{2}$$

Hence, option C is the correct answer.