MAT 2012 Question Paper

Work done by A in one day = $$\frac{1}{20}$$ and work done by B in one day = $$\frac{1}{15}$$

Work done by both A and B in one day = $$\frac{1}{20} + \frac{1}{15} = \frac{7}{60}$$

Work done by A and B in 6 days = $$\frac{7}{60} \times 6 = \frac{7}{10}$$

Remaining work = $$1 - \frac{7}{10} = \frac{3}{10}$$

Now, Time taken by A and B to complete the work in next 4 days is given by

$$\Rightarrow$$ (Remaining work / (A + C)'s effeciency) = Remaining time to complete the work.

$$\Rightarrow \frac{\frac{3}{10}}{\frac{1}{20} + \frac{1}{C}} = 4$$ days

$$\Rightarrow \frac{3}{10} = 4(\frac{1}{20} + \frac{1}{C})$$

$$\Rightarrow \frac{3}{10} = \frac{4}{20} + \frac{4}{C} \Rightarrow \frac{4}{C} = \frac{3}{10} - \frac{1}{5}$$

$$\Rightarrow \frac{1}{C} = \frac{1}{40}$$ (or) C = 40 days

Hence, option B is the correct answer.

Work done by B in one hour = $$\frac{1}{6}$$

Work done by B and C together in one hour = $$\frac{1}{B} + \frac{1}{C} = \frac{1}{6} + \frac{1}{12} = \frac{1}{12}$$

Work done by A, B and C together in one hour,

$$\frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{3}{8}$$

$$\frac{1}{A} + \frac{1}{6} + \frac{1}{12} = \frac{3}{8}$$

$$\frac{1}{A} = \frac{3}{8} - \frac{1}{6} - \frac{1}{12}$$ 

$$\frac{1}{A} = \frac{9 - 4 - 2}{24} = \frac{1}{8}$$

Work done by A and B together in one hour = $$\frac{1}{A} + \frac{1}{B} = \frac{1}{8} + \frac{1}{6} = \frac{7}{24}$$

Total work done by A and B together = $$\frac{24}{7}$$ (or) $$3\frac{3}{7}$$ hours

Hence, option D is the correct answer. 

Given, X can complete $$\frac{1}{4}$$ of the work in 10 days i.e he can complete the whole work in 10 x 4 = 40 days

Y can do 40% of the work in 40 days i.e he can complete the whole work in 40 x $$\frac{1}{0.4}$$ = 100 days

Z can do $$\frac{1}{3}$$rd of the work in 13 days i.e he can complete the whole work in 13 x 3 = 39 days

Z will complete the work first.

Hence, option C is the correct answer.

Work done by A and B in 1 hr per day = $$\frac{1}{63}+\frac{1}{42}=\frac{5}{126}$$

Number of hours they are required to work = $$\frac{42}{5}$$ hours

Number of days required = $$(\frac{1}{\frac{5}{126}})\div(\frac{42}{5})$$

= $$\frac{126}{5}\times\frac{5}{42}=3$$ days

=> Ans - (B)

Given, when 5 percent is lost in grinding wheat, the country has to import 20 million bags to make up of the loss,

5% of wheat = 20 million bags ......(1)

When only 2 percent is lost, it has to import 15 million bags,

2% of wheat = 15 million bags ......(2)

From equation (1) and (2) 

3% of wheat = 5 million and 100% = $$\frac{500}{3} = 166 \frac{2}{3}$$

Hence, option D is the correct answer.

Let the total number of individuals who had been interviewed for the survey = 'x'

Number of respondents who had opted for product A = 20% of x 

Number of respondents who had opted for product B = 60% of x

Number of respondents who had opted for none of the products = 20% of x ( remaining part out of 100%)

Difference between those who opted for product B and those who were undecided is 720.

$$\Rightarrow$$ (60% - 20%)x = 720 

$$\Rightarrow$$ 40% of x = 720 (or) x = 720 ($$\frac{100}{40}$$) (or) x = 1800

Hence, option B is the correct answer.

Cost price of items bought by Gauri in the stationers = 25

Price after deducting sales tax = 25 - 0.30 = 24.70

Let the amount of sales tax on taxable purchases be x

Then, 6% of x = 0.3 rs

x = (0.3)($$\frac{100}{6}$$)

x = 5

Cost price of the tax free items = 24.70 - 5 = 19.70

Hence, option C is the correct answer.

Total number of books shyam had,

X = 120, Y = (1.1 x 120) = 132, Z = (0.9 x 120) = 108

Total number of books he tore,

X = 5 % of X = (0.05 x 120) =  6 

Y = 10% of Y = (0.1 X 132) = 13.2

Z = 15% of Z = (0.15 x 108) = 16.2

Percentage of books he tore,

$$\Rightarrow \frac{6 + 13.2 + 16.2}{120 + 132 + 108}$$ x 100

$$\Rightarrow \frac{35.4}{360}$$ x 100 = 9.83%

Hence, option D is the correct answer.

Total number of coins in the bag = 4 + 3 + 3 = 10

If 6 coins are drawn from the bag at random, then the probability that the draw yields maximum amount is given by,

= $$\frac{4_{C_{4}} \times 3_{C_{2}}}{10_{C_{6}}}$$

= $$\frac{1 \times 3}{210}$$

= $$\frac{1}{70}$$

Odds in favour of the draw = $$1 - \frac{1}{70} = \frac{69}{70}$$

Hence, option C is the correct answer.

The number of possibilities where the sum is less than '7' is,

(1,1), (1,2), (1,3), (1,4), (1,5), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (4,1), (4,2), (5,1)

There are 15 possibilities out of 36(6 x 6) possible outcomes.

$$\therefore$$ The probability that the sum is less than 7 is,

$$\Rightarrow \frac{15}{36} = \frac{5}{12}$$

Hence, option D is the correct answer.