MAT 2012 Question Paper

Quantity of milk in the mixture,

= $$\frac{13}{20} \times 48 + \frac{18}{35} \times 42$$

= $$\frac{156}{5} + \frac{108}{5} = \frac{264}{5}$$L

Quantity of water in the mixture,

= $$\frac{7}{20} \times 48 + \frac{17}{35} \times 42$$

= $$\frac{84}{5} + \frac{102}{5} = \frac{186}{5}$$L

20L of mixture is added to the mixture,

= $$\frac{186}{5} + 20 = \frac{186 + 100}{5}$$

= $$\frac{286}{5}$$L

Required ratio = $$\frac{264}{5} : \frac{286}{5}$$

= 12 : 13

Hence, option C is the correct answer.

Let the quotes of A and B be $$7x$$ and $$4x$$

According to the question,

= $$\frac{4x - 1}{4x} = \frac{3}{4}$$

= $$4x - 1 = 3x$$ (or) $$x = 1$$ Lakh

Amount decreased by A in price to win the bid is given by,

= $$7x - (4x - 1)$$

= $$3x + 1 = 4$$ Lakh

Hence, option B is the correct answer.

Part filled by 3 pipes in 3 minutes = $$\frac{1}{20} + \frac{1}{15} + \frac{1}{12} = \frac{3 + 4 + 5}{60} = \frac{1}{5}$$

1/5th of the tank is filled in 3 minutes then whole tank is filled in 5 x 3 = 15min

Hence, option D is the correct answer.

Time taken by the hot and cold water pipes to fill the bath,

= $$\frac{10 \times 15}{10 + 15} = 6$$ min

Total time taken to fill the bath including the empty pipe is also open = (6 + 4) = 10 min

Let '$$x$$' be the time taken by empty pipe to empty the bath.

$$\therefore \frac{1}{10} + \frac{1}{15} + \frac{1}{x} = \frac{1}{10}$$ 

$$\Rightarrow \frac{1}{x} = \frac{1}{15}$$

$$\Rightarrow x = 15$$ min

Hence, option D is the correct answer.

Part filled by A, B and emptied by C in one minute = $$\frac{1}{20}, \frac{1}{30}$$ and $$\frac{1}{15}$$ 

Part filled by three pipes in 3 minutes = $$\frac{1}{20} + \frac{1}{30}$$ - $$\frac{1}{15}$$ 

$$\Rightarrow \frac{3 + 2 - 4}{60} = \frac{1}{60}$$

Cistern can be filled in 60 x 3 = 180 minutes.

Hence, option A is the correct answer.

Let, the time taken by waste pipe to empty the cistern = '$$x$$'

Part filled by two pipes in 1 minute is 1/10 and 1/15 respectively and the cistern gets filled in 18 minutes (given).

Net part filled in 1 hour is given by,

$$\frac{1}{10} + \frac{1}{15} - \frac{1}{x} = \frac{1}{18}$$

$$\frac{1}{10} + \frac{1}{15} - \frac{1}{18} = \frac{1}{x}$$

$$\frac{9 + 6 - 5}{90} = \frac{1}{x}$$

$$\frac{10}{90} = \frac{1}{x}$$ or $$x = 9$$ minutes

Hence, option D is the correct answer.

Let, the whole distance travelled be 'x',

Raghu travelled 1200 km by air which formed $$\frac{2}{5}^{th}$$ of his trip,

$$\frac{2}{5}^{th}$$ of x = 1200

x = 3000 km

One - third of the whole trip he travelled by car,

$$\frac{1}{3}$$ x 3000 = 1000 km

The rest of the journey he did by train,

$$\therefore$$ The distance travelled by train = 3000 - (1200 + 1000) 

$$\Rightarrow$$ 800 km

Hence, option B is the correct answer.

Remaining Distance to be covered = 6 - $$\frac{1}{2}$$ x 6 = 3 km

Remaining time = 45 - $$\frac{2}{3}$$ x 45 = 15 minutes (or) $$\frac{1}{4}$$hour

Required speed to cover the remaining distance in the remaining time,

= $$\frac{3}{(\frac{1}{4})} = 12$$ km/h

Hence, option A is the correct answer.

Distance travelled by A and B is same.

Time taken by A to complete the journey = $$\frac{5}{3}$$ h

Speed at which A travelled = 16 km/h

Distance travelled by A = $$\frac{5}{3} \times 16 = \frac{80}{3}$$ km

Now, speed of the second man is given by,

S = $$\frac{80/3}{12/5}$$

S = $$\frac{80 \times 5}{3 \times 12}$$

S = $$\frac{100}{9} = 11\frac{1}{9}$$ km/h

Hence, option C is the correct answer.

Relative speed between A and B is given by,

$$\frac{23}{4} - \frac{9}{2} = \frac{23 - 18}{4} = \frac{5}{4}$$ km/h

Total number of km's away is given by,

D = $$\frac{5}{4} \times \frac{7}{2}$$

D = $$\frac{35}{8}$$ (or) $$4\frac{3}{8}$$ km

Hence, option C is the correct answer.