MAT 2008 Question Paper

Let the volume of the tank be v litres.

Pipe 1 : 1 min = v/20 lt fill 

Pipe 2 : 1 min = v/30 lt fill

Pipe 3: 1 min = v/50 lt empty

Both inlet pipes can together fill the tank in 20*30/20+30 = 600/50 = 12 minutes, half a tank in  6 minutes.

So, for 6 minutes, both inlet pipe work and for (24-6) = 18 minutes both inlet and one outlet pipes work.

1/2 {6v/20 + 6v/30} +1/2{ 18v/20 + 18v/30 -18v/ 50} = v

v = 1800 litres

Time taken including stoppages be t and without stoppages be x.

d = 54 *x

d = 45*t

54x = 45t

x/t = 5/6

Let t be 1 hr = 60 minutes, x = 50 minutes (from above equation)

So, stoppage per hour = 10 minutes.

Bharat express:

3:00- start from Rampur

3:20 reach stoppage

3:35 start for Sitapur

4:30 30*55/60 = 330/12 = 27.5 km travel since 3:35

Total travel till 4:30 = 27.5 + 30/3 = 37.5 km

Laxman Mail:

4:00 at Sitapur

4:15 reach stopage

4:30 start for Rampur

Total travel till 4:30 = 10 kms

Time after two trains meet = distance between them/Relative speed

t = 90-37.5/70 = 0.75 hrs = 45 min

5:15 meeting of 2 trains

At 5:15. 

Bharat express from Rampur = 37.5 + 45/60 * 30 = 60 kms

Laxman Mail from Rampur = 100 - {10+45/60 * 40} = 100 - 40 = 60 kms

So both trains are equidistant from Rampur.

There is an Arithmetic Progression here,

40 + (40+5)+ (40+5+5) .....n terms = 385

So, sum of terms = 385

first term = 40 and common difference = 5

S = n/2[2a + (n-1)*d]

385 = n/2[80+5n-5)

770 = n[75+5n]

154 = n(15+n)

n = 7

So it takes 7 hours to cover a distance of 385 kms.

Cost price of the mixture $$\times$$ (100 + Gain)% = Selling price of the mixture

Cost price of the mixture $$\times$$ (100 + 10)% = 30

Cost price of the mixture $$\times$$ 110% = 30

Cost price of the mixture = $$\frac{30}{1.10}$$ = Rs. 27.27

By the rule of mixture and Alligation,

 Let the quantity of tea = x

Type A                                                     Type B

x                                                                        30

25                                                                      30

                                27.27

2.73                                                                    2.27

$$\frac{x}{2.73}$$ = $$\frac{30}{2.27}$$

x = $$\frac{30\times2.73}{2.27}=36 kg$$

 Let the quantity of sugar = x

By the rule of mixture and Alligation,

Type A                                                            Type B

x                                                                        75

5.75                                                                   4.50

                                5.50

1.00                                                                   0.25

$$\frac{x}{1.00}$$ = $$\frac{75}{0.25}$$

x = $$\frac{75  \times 1}{0.25}$$

x = 300 Kg

Let the number of officers be x.

And number of workers be = (400 - x).

3000$$ \times $$400 = 10,000$$ \times $$x + 2000$$ \times $$(400 - x)

1200000 = 10000x + 800000 - 2000x

1200000 - 800000 = 8000x

400000 = 8000x

x = $$\frac{400000}{8000}$$

x = 50

Number of officers = 50

Number of workers = 400 - 50 = 350

Let the person travel be train = x hours

Person travel by bus = (6 - x)

The distance traveled by train = 55$$ \times $$x

The distance traveled by bus = 40$$ \times $$(6 - x)

40 (6 - x) + 55x = 285

240 - 40x + 55x = 285

15x = 285 - 240

15x = 45

x = $$\frac{45}{15}$$

x = 3 hours

Distance traveled by train = 55 $$ \times $$ 3 = 165 Km.

Amount of salt in 30kg solution = $$\frac{2 \times 30} {100}$$ = 0.6 kg

Let x kg of pure salt is added:

Then, $$\frac{0.6 + x} {30 + x}$$ = $$\frac{10} {100}$$

60 + 100x = 300 + 10x

100x - 10x = 300 - 60

90x = 240

x = $$\frac{240} {90}$$

x = $$\frac{8} {3}$$

x = 2$$\frac{2} {3}$$ kg

Let the speed of train = S km/hr

Let the length of train = x meters

When a train crosses a man, it travels its own distance.

According to question,

Distance = Speed$$ \times $$Time

x = (S - 3)$$ \times \frac{5}{18} \times 9$$      (1)

Also,

x = (S - 6)$$ \times \frac{5}{18} \times 10$$      (2)

From equation (1) and (2)

(S - 3)$$ \times \frac{5}{18} \times 9$$ = (S - 6)$$ \times \frac{5}{18} \times 10$$

(S - 3)$$ \times $$9 = (S - 6)$$ \times $$10

9S - 27 = 10S - 60

10S - 9S = 60 - 27

S = 33 km/hr