MAT 2006 Question Paper

As per the given question,

Let, the height of first plane be 'AC' which is equal to 300m. Height of second plane be 'BC'.

From triangle ACD,

tan 60$$^{\circ} = \frac{300}{CD}$$

$$\sqrt{3} = \frac{300}{CD}$$

CD = $$\frac{300}{\sqrt{3}}$$......(1)

From triangle BCD,

tan 45$$^{\circ} = \frac{BC}{CD}$$

CD = BC ...........(2)

From equations (1) and (2) 

BC = $$\frac{300}{\sqrt{3}}$$

BC = 100$$\sqrt{3}$$

Hence, option A is the correct answer.

As per the given question,

Let C be the point where the man is and D be the point when retires 40m. Let AB be the length of the tree.

We need to find the breadth of river that is 'BC' 

From triangle ABD,

Tan 30$$^{\circ}$$ = $$\frac{AB}{40 + x}$$

$$\frac{1}{\sqrt{3}} = \frac{AB}{40 + x}$$

AB = $$\frac{40 + x}{\sqrt{3}}$$ ......(1)

From triangle ABC,

Tan 60$$^{\circ} = \frac{AB}{x}$$

$$\sqrt{3} = \frac{AB}{x}$$

AB = $$\sqrt{3}x$$ ..................(2)

From equations (1) and (2)

$$\frac{40 + x}{\sqrt{3}}$$ = $$\sqrt{3}x$$

40 + x = 3x 

x = 20 m

Hence, option B is the correct answer.

The probability of picking up a defective lamp in the first attempt = $$\frac{6}{10}$$

The probability of picking up a defective lamp in the second attempt = $$\frac{5}{9}$$

The probability of picking up a defective lamp in the second attempt = $$\frac{4}{8}$$

The probability that he will have light = 1 - {$$\frac{6}{10} \times \frac{5}{9} \times \frac{4}{8}$$}

$$\Rightarrow$$ 1 - $$\frac{120}{720}$$ = 1 - $$\frac{1}{6}$$ = $$\frac{5}{6}$$

Hence, option A is the correct answer.

Two persons are to be selected from 5 women and 8 men for the job

The probability that no women is selected = $$\frac{8}{13}$$ and

The probability that atleast one person is women = 1 - $$\frac{8}{13}$$ = $$\frac{5}{13}$$

Hence, option B is the correct answer.

The probability that the component fails when used for the first time = 0.1 (or)

The probability that the component does not fail when used for the first time = 1 - 0.1 = 0.9

The probability that the component lasts for one year = 0.99

Hence, the probability that the component will last for one year = 0.9 x 0.99 = 0.891

Hence, option A is the correct answer.

As per the given question,

The area of the outer path = Area of outer circle - Area of inner circle

$$\Rightarrow \pi r_{1}^{2} - \pi r_{2}^{2}$$

$$\Rightarrow \pi (24.5^{2} - 21^{2})$$ 

$$\Rightarrow \frac{22}{7} (600.25 - 441)$$

$$\Rightarrow \frac{22}{7} (159.25) = 500.5$$

The cost of gravelling the path at Rs 4 per square metre = $$500.5 \times 4 = 2002$$

Hence, option C is the correct answer.

Volume of the plot = 240 m x 180 m x 0.25 m = 10800 m$$^{3}$$

($$\because$$ The earth dug out is evenly spread over the plot, increasing its surface level by 25 cm)

Total volume of the drainlet is given by,

= (2 x 260 x 10 x h) + (2 x 180 x 10 x h) m$$^{3}$$

Volume of the plot is equal to the volume of the drainlet,

10800 m$$^{3}$$ = (2 x 260 x 10 x h) + (2 x 180 x 10 x h) m$$^{3}$$

10800 = h(5200 + 3600)

10800 =  h(8800)

h = 1.227

Hence, option C is the correct answer.

Given, Length of the rectangular tank = 225 m

Breadth of the rectangular tank = 162 m 

height of rectangular tank = 0.2 m

Volume of the water to be filled in tank = 225 x 162 x 0.2 = 7290 m$$^{3}$$

Speed of the flowing water in the aperture per hour is given by,

= $$\frac{7290}{5 \times (60/100) \times (45/100)}$$

= $$\frac{7290 \times 100 \times 100}{5 \times 60 \times 45}$$

= $$\frac{72900000}{13500}$$

= $$5400$$m/hr

Hence, option B is the correct answer.

Given,

The length of iron pipe (h) = 20 cm

External diameter = 25 cm, then external radius will be (R) = 12.5 cm

Internal diameter = 25 - (2) = 23 cm (as thickness of the pipe is 1cm), then internal radius will be (r) = 11.5cm

Total surface area of the pipe = External curved surface area + Internal curved surface area + 2 (Base area)

$$\Rightarrow$$ 2$$\pi$$Rh + 2$$\pi$$rh + 2$$\pi$$(R$$^{2}$$ - r$$^{2}$$)

$$\Rightarrow$$ 2$$\pi$$h(R + r) + 2$$\pi$$(R$$^{2}$$ - r$$^{2}$$)

$$\Rightarrow$$ 2($$\frac{22}{7}$$)(20)(24) + 2($$\frac{22}{7}$$)(24)

$$\Rightarrow \frac{44 \times 24 \times 21}{7}$$ = 3168 cm$$^{2}$$

Hence, option C is the correct answer.

According to the question,

Let 'h' be the height of the vertical tower. And

$$\angle$$ADC = P, $$\angle$$ACB = Q, CB = $$x$$ 

In triangle ABC,

tan Q = $$\frac{AB}{BC}$$

$$\frac{3}{4} = \frac{h}{x}$$

$$x = \frac{4h}{3}$$................(1)

In triangle ABD,

tan P = $$\frac{h}{192 + x}$$

$$\frac{5}{12} = \frac{h}{192 + (4h/3)}$$

$$\frac{5}{12} = \frac{3h}{576 + 4h}$$

$$5(576 + 4h) = 36h$$

$$16h = 2880$$ (or) $$h = 180 $$m

Hence, option B is the correct answer.