MAT 2006 Question Paper

Each men can vote for any of the 4 candidates i.e number of ways each men can vote is '4'.

Total number of ways = 4 x 4 x 4 x 4 x 4 = 1024.

Hence, option C is the correct answer.

As no two women sit together, 

6 men are arranged alternately around the table as shown

These 6 men can be arranged in (6 - 1)! ways (or) 5! ways.

Now, 5 women are arranged in remaining 6 places in $$6_{P_{5}}$$ ways (or) 6! ways.

$$\therefore$$ Total number of ways = 6! × 5! 

Hence, option A is the correct answer.

Given, at least 4 questions are to be solved from the first 5 questions. Hence, there are two conditions where he chooses 4 questions from 

the first 5 (or) all 5 questions from the first 5.

(1) If he chooses to write 4 questions from the first 5 and 6 questions from remaining 8,

Then number of ways = $$5_{C_{4}}$$ x $$8_{C_{6}}$$ = 5 x 28 = 140.

(2) If he chooses to write 5 questions from the first 5 and 5 questions from the remaining 8,

Then number of ways = $$5_{C_{5}}$$ x $$8_{C_{5}}$$ = 1 x 56 = 56.

$$\therefore$$ Total number of ways = 140 + 56 =196.

Hence, option C is the correct answer.

For arithmetic progression,

S(n) = $$\frac{n}{2}[2a + (n - 1)d]$$

Given, S(n) = 250 years; d = 3 months (or) 0.25 years; a = 7 years

Substitute the values in the above formula,

$$250 = \frac{n}{2}[2(7) + (n - 1)0.25]$$

$$500 = 14n + 0.25 n^{2} - 0.25n$$

$$ 0.25 ^{2} + 13.75 n - 500 = 0$$

After solving, we get the roots as '-80' and '25'.

we will consider only '25' as '-80' is an imaginary value.

Hence, option C is the correct answer.

Edge of the cube before melting = $$12\ cm$$

Volume of this cube $$(V)= 12^{3} = 1728\ cm$$

Volume of two smaller cubes will be $$6^{3}$$ and $$8^{3}$$ respectively i.e 

Volume of first smaller cube $$(V1) = 216\ cm$$; 

Volume of second smaller cube $$(V2)= 512\ cm$$

As the bigger cube is melted into three smaller cubes. The volume of the bigger cube is equal to volume of the smaller cubes taken together.

$$V = V1 + V2 + V3$$

$$1728 = 216 + 512 + V3\ (or)\ V3 = 1000$$

Now edge of the third smaller cube = $$\sqrt[3]{1000} = 10$$.

Hence, option A is the correct answer.

Given, 

Height (h) = 22.5, Diameter(d) = 7m and radius(r) = 3.5m

Curved surface area of a cylinder = 2 $$\pi$$ r h ..........(1)

By substituting the above values in equation (1) we get,

$$\Rightarrow$$ 2 x $$\frac{22}{7}$$ x 3.5 x 22.5

$$\Rightarrow$$ 2 x 22 x 0.5 x 22.5 = 495

Curved surface area of the well = 495 m$$^{2}$$

Now, cost of plastering its inner curved surface at 3 per sq meter = 495 x 3 = 1485 rs

Hence, option B is the correct answer.

Rate of flow of water in in the pipe = $$\pi r^{2} h_{1}$$

Where, r = 1 cm, h = 6 m (or) 600 cm

$$\therefore$$ Rate of flow of water in the pipe = 600 $$\pi$$ cm$$^{3}$$/second (or) 600 x 60 $$\pi$$ cm$$^{3}$$/minute

And Water after 30 minutes = 600 x 60 x 30 $$\pi$$ cm$$^{3}$$/minute.........(1) 

Volume of water in the tank = $$\pi r^{2} h$$

Where, r = 60 cm, h = ?

$$\therefore$$ Volume of water flown in the tank =  3,600 $$\pi$$ h cm$$^{3}$$......(2)

Equate (1) and (2) as the volume of water in the pipe will be equal to volume of water in tank

600 x 60 x 30 $$\pi$$ = 3,600 $$\pi$$ h

h = 300 cm (or) 3 m

Hence, option C is the correct answer.

As per the given question,

Using pythagoras theorem,

$$BC^{2} = AC^{2} - AB^{2}$$

$$BC^{2} = 15^{2} - 12^{2} = 225 - 144 = 81$$

$$BC = 9\ m $$

$$CD^{2} = EC^{2} - ED^{2}$$

$$CD^{2} = 15^{2} - 9^{2} = 225 - 81 = 144$$

$$CD = 12\ m $$

Width of the street = $$12 + 9 = 21\ m$$

Hence, option B is the correct answer.

As per the given question,

Let, 'EC' be the first tree and 'AB' be the second tree.

Height of first tree is '$$80\ m$$' and let height of second tree be '$$x$$'

Then, $$AB = x$$ and $$ED = (80 - x)$$ 

In $$\triangle EAD$$,

tan $$45^{\circ} = \frac{ED}{AD}$$

$$1 = \frac{80-x}{60}$$

$$x = 20\ m$$ 

Hence, option A is the correct answer.

There are two conditions here,

(i) If Mr Y is in the committee then Mrs X will not be selected (or)

(ii) If Mr Y is not selected in the committee then Mrs X will be selected 

$$\therefore$$ Total number of ways = $$8_{C_{3}} \times 6_{C_{4}} + 7_{C_{3}}\times 7_{C_{4}}$$

$$\Rightarrow$$ 56 x 15 + 35 x 35 = 2065

Hence, option D is the correct answer.