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Q:

Find the area of a triangle whose sides measure 13 cm, 14 cm and 15 cm.

A) 84	B) 64
C) 44	D) 22

Answer & Explanation Answer: A) 84

Explanation:

Let a = 13, b = 14 and c = 15. Then, $s = \frac{1}{2} (a + b + c)$ =21

(s- a) = 8, (s - b) = 7 and (s - c) = 6.

Area = $\sqrt{[s (s - a) (s - b) (s - c)]}$ = $\sqrt{(21 \times 8 \times 7 \times 6)}$ = 84 sq.cm

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Q:

A room is half as long again as it is broad. The cost of carpeting the at Rs. 5 per sq.m is Rs. 270 and the cost of papering the four walls at Rs. 10 per sq.m is Rs. 1720. If a door and 2 windows occupy 8 sq. m, find the dimensions of the room.

A) b=6; l=18; H=6	B) b=5; l=6; H=18
C) l=6; b=18; H=15	D) l=5; b=18; H=18

Answer & Explanation Answer: A) b=6; l=18; H=6

Explanation:

Let breadth = x metres, length = 3x metres, height = H metres.

Area of the floor=(Total cost of carpeting)/(Rate) = (270/5) sq.m = 54 sq.m

$x \times \frac{3 x}{2} = 54 \Rightarrow x^{2} = 54 \times 2 \Rightarrow x = 6$

So, breadth = 6 m and length = $\frac{3 (6)}{2}$ = 9 m.

Now, papered area = (1720/10) = 172 sq.m

Area of 1 door and 2 windows = 8 sq.m

Total area of 4 walls = (172 + 8) sq.m = 180 sq.m

$2 \times (9 + 6) H = 180 \Rightarrow H = 6$

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Q:

If the length of a certain rectangle is decreased by 4 cm and the width is increased by 3 cm, a square with the same area as the original rectangle would result. Find the perimeter of the original rectangle.

A) 20	B) 30
C) 40	D) 50

Answer & Explanation Answer: D) 50

Explanation:

Let x and y be the length and breadth of the rectangle respectively.
Then, x - 4 = y + 3 or x - y = 7 ----(i)
Area of the rectangle =xy; Area of the square = (x - 4) (y + 3)
(x - 4) (y + 3) =xy <=> 3x - 4y = 12 ----(ii)
Solving (i) and (ii), we get x = 16 and y = 9.
Perimeter of the rectangle = 2 (x + y) = [2 (16 + 9)] cm = 50 cm.

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Q:

If each side of a square is increased by 25%, find the percentage change in its area.

A) 56.25%	B) 36.25%
C) 16.25%	D) 12.25%

Answer & Explanation Answer: A) 56.25%

Explanation:

Let each side of the square be a. Then, area = . $a^{2}$
New side = $\frac{125 a}{100} = \frac{5 a}{4}$ . New area = ${(\frac{5 a}{4})}^{2}$ = $(\frac{25 a^{2}}{16})$

Increase in area = $\frac{25 a^{2}}{16} - a^{2} = \frac{9 a^{2}}{16}$
Increase% = $[\frac{9 a^{2}}{16} * \frac{1}{a^{2}} * 100]$ % = 56.25%.

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Q:

The diagonals of two squares are in the ratio of 2 : 5. Find the ratio of their areas.

A) 4:25	B) 3:25
C) 3:15	D) 5:25

Answer & Explanation Answer: A) 4:25

Explanation:

Let the diagonals of the squares be 2x and 5x respectively.
Ratio of their areas = $\frac{1}{2} * {(2 x)}^{2} : \frac{1}{2} * {(5 x)}^{2} = 4 : 25$

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Q:

First,second and third prizes are to be awarded at an engineering fair in which 13 exhibits have been entered.In how many different ways can the prizes be awarded?

A) 1736	B) 1716
C) 1216	D) 1346

Answer & Explanation Answer: B) 1716

Explanation:

$13 P_{3}$ = 13!/10! = 1716

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Q:

Find the area of a square, one of whose diagonals is 3.8 m long

A) 7.22	B) 6.22
C) 4.22	D) 3.22

Answer & Explanation Answer: A) 7.22

Explanation:

Area of the square = $\frac{1}{2} * {(d i a g o n a l)}^{2} = \frac{1}{2} * 3.8 * 3.8 = 7.22 m^{2}$

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Q:

A tank is 25 m long, 12 m wide and 6 m deep. The cost of plastering its walls and bottom at 75 paise per sq. m, is:

A) rs.458	B) rs,558
C) rs.658	D) rs.758

Answer & Explanation Answer: B) rs,558

Explanation:

Area to be plastered= [2(l + b) x h] + (l x b)

= [2(25 + 12) x 6] + (25 x 12)

= (444 + 300)

= 744 sq.m

Cost of plastering = Rs.744 x (75/100)

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