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Q:

A room of 5m 44cm long and 3m 74cm broad is to be paved with squre tiles. Find the least number of squre tiles required to cover the floor.

A) 136	B) 146
C) 166	D) 176

Answer & Explanation Answer: D) 176

Explanation:

area of the room = 544 x 374 sq.cm

size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm

area of 1 tile = 34x34 sq cm

no. of tiles required = (544 x 374) / (34 x 34) = 176

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Q:

The perimeters of two squares are 40cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.

A) 12	B) 24
C) 36	D) 48

Answer & Explanation Answer: B) 24

Explanation:

side of first square = 40/4 =10cm

side of second square = 32/4 = 8cm

area of third squre = 10² - 8² = 36 sq cm

side of third square = $\sqrt{36}$ = 6 cm

required perimeter = 6*4 = 24cm

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Q:

A rectangular grassy plot 110m by 65cm has a gravel path .5cm wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq.mt

A) 680rs	B) 480rs
C) 340rs	D) 320rs

Answer & Explanation Answer: A) 680rs

Explanation:

area of theplot = 110 * 65 = 7150 sq.m
area of the plot excluding the path = (110-5)* (65-5) = 6300 sq.m
area of the path = 7150 - 6300 =850 sq.m
cost of gravelling the path = 850 x (80/100) = 680 Rs

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Q:

Two concentric circles form a ring. The inner and outer circumference of the ring are 352/7 m and 528/7m respectively.Find the width of the ring.

A) 5	B) 4
C) 3	D) 2

Answer & Explanation Answer: B) 4

Explanation:

let the inner and outer radii be r and R meters
then, $2 πr$ = 352/7

=> r = (352/7) * (7/22) * (1/2) = 8m
$2 πR$ = 528/7

=> R= (528/7) * (7/22) * (1/2)= 12m
width of the ring = R-r = 12-8 = 4m

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Q:

If the radius of a circle is decreased by 50% , find the percentage decrease in its area.

A) 75%	B) 65%
C) 35%	D) 25%

Answer & Explanation Answer: A) 75%

Explanation:

let original radius = r and new radius = (50/100) r = r/2

original area = ${πr}^{2}$ and new area = $π {(r / 2)}^{2}$

decrease in area = $3 {πr}^{2} / 4$ * $\frac{1}{{πr}^{2}}$ *100 = 75%

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Q:

There are eight boxes of chocolates, each box containing distinct number of chocolates from 1 to 8. In how many ways four of these boxes can be given to four persons (one boxes to each) such that the first person gets more chocolates than each of the three, the second person gets more chocolates than the third as well as the fourth persons and the third person gets more chocolates than fourth person?

A) 70	B) 40
C) 72	D) 80

Answer & Explanation Answer: A) 70

Explanation:

All the boxes contain distinct number of chocolates.
For each combination of 4 out of 8 boxes, the box with the greatest number has to be given to the first person, the box with the second highest to the second person and so on.

The number of ways of giving 4 boxes to the 4 person is: 8 $C_{4}$ = 70

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Q:

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

A) 65.25	B) 56.25
C) 65	D) 56

Answer & Explanation Answer: B) 56.25

Explanation:

let each side of the square be a , then area = $a^{2}$

As given that The side is increased by 25%, then

New side = 125a/100 = 5a/4

New area = ${(\frac{5 a}{4})}^{2}$

Increased area= $\frac{25 a^{2}}{16} - a^{2}$

Increase %= $\frac{[9 a^{2} / 16]}{a^{2}} * 100$ % = 56.25%

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Q:

The altitude drawn to the base of an isosceles triangle is 8cm and the perimeter is 32cm. Find the area of the triangle?

A) 50	B) 60
C) 70	D) 80

Answer & Explanation Answer: B) 60

Explanation:

let ABC be the isosceles triangle, the AD be the altitude

Let AB = AC = x then BC= 32-2x [because parameter = 2 (side) + Base]

since in an isoceles triange the altitude bisects the base so

BD = DC = 16-x

In a triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

$x^{2} = 8^{2} + {(16 - x)}^{2}$ $\Rightarrow x = 10$

BC = 32-2x = 32-20 = 12 cm

Hence, required area = $\frac{1}{2} * B C * A D$ = $\frac{1}{2} * 12 * 10$ = 60 sq cm

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