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Q:

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

 A) 55% B) 65% C) 75% D) 85%

Answer:   C) 75%

Explanation:

Let original radius = R.

New radius = $\inline \fn_cm \frac{50}{100}R$$\inline \fn_cm \frac{R}{2}$

Original area =$\inline \fn_cm \tiny \prod R^{2}$  and new area = $\inline \fn_cm \tiny \prod(\frac{R}{2})^{2}=\frac{\prod R^{2}}{4}$

Decrease in area = $\inline \fn_cm \tiny \frac{3\prod R^{2}}{4}\times \frac{1}{\prod R^{2}}\times 100$ = 75%

Q:

Calculate the number of bricks, each measuring 25 cm x 15 cm x 8 cm required to construct a wall of dimensions 10 m x 4 cm x 6 m when 10% of its volume is occupied by mortar  ?

 A) 720 B) 600 C) 660 D) 6000

Answer & Explanation Answer: A) 720

Explanation:

Let the number of bricks be 'N'

10 x 4/100 x 6 x 90/100 = 25/100 x 15/100 x 8/100 x N

10 x 4 x 6 x 90 = 15 x 2 x N => N = 720.

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Q:

If the sides of a triangle are 26 cm, 24 cm and 10 cm, what is its area ?

 A) 108 sq.cm B) 112 sq.cm C) 116 sq.cm D) 120 sq.cm

Answer & Explanation Answer: D) 120 sq.cm

Explanation:

The triangle with sides 26 cm, 24 cm and 10 cm is right angled, where the hypotenuse is 26 cm.

Area of the triangle =1/2 x base x height => 1/2 x 24 x 10 = 120 sq.cm

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Q:

An order was placed for the supply of a carpet whose breadth was 6 m and length was 1.44 times the breadth. What be the cost of a carpet whose length and breadth are 40% more and 25% more respectively than the first carpet. Given that the ratio of carpet is Rs. 45 per sq. m ?

 A) Rs. 4082.40 B) Rs. 1024.21 C) Rs. 2810.6 D) Rs. 3214

Answer & Explanation Answer: A) Rs. 4082.40

Explanation:

Length of the first carpet = (1.44)(6) = 8.64 cm

Area of the second carpet = 8.64(1 + 40/100) 6 (1 + 25/100)

= 51.84(1.4)(5/4) sq m = (12.96)(7) sq m

Cost of the second carpet = (45)(12.96 x 7) = 315 (13 - 0.04) = 4095 - 12.6 = Rs. 4082.40

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Q:

What is the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high  ?

 A) 13 m B) 14 m C) 15 m D) 16 m

Answer & Explanation Answer: A) 13 m

Explanation:

the length of the longest pole which can be kept in a room 12 m long, 4 m broad and 3 m high is

=> $\inline \fn_jvn \small d^{2}=12^{2}+4^{2}+3^{2} = 13$ mts.

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Q:

A class is 6 meters 24 centimeters in length and 4 meters 32 centimeters in width. Find the least number of square tiles of equal size required to cover the entire floor of the class room ?

 A) 115 B) 117 C) 116 D) 114

Answer & Explanation Answer: B) 117

Explanation:

Length = 6 m 24 cm = 624 cm
Width = 4 m 32 cm = 432 cm
HCF of 624 and 432 = 48
Number of square tiles required = (624 x 432)/(48 x 48) = 13 x 9 = 117.

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