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%


Let original radius = R.

New radius = 

Original area =  and new area = 

Decrease in area =  = 75%


Find the ratio of the areas of the incircle and circumcircle of a square ?

A) 1 : 4 B) 4 : 1
C) 2 : 1 D) 1 : 2
Answer & Explanation Answer: D) 1 : 2


Let the side of the square be x. Then, its diagonal = .


Radius of incircle =  and radius of circumcircle = . k11482389216.png image


Therefore, required ratio =  = 

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Filed Under: Area - Quantitative Aptitude - Arithmetic Ability
Exam Prep: CAT
Job Role: Bank Clerk

10 218

The base of a parallelogram is twice its height. If the area of the parallelogram is 72 sq.cm. Find its height?

A) 3 cm B) 4 cm
C) 6 cm D) 8 cm
Answer & Explanation Answer: C) 6 cm


Let height =x Then,  base =2x

2x * x = 72      


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51 1930

A room 5m 55cm long and 3m 74cm broad is to be paved with square tiles.Find the least number of square tiles required to cover the floor?

A) 156 B) 166
C) 176 D) 186
Answer & Explanation Answer: C) 176


Area of the room=(544 * 374)

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

Area of 1 tile = (34  x  34) 

Number of tiles required== [(544 x 374) / (34 x 34)] = 176

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35 2256

The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. Find the length of the rectangle?

A) 10cm B) 20cm
C) 30cm D) 40cm
Answer & Explanation Answer: B) 20cm


Let the breadth of the given rectangle be x then length is 2x.
thus area of the given rect is \fn_jvn {\color{Black} 2x^{2}}


after dec 5cm from length and inc 5cm breadth , new lenght becomes 2x-5 and breadth is x+5.thus new area =(2x-5)(x+5)=\fn_jvn {\color{Black} 2x^{2}+5x-25}

since new area is 75 units greater than original area thus
\fn_jvn {\color{Black}2x^{2}+75= 2x^{2}+5x-25}
therefore x=20
hence length of the rectangle is 40 cm.

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16 1693

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

A) 56m B) 65m
C) 34m D) 36m
Answer & Explanation Answer: A) 56m


Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres


Two poles will be kept 5 metres apart. Also remember that the poles will be placed 

along the perimeter of the rectangular plot, not in a single straight line which is

very important.


Hence number of poles required = 280 / 5 = 56

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14 1877