30
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%

Explanation:

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:

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

Explanation:

Let the side of the square be x. Then, its diagonal = $\inline&space;\fn_jvn&space;\small&space;\sqrt{2}&space;x$.

Radius of incircle = $\inline&space;\fn_jvn&space;\small&space;\frac{x}{2}$ and radius of circumcircle = $\inline&space;\fn_jvn&space;\small&space;\frac{\sqrt{2}x}{2}=&space;\frac{x}{\sqrt{2}}$.

Therefore, required ratio = $\inline&space;\fn_jvn&space;\small&space;\left&space;(&space;\frac{\prod&space;x^{2}}{4}&space;:&space;\frac{\prod&space;x^{2}}{2}&space;\right)$ = $\inline&space;\fn_jvn&space;\small&space;\frac{1}{4}&space;:&space;\frac{1}{2}&space;=&space;1&space;:&space;2$

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

10 218
Q:

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

Explanation:

Let height =x Then,  base =2x

2x * x = 72

=>x=6

51 1930
Q:

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

Explanation:

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) $\inline cm^{2}$

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

35 2256
Q:

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

Explanation:

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

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)=$\fn_jvn&space;{\color{Black}&space;2x^{2}+5x-25}$

since new area is 75 units greater than original area thus
$\fn_jvn&space;{\color{Black}2x^{2}+75=&space;2x^{2}+5x-25}$
5x=75+25
5x=100
therefore x=20
hence length of the rectangle is 40 cm.

16 1693
Q:

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

Explanation:

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