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Q:

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

A) 1:1	B) 1:2
C) 1:3	D) 1:4

Answer & Explanation Answer: B) 1:2

Explanation:

Let the side of the square be x. Then, its diagonal = $\sqrt{2 x^{2}} = \sqrt{2 x}$

Radius of incircle = $\frac{x}{2}$

Radius of circum circle= $\sqrt{2} \times \frac{x}{2} = \frac{x}{\sqrt{2}}$

Required ratio = $\frac{{πx}^{2}}{4} : \frac{{πx}^{2}}{2} = \frac{1}{4} : \frac{1}{2} = 1 : 2$

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Q:

Two concentric circles form a ring. The inner and outer circumferences of ring are $\frac{518}{7}$ m and $\frac{352}{7}$ m respectively. Find the width of the ring.

A) 4	B) 5
C) 6	D) 7

Answer & Explanation Answer: A) 4

Explanation:

Let the inner and outer radii be r and R metres.

Then $2 πR = \frac{352}{7} = > R = \frac{352}{7} * \frac{7}{22} * \frac{1}{2} = 8 m$

$2 πR = \frac{528}{7} = > R = \frac{528}{7} * \frac{7}{22} * \frac{1}{2} = 12 m$

=> Width of the ring = (R - r) = (12 - 8) m = 4 m.

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Q:

If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM ?

A) 32	B) 24
C) 72	D) 36

Answer & Explanation Answer: A) 32

Explanation:

The 5 letter word can be rearranged in 5!=120 Ways without any of the letters repeating.

The first 24 of these words will start with A.

Then the 25th word will start will CA _ _ _.
The remaining 3 letters can be rearranged in 3!=6 Ways. i.e. 6 words exist that start with CA.

The next word starts with CH and then A, i.e., CHA _ _.
The first of the words will be CHAMS. The next word will be CHASM.

Therefore, the rank of CHASM will be 24+6+2= 32

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Q:

In two triangles, the ratio of the areas is 4 : 3 and the ratio of their heights is 3 : 4. Find the ratio of their bases.

A) 13:9	B) 16:9
C) 15:9	D) 14:9

Answer & Explanation Answer: B) 16:9

Explanation:

Let the bases of the two triangles be x and y and their heights be 3h and 4h respectively.
Then, $\frac{\frac{1}{2} \times x \times 3 h}{\frac{1}{2} \times y \times 4 h} = \frac{4}{3} \Rightarrow \frac{x}{y} = \frac{4}{3} \times \frac{4}{3} = \frac{16}{9}$

Required ratio = 16 : 9.

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Q:

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

A) 24	B) 48
C) 60	D) 72

Answer & Explanation Answer: B) 48

Explanation:

Let ABC be the isosceles triangle and AD be the altitude

Let AB = AC = x. Then, BC = (32 - 2x).

Since, in an isosceles triangle, the altitude bisects the base,
so BD = DC = (16 - x).
In triangle ADC, ${(A C)}^{2} = {(A D)}^{2} + {(D C)}^{2}$

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

BC = (32- 2x) = (32 - 20) cm = 12 cm.

Hence, required area = $\frac{1}{2} * B C * A D = \frac{1}{2} * 12 * 8 = 48 c m^{2}$

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Q:

The base of a triangular field is three times its altitude. If the cost of cultivating the field at Rs. 24.68 per hectare be Rs. 333.18, find its base and height.

A) B=900;H=300	B) B=300;H=900
C) B=600;H=700	D) B=500;H=900

Answer & Explanation Answer: A) B=900;H=300

Explanation:

Area of the field = Total cost/rate = (333.18/25.6) = 13.5 hectares
$13.5 * 10000 m^{2} = 135000 m^{2}$
Let altitude = x metres and base = 3x metres.
Then, $\frac{1}{2} * 3 x * x = 135000 \Rightarrow x^{2} = 90000 \Rightarrow x = 300$

Base = 900 m and Altitude = 300 m.

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Q:

Find the area of a right-angled triangle whose base is 12 cm and hypotenuse is 13cm.

A) 30	B) 40
C) 50	D) 60

Answer & Explanation Answer: A) 30

Explanation:

Height of the triangle = $\sqrt{(13^{2} - 12^{2})} = \sqrt{25}$ = 5 cm.

Its area = $\frac{1}{2} * b a s e * h e i g h t$ = $\frac{1}{2} * 12 * 5$ = $30 c m^{2}$ .

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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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