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Q:

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

A) 6	B) 7
C) 8	D) 9

Answer & Explanation Answer: A) 6

Explanation:

Let the height of the parallelogram be x. cm. Then, base = (2x) cm.

$2 x \times x = 72 = > 2 x^{2} = 72 = > x = 6$

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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:

When John arrives in New York,he has eight stops to see, but he has time only to visit six of them.In how many different ways can he arrange his schedule in New York?

A) 20610	B) 24000
C) 20160	D) 21000

Answer & Explanation Answer: C) 20160

Explanation:

He can arrange his schedule in $8 P_{6}$ = 20160 ways

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Q:

Find the length of the altitude of an equilateral triangle of side $3 \sqrt{3}$ cm.

A) 4.5	B) 3.5
C) 2.5	D) 6.5

Answer & Explanation Answer: A) 4.5

Explanation:

Let the length of the altitude be h.
Then, h = $a \frac{\sqrt{3}}{2}$ = $3 \sqrt{3} \times \frac{\sqrt{3}}{2}$ = 4.5

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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:

Compute the sum of 4 digit numbers which can be formed with the four digits 1,3,5,7, if each digit is used only once in each arrangement.

A) 105555	B) 106665
C) 106656	D) 108333

Answer & Explanation Answer: C) 106656

Explanation:

The number of arrangements of 4 different digits taken 4 at a time is given by $4 P_{4}$ = 4! = 24.All the four digits will occur equal number of times at each of the position,namely ones,tens,hundreds,thousands.

Thus,each digit will occur 24/4 = 6 times in each of the position.The sum of digits in one's position will be 6 x (1+3+5+7) = 96.Similar is the case in ten's,hundred's and thousand's places.

Therefore,the sum will be 96 + 96 x 10 + 96 x 100 + 96 x 100 = 106656

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