Question Detail

Find the circumference of a circle, whose area is 24.64 meter sqaure

17.90m
17.80m
17.60m
17.40m

Answer: Option C

Explanation:

$\begin{aligned} \text{Area of Square =} \pi*r^2\\ => \pi*r^2 = 24.64 \\ => r^2 = \frac{24.64}{22}*7 \\ => r^2 = 7.84 \\ => r = \sqrt{7.84} \\ => r = 2.8 \\ \text{Circumference =}2\pi*r\\ = 2*\frac{22}{7}*2.8 \\ = 17.60m \end{aligned}$

1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

$\begin{aligned} 120m^2 \end{aligned}$
$\begin{aligned} 130m^2 \end{aligned}$
$\begin{aligned} 140m^2 \end{aligned}$
$\begin{aligned} 150m^2 \end{aligned}$

Answer: Option A

Explanation:

$\begin{aligned} \text{We know }h^2 = b^2+h^2 \\ =>\text{Other side }= \sqrt{(17)^2-(15)^2} \\ = \sqrt{289-225} = \sqrt{64} \\ = 8 meter \\ Area = Length \times Breadth \\ = 15\times8 m^2 = 120 m^2 \end{aligned}$

2. If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :

15:12
15:14
15:16
15:22

Answer: Option C

Explanation:

$\begin{aligned} \frac{a^2}{b^2} = \frac{225}{256} \\ \frac{15}{16} \\ <=> \frac{4a}{4b} = \frac{4*15}{4*16} \\ = \frac{15}{16} = 15:16 \end{aligned}$

3. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then

$\begin{aligned} \text{Area of triangle =}\frac{1}{2}*b*h1\\ \text{Area of rectangle =}b*h2\\ \text{As per question }\\ \frac{1}{2}*b*h1 = b*h2 \\ \frac{1}{2}*b*h1 = b*100 \\ h1 = 100*2 = 200 m \\ \end{aligned}$

4. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

22 cm
20 cm
18 cm
10 cm

Answer: Option C

Explanation:

$\begin{aligned} \text{Area of triangle, A1 = }\frac{1}{2}*base*height \\ = \frac{1}{2}*15*12 = 90 cm^2 \\ \text{Area of second triangle =} 2*A1 \\ = 180 cm^2 \\ \frac{1}{2}*20*height = 180 \\ => height = 18 cm \end{aligned}$

5. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

16000
18000
20000
22000

Answer: Option C

Explanation:

$\begin{aligned} \text{Number of bricks =}\frac{\text{Courtyard area}}{\text{1 brick area}} \\ = \left( \frac{2500 \times 1600}{20 \times 10} \right) \\ = 20000 \end{aligned}$

Question Detail

Find the circumference of a circle, whose area is 24.64 meter sqaure

1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

2. If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :

3. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

4. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

5. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

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Question Detail

Find the circumference of a circle, whose area is 24.64 meter sqaure

1. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

2. If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :

3. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

4. The base of a triangle is 15 cm and height is 12 cm. The height of another triangle of double the area having the base 20 cm is :

5. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :

Thanks ! Your comment will be approved shortly !

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