Question Detail

The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares .

22 cm
24 cm
26 cm
28 cm

Answer: Option B

Explanation:

We know perimeter of square = 4(side)
So Side of first square = 40/4 = 10 cm
Side of second square = 32/4 = 8 cm

Area of third Square = 10*10 - 8*8
= 36 cm

So side of third square = 6 [because area of square = side*side]
Perimeter = 4*Side = 4*6 = 24 cm

1. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?

Area is quadrupled
Area is tripled
Area is doubles
Area become half

Answer: Option A

Explanation:

\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}

So the area quadruples.

2. 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}

3. What will be the cost of gardening 1 meter boundary around a rectangular plot having perimeter of 340 meters at the rate of Rs. 10 per square meter ?

Rs. 3430
Rs. 3440
Rs. 3450
Rs. 3460

Answer: Option B

Explanation:

In this question, we are having perimeter.
We know Perimeter = 2(l+b), right
So,
2(l+b) = 340
As we have to make 1 meter boundary around this, so
Area of boundary = ((l+2)+(b+2)-lb)
= 2(l+b)+4 = 340+4 = 344

So required cost will be = 344 * 10 = 3440

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. 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}