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 :

Answer: Option C

Explanation:

Let Breadth = x cm,
then, Length = 3x cm

$$\begin{aligned} x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\ => 10x^2 = 640 \\ => x = 8 \\ \end{aligned}$$
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm

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

Answer: Option D

Explanation:

Let original length = x metres and original breadth = y metres.

$$\begin{aligned} \text{Original area } = \text{xy } m^2 \\ \text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\ \text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\ =>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\ =>\text{New Area }= \frac{36}{25}xy \\ \text{Area Difference} = \frac{36}{25}xy - xy \\ = \frac{11}{25}xy \\ Increase \% = \frac{Differnce}{Actual}*100 \\ = \frac{11xy}{25}*\frac{1}{xy}*100 = 44\% \end{aligned}$$

Answer: Option C

Explanation:

Let each side be a cm, then

$$\begin{aligned} \left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\ <=>\left(a^2-\frac{a^2}{4}\right) = 100 \\ <=> \frac{3a^2}{4} = 100 \\ a^2 = \frac{400}{3} \\ Area = \frac{\sqrt{3}}{4}*a^2 \\ = \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\ = \frac{100}{\sqrt{3}}cm^2 \end{aligned}$$

Answer: Option C

Explanation:

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