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 :

Answer: Option C

Explanation:

$$\begin{aligned} \text{We know area of square =}a^2 \\ \text{Area of triangle =}\frac{1}{2}*a*h \\ => \frac{1}{2}*a*h = a^2 \\ => h = 2a \end{aligned}$$

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:

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

Answer: Option B

Explanation:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =
$$\begin{aligned} \frac{\text{Area of floor}}{\text{Area of tile}} \\ = \left(\frac{1517\times902}{41 \times 41}\right)\\ = 814 \end{aligned}$$

Answer: Option C

Explanation:

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