A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?

Answer: Option D

Explanation:

Let original length = x
and original width = y
Decrease in area will be
$$\begin{aligned} = xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\ = \left(xy- \frac{18}{25}xy\right) \\ = \frac{7}{25}xy \\ \text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\ = 28\% \end{aligned}$$

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 D

Explanation:

$$\begin{aligned} l*b = 460 m^2 \end{aligned}$$ .. (i)
Lets suppose breadth of plot is = b rnLength will be, $$\begin{aligned}l=b*\frac{100+15}{100}\\=\frac{115b}{100}\end{aligned}$$ .. (ii)

From (i) and (ii), we get :
$$\begin{aligned} \frac{115b}{100}*b = 460 \\ b^2 = \frac{46000}{115} = 400 \\ => b = \sqrt{400} = 20 meter \end{aligned}$$

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 A

Explanation:

Let the diagonals of the squares be 2x and 5x.
Then ratio of their areas will be
$$\begin{aligned} \text{Area of square} = \frac{1}{2}*{Diagonal}^2 \\ \frac{1}{2}*{2x}^2:\frac{1}{2}*{5x}^2 \\ 4x^2:25x^2 = 4:25 \end{aligned}$$