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Question Detail
The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:
 32%
 34%
 42%
 44%
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}
1. 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 ?
 25%
 26%
 27%
 28%
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}
2. The height of an equilateral triangle is 10 cm. find its area.
 \begin{aligned} \frac{120}{\sqrt{3}} cm^2 \end{aligned}
 \begin{aligned} \frac{110}{\sqrt{3}} cm^2 \end{aligned}
 \begin{aligned} \frac{100}{\sqrt{3}} cm^2 \end{aligned}
 \begin{aligned} \frac{90}{\sqrt{3}} cm^2 \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}
3. The area of a square is 69696 cm square. What will be its diagonal ?
 373.196 cm
 373.110 cm
 373.290 cm
 373.296 cm
Answer: Option D
Explanation:
If area is given then we can easily find side of a square as,
\begin{aligned}Side = \sqrt{69696} \\
= 264 cm \\
\text{we know diagonal =}\sqrt{2}\times side \\
= \sqrt{2}\times 264 \\
= 1.414 \times 264 \\
= 373.296 cm \end{aligned}
4. 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
5. 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{289225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2
\end{aligned}
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