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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 rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?
 88 feet
 86 feet
 84 feet
 82 feet
Answer: Option A
Explanation:
We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet
Area to be fenced = 2B + L = 2*34 + 20
= 88 feet
2. If the radius of a circle is diminished by 10%, then the area is diminished by:
 200%
 210%
 300%
 310%
Answer: Option C
Explanation:
Let the original radius be R cm. New radius = 2R
\begin{aligned}
Area = \pi R^2 \\
\text{New Area =} \pi {2R}^2 \\
= 4\pi R^2 \\
\text{Increase in area =}(4\pi R^2  \pi R^2) \\
= 3\pi R^2 \\
\text{Increase percent =} \frac{3\pi R^2}{\pi R^2}*100 \\
= 300 \%
\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 area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?
 18 meter
 20 meter
 22 meter
 25 meter
Answer: Option B
Explanation:
Let breadth =x metres.
Then length =(115x/100)metres.
\begin{aligned}
=x*\frac{115x}{100}= 460\\
x^2=(460 x 100/115) \\
x^2=400 \\
x= 20 \\
\end{aligned}
5. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.
 200 m
 150 m
 148 m
 140 m
Answer: Option A
Explanation:
Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\
\frac{1}{2}*b*h1 = b*100 \\
h1 = 100*2 = 200 m \\
\end{aligned}
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