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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. If the ratio of the areas of two squares is 225:256, then the ratio of their perimeters is :
 15:12
 15:14
 15:16
 15:22
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}
2. 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
3. The Diagonals of two squares are in the ratio of 2:5. find the ratio of their areas.
 4:25
 4:15
 3:25
 3:15
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}
4. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?
 714
 814
 850
 866
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}
5. If the circumference of a circle increases from 4pi to 8 pi, what change occurs in the area ?
 Area is quadrupled
 Area is tripled
 Area is doubles
 Area become half
Answer: Option A
Explanation:
\begin{aligned}
2\pi R1 = 4 \pi \\
=> R1 = 2 \\
2\pi R2 = 8 \pi \\
=> R2 = 4 \\
\text{Original Area =} 4\pi * 2^2 \\
= 16 \pi \\
\text{New Area =} 4\pi * 4^2 \\
= 64 \pi
\end{aligned}
So the area quadruples.
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