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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. The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:
 15 cm
 20 cm
 25 cm
 30 cm
Answer: Option D
Explanation:
We know the product of diagonals is 1/2*(product of diagonals)
Let one diagonal be d1 and d2
So as per question
\begin{aligned}
\frac{1}{2}*d1*d2 = 150 \\
\frac{1}{2}*10*d2 = 150 \\
d2 = \frac{150}{5} = 30 \\
\end{aligned}
2. 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}
3. A courtyard is 25 meter long and 16 meter board is to be paved with bricks of dimensions 20 cm by 10 cm. The total number of bricks required is :
 16000
 18000
 20000
 22000
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}
4. 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 :
 22 cm
 20 cm
 18 cm
 10 cm
Answer: Option C
Explanation:
\begin{aligned}
\text{Area of triangle, A1 = }\frac{1}{2}*base*height \\
= \frac{1}{2}*15*12 = 90 cm^2 \\
\text{Area of second triangle =} 2*A1 \\
= 180 cm^2 \\
\frac{1}{2}*20*height = 180 \\
=> height = 18 cm
\end{aligned}
5. The area of incircle of an equilateral triangle of side 42 cm is :
 \begin{aligned} 462 cm^2 \end{aligned}
 \begin{aligned} 452 cm^2 \end{aligned}
 \begin{aligned} 442 cm^2 \end{aligned}
 \begin{aligned} 432 cm^2 \end{aligned}
Answer: Option A
Explanation:
\begin{aligned}
\text{Radius of incircle} = \frac{a}{2\sqrt3} \\
= \frac{42}{2\sqrt3} \\
= 7\sqrt{3} \\
\text{Area of incircle =} \\
\frac{22}{7}*49*3 = 462 cm^2
\end{aligned}
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