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Question Detail
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}
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. The length of a rectangle is three times of its width. If the length of the diagonal is \begin{aligned}8\sqrt{10}\end{aligned}, then find the perimeter of the rectangle.
- 60 cm
- 62 cm
- 64 cm
- 66 cm
Answer: Option C
Explanation:
Let Breadth = x cm,
then, Length = 3x cm
\begin{aligned}
x^2+{(3x)}^2 = {(8\sqrt{10})}^2 \\
=> 10x^2 = 640 \\
=> x = 8 \\
\end{aligned}
So, length = 24 cm and breadth = 8 cm
Perimeter = 2(l+b)
= 2(24+8) = 64 cm
3. 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}
4. There is a plot which is rectangular in shape and it area is 460 square metres. What will be breadth of the plot if length of plot is 15% more than the breadth of plot ?
- 8 mtr
- 10 mtr
- 15 mtr
- 20 mtr
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}
5. 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
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