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Question Detail
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}
1. 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}
2. 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}
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. 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}
5. A farmer wishes to start a 100 sq. m. rectangular vegetable garden. Since he has only 30 meter barbed wire, he fences three sides of the garden letting his house compound wall act as the fourth side fencing. Then find the dimension of the garden.
- 10 m * 5 m
- 15 m * 5 m
- 20 m * 5 m
- 25 m * 5 m
Answer: Option C
Explanation:
From the question, 2b+l = 30
=> l = 30-2b
\begin{aligned}
Area = 100m^2 \\
=> l \times b = 100 \\
=> b(30-2b) = 100 \\
b^2 - 15b + 50 = 0 \\
=>(b-10)(b-5)=0 \\
\end{aligned}
b = 10 or b = 5
when b = 10 then l = 10
when b = 5 then l = 20
Since the garden is rectangular so we will take value of breadth 5.
So its dimensions are 20 m * 5 m
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