Question Detail

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

1. Find the circumference of a circle, whose area is 24.64 meter sqaure

17.90m
17.80m
17.60m
17.40m

Answer: Option C

Explanation:

\begin{aligned}
\text{Area of Square =} \pi*r^2\\
=> \pi*r^2 = 24.64 \\
=> r^2 = \frac{24.64}{22}*7 \\
=> r^2 = 7.84 \\
=> r = \sqrt{7.84} \\
=> r = 2.8 \\
\text{Circumference =}2\pi*r\\
= 2*\frac{22}{7}*2.8 \\
= 17.60m

\end{aligned}

2. 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.

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