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.

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

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

$$\begin{aligned} \frac{1}{2}*d1*d2 = 150 \\ \frac{1}{2}*10*d2 = 150 \\ d2 = \frac{150}{5} = 30 \\ \end{aligned}$$

Answer: Option D

Explanation:

Let original length = x metres and original breadth = y metres.

$$\begin{aligned} \text{Original area } = \text{xy } m^2 \\ \text{New Length }= \frac{120}{100}x = \frac{6}{5}x \\ \text{New Breadth }= \frac{120}{100}y = \frac{6}{5}y \\ =>\text{New Area }= \frac{6}{5}x * \frac{6}{5}y \\ =>\text{New Area }= \frac{36}{25}xy \\ \text{Area Difference} = \frac{36}{25}xy - xy \\ = \frac{11}{25}xy \\ Increase \% = \frac{Differnce}{Actual}*100 \\ = \frac{11xy}{25}*\frac{1}{xy}*100 = 44\% \end{aligned}$$

Answer: Option A

Explanation:

$$\begin{aligned} \text{Ratio of sides =}\frac{1}{2}:\frac{1}{3}:\frac{1}{4} \\ =6:4:3\\ Perimeter = 52 cm \\ \text{So sides are =} \\ \left( 52*\frac{6}{13}\right)cm,\left( 52*\frac{4}{13}\right)cm, \left( 52*\frac{3}{13}\right)cm \end{aligned}$$
a = 24 cm, b = 16 cm and c = 12 cm
Length of the smallest side = 12 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