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 ?

Answer: Option D

Explanation:

Area of each slab =
\begin{aligned}
\frac{72}{50}m^2 = 1.44 m^2\\
\text{Length of each slab =}\sqrt{1.44} \\
= 1.2m = 120 cm

\end{aligned}

Answer: Option B

Explanation:

In this type of question, we will first calculate the distance covered in given time.
Distance covered will be, Number of revolutions * Circumference

So we will be having distance and time, from which we can calculate the speed. So let solve.

Radius of wheel = 70/2 = 35 cm
Distance covered in 40 revolutions will be

\begin{aligned}
\text{40 * Circumference } \\
= \text{40 * 2*\pi*r } \\
= 40 * 2* \frac{22}{7}* 35 \\
= 8800 cm \\
= \frac{8800}{100} m = 88 m\\

\text{Distance covered in 1 sec =}\\
\frac{88}{10} \\
= 8.8 m \\

Speed = 8.8 m/s \\
= 8.8*\frac{18}{5} = 31.68 km/hr

\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 C

Explanation:

\begin{aligned}
\text{We know area of square =}a^2 \\
\text{Area of triangle =}\frac{1}{2}*a*h \\
=> \frac{1}{2}*a*h = a^2 \\
=> h = 2a

\end{aligned}

Answer: Option C

Explanation:

Let each side be a cm, then
\begin{aligned}
\left(\frac{a}{2}\right)^2+{10}^2 = a^2 \\
<=>\left(a^2-\frac{a^2}{4}\right) = 100 \\
<=> \frac{3a^2}{4} = 100 \\
a^2 = \frac{400}{3} \\
Area = \frac{\sqrt{3}}{4}*a^2 \\
= \left(\frac{\sqrt{3}}{4}*\frac{400}{3}\right)cm^2 \\
= \frac{100}{\sqrt{3}}cm^2
\end{aligned}