Question Detail

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}

1. A rectangular field is to be fenced on three sides leaving a side of 20 feet uncovered.If the area of the field is 680 sq.ft, how many feet of fencing will be required ?

88 feet
86 feet
84 feet
82 feet

Answer: Option A

Explanation:

We are given with length and area, so we can find the breadth.
as Length * Breadth = Area
=> 20 * Breadth = 680
=> Breadth = 34 feet

Area to be fenced = 2B + L = 2*34 + 20
= 88 feet

2. The wheel of a motorcycle, 70 cm in diameter makes 40 revolutions in every 10 seconds. What is the speed of the motorcycle in km/hr

30.68 km/hr
31.68 km/hr
32.68 km/hr
33.68 km/hr

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}

3. The percentage increase in the area of a rectangle, if each of its sides is increased by 20% is:

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}

4. What are the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad ?

Answer: Option B

Explanation:

In this type of questions, first we need to calculate the area of tiles. With we can get by obtaining the length of largest tile.
Length of largest tile can be obtained from HCF of length and breadth.
So lets solve this,

Length of largest tile = HCF of (1517 cm and 902 cm)
= 41 cm

Required number of tiles =
\begin{aligned}
\frac{\text{Area of floor}}{\text{Area of tile}} \\
= \left(\frac{1517\times902}{41 \times 41}\right)\\
= 814

\end{aligned}

5. The height of an equilateral triangle is 10 cm. find its area.

\begin{aligned} \frac{120}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{110}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{100}{\sqrt{3}} cm^2 \end{aligned}
\begin{aligned} \frac{90}{\sqrt{3}} cm^2 \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}