Question Detail

The area of rhombus is 150 cm square. The length of one of the its diagonals is 10 cm. The length of the other diagonal is:

15 cm
20 cm
25 cm
30 cm

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}

1. The perimeters of 5 squares are 24 cm, 32 cm, 40 cm, 76 cm and 80 cm respectively. The perimeter of another square equal in area to the sum of the area of these square is:

124 cm
120 cm
64 cm
56 cm

Answer: Option A

Explanation:

Clearly first we need to find the areas of the given squares, for that we need its side.

Side of sqaure = Perimeter/4

So sides are,

\begin{aligned}
\left(\frac{24}{4}\right),\left(\frac{32}{4}\right),\left(\frac{40}{4}\right),\left(\frac{76}{4}\right),\left(\frac{80}{4}\right) \\
= 6,8,10,19,20 \\
\text{Area of new square will be }\\
= [6^2+8^2+10^2+19^2+20^2] \\
= 36+64+100+361+400 \\
= 961 cm^2 \\
\text{Side of new Sqaure =}\sqrt{961} \\
= 31 cm \\
\text{Required perimeter =}(4\times31) \\
= 124 cm

\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. The difference of the areas of two squares drawn on two line segments in 32 sq. cm. Find the length of the greater line segment if one is longer than the other by 2 cm.

9 cm
8 cm
7 cm
6 cm

Answer: Option C

Explanation:

Let the lengths of the line segments be x and x+2 cm
then,
\begin{aligned}
{(x+2)}^2 - x^2 = 32 \\
x^2 + 4x + 4 - x^2 = 32 \\
4x = 28 \\
x = 7 cm
\end{aligned}

4. One side of rectangular field is 15 meter and one of its diagonals is 17 meter. Then find the area of the field.

\begin{aligned} 120m^2 \end{aligned}
\begin{aligned} 130m^2 \end{aligned}
\begin{aligned} 140m^2 \end{aligned}
\begin{aligned} 150m^2 \end{aligned}

Answer: Option A

Explanation:

\begin{aligned}
\text{We know }h^2 = b^2+h^2 \\
=>\text{Other side }= \sqrt{(17)^2-(15)^2} \\
= \sqrt{289-225} = \sqrt{64} \\
= 8 meter \\
Area = Length \times Breadth \\
= 15\times8 m^2 = 120 m^2

\end{aligned}

5. 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}