Question Detail

A towel, when bleached, was found to have lost 20% of its length and 10% of its breadth. The percentage of decrease in area is ?

Answer: Option D

Explanation:

Let original length = x
and original width = y
Decrease in area will be
\begin{aligned}
= xy-\left( \frac{80x}{100}\times\frac{90y}{100}\right) \\
= \left(xy- \frac{18}{25}xy\right) \\
= \frac{7}{25}xy \\
\text{Decrease = }\left(\frac{7xy}{25xy} \times100\right) \% \\
= 28\%
\end{aligned}

1. What will be the ratio between the area of a rectangle and the area of a triangle with one of the sides of the rectangle as base and a vertex on the opposite side of the rectangle ?

Answer: Option D

Explanation:

As far as questions of Area or Volume and Surface area are concerned, it is all about formulas and very little logic. So its a sincere advice to get all formulas remembered before solving these questions.

Lets solve this,

\begin{aligned}
\text{Area of rectangle =}l*b\\
\text{Area of triangle =}\frac{1}{2}l*b\\
\text{Ratio =}l*b:\frac{1}{2}l*b \\
= 1:\frac{1}{2} \\
= 2:1

\end{aligned}

One little thing which should be taken care in this type of question is, be sure you are calculating ration in the given order of the question.
If it is ratio of triangle and rectangle then we have to write triangle formula first. cheers :)

2. 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}

3. A triangle and a parallelogram are constructed on the same base such that their areas are equal. If the altitude of the parallelogram is 100 m , then the altitude of the triangle is.

200 m
150 m
148 m
140 m

Answer: Option A

Explanation:

Let the triangle and parallelogram have common base b,
let the Altitude of triangle is h1 and of parallelogram is h2(which is equal to 100 m), then
\begin{aligned}
\text{Area of triangle =}\frac{1}{2}*b*h1\\
\text{Area of rectangle =}b*h2\\
\text{As per question }\\
\frac{1}{2}*b*h1 = b*h2 \\

\frac{1}{2}*b*h1 = b*100 \\

h1 = 100*2 = 200 m \\

\end{aligned}

4. 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

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}