# Area

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**: (d)

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

Decrease in area = \(xy - \left ( \frac{80}{100}x \times \frac{90}{100}y \right )\)

= \(xy - \frac{18}{25}xy\)

= \(\frac{7}{25}xy\)

∴ Decrease % = \(\left ( \frac{7}{25}xy \times \frac{1}{xy} \times 100 \right )\) % = 28 %

The length of a rectangle is 18 cm and its breadth is 10 cm. When the length is increased to 25 cm, what will be the breadth of the rectangle if the area remains the same ?

**Answer**: (c)

Let the breadth be b. Then,

⇒ 25b = 18 × 10

⇒ b = \(\left ( \frac{18 \times 10 }{25} \right )\) cm = 7.2 cm

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed ?

**Answer**: (b)

Perimeter of the plot

= 2 (90 + 50)

= 280 m.

∴ Number of poles = \(\frac{280}{5}\) = 56 m

The length of a room is 5.5 m and width is 3.75 m. Find the cost of paving the floor by slabs at the rate of Rs. 800 per sq. metre.

**Answer**: (d)

Area of the floor = (5.5 × 3.75) m²

= 20.625 m²

∴ Cost of paving = Rs. (800 × 20.625)

= 16500.

The length of a rectangle hall is 5 m more than its breadth. The area of the hall is 750 m². The length of the hall is :

**Answer**: (d)

Let breadth = *x* metres. Then, length = (*x* + 5) metres.

Then *x *(*x*+5) = 750

⇔ *x*^{2} + 5*x* − 750 = 0

⇔ (*x* + 30) (x + 25) = 0

⇔ *x* = 25.

∴ Length = (*x* + 5) = 30 m.

The difference between the length and breadth of a rectangle is 23 m. If its perimeter is 206 m, then its area is :

**Answer**: (d)

We have : (*l* - b) = 23 and 2(*l* + b) = 206 or (*l* + b) = 103.

Solving the two equations, we get, *l* = 63 and b = 40.

∴ Area = (*l* × b) = (63 × 40) m² = 2520 m².

The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres ?

**Answer**: (d)

Let breadth = *x* metres. Then, length = (*x* + 20) metres.

Perimeter = \(\left ( \frac{5300}{26.50} \right )\) = 200 m

∴ 2[(*x* + 20) + *x*] = 200

⇒ 2*x* + 20 = 100

⇒ 2*x* = 80

⇒ *x* = 40.

Hence, length = *x* + 20 = 60 m.

The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, what is the area of the field ?

**Answer**: (b)

Let length = *x* metres. Then, breadth =\(\left ( \frac{60}{100}x \right )\) metres = \(\left ( \frac{3x}{5} \right )\) metres.

Perimeter = \(\left [ 2\left ( x + \frac{3x}{5} \right ) \right ] = \left ( \frac{16x}{5} \right )\) m

∴ \(\left ( \frac{16x}{5} \right )\) = 800

⇒ x = \(\left ( \frac{800 \times 5}{60} \right )\) = 250.

So, length = 250 m; breadth = 150 m.

∴ Area = (250 × 150 ) m^{2} = 37500 m^{2}

A courtyard 25 m long and 16 m broad is to be paved with bricks of dimensions 20 cm by 10 cm. the total number of bricks required is :

**Answer**: (b)

Number of bricks = Area of Courtyard/Area of 1 brick

⇒ \(\left ( \frac{2500 \times 1600}{20 \times 10} \right )\) = 20000.

A rectangular carpet has an area of sq. m. if its diagonal and longer side together equal 5 times the shorter side, the length of the carpet is :

**Answer**: (b)

We have *l*b = 60 and \(\sqrt{l^{2} + b^{2}} = 5b\)

Now, *l*^{2} + b^{2} = (5b - *l*)^{2}

⇒ 24b^{2} - 10*l*b = 0

⇒ 24b^{2} - 600 = 0

⇒ b2 = 25 ⇒ b = 5.

∴ *l* = \(\frac{60}{5}\) m = 12m.

So, length of the carpet = 12 m.

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