# Area

The cost price of 20 articles is the same as the selling price of *x* articles. If the profit is 25%, then the value of *x* is:

**Answer**: (b)

Let C.P. of each article be Rs. 1 C.P. of *x* articles = Rs. *x*.

S.P. of x articles = Rs. 20.

Profit = Rs. (20 - x).

∴ \(\frac{20 - x}{x} \times 100 = 25\)

⇒ 2000 - 100*x* = 25*x*

⇒ 125*x *= 2000

⇒ *x *= 16

Jo’s collection contains US, Indian and British stamps. If the ratio of US to Indian stamps is 5 to 2 and the ratio of Indian to British stamps is 5 to 1, what is the ratio of US to British stamps?

**Answer**: (d)

Indian stamps are common to both ratios. Multiply both ratios by factors such that the Indian stamps are represented by the same number.

US : Indian = 5 : 2, and

Indian : British = 5 : 1.

Multiply the first by 5, and the second by 2.

Hence the two ratios can be combined and

US : British = 25 : 2

Now,

US : Indian = 25 : 10, and

Indian : British = 10 : 2

A bag contains 50 P, 25 P and 10 P coins in the ratio 5 : 9 : 4, amounting to Rs. 206. Find the number of coins of each type respectively.

**Answer**: (c)

Let ratio be x.Hence no. of coins be 5*x* ,9*x* , 4*x* respectively.

Now given total amount

= Rs. 206 ⇒ (.50)(5*x*) + (.25)(9*x*) + (.10)(4*x*)

= 206 we get *x* = 40 ⇒ No. of 50p coins

= 200

⇒ No. of 25p coins = 360

⇒ No. of 10p coins = 160

Fresh fruit contains 68% water and dry fruit contains 20% water. How much dry fruit can be obtained from 100 kg of fresh fruits ?

**Answer**: (c)

The fruit content in both the fresh fruit and dry fruit is the same.

Given, fresh fruit has 68% water. so remaining 32% is fruit content. weight of fresh fruits is 100 kg.

Dry fruit has 20% water. so remaining 80% is fruit content.

Let weight if dry fruit be y kg.

fruit % in fresh fruit = fruit % in dryfruit

(32/100) × 100 = (80/100 ) × y

we get, y = 40 kg

The area of a rectangle is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the rectangular field ?

**Answer**: (b)

Let breadth = *x* metres.

Then length =(115*x*/100) metres.

*x* × (115*x*/100) = 460

⇒ x^{2} = (460 × 100/115)

⇒ x^{2} = 400

⇒ x = 20

A man walked diagonally across a square lot. Approximately, what was the percent saved by not walking along the edges?

**Answer**: (a)

Let the side of the square(ABCD) be *x* meters.

Then, AB + BC = 2*x* metres.

AC = √2*x* = (1.41x) m.

Saving on 2*x* metres = (0.59*x*) m.

Saving % = (0.59*x*/2*x)* × 100 % =30% (approx)

A tank is 25m long 12m wide and 6m deep. The cost of plastering its walls and bottom at 75 paise per sq. m is

**Answer**: (d)

Area to be plastered

= [2 (*l* + b) × h] + (*l* × b)

= [2 (25 + 12) × 6] + (25 × 12)

= 744 sq. m

Cost of plastering = 744 × (75/100)

= Rs. 558

If the radius of a circle is decreased by 50%, find the percentage decrease in its area.

**Answer**: (c)

Let original radius = R.

New radius =b(50/100) R = (R/2)

Original area ΠR^{2} and new area= \(\pi \left ( \frac{R}{2} \right )^{2} = \frac{\pi R^{2}}{4}\)

Decrease in area

= \(\frac{3\pi R^{2}}{4} \times \frac{1}{\pi R^{2}} \times 100\)

= 75%

If each side of a square is increased by 25%, find the percentage change in its area?

**Answer**: (b)

Let each side of the square be a,

then area = a × a

New side = 125a/100 = 5a/4

New area = (5a × 5a) / (4 × 4)

= (25a²/16)

increased area== (25a²/16) - a²

Increase % = [(9a²/16 ) × (1/a² ) × 100] %

= 56.25%

The diagonal of a rectangle is √41 cm and its area is 20 sq. cm. The perimeter of the rectangle must be:

**Answer**: (a)

Let \(\sqrt{l^{2} + b^{2}} = \sqrt{41}\)

Also, *l*b = 20

(*l* + b)^{2} = *l*^{2} + b^{2} + 2*l*b

= 41 + 40 = 81

⇒ (*l* + b) = 9

∴ Perimeter = 2(*l* + b) = 18 cm.

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