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Parin Sir > Quantitative Aptitude (Maths) > Percentage > Examples 16 to 22 > Examples 16 to 22

Examples 16 to 22

June 13, 2020
Category: Examples 16 to 22

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(16) (i) If the diameter of a circle increases by 30%, its circumference will increase by how many percentage ?

(ii) In above example area of the circle will increase by how many percent ?

Soln.

(i) In circumference, the increment is 60 $\large \pi$ from 200 $\large \pi$

Means 30 %

(ii)

In Area, the increment is 6900 $\large \pi$ from 10000 $\large \pi$

Means 69 %

(17) In an election where only A and B contested, A got 58 % of total votes and won by 1440 votes. Find the total number of votes polled assuming that there was no invalid vote ?

Soln. A got 58 % of total votes and B got ( 100 – 58 ) % = 42 % of total votes.

$\therefore$ A – B = 16 % = 1440 votes.

$\therefore$ 100 % = 9000 votes.

$\therefore$ Total votes = 9000 votes.

(18) A sum of Rs. 890 was distributed among 4 people, such that A gets 20% less than C and C gets 50 % more than B. If D gets Rs. 150, how much does C get?

Soln. Amount of A + B + C get = Rs. 890 – Rs. 150 = Rs. 740

Let, B gets Rs. 100

$\therefore$ C gets Rs. 150 ( 50 % more than B. )

$\therefore$ A gets Rs. 120 ( 20 % less than C. )

i.e. If total amount get by A + B + C = Rs. 370, then C gets Rs. 150

$\therefore$ If total amount get by A + B + C = Rs. 740, then C gets (?)

Hence, C get Rs. 300.

(19) If the sales tax be reduced from 5 $\frac{1}{2}$ % to 5 $\frac{1}{3}$ %, then what difference does it make to a person who purchases an article with marked price of Rs. 5400 ?

Soln.

Required difference

= $\left (5\frac{1}{2}-5\frac{1}{3} \right )$ % of Rs. 5400

= $\left (\frac{1}{6} \right )$ % of Rs. 5400

= $\frac{1}{6}\times \frac{1}{100}\times$ 5400

= Rs. 9

(20) If 9 % of ( x – y ) = 6 % of ( x + y ), then what percent of x is y ?

Soln. 9 % of (x – y) = 6 % of (x + y)

9 ( x – y ) = 6 ( x + y ) $\left [\because \frac{9}{100}\left (x-y \right ) = \frac{6}{100}\left (x+y \right ) \right ]$

$\therefore$ x = 5 y

$\therefore$ y = $\frac{1}{5}$ x

$\therefore$ y = $\frac{20}{100}$ x = 20 % of x

(21) A salesman’s commission is 5 % on all sales upto Rs.10000 and 10 % on all sales exceeding this. He remits ( submits ) Rs. 25700 to his parent company after deducting his commission. Find the total sales.

Soln. Let his total sales be Rs. x. Now,

(Total Sales) – (Commission) = Rs.25700.

$\therefore$ x – [ 5 % of 10000 + 10 % of (x – 10000) ] = 25700

$\therefore$ x – $\left [\frac{5}{100} \times 10000+\frac{10}{100}\left (x-10000 \right ) \right ]$ = 25700

$\therefore$ x – 500 – $\frac{\left (x-10000 \right )}{10}$ = 25700

$\therefore$ x = 28000 Rs. = Total sales.

(22) How many kg of pure salt must be added to 75 kg of 20 % solution of salt and water to increase it to a 25 % solution?

Soln. 20 % solution of salt means 20 % salt and 80 % water.

$\therefore$ Salt = 15 kg and Water = 60 kg

Let, x kg of pure salt is added.

So, new quantity of salt = 15 + x & Total quantity = 75 + x

$\therefore$ 25 % of (75 + x) = 15 + x

$\therefore$ $\frac{25}{100} \times$ (75 + x) = 15 + x

$\therefore$ 75 + x = 4 (15 + x)

$\therefore$ x = 5

So, pure salt must be added = 5 kg

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Examples 16 to 22

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