Percentages

Percentages

Percent (%)

Percent means for every hundred or per hundred. The numerator of “per hundred” is called the rate percent. Example 12/100 can be called as 12%, and 12% is the rate percent, and 12 is the rate. The other way to look at it is if some makes a profit of 20%, then one has gained 20/100 of the value invested.

Important notes
1. To express percent as a fraction divide it by 100,
12 % =
2. To express a fraction as a percent multiply it by 100,
½ =
3. Increase/Decrease Percent
=
Here Increase/Decrease = (Final value – initial value)

Important relations in percentage

1. If the price of a commodity increases by r%, then percentage reduction in consumption, so as not to increase expenditure is

Example: If the cost of petrol increases by 40%, by what percent the person should reduce his consumption considering expenditure on petrol remains the same.
Increase in price = 40%, by the formula, decrease in consumption is = = 28.57%

2. If the price of a commodity decreases by r%, then increase in consumption, so as not to decrease expenditure is

Example: If the cost of petrol decreases by 10 %, by what percent can a person increase his consumption considering expenditure on petrol remains the same?
Decrease in price = 10%, by the formula, increase in consumption is = 11.11%

3. If A’s income is r% more than B’s then B’s income is % less than A’s.

Example: If A’s income is 20% more than B’s, then what percent is B’s income lesser than A?
A’s income more than B = 20%, by the formula, B’s income is = × 100 = 16.66 % less of A

4. If A’s income is r % less than B’s then B’s income is
% more than A’s

Example: If A’s income is 20 % less than B’s, then what percent is B’s income more than A?
A’s income less than B = 20%, by the formula, B’s income is = × 100 = 25% more of A

5. If the present population of a town is p and let there be an increase of X % per annum. Then:
(i) Population after n years =
(ii) Population n years ago =
This is the compound interest formula, which we will study in detail later. If the decrease or depreciation is r%, then population or value of a machine (after depreciation) after n years
=
CALCULATIONS IN PERCENTAGES

Let’s start with a number A (= 1A)
1. A increased by 10% would become A + 0.1A = 1.1A
2. A decreased by 10% would become A – 0.1A = 0.9 A
3. A increased by 200% would become A + 2A = 3A
4. A decreased by 50 % would become = 0.5A
5. Use decimal fractions while adding and subtracting and normal fractions while multiplying.

SOLVED EXAMPLES
Q1. Express as percentages: .
Ans. To convert to percent, multiply each by 100
× 100 = 100/3 %, do the others similarly.

Q2. If a pipe, A is 30 meters and 45% longer than another pipe, B find the length of the pipe B.
Ans. A = 30m, A = B + B => A = B => A = 1.45B Therefore B = = = 20.68.

Q3. On my sister’s 15th birthday, she was 159 cm in height, having grown 6% since the year before. How tall was she the previous year?
Ans. Height this year = 159 cm, growth = 6 %, let last year height be A
Now A = 159 A = = 150
Last year height = 150

Q4. Arun spent 25 % of his pocket money, and has Rs 125 left. How much had he at first?
Ans. Pocket money spent = 25%, left = 75%, Let original pocket money be A
Therefore A = 125 A = 125 ×
(By now students should be able to do this in single step), A = 166.66

Q5. If the cost of electricity increases by 30%, by what percent one should reduce his spend in order that spent on electricity stays the same?
Ans. As per the formula, × 100 The reduced percentage spent = × 100 = 23.07%

Q6. If the price of petrol increases by 25% and Rajesh intends to spend only 15% more on petrol, by how much % should he reduce the quantity of petrol that he buys?
Ans. Let the initial cost of 1 litre be A
Cost after increase = 1.25A (25% increase)
Let Rajesh’s initially buy be ‘B’ litres of petrol
Initial spent = AB
Increase in spent = 15%,
Current spent = 1.15 AB (15% increase)
Let the number of litres he is buying now is C.
Therefore 1.25AC = 1.15 AB, 1.25 C = 1.15 B, C = 0.92B, which means that current consumption is 92% of earlier consumption, therefore Rajesh has reduced his consumption by 8 %

Q7. In an election, Congress secured 10% of the total votes more than BJP (consider only two parties in the election and everyone voting). If BJP got 126000 votes, by how many votes did it lose the election?
Ans. Let congress secured X % of the total votes, therefore BJP had secured (X – 10) % of votes, being a two party election:
X + X – 10 = 100
2X = 110
X = 55
Therefore Congress has 55% of vote and BJP has 45%, since BJP got 126000 votes
of total votes = 126000
Total votes = 280000
Congress votes = 55/100 × 280000 = 154000
Difference = 28000, which is the victory margin.

Q8. If the population is1500000 and the expected birth rate is 50%, while the expected death rate is 31%, What will be the net change in the in the population at the end of the one year.
Ans. The current population is 1500000
Number of births will be 50/100 × 1500000 = 750000
Number of deaths was 31/100 × 1500000 = 465000
Net change = 750000 – 465000 = 285000

Q9. What is the % change in the area of a square (which will become rectangle) if its length side is increased by 10% and its width side is decreased by 10%?
Ans. In these types of problems, assume a percent (100) base and then move forward.
Let the side of square be 100
So length side will become
1.1(10% increase) × 100 = 110
And the width side will become
0.9(10% decrease) × 100 = 90
Old Area = 100 × 100 = 10000
New Area = 110 × 90 = 9900
Difference = 10000 – 9900 = 100
Difference percent = 100/10000 × 100
= 1% decrease in the area

Q10. Ram obtains 40 % of the marks in a paper of 200 marks. Shyam is ahead of Ram by 25 % of Ram’s marks, while Bhuvan is ahead of Shyam by one ninth of his own marks. How many marks does Bhuvan get?
Ans. Ram’s marks = 40/100 × 200 = 80
Shyam’s Marks = 1.25 (25% ahead) × 80 = 100
Let Bhuvan’s Marks be A, therefore
A = 100 (Shyam’s marks) + 1/9A
8/9 A = 100, A = 112.5
After grasping the concept of percentages, let us move to its biggest application, problems of profit and losses.