## Quantitative Aptitude Notes - Percentages

### Quantitative Aptitude Notes - Percentages

What is Percentage: A fraction with its denominator as ‘100’ is called a percentage. Percentage means per hundred. So it is a fraction of the form 6/100 , 37/100, 151/100 and these fractions can be expressed as 6%, 37% and 151% respectively. By a certain percent ,we mean that many hundredths.
Thus x percent means x hundredths, written as x%.
To express x% as a fraction: We have, x% = x/100.
Thus, 20% =20/100 =1/5; 48% =48/100 =12/25, etc.
To express a/b as a percent: We have, a/b = ((a/b)*100)%
Thus, ¼ =[(1/4)*100] = 25%; 0.6 =6/10 =3/5 =[(3/5)*100]% =60%
Why Percentage: Percentage is a concept evolved so that there can be a uniform platform for comparison of various things. (Since each value is taken to a common platform of 100)
Example: To compare three different students depending on the marks they scored we cannot directly compare their marks until we know the maximum marks for which they took the test. But by calculating percentages they can directly be compared with one another.
Important Points to Remember:
a) If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is
[R / (100+R))*100] %
b) If the price of the commodity decreases by R%, then the increase in consumption so as to decrease the expenditure is
[(R / (100-R)*100] %
c) If A is R% more than B, then B is less than A by
[(R/(100+R))*100]%
d) If A is R% less than B, then B is more than A by
[(R/(100-R))*100]%
Results on Population: Let the population of the town be P now and suppose it increases at the rate of R% per annum, then:
1. Population after n years = P [1+(R/100)]n
2. Population n years ago = P / [1+(R/100)]n
Results on Depreciation: Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1. Value of the machine after n years = P [1-(R/100)]n
2. Value of the machine n years ago = P / [1-(R/100)]n

MEMORY BASED SOLVED EXAMPLES BASED ON VARIOUS TYPES:
Type 1: Problems based on partial percentage:
1. Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was 56% of the sum of their marks. The marks obtained by them are:
Solution:
Let the marks be (x + 9) and x.
Then, (x + 9) = (56/100) x (x + 9 + x)
=> 25(x + 9) = 14(2x + 9) => x = 33
So there marks are 42 and 33.
2. In an election between two candidates, one got 55% of the total valid votes, 20% of the votes were invalid. If the total number of votes were 7500, the number of valid votes that the other candidate got, was:
Solution:
Number of valid votes = 80% of 7500 = 6000
Valid votes polled by other candidate = 45% of 6000 = 2700.
3. Teacher took exam for English, average for the entire class was 80 marks. If we say that 10% of the students scored 95 marks and 20% scored 90 marks then calculate average marks of the remaining students of the class
Solution:
Let’s assume that total number of students in class is 100 and required average be x.
Then from the given statement we can calculate=(10 * 95)+(20 * 90)+(70 * x)=(100 * 80)
=> 70x = 8000 - (950 + 1800) = 5250 => x = 75.
4. In a hotel, 60% had vegetarian lunch while 30% had non-vegetarian lunch and 15% had both type of lunch. If 96 people were present, how many did not eat either type of lunch?
Solution:
n(A) = (60/100 * 96) = 288/5; n(B) = (30/100 * 96) = 144/5
n(A n B) = (15/100 * 96) = 72/5
People who have either or both lunch
n(A U B) = 288/5 + 144/5 – 72/5 = 360/5 = 72
So People who do not have either lunch were = 96 – 72 = 24
5. A candidate who gets 20% marks fails by 10 marks but another candidate who gets 42% marks gets 12% more than the passing marks. Find the maximum marks.
Solution:
From the given statement pass percentage = 42% - 12% = 30%
By this 30% of x – 20% of x = 10 marks => 10% of x = 10marks => 100% = 100 marks
6. In an election, three candidates A, B and C contested. A got 42% of the votes polled, B got 28% of the votes polled and C got the rest. A got 3096 more votes than C. How many votes were polled in total?
Solution:
In 100% votes, A got 42% of the votes polled, B got 28% of the votes polled, now the remaining 30% of votes would be polled by C, i.e., 30%
Difference between A and C =3096
(42% - 30%) * x = 4096 => 12% x = 3096 => x= 25800.
7. In a test consisting of 75 questions carrying one mark each, Samir answered 75% of the first 40 questions correctly. What approximate percent of the other 35 questions does he need to answer correctly to score 80% on the entire test?
Solution:
Total marks obtained = 75 * 80% =60
Marks obtained in first 40 questions = 40 *75% =30
Remaining marks to secure = 60-30= 30 out of remaining 75-40 = 35 questions.
Therefore, Required percentage = 30/35 * 100 ≈ 86%
Type 2: Percentage based on comparison
1. One student secured 14 marks more than the other and his marks was 60% of the sum of their marks. Find the marks obtained by them?
Solution:
x +14 = 3/5*(x + 14+ x) => 5x + 70 = 6x + 42 => x = 28.
Therefore, marks obtained by them 28 and 42.
2. In a competitive examination in State A, 6% candidates got selected from the total appeared candidates. State B had an equal number of candidates appeared and 7% candidates got selected with 80 more candidates got selected than A. What was the number of candidates appeared from each State?
Solution:
State A and State B had an equal number of candidates appeared.
In state A, 6% candidates got selected from the total appeared candidates
In state B, 7% candidates got selected from the total appeared candidates
But in State B, 80 more candidates got selected than State A
From these, it is clear that 1% of the total appeared candidates in State B = 80
=> Total appeared candidates in State B = 80 x 100 = 8000
=> Total appeared candidates in State A = total appeared candidates in State B = 8000
3. In a certain school, 20% of students are below 8 years of age. The number of students above 8 years of age is 2/3 of the number of students of 8 years of age which is 48. What is the total number of students in the school?
Solution:
Let the total number of students = x
20% of students are below 8 years of age
=> number of students whose age ≥ 8 years = 80% of x ______(i)
Number of students whose age is 8 years =48 _________(ii)
Number of students whose age is greater than 8 years = 48 x 2/3 = 32 _______ (iii)
From (i), (ii), (iii)
80% of x = 42 + 32 = 80
=> 100% of x = 80 x 100/80 = 100 => x = 100
4. 30% of the men are more than 25 years old and 80% of the men are less than or equal to 50 years old. 20% of all men play football. If 20% of the men above the age of 50 play football, what percentage of the football players are less than or equal to 50 years?
Solution:
Let total number of men = 100
Then, 20 men play football. 80 men are less than or equal to 50 years old. Remaining 20 men are above 50 years old.
Number of football players above 50 years old = 20 x 20/100 = 4
Number of football players less than or equal to 50 years old = 20 – 4 = 16
Required percentage = 16/20 x 100 = 80%
5. 605 sweets were distributed equally among in such a way that the no. Of sweets received by each child is 20% of the total no. of children. How many sweets did each child receive?
Solution:
Let no. Of students = x;
Then each child got = 20% of X;
20% of x * x = 605 => x2 = 605* 5 = 3025 => x =55
Each child got = 20% of 55 = 11 sweets
Type 3: Percentage based on Consumption
1. Due to a 25% increase in the price of rice per kilogram, a person is able to purchase 20 kg less for Rs. 400. What is the increased price of rice per kilogram?
Solution:
20kg = 25% of 400 = Rs.100 => 1kg = Rs.5
2. Wheat is now being sold at Rs. 27 per kg. During last month, its cost was Rs. 24 per kg. Find by how much per cent a family reduces its consumption, so as to keep the expenditure fixed.
Solution:
Percentage increase in cost = (27 – 24) x 100/24 = 12.5
Required decrease = (12.5%) x 100/(100 + 12.5) = 11.11
Type 4: Problems based on Depreciation
1. The price of a car is Rs. 3,25,000. It was insured to 85% of its price. The car was damaged completely in an accident and the insurance company paid 90% of the insurance. What was the difference between the price of the car and the amount received?
Solution:
Price of the car = Rs.3,25,000
Car was insured to 85% of its price
Insured price = 325000 x 85/100
Insurance company paid 90% of the insurance.
Amount paid by insurance company = 325000 x 85/100 x 90/100 = 325 x 85 x 9 = 248625
Difference between the price of the car and the amount received = 325000 – 248625 = Rs.76375
Type 5: Problems based on Population
1. The population of a village is 5500. If the number of males increases by 11% and females by 20%, then the population becomes 6330. Find the population of females in the town.
Solution:
Let the male and female of the village be x and y.
Therefore, x+ y = 5500 ---- (1)
1.11 x + 1.2 y = 6330 ------ (2)
(2) - (1) * 1.11 => 0.09 y = 225 => y = 2500
Therefore, the population of females in the town = 2500.
2. The population of a city has increased continually from 1,71,000 to 2,56,500 in a decade. Find the average percentage increases of that city yearly.
Solution:
( 2,56,500 – 1,71,000 ) = 85,500.
% = 85,500 x 100 / 1,71,000 = 50%.
Average is (50 / 10)% = 5%
Type 6: Miscellaneous
1. A mixture of 20 litres of milk and water contains 20% of water. The new mixture is formed by adding 5 lit of water. What is the percentage of milk in the new mixture?
Solution:
20 lit = 4 lit water 16 lit milk
By adding 5 lit water
Total = 9 lit water 16 lit milk (new)
16/25×100=64%
2. A batsman scored 110 runs which included 3 boundaries and 8 sixes. What percent of his total score did he make by running between the wickets?
Solution:
Number of runs made by running = 110 – (3 × 4 + 8 × 6) = 110 – 60 = 50
Therefore, required percentage = 50/110 × 100 = 500/11 = 45 5/11%
3. From the salary of Roja, 20% is deducted as house rent, 10% of rest she spends on children's education and 20% of balance she spends on watching movies. If her savings are Rs.5760/- then hers total salary is:
Solution:
First value = last value× [100/(100-p1)] × [100/(100-p2)]
= 5760× [100/(100 – 20)] × [100/(100 – 10)] × [100/(100 – 20)] = 5760 × 100/80× 100/90× 100/80 = 10000/-
4. Ten percent of Ram's monthly salary is equal to eight percent of Shyam's monthly salary. Shyam's monthly salary is twice Abhinav's monthly salary. If Abhinav's annual salary is Rs. 1.92 lakhs, find Ram's monthly salary?
Solution:
Let the monthly salaries of Ram and Shyam be Rs. r and Rs. s respectively.
10/100 r = 8/100 s => r = 4/5 s
Monthly salary of Abhinav = (1.92 lakhs)/12 = Rs. 0.16 lakhs
s = 2(0.16 lakhs) = 0.32 lakhs
r = 4/5(0.32 lakhs) = Rs. 25600
5. A man spends 40% of his monthly salary on food and one – third of the remaining on Transport. If he saves Rs. 4500 per month which is equal to half the balance after spending on food and transport, his monthly salary is:
Solution:
Let salary = x;
Expenditure of food = x * 40/100 = 2x /5
Remaining Salary = x – (2x /5) = 3x/5
Expenditure of transport = 3x/5 * 1/3 = x/5
Total expenditure of Food and transport = x/5 + 2x/ 5 = 3x/ 5
Remaining Expenditure = 2x/5
According to the question, 2x/5 * ½ = 4500 => x = 22500
6. Ms. Sujata invests Rs. 2170 which is 7% of her monthly salary in mutual funds. Later she invests 18% of her monthly salary in recurring deposit also, she invests 6% of her salary in NSCs. What is the total annual amount invested by Ms. Sujata?
Solution:
7% of monthly salary = Rs. 2170
Total monthly investment= 75 + 18 % + 6% = 31% => 2170*31/7= 9610
Therefore, Annual instalment = 9610 * 12 = Rs. 115320.