The concept of percentage is important not only for the Quantitative Aptitude section but also for the Data Interpretation (DI) and Data Sufficiency (DS) topics of the SSC CGL Exam.
So let us understand the basic Concepts of Percentage.
Percentage can be divided into per-cent-age that means ‘per every hundred’. It is denoted by the symbol %.
For instance if a student scores 20 out of 25, his score out of 100 is given by multiplying the fraction by 100.
That is, 20/25 × 100 = 80%
Thus any given ratio can be converted into percentage by multiplying the ratio with hundred, and the vice versa to convert percentage into decimal or fraction.
331⁄3% or 33.33
162⁄3% or 16.66%
142⁄7% or 14.28%
121⁄2% or 12.5%;
111⁄9% or 11.11%
91⁄11% or 9.09%
Now we will look at the Concept of Percentage change. There are two changes when a quantity gets modified. They are:
i. Actual change
ii. Percentage change
For example, when the price of a product goes up from Rs 100 to Rs 120, the actual change is Rs 20. This is given by Rs 120-Rs 100.
The Percent change can be calculated by,
(Actual change)/(Original Value) × 100
Thus, the percentage rise in value can be given by
20/100 × 100 = 20%
There is a direct relationship between the numerator of a ratio and the ratio itself. If the numerator increases by a certain percentage, the ratio also increases by the same percentage.
Note: if the denominator remains constant.
For example, 25/40 is 25% more than 20/40. (As the rise in numerator (5) is 25% of its initial value (20))
If there is a percentage rise then the effect of its change can be nullified by 100x/(100 + x) %
For Example: If the price of a product went up by 25% by what percentage should the price be reduced to make the price even?
Percentage reduction required=
(100 × 25)/( 100 + 25) = 2500/125 = 20%
If a value A increases by a%, b%, c% and so on upto n%, the total rise in % is given by the formula,
Final output = A(1 + a/100)(1 + b/100)(1 + c/100)……….(1 + n/100)
If a value A decreases by a%, b%, c% and so on upto n%, the total decline in % is given by the formula,
Final output = A(1 - a/100)(1 - b/100)(1 - c/100)……….(1 - n/100)
1. An electric iron is offered at a discount of 10%. It is sold during clearance sale at 6% discount over the already discounted price at Rs. 1692/- . The original marked price of the electric iron is:
a) Rs. 2000/-
b) Rs. 1896/-
c) Rs. 1900/-
d) Rs. 1946/-
Explanation: SP before 6% discount = 1692 * 100/94
So, original MP = 1692 x 100/94 * 100/90 = 2000
2. In a test, P scored 10% more than Q and Q scored 5% more than R. If R scored 600 marks out of 800, then P’s marks are:
Explanation: Given, R's marks = 600
R's % marks = 600*100/800 = 75%
Q’s% marks = 80%
P’s% marks = 90%
P’s marks = 90% of 800 = 720
3. A person who has a certain amount with him goes to market. He can buy 50 apples or 40 bananas. He retains 10% of the amount for taxi fares and buys 20 bananas and of the balance, he purchases apples. Number of apples he can purchase is:
Explanation: Suppose the person has Rs. 100 with him.
∴ Price per apple is Rs. 2 and that of a banana is Rs. 2.50. After keeping Rs. 10 for taxi, he is left with Rs. 90.
Price of 20 bananas = Rs. 40. Remaining money = (90-40) = Rs. 50
So, he can buy 50/2.5 = 20 apples for this amount.
4. A student took five papers in an examination, where the full marks were the same for each paper. His marks in these papers were in the proportion of 3: 4: 9: 11: 13. In all papers together, the candidate obtained 60% of the total marks. Then, the number of papers in which he got less than 70% marks is:
Explanation: Let the marks scored in five subjects be 3x, 4x, 9x, 11x and 13x
Total marks in all the five subjects = 40x
Max marks of the five subjects = 40x/0.6
(∵ 40x is 60% of total marks)
∴ Max marks in each subject = 40x/(0.6 × 5) = 13.3x
Hence, percentage in each subject = 3x/13.33x × 100,
4x/13.33 × 100, 9x/13.33x × 100, (11x × 100)/13.33x and (16x × 100)/13.33x
Or 22.50%, 30 %, 67.51%, 82.52% and 97.52%
∴ Number of papers in which he got less than 70% marks is 3.
5. 2/5 of the voters promise to vote for A and the rest promised to vote for B. Of these, on the last day 15% of the voters went back their promise to vote for A and 25% of voters went back of their promise to vote for B, and A lost by 4 votes. Then, the total number of voters is:
Explanation: Let total number of votes polled by 100%, then votes polled in favour of P = 40 - 6 + 15 = 49%
Voters polled in favour of Q = 60 -15 + 6 = 51%
Difference = 51 - 49 = 2%
It is already given that a lost by 4 votes, hence total number of votes polled = 200.
6. A dishonest shopkeeper sells his rice using weights 15% less than the true weights and makes a profit of 20%. Find his total gain percentage.
d) None of these
Explanation: Let us consider 1 kg of rice. Its actual weight is 85% of 1000 gm = 850 gm.
Let the cost price of each gram be Re. .1. Then the CP of each bag = Rs. 85.
SP of 1 kg of bag = 120% of the true CP
Therefore, SP = 120/100 * 100 = Rs. 120
Gain = 120 – 85 = 35
Hence, Gain % = 35/850* 100 = 41.17%
7. A man bought two books for Rs. 250 each. If he sells one at a profit of 5%, then how much should he sell the other so that he makes a profit of 20% on the whole?
d) None of these
Explanation: Before we start, it’s important to note here that it is not 15% to be added to 5% to make it a total of 20%.
Let the other profit percent be x.
Then, our equation looks like this.
105/100 * 250 + [(100+x)/100] * 250 = 120/100 * 500 → x= 35.
Hence, if he makes a profit of 35% on the second, it comes to a total of 20% profit on the whole.
8. The numerator and denominator of a fraction is increased and decreased by 12% & and 10% respectively and the fraction becomes 224/270. What is the original fraction?
Explanation: Let the fraction be a/b
4% less than 40% is 522
So, 36% of the total marks=522
9. In an examination it is required to get 40% of the aggregate marks to pass. A student gets 522 marks and is declared failed by 4% marks. What are the maximum aggregate marks a student can get?
Explanation: Let the total marks be x.
4% less than 40% is 522
So, 36% of the total marks = 522
36/100 × x = 522
x = 522 × 100/36
x = 1450
10. A town’s population is 125000 and increases at a rate of 2%. What is the town’s population after 3 years and before 3 years?
a) 132651 & 117649
b) 152361 & 171469
c) 136251 & 114679
d) None of these
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