Profit-Loss, Simple and Compound Interest and Ages Notes for UPSC

Profit and loss, simple and compound interest and the problems on ages are the commonest topics of arithmetic. They are also the most useful: the same ideas of percentage and ratio appear in everyday life, in the budget and in the data interpretation questions. In the UPSC CSAT, they come as part of the basic numeracy at the Class X level. A student with a clear idea of the few formulas and the right method can solve these questions in a minute. This note explains the formulas, the method and the typical solved examples, in simple language for students from any background.

Profit and Loss: The Basics

The cost price (CP) is the price at which an article is bought, and the selling price (SP) is the price at which it is sold. If SP is more than CP, there is a profit (SP - CP); if it is less, there is a loss (CP - SP). The profit and the loss percentages are always calculated on the cost price unless the question says otherwise.

Important Rules

• If two articles are sold at the same price, one at a profit of x per cent and the other at a loss of x per cent, there is always a loss of (x squared / 100) per cent.
• A shopkeeper who uses a false weight gains by the difference between the real and the false weight, divided by the false weight.
• When the marked price is raised by a per cent and then a discount of b per cent is given, the final change is calculated by multiplying the factors: (1 + a/100) x (1 - b/100).
• Profit and loss are about money, so always compare the SP and the CP; the percentage depends on the base used.

Solved Examples on Profit and Loss

1. An article bought for Rs 400 is sold at a gain of 25 per cent. Find the SP. SP = 400 x 125 / 100 = Rs 500.
2. A product costing Rs 500 is sold for Rs 450. Find the loss per cent. Loss = Rs 50, and the loss per cent = 50 / 500 x 100 = 10 per cent.
3. Two successive discounts of 20 per cent and 10 per cent are given. Find the net discount. Net = 20 + 10 - (20 x 10 / 100) = 30 - 2 = 28 per cent.
4. The marked price of an article is Rs 1,000, and a discount of 20 per cent is given. Find the selling price. SP = 1,000 x 80 / 100 = Rs 800.
5. Two articles are sold at Rs 1,200 each, one at a gain of 20 per cent and the other at a loss of 20 per cent. Find the overall result. There is a loss of (20 x 20) / 100 = 4 per cent on the total cost.

Simple Interest (SI)

Interest is the money paid for using the money of another person. The amount borrowed or lent is the principal (P), the yearly interest percentage is the rate (R), and the period is the time (T) in years. In simple interest, the interest is calculated only on the original principal.

• SI = (P x R x T) / 100
• Amount (A) = P + SI
• If a sum becomes n times itself in T years at simple interest, the rate = [(n - 1) x 100] / T per cent. A sum that doubles in 5 years has a rate of 20 per cent per year.

Compound Interest (CI)

In compound interest, the interest of each period is added to the principal, and the next interest is calculated on the increased amount ("interest on interest").

The same formula is used for the growth of population and the depreciation (fall in value) of machines, with the positive or negative sign. For population growth, A = P x (1 + R/100) to the power n; for the fall in value, A = P x (1 - R/100) to the power n.

Solved Examples on Interest

1. Find the simple interest on Rs 5,000 at 8 per cent for 3 years. SI = 5,000 x 8 x 3 / 100 = Rs 1,200.
2. Find the compound interest on Rs 10,000 at 10 per cent for 2 years. Amount = 10,000 x 1.1 x 1.1 = Rs 12,100; CI = Rs 2,100.
3. Find the difference between the CI and the SI on Rs 5,000 for 2 years at 10 per cent. Difference = 5,000 x (0.1) squared = Rs 50.
4. At what rate does a sum of money double itself in 5 years at simple interest? Rate = 100 / 5 = 20 per cent.
5. The population of a town is 20,000 and grows at 10 per cent per year. What will it be after 2 years? 20,000 x 1.1 x 1.1 = 24,200.

Problems on Ages

Age problems are word problems that are solved by making equations. The key rules are: (1) the difference between the ages of two persons stays the same at all times; (2) after n years, each person's age increases by n; and (3) n years ago, each person's age was less by n. Let the unknown age be x, and write the facts as an equation.

• Ratio type: If the present ages are in the ratio a:b, take them as ax and bx, and use the information of the future or the past to find x.
• Sum and difference type: Use the sum of the ages and the difference to find each age.

Solved Examples on Ages

1. A father is three times as old as his son. After 10 years, he will be twice as old. Find their present ages. Let the son be S and the father 3S. Then 3S + 10 = 2(S + 10), so S = 10 and the father is 30.
2. The ages of A and B are in the ratio 3:4, and the sum of their ages is 56. Find the age of A. 7 parts = 56, so one part is 8, and A = 3 x 8 = 24 years.
3. Five years ago, the age of a mother was 4 times her daughter. The mother is now 45. Find the daughter's age. Mother 5 years ago = 40, daughter = 10, so the daughter is now 15 years.

Tips for the CSAT Exam

• Convert percentage problems into multiplication factors, such as 1.25 for a gain of 25 per cent and 0.8 for a loss of 20 per cent.
• For CI of 2 years, calculate step by step and do not use the long formula unless the rate is a hard number.
• Check the options first: in many questions, only one option fits a quick estimate.
• In the age problems, write the unknown clearly and check the answer by putting the values back.
• Do not spend more than a minute on any single question in the exam.

Practice Questions With Answers

• Q1. A shopkeeper buys an item for Rs 800 and sells it at a profit of 15 per cent. Find the selling price. Answer: Rs 920.
• Q2. Find the SI on Rs 2,000 at 5 per cent for 4 years. Answer: Rs 400.
• Q3. Find the CI on Rs 4,000 at 5 per cent for 2 years. Answer: Rs 410.
• Q4. The ages of two friends are in the ratio 2:3, and the difference is 6 years. Find the older age. Answer: 18 years.

Read it together with our notes on profit, loss and interest and the CSAT notes section.

Quick Revision

• SP = CP x (100 + gain per cent) / 100; successive discount = a + b - ab/100.
• SI = PRT / 100; CI = P(1 + R/100)^n - P; the CI - SI for 2 years = P(R/100) squared.
• Age problems: the difference remains constant; write equations.

Frequently Asked Questions (FAQ)

What is the formula for profit per cent?

Profit per cent equals profit divided by the cost price, multiplied by 100.

How do I find the selling price with a gain of x per cent?

Selling price equals the cost price multiplied by (100 + x) and divided by 100.

What is the net discount for two successive discounts a and b?

The net discount is a + b - (ab / 100) per cent.

What is the formula for simple interest?

Simple interest equals principal times rate times time, divided by 100.

What is the formula of the amount in compound interest?

Amount equals the principal multiplied by (1 + R/100) to the power n.

What is the difference between CI and SI for 2 years?

The difference is the principal multiplied by (R/100) squared.

How do I solve half-yearly compound interest?

Use half the rate and double the number of years.

When is there a loss if two items are sold at the same price at x per cent profit and loss?

There is always a loss of (x squared / 100) per cent on the total cost.

What are the key rules of age problems?

The difference between two ages stays constant, and after n years each age increases by n.

How can I solve ratio-based age problems?

Take the ages as ax and bx, form an equation using the given future or past information, and solve for x.

Are profit and loss questions asked in CSAT?

Yes. They come as basic numeracy questions at Class X level.

How much time should I spend on a CSAT arithmetic question?

About one minute, using the options and the estimation to save time.