Quantitative Aptitude for JKSSB Exams: Formulas and Worked Examples for Every Topic

The mathematics in most JKSSB papers is arithmetic, pitched at school level. You do not need advanced topics, but you do need speed and accuracy on a fixed set of chapters. This guide lists the main topics, the formulas to remember, and a worked example for each. Practise the examples by hand before you read the solutions.

1. Percentage

A percentage is a fraction out of 100. To find x% of a number, multiply the number by x and divide by 100. If a value increases by a% and then decreases by the same fraction needed to return, the decrease is smaller in percentage terms than the increase.

Example: What is 20% of 450? 450 × 20 ÷ 100 = 90.

Example: A price rises by 25%. By what percentage must it fall to return to the original price? If the price is 100, it becomes 125. The fall needed is 25 out of 125, which is 20%.

2. Profit, loss and discount

Profit = selling price − cost price. Profit percentage = profit ÷ cost price × 100. Discount is taken on the marked price. For two successive discounts of a% and b%, the final price is the marked price × (1 − a/100) × (1 − b/100).

Example: A trader buys an article for ₹400 and sells it at a 20% profit. The selling price is 400 × 1.2 = ₹480.

Example: An article is marked at ₹1,000 and sold with successive discounts of 10% and 5%. The price is 1,000 × 0.9 × 0.95 = ₹855, which is the same as a single discount of 14.5%.

3. Ratio and proportion

To divide a quantity in the ratio a : b : c, add the parts, divide the quantity by the total, and multiply by each part.

Example: Divide ₹1,800 in the ratio 2 : 3 : 4. The total parts are 9, and one part is 1,800 ÷ 9 = 200. The shares are 400, 600 and 800.

4. Average

Average = sum of items ÷ number of items. If you know the average and the number of items, the sum is average × number.

Example: The average of five numbers is 24. If one number, 30, is removed, what is the new average? The sum is 24 × 5 = 120. After removing 30, the sum is 90 for 4 numbers, so the average is 22.5.

5. Time and work

If A can do a job in x days, A's one-day work is 1/x. When two people work together, add their one-day work and take the reciprocal.

Example: A can finish a job in 10 days and B in 15 days. Working together, one day's work is 1/10 + 1/15 = 3/30 + 2/30 = 5/30 = 1/6. They finish in 6 days.

6. Speed, distance and time

Speed = distance ÷ time. To convert km/h to m/s, multiply by 5/18. To convert m/s to km/h, multiply by 18/5. A train crossing a pole travels its own length, and crossing a platform travels its own length plus the platform's length.

Example: Convert 54 km/h to m/s. 54 × 5/18 = 15 m/s.

Example: A train 150 m long crosses a pole in 10 seconds. Its speed is 150 ÷ 10 = 15 m/s, which is 54 km/h.

7. Simple and compound interest

Simple interest = P × R × T ÷ 100. Compound amount = P × (1 + R/100)^T, and compound interest is the amount minus the principal.

Example: Find the simple interest on ₹5,000 at 8% a year for 3 years. 5,000 × 8 × 3 ÷ 100 = ₹1,200.

Example: Find the compound interest on ₹10,000 at 10% a year for 2 years. The amount is 10,000 × 1.1 × 1.1 = 12,100. The compound interest is 12,100 − 10,000 = ₹2,100.

8. LCM and HCF

For two numbers, the product of the numbers equals the product of their HCF and LCM. Use prime factorisation to find each.

Example: Find the HCF and LCM of 12 and 18. The prime factors are 12 = 2² × 3 and 18 = 2 × 3². The HCF is 2 × 3 = 6, and the LCM is 2² × 3² = 36. Check: 6 × 36 = 216 = 12 × 18.

9. Problems on ages

Let the younger person's present age be x. Write the other age in terms of x, then use the change after or before some years to form an equation.

Example: A father is three times as old as his son. After 12 years, he will be twice as old. Find their present ages. Let the son's age be x, so the father's is 3x. After 12 years: 3x + 12 = 2(x + 12), so 3x + 12 = 2x + 24 and x = 12. The son is 12 and the father is 36.

10. Mensuration

With π = 22/7: circle area = πr², circumference = 2πr; cylinder volume = πr²h; rectangle area = l × b; cuboid volume = l × b × h.

Example: A circle has radius 7 cm. Area = 22/7 × 7 × 7 = 154 cm², and circumference = 2 × 22/7 × 7 = 44 cm.

Example: A cylinder has radius 7 cm and height 10 cm. Volume = 22/7 × 7 × 7 × 10 = 1,540 cm³.

11. Number system and simplification

The sum of the first n natural numbers is n(n + 1) ÷ 2. For simplification, apply BODMAS: brackets, orders, division and multiplication, then addition and subtraction.

Example: Find 1 + 2 + 3 + … + 50. 50 × 51 ÷ 2 = 1,275.

Example: Simplify 25% of 80 + 15% of 200. 20 + 30 = 50.

How to practise

• Learn one chapter at a time and finish 20 questions at the Easy level before moving on
• Memorise the squares up to 25, cubes up to 10 and common fraction-percentage pairs (1/8 = 12.5%, 1/6 = 16.67%)
• Time yourself: most arithmetic questions should take about a minute
• Keep a formula sheet of one page and revise it weekly

Practise by topic in the topic sets, and read the guides on HCF and LCM, problems on ages and ratio, percentage and profit-loss for Class-IV.