Option elimination and value-putting for SSC CHSL

Every SSC CHSL maths question has four options, and exactly one is correct. That structure is information. A candidate who uses it well can test options instead of forming equations, put in easy numbers instead of expanding algebra, and reject three options by their last digit alone.

These techniques don't replace knowing the method. They speed it up and act as a check. In Tier 1 you have 15 minutes for 25 maths questions, about 36 seconds each, so shaving 15 seconds off a question matters. In Tier 2, where a wrong answer costs a full mark, the checking value matters even more.

1. Work backwards from the options

Put each option into the conditions of the question. When the options are in order, start with a middle one. If it is too big, the answer is smaller; if too small, it is larger. You rarely need more than two tests.

Why it works: checking an answer is almost always easier than finding it. Equations in ages, numbers and percentages become simple substitutions.

2. Put simple values

If an expression must hold for all values, or for all values satisfying a condition, any valid numbers will do. Choose values that make the condition true and the arithmetic easy.

For example, (a + b)² − (a − b)² equals what? Put a = 2 and b = 1. You get 9 − 1 = 8, and only 4ab gives 8.

Avoid 0 and 1 when they make several options equal. If two options tie, try a second set of values.

3. Assume a convenient total

In percentage, profit and ratio questions, the actual amount often doesn't matter. Take 100 for percentages, or the LCM of the numbers for time and work. A price that rises 25% and then falls 20% goes 100 → 125 → 100, so there is no net change.

4. Unit-digit check

If the options differ in their last digit, compute only the last digit. The unit digit of a product depends only on the unit digits of the numbers multiplied. For example, 347 × 23 must end in 7 × 3 = 21, so the answer ends in 1.

5. Digit-sum check (casting out nines)

When all options end in the same digit, use the digit sum. Reduce each number to a single digit by adding its digits repeatedly. The digit sum of a product equals the digit sum of the product of the digit sums.

Why: every number leaves the same remainder when divided by 9 as its digit sum does, and remainders multiply consistently. The check can't prove an answer right, but it very often proves the others wrong.

6. Ratio multiples and divisibility

If ₹X is shared in the ratio 2 : 3 : 6, the second share must be a multiple of 3 (it is 3 × some unit). An age in the ratio 5 : 3 must be a multiple of 5. A number divisible by 36 must pass the tests for both 4 and 9. These checks strike out options before you calculate.

7. Estimate and eliminate by range

Round the numbers, find a rough answer, and pick the option nearest to it. This works when options are spread out, not when they are bunched.

Six worked questions

Q1. A number increased by 20% becomes 90. Find the number. Options: 70, 72, 75, 80.
Start in the middle: 75 × 1.2 = 90. Answer: 75. One test, no equation.

Q2. The present ages of two siblings are in the ratio 5 : 3. After 6 years the ratio will be 7 : 5. Find the elder one's present age. Options: 12, 15, 18, 20.
The age must be a multiple of 5: only 15 or 20. Test 15: the other is 9, and after 6 years 21 : 15 = 7 : 5. Answer: 15. (Test 20: 12, then 26 : 18 = 13 : 9, which fails.)

Q3. If a + b + c = 0, find (a² + b² + c²)/(ab + bc + ca).
Put a = 1, b = 1, c = −2. Numerator = 1 + 1 + 4 = 6. Denominator = 1 − 2 − 2 = −3. Answer: −2.

Q4. 457 × 38 = ? Options: 17,346; 17,366; 17,376; 17,386.
All end in 6, so use digit sums. 457 → 7 and 38 → 2, so the product → 14 → 5. The options give 3, 5, 6 and 7. Answer: 17,366.

Q5. ₹1,540 is divided among three people in the ratio 2 : 3 : 6. Find the second share. Options: 400, 420, 440, 460.
It must be a multiple of 3. Only 420 has a digit sum divisible by 3. Answer: ₹420 (1,540 ÷ 11 = 140, and 3 × 140 = 420).

Q6. 18.2% of 596 ≈ ? Options: 84, 108, 126, 150.
18% of 600 = 108. The options are far apart, so the answer is 108 (exact value 108.47).

When not to use shortcuts

• Options close together. Estimation can mislead when the options are 24.2 and 24.8.
• Questions that can't be tested, such as "which of these statements is true?".
• When the direct method is already fast. Testing four options can take longer than a two-line solution.
• Value-putting that ties. If two options give the same value, choose new values rather than guessing.

Common mistakes

Practice

1. A number decreased by 15% becomes 170. Find the number. Options: 190, 195, 200, 210.
2. The present ages of two people are in the ratio 4 : 5. Five years ago the ratio was 3 : 4. Find the younger one's present age. Options: 16, 18, 20, 24.
3. If a + b = 1, find a³ + b³ + 3ab.
4. What is the unit digit of 123 × 456 × 789?
5. Which of these is divisible by 24: 1,044; 1,056; 1,062; 1,070?
6. Estimate 49.8% of 402. Options: 150, 175, 200, 225.
7. 239 × 47 = ? Options: 11,223; 11,233; 11,243; 11,253.
8. ₹2,600 is divided in the ratio 3 : 4 : 6. Find the largest share. Options: 1,100; 1,150; 1,200; 1,250.

Answers:

1. 200. 200 × 0.85 = 170.
2. 20. Multiples of 4 are 16, 20 and 24; only 20 and 25 give 15 : 20 = 3 : 4 five years ago.
3. 1. Put a = 1, b = 0. Or (a + b)³ = a³ + b³ + 3ab(a + b).
4. 2. 3 × 6 = 18 → 8; 8 × 9 = 72 → 2.
5. 1,056. Last three digits 056 = 56, divisible by 8; digit sum 12, divisible by 3. 1,056 = 24 × 44.
6. 200. Half of 400.
7. 11,233. Digit sums: 239 → 5, 47 → 2, product → 10 → 1. Only 11,233 → 10 → 1.
8. ₹1,200. It must be a multiple of 6; only 1,200 is. 2,600 ÷ 13 × 6 = 1,200.

What to do next

• In your next mock, mark every question where an option test or check would have been faster.
• Practise digit-sum checks on 20 multiplications until they take five seconds each.
• Before each question, glance at the options and decide: solve, test or estimate.
• Revise percentages and algebra, where value-putting saves the most time, and the number system for divisibility rules.