Here’s a secret: in multiple-choice aptitude tests, we often don’t need to find the exact answer. We just need to find which option is correct. That’s a completely different game. This topic is about playing that game — using the options themselves as our biggest advantage.
This isn’t about math. It’s about exam META-strategy.
When to Approximate vs Compute Exactly
Look at the options BEFORE we start computing. If they’re spread out, we can round aggressively and still get the right answer. If they’re bunched together, we need precision.
Rounding Strategy
The key to good approximation: round in opposite directions so errors cancel out.
Example 1: Compute 197 × 203 ÷ 401
Options: (a) 80 (b) 100 (c) 120 (d) 150
Options are far apart. Let’s approximate:
- 197 ≈ 200 (rounded up by 3)
- 203 ≈ 200 (rounded down by 3)
- 401 ≈ 400
Answer ≈ 200 × 200 / 400 = 40000/400 = 100
Answer: (b) 100 — didn’t need any actual computation!
Example 2: What is 17.8% of 493?
Options: (a) 72.3 (b) 87.8 (c) 94.2 (d) 101.6
Let’s approximate: 17.8% ≈ 18%, and 493 ≈ 500.
18% of 500 = 90. The actual answer will be slightly less (since we rounded both up). Looking at options, 87.8 is the closest below our estimate.
Answer: (b) 87.8
The exact calculation: 17.8% of 493 = 0.178 × 493 = 87.754 ≈ 87.8 ✓
Unit Digit Elimination
This is one of the most powerful tricks. The unit digit of the answer depends only on the unit digits of the operands. We can eliminate options without doing the full calculation.
Example 3: What is 347 × 23?
Options: (a) 7871 (b) 7981 (c) 7991 (d) 8011
Unit digit of 347 × 23: unit digit of 7 × 3 = 21 → unit digit = 1
All four options end in 1! Unit digit doesn’t help here. But wait — let’s approximate: 347 × 23 ≈ 350 × 23 = 8050. The actual answer should be slightly less. Looking at options, 7981 is the closest below 8050.
More precisely: 347 × 23 = 347 × 20 + 347 × 3 = 6940 + 1041 = 7981 ✓
Answer: (b) 7981
Example 4: What is 143 × 37?
Options: (a) 5271 (b) 5281 (c) 5291 (d) 5301
Unit digit: 3 × 7 = 21 → unit digit = 1. Options (a) and (b) end in 1. Eliminates (c) and (d).
Now between 5271 and 5281: 143 × 37 = 143 × 40 - 143 × 3 = 5720 - 429 = 5291
Wait, that’s option (c)! But (c) ends in 1 too. Let me recheck: 5291 ends in 1. Ah, all of (a), (b), (c) end in 1. Only (d) ends in 1 too. So they ALL end in 1. Unit digit elimination doesn’t help in this case.
Correct answer: 143 × 37 = 5291 → (c)
The lesson: unit digit elimination is powerful when options have different unit digits. When they all share the same unit digit, we need other techniques.
Back-Substitution
This is the secret weapon for equation-solving problems. Instead of solving the equation algebraically, we plug each option into the equation and see which one works.
Example 5: If 3x + 7 = 22, find x.
Options: (a) 3 (b) 5 (c) 7 (d) 9
Instead of solving: plug in each option.
- (a) 3(3) + 7 = 16 ✗
- (b) 3(5) + 7 = 22 ✓
Answer: (b) 5 — faster than solving algebraically for simple equations.
When back-substitution shines:
- Quadratic equations (plug options instead of using the formula)
- Word problems (plug in and verify the condition)
- “Which of the following satisfies…” type questions
Parity Check (Odd/Even)
If we know the answer must be even, we can eliminate all odd options (and vice versa).
Example: What is 124 × 37 + 63?
We don’t need to compute: Even × Odd = Even. Even + Odd = Odd. So the answer is odd.
If options are (a) 4650 (b) 4651 (c) 4652 (d) 4653 — we know the answer must be odd, eliminating (a) and (c).
Order of Magnitude Check
Sometimes we can tell at a glance that an answer is way too big or too small.
Example: A shopkeeper bought 15 items at Rs 230 each. What is the total cost?
Options: (a) 3,450 (b) 34,500 (c) 345 (d) 3,500
Quick estimate: 15 × 230 ≈ 15 × 200 = 3000. The answer is around 3000-4000. Only (a) 3,450 and (d) 3,500 are in range.
Exact: 15 × 230 = 15 × 23 × 10 = 345 × 10 = 3,450
Answer: (a) 3,450
The “Work Backward from Options” Strategy
For some problems, it’s faster to start from the options and work backward.
Example: The sum of three consecutive even numbers is 42. What is the smallest?
Options: (a) 10 (b) 12 (c) 14 (d) 16
Start from option (b): 12, 14, 16 → sum = 42 ✓
Answer: (b) 12
We could set up an equation (x + x+2 + x+4 = 42 → 3x+6 = 42 → x = 12), but plugging in was faster.
Combining Techniques
The real power comes from combining multiple elimination techniques.
Example: What is 387 × 113?
Options: (a) 43,621 (b) 43,731 (c) 43,831 (d) 43,931
Step 1 - Unit digit: 7 × 3 = 21 → unit digit = 1. All options end in 1. No help.
Step 2 - Approximation: 387 × 113 ≈ 400 × 110 = 44,000. Answer is close to 44,000 but slightly less.
Step 3 - Narrow down: 387 × 113 = 387 × 100 + 387 × 13 = 38,700 + 5,031 = 43,731
Answer: (b) 43,731
Strategy Summary
Practice Problems
Problem 1: Approximate: 4987 × 51 ÷ 99. Options: (a) 2400 (b) 2550 (c) 2700 (d) 2850
Problem 2: What is the unit digit of 347^23? Options: (a) 1 (b) 3 (c) 7 (d) 9
Problem 3: If x² - 5x + 6 = 0, find x. Options: (a) 1, 6 (b) 2, 3 (c) -2, -3 (d) 3, 4. Use back-substitution.
Answers
Problem 1: 4987 ≈ 5000, 51 ≈ 50, 99 ≈ 100. Answer ≈ 5000 × 50 / 100 = 250000/100 = 2500. Closest option: (b) 2550. (Actual: 4987 × 51 / 99 ≈ 2568.8)
Problem 2: Unit digit of 7 cycles every 4 powers: 7¹=7, 7²=9, 7³=3, 7⁴=1, 7⁵=7… 23 ÷ 4 = 5 remainder 3. So unit digit = same as 7³ = 3. Answer: (b)
Problem 3: Try option (b): x=2 → 4-10+6=0 ✓. x=3 → 9-15+6=0 ✓. Answer: (b) 2, 3