How to Solve Percentage Problems in Aptitude Tests

Percentage problems appear in every campus aptitude test, and two shortcuts cover the fastest paths: the commutative swap rule and the successive-change formula.

Understanding why these shortcuts work, not just that they work, is what lets you adapt them under time pressure. Each section below derives the method from scratch and then applies it to verified worked examples.

The Swap Rule: a% of b = b% of a

Why It Works

The rule follows directly from multiplication being commutative:

• (a/100) times b = a times b / 100
• (b/100) times a = b times a / 100
• Since a times b = b times a, both sides are equal.

Use the swap whenever one direction gives a round calculation. Choose the easier side; the result is the same.

Applying the Swap to a Single Term

• Q: Find 32% of 50.
• Swap: 32% of 50 = 50% of 32 (valid because both equal 32 times 50 / 100).
• Compute: 50% of 32 = 32 divided by 2 = 16.

The original form (0.32 times 50) requires multiplication. The swapped form is mental arithmetic.

Applying the Swap to a Sum of Two Terms

For a sum of two percentage terms, check whether the two expressions are numerically equal before adding. If they are, the sum simplifies to twice one term.

• Q: Find 45% of 280 + 28% of 450.
• Step 1: 45% of 280 = (45 / 100) times 280 = 45 times 2.8 = 126.
• Step 2: 28% of 450 = (28 / 100) times 450 = 28 times 4.5 = 126.
• Step 3: Both terms are 126, because 45 times 280 = 28 times 450 = 12,600.
• Step 4: Sum = 126 + 126 = 252. Alternatively, 90% of 280 = 280 minus 28 = 252.

Note: the equality at Step 3 holds because the products 45 times 280 and 28 times 450 happen to be the same number. It is not a direct consequence of the swap rule; it is a property of this specific pair of values. Check numerically before using this simplification.

Cricket Score Problem (Verified)

• Q: A batsman scored 110 runs, including 3 boundaries (4 runs each) and 8 sixes (6 runs each). What percentage of his total runs came from running between the wickets?
• Boundary runs: 3 times 4 = 12.
• Six runs: 8 times 6 = 48.
• Non-running total: 12 + 48 = 60.
• Runs from running: 110 minus 60 = 50.
• Percentage: (50 / 110) times 100 = 5,000 / 110 = 500 / 11 = 45 and 5/11 percent.

Successive Percentage Changes

When a value changes by x% and then by y%, the net effect is not simply x plus y. The second percentage is applied to a different base than the first.

Deriving the Net-Change Formula

Starting from first principles, with initial value V:

• After first change of x%: new value = V times (1 + x/100).
• After second change of y%: final value = V times (1 + x/100) times (1 + y/100).
• Expanding the product: V times (1 + x/100 + y/100 + xy/10,000).
• Net percentage change = x + y + xy/100.

Use positive x or y for an increase; use negative x or y for a decrease.

The Classic Trap: Equal Increase and Decrease

• Q: A value increases by 25%, then decreases by 25%. What is the net change?
• x = 25, y = -25
• Net = 25 + (-25) + (25 times -25) / 100 = 0 - 6.25 = -6.25%
• The value is 6.25% below the start, not back at the start.
• Verification with 100: 100 after +25% = 125. 125 after -25% = 93.75. Change = -6.25. Formula confirmed.

Worked Example: Price Change

• Q: A laptop price increases by 15% in year one and decreases by 10% in year two. What is the net percentage change?
• x = 15, y = -10
• Net = 15 + (-10) + (15 times -10) / 100 = 5 - 1.5 = 3.5%
• Answer: Net increase of 3.5%.

Election and Vote Problems

The Two-Step Method

Every election-type percentage problem follows the same structure:

1. Find valid votes: Valid = Total votes times (1 minus invalid-vote fraction).
2. Apply each candidate’s percentage to valid votes only.

Applying candidate percentages to the total vote count gives an inflated denominator and a wrong answer. Step 1 is non-negotiable.

Worked Example (Re-Derived)

• Q: Candidate A received 55% of valid votes. Invalid votes were 20% of 7,500 total votes. How many valid votes did Candidate B receive?
• Step 1: Valid votes = 7,500 times (1 - 0.20) = 7,500 times 0.80 = 6,000.
• Step 2: Candidate B’s share = 100% - 55% = 45% of valid votes.
• Step 3: 45% of 6,000 = 0.45 times 6,000 = 2,700 votes.

Quick Calculation Methods

Benchmark Fractions

Memorising the fraction equivalents below lets you skip the division step for the most common percentages:

Fraction	Percentage
1/2	50%
1/4	25%
3/4	75%
1/5	20%
2/5	40%
1/8	12.5%
3/8	37.5%
1/3	33.33%
2/3	66.67%
1/6	16.67%

The Decomposition Method

Split the target percentage into two benchmark-friendly parts, compute each, then add:

• 42% of 250: (40% + 2%) of 250 = (100 + 5) = 105.
• 17.5% of 400: (10% + 5% + 2.5%) of 400 = (40 + 20 + 10) = 70.
• 37.5% of 960: (1/8) times 960 = 360 (using benchmark fraction directly).

Shifting the decimal one place left gives a tenth of the value; doubling, halving, or adding that anchor covers most targets without multiplication.

Practice sets using both methods are available at IndiaBix’s percentage section, which organises problems by the format campus tests use.

Practice Problems with Worked Solutions

These three problems cover the successive-change, election-type, and equal-product patterns. Re-derive each answer before checking.

Problem 1: Successive Discount

• Q: A retailer applies successive discounts of 20% and 10% on a product. What is the net discount?
• x = -20, y = -10 (both are decreases, so both are negative).
• Net = (-20) + (-10) + ((-20) times (-10)) / 100 = -30 + 2 = -28%
• Answer: Net discount of 28%.

Problem 2: Salary Change

• Q: An employee’s salary increases by 30% in April and decreases by 15% in October. What is the net percentage change?
• x = 30, y = -15
• Net = 30 + (-15) + (30 times (-15)) / 100 = 15 - 4.5 = 10.5%
• Answer: Net increase of 10.5%.

Problem 3: Equal-Product Sum

• Q: Find 35% of 160 + 16% of 350.
• Step 1: 35% of 160 = 35 times 1.6 = 56.
• Step 2: 16% of 350 = 16 times 3.5 = 56.
• Step 3: 35 times 160 = 16 times 350 = 5,600. Both terms are equal.
• Answer: 56 + 56 = 112.

The NCERT Class 8 Comparing Quantities chapter provides the algebraic proofs behind these rules for students who want to trace each formula to its basis.

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The same pattern recognition that shortens these problems, identifying equal products and applying the net-change formula, appears in the quantitative sections of placement tests at companies with structured aptitude rounds. FACE Prep’s guides on the Mu Sigma aptitude test and the ZS Associates recruitment process cover those specific formats in detail. Both appear in nearly every campus online test alongside percentages. The campus placement evaluation test guide and the time and work aptitude questions guide complete the core aptitude framework.

The successive-change formula (net = x + y + xy/100) shows up in AI output metrics. Precision, recall, and accuracy tradeoffs across evaluation runs follow the same compounding logic. If that connection interests you after clearing the placement round, TinkerLLM covers those applied AI concepts at ₹299.