Squaring Numbers Fast: Three Techniques for Aptitude Tests

Squaring a number mentally comes down to picking the right algebraic identity for the number’s position relative to a round anchor.

Three identities cover almost every case that appears in Indian placement aptitude tests. This article derives each one, verifies every worked example, and ends with a practice set to test recall under time pressure.

Why Squaring Speed Matters in Placement Tests

Quantitative aptitude rounds in TCS NQT and AMCAT include 25 to 35 quantitative questions in a 40 to 60-minute window, which works out to roughly 80 to 100 seconds per question. Squares appear in number-system questions, geometry, and series completion. Solving a squares question in 12 seconds instead of 30 seconds returns 18 seconds to your buffer. Across five such questions, that reclaimed time goes to the harder problems.

Speed techniques are not about showing off arithmetic. They are a resource-allocation strategy.

The three techniques below cover:

• Numbers within ±15 of 50 or 100 (base method)
• Any number whose units digit is 5 (ends-in-5 shortcut)
• Numbers within ±5 of any multiple of 10 (binomial expansion)

The Base Method (Difference of Squares)

Derivation

The algebraic identity N² − B² = (N−B)(N+B) rearranges to:

N² = B² + (N−B)(N+B)

Pick B as a round number close to N, and B² is either memorised or computed in one step. The product (N−B)(N+B) then involves two numbers that are both easy to handle.

Suitable bases: 10, 20, 30, 40, 50, 60, 70, 80, 90, 100.

Worked Examples

• 48² with base 50:

  • Step 1: difference = 48 − 50 = −2
  • Step 2: (N−B)(N+B) = (−2)(48+50) = (−2)(98) = −196
  • Step 3: N² = 50² + (−196) = 2500 − 196 = 2304
  • Verify: 48 × 48 = 48×40 + 48×8 = 1920 + 384 = 2304 ✓

• 47² with base 50:

  • Step 1: difference = 47 − 50 = −3
  • Step 2: (N−B)(N+B) = (−3)(47+50) = (−3)(97) = −291
  • Step 3: N² = 2500 − 291 = 2209
  • Verify: (50−3)² = 2500 − 300 + 9 = 2209 ✓

• 98² with base 100:

  • Step 1: difference = 98 − 100 = −2
  • Step 2: (N−B)(N+B) = (−2)(98+100) = (−2)(198) = −396
  • Step 3: N² = 10000 − 396 = 9604
  • Verify: (100−2)² = 10000 − 400 + 4 = 9604 ✓

• 103² with base 100:

  • Step 1: difference = 103 − 100 = 3
  • Step 2: (N−B)(N+B) = 3 × (103+100) = 3 × 203 = 609
  • Step 3: N² = 10000 + 609 = 10609
  • Verify: 103 × 103 = 10300 + 309 = 10609 ✓

When to Use

Use the base method whenever N is within 15 of 50 or 100. If N is 62, the difference from 50 is 12 and from 60 is only 2. At that point, binomial expansion (below) handles it faster.

The Ends-in-5 Shortcut

Derivation

Any number ending in 5 can be written as 10x + 5, where x represents the digits before the 5.

(10x + 5)² = 100x² + 100x + 25 = 100x(x+1) + 25

This means:

• Left part of the result: x × (x+1)
• Right part of the result: always 25

The right part never varies. Only x changes.

Worked Examples

• 25²: x = 2, x(x+1) = 2×3 = 6, result = 625
• 65²: x = 6, x(x+1) = 6×7 = 42, result = 4225
• 75²: x = 7, x(x+1) = 7×8 = 56, result = 5625
• 85²: x = 8, x(x+1) = 8×9 = 72, result = 7225

Verification for 85²: 85×80 + 85×5 = 6800 + 425 = 7225 ✓

The shortcut extends to 3-digit numbers ending in 5. For 125, x = 12, x(x+1) = 12×13 = 156, so 125² = 15625.

What This Technique Demands

Three steps, no exceptions:

• Step 1: Identify x (all digits before the 5).
• Step 2: Multiply x by (x+1).
• Step 3: Write the product, then append 25.

The only variable is the multiplication in Step 2. For single-digit x (numbers 15 through 95), that multiplication uses one of the simplest products in the table: 1×2, 2×3, up to 9×10. No carrying required.

Binomial Expansion for Nearest Multiple of 10

Derivation

For a number N near a multiple of 10, write N = a + b, where a is the nearest multiple of 10 and b is the small remainder (positive or negative, typically in the range −5 to +5).

(a + b)² = a² + 2ab + b²

With small b, the computation has three parts:

• a²: square of a round number (trivial)
• 2ab: double the product of two manageable numbers
• b²: square of a single digit

Worked Examples

• 37²: a = 40, b = −3

  • a² = 1600
  • 2ab = 2 × 40 × (−3) = −240
  • b² = 9
  • N² = 1600 − 240 + 9 = 1369
  • Verify: 37×30 + 37×7 = 1110 + 259 = 1369 ✓

• 63²: a = 60, b = 3

  • a² = 3600
  • 2ab = 2 × 60 × 3 = 360
  • b² = 9
  • N² = 3600 + 360 + 9 = 3969
  • Verify: 63×60 + 63×3 = 3780 + 189 = 3969 ✓

When to Use

Binomial expansion wins when N is within ±5 of a multiple of 10, because b is small and b² is a single-digit-times-single-digit operation. For N = 62, using a = 60 and b = 2 is faster than the base method with base 50 (where the difference is 12).

The base method and binomial expansion are algebraically the same identity when the base is a multiple of 10. The practical difference is which arithmetic path is easier to execute under time pressure.

Practice Set

Work each problem using the appropriate technique. Check your answer before looking at the result.

Problem	Technique	Answer
45²	Ends-in-5	2025
52²	Binomial (a=50, b=2)	2704
97²	Base method (B=100)	9409
105²	Base method (B=100)	11025
35²	Ends-in-5	1225
67²	Binomial (a=70, b=−3)	4489

Verification for the non-obvious answers:

• 52²: (50+2)² = 2500 + 200 + 4 = 2704 ✓
• 97²: 100² + (−3)(97+100) = 10000 − 591 = 9409 ✓
• 105²: 100² + 5×(105+100) = 10000 + 5×205 = 10000 + 1025 = 11025 ✓
• 67²: (70−3)² = 4900 − 420 + 9 = 4489 ✓

For the same carry-chain discipline applied to multiplication, see the FACE Prep guide to multiplying any number by 111. If you’re building a full quantitative aptitude sprint, the guides on solving clock problems and calendar problems cover two more high-frequency sub-topics in TCS NQT and AMCAT.

The ends-in-5 shortcut reduces squaring to a fixed three-step rule with one variable: look at x, compute x(x+1), append 25. That architecture, a fixed template with one input slot and a deterministic output, is what good prompts look like too. TinkerLLM is where you move from understanding that pattern to testing whether it holds across edge cases. Entry is ₹299, and the discipline carries over directly from aptitude prep to building with AI models.