Squaring Numbers 40 to 60: Aptitude Shortcuts for Placement Tests

Squaring numbers between 40 and 60 without a calculator takes under 5 seconds once you know which of three shortcuts applies. Each shortcut works because 50 is a convenient base for this range.

Where Squaring 40–60 Shows Up in Placement Tests

Quantitative aptitude sections test squares and square roots in two ways: direct computation (“find 47 squared”) and embedded application (distance problems, profit-and-loss, area questions where an intermediate step requires a square). The 40–60 range appears often because these values produce four-digit results that are non-trivial to compute conventionally but are fast with a formula.

Placement tests where this range is relevant:

• TCS NQT Numerical Ability section: squares, cubes, and surds appear across difficulty bands. Each numerical ability question is typically allocated roughly 60 seconds.
• AMCAT Quantitative: speed-and-accuracy format. Squaring identities are a standard shortcut topic.
• Banking sector exams (IBPS PO, SBI PO): quantitative aptitude papers include squares and cubes directly. Students considering banking careers after engineering, discussed in why choose the banking sector after engineering, encounter these in every mock test.

These shortcuts reduce a 20-second pencil calculation to a 5-second mental one, a meaningful edge when 60 questions share a 70-minute window.

Method 1: The Base-50 Addition Formula

Every number between 40 and 60 can be written as 50 + d, where d is the signed difference from 50. Expanding (50 + d)²:

(50 + d)² = 50² + 2 × 50 × d + d² = 2500 + 100d + d²

The formula becomes:

x² = 2500 + 100d + d² where d = x – 50

Three steps: find d, add 100d to 2500, add d².

• Example 1 — 47²:

  • Step 1: d = 47 – 50 = –3
  • Step 2: 2500 + 100 × (–3) = 2500 – 300 = 2200
  • Step 3: 2200 + (–3)² = 2200 + 9 = 2209

• Example 2 — 53²:

  • Step 1: d = 53 – 50 = 3
  • Step 2: 2500 + 300 = 2800
  • Step 3: 2800 + 9 = 2809

• Example 3 — 58²:

  • Step 1: d = 58 – 50 = 8
  • Step 2: 2500 + 800 = 3300
  • Step 3: 3300 + 64 = 3364

Note the sign discipline in Example 1: d is negative when x is below 50, so 100d subtracts from 2500. The formula handles this automatically; the only error risk is dropping the negative sign on d.

Method 2: The Two-Part Visual Split

The same formula rearranges into a pattern that is easier to read off mentally. Notice that 2500 + 100d = 100 × (25 + d) = 100 × (x – 25). So:

x² = 100 × (x – 25) + d²

In plain terms: the first two digits of the answer equal (x – 25), and the last two digits equal d² (zero-padded to two digits if needed).

• Example — 44²:

  • First two digits: 44 – 25 = 19
  • d = 44 – 50 = –6; d² = 36
  • Answer: 19 | 36 = 1936

• Example — 55²:

  • First two digits: 55 – 25 = 30
  • d = 55 – 50 = 5; d² = 25
  • Answer: 30 | 25 = 3025

• Example — 48²:

  • First two digits: 48 – 25 = 23
  • d = –2; d² = 4, zero-padded to 04
  • Answer: 23 | 04 = 2304

The edge case at 40 and 60

At x = 40 and x = 60, d = –10 or +10 and d² = 100, which is three digits. The formula still works: carry 1 into the first-two-digits portion.

• 40²: first two digits = 40 – 25 = 15; d² = 100. Add the carry: 15 + 1 = 16, remainder 00. Answer: 1600.
• 60²: first two digits = 60 – 25 = 35; d² = 100. Add carry: 35 + 1 = 36, remainder 00. Answer: 3600.

For any number from 41 through 59, d² is at most 81 (at d = ±9), so no carry is ever needed.

Method 3: The Difference-of-Squares Approach

The algebraic identity a² – b² = (a – b)(a + b), described formally in Wikipedia’s article on the difference of two squares, rearranges into a squaring shortcut:

x² = (x – d)(x + d) + d²

Pick d so that one of the factors becomes a round number you can multiply quickly.

• Example — 47²: pick d = 3 (lands on 44 and 50)

  • (47 – 3)(47 + 3) + 3² = 44 × 50 + 9 = 2200 + 9 = 2209

• Example — 42²: pick d = 2 (lands on 40 and 44)

  • (42 – 2)(42 + 2) + 2² = 40 × 44 + 4 = 1760 + 4 = 1764
  • Verify: 40 × 44 = 40 × 40 + 40 × 4 = 1600 + 160 = 1760 ✓

• Example — 57²: pick d = 3 (lands on 54 and 60)

  • (57 – 3)(57 + 3) + 3² = 54 × 60 + 9 = 3240 + 9 = 3249

This method is fastest when x is within 3 of a round number (40, 50, or 60), so the factor product is trivial. For d ≥ 5, the two-part split from Method 2 is usually quicker because the factor multiplication gets messy.

IndiaBix’s square and cube roots section has timed practice problems in the same question format used in placement tests, including mixed square-and-root combinations.

Practice Problems and Method Guide

Try these before moving on

Each of the following can be solved mentally in under 20 seconds with the methods above. Work each one, then check the answers below.

• P1: 52² — two-part split (d = 2, d² = 4)
• P2: 44² — already worked above; confirm: 1936
• P3: 57² — difference-of-squares with d = 3
• P4: 41² — two-part split (d = –9, d² = 81)
• P5: 45² — two-part split (d = –5, d² = 25)

Answers: P1 = 2704, P2 = 1936, P3 = 3249, P4 = 1681, P5 = 2025.

Which method to use

Situation	Best method
x is between 41 and 59, no special feature	Two-part split (Method 2)
x is within 3 of 50 (that is, 47 to 53)	Difference-of-squares, pick d = 3
x ends in 5 (45, 55)	(n5)² = n × (n+1) | 25 shortcut
x = 40 or x = 60	Method 1 or Method 2 with carry
First exposure to any of these	Method 1, since the formula is fully explicit

The 10 Technical Interview Aptitude Questions solved set in this cluster covers number series, C programming questions, and speed problems in the same placement-test format.

The three-step decomposition from Method 1 maps onto AI-output verification: break the answer into components, check each one, reassemble. When an LLM produces a multi-part answer, checking each component before accepting the whole is that same instinct applied differently. TinkerLLM (₹299 at tinkerllm.com) is where that habit meets live model calls, applied to real engineering problems rather than aptitude shortcuts.