---
title: "Round off and compute errors"
subject: "Numerical Method"
chapter: 1
key: "semester_4/numerical_method:round-off-significant-figures"
source: https://www.bctengineer.com/q/semester_4/numerical_method/round-off-significant-figures
site: BCT Engineering
---

# Round off and compute errors

Numerical Method · Chapter 1: Introduction, Approximation and errors of computation

- Asked in: 2068 Magh, 4 marks
- Video: [Chopping & Rounding off | Approximation | Numerical Methods - 1 | Expert Tutor](https://www.youtube.com/watch?v=Ey92xybNC8Q) — Expert Tutor
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_4/numerical_method/round-off-significant-figures
- Whole chapter: https://www.bctengineer.com/md/semester_4/numerical_method/chapter-1

## Question

Round off the numbers 865250 and 37.46235 to four significant figures and compute relative, absolute and percentage errors.

## Short answer

Rounding to four significant figures: 865250 becomes 865200 (absolute error 50, relative 5.779×10⁻⁵, percentage 0.00578%); 37.46235 becomes 37.46 (absolute error 0.00235, relative 6.273×10⁻⁵, percentage 0.00627%).

## Answer

Rounding to *n* significant figures keeps the first n digits, counting from the first non-zero digit. The discarded part decides what happens to the last kept digit:
- less than half a unit in the nth place: leave the nth digit unchanged;
- more than half a unit: increase the nth digit by 1;
- exactly half a unit: increase the nth digit by 1 if it is odd, and leave it unchanged if it is even (the round-to-even rule of standard numerical methods texts).

Error formulas (X = given value, X′ = rounded value):
- Absolute error = |X − X′|
- Relative error = |X − X′| / |X|
- Percentage error = relative error × 100

#### (i) 865250
First four significant figures: 8 6 5 2. The discarded part is 50, which is exactly half a unit of the fourth place (100). The fourth digit 2 is even, so it is left unchanged: **865200**.
(With the simpler rule "5 or more, round up", the result would be 865300. The absolute error is 50 either way, so the errors below are the same.)

- Rounded value = 865200
- Absolute error = |865250 − 865200| = 50
- Relative error = 50 / 865250 = 5.779 × 10⁻⁵
- Percentage error = 0.00578 %

#### (ii) 37.46235
First four significant figures: 3 7 4 6. The next digit is 2 (less than half a unit), so no change: **37.46**.

- Rounded value = 37.46
- Absolute error = |37.46235 − 37.46| = 0.00235
- Relative error = 0.00235 / 37.46235 = 6.273 × 10⁻⁵
- Percentage error = 0.00627 %

**Answer:** 865250 → 865200 (865300 if halves are always rounded up) with Ea = 50, Er = 5.779 × 10⁻⁵, 0.00578 %; 37.46235 → 37.46 with Ea = 0.00235, Er = 6.273 × 10⁻⁵, 0.00627 %.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Rounding 865250 to 4 significant figures gives:
   A. 865300
   B. 865200
   C. 865250
   D. 8653
   Answer: A. First four digits 8652, next digit 5 so round up to 8653, restore place value: 865300. (Key point: Rounding rule)

2. Absolute error when 865250 is rounded to 865300 is:
   A. 0.00578
   B. 50
   C. 5.78e-5
   D. 865300
   Answer: B. |865250 - 865300| = 50. (Key point: Absolute error)

3. Relative error for 865250 -> 865300 is:
   A. 50/865300 * 100
   B. 865250/50
   C. 50/865250 = 5.78e-5
   D. 0.00578 (as a plain ratio)
   Answer: C. Relative error is Ea divided by the given value; 0.00578 is its percentage. (Key point: Relative error)

4. Why is 37.46235 rounded to 37.46 (not 37.47)?
   A. The last digit is odd
   B. It has a decimal point
   C. Four digits are always truncated
   D. The next digit 2 is less than 5
   Answer: D. A digit under 5 leaves the last kept digit unchanged. (Key point: No round up)

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