To round off a number, look at the first digit you must drop. If it is more than 5, raise the digit before it by 1. If it is less than 5, keep the digit before it unchanged. If the dropped digit is exactly 5 (with nothing after it), use the "even rule": raise the digit before it only if that digit is odd, otherwise leave it. Memory hook: "5 sits on the fence, so the digit before it must become EVEN."
Decision chart for rounding off the last uncertain digit: look at the digit being dropped, then apply the less-than-5, exactly-5 (even rule), or more-than-5 case.
Your doubts, answered
What do I do when the digit I have to drop is more than 5?
Raise the preceding digit by 1. Example: 2.746 rounded to 3 significant figures. The digit to drop is 6, which is more than 5, so the 4 before it becomes 5. Answer: 2.75. This is the simplest case and most NEET numbers fall here.
What if the digit to be dropped is less than 5?
Leave the preceding digit unchanged. Example: 1.743 rounded to 3 significant figures. The digit to drop is 3, which is less than 5, so the 4 stays as it is. Answer: 1.74. Do not touch the kept digits at all.
What happens when the digit to drop is exactly 5?
NCERT uses the round-half-to-even rule. If the preceding digit is EVEN, simply drop the 5 (no change). If the preceding digit is ODD, raise it by 1. So 2.745 becomes 2.74 (4 is even, drop the 5) and 2.735 becomes 2.74 (3 is odd, raise it to 4). This rule stops errors from always rounding 5 upward.
Do I round off after every step or only at the end?
Round only the FINAL answer. If you round in the middle of a calculation, small errors add up and your final answer can be wrong by a full digit. Carry one or two extra digits through the working, then round once at the end to the correct number of significant figures.
How many digits should the rounded answer have?
The answer must match the significant figure rule of the operation. For multiplication or division, keep as many significant figures as the number with the fewest. For addition or subtraction, keep as many decimal places as the number with the fewest decimal places. Round only after you know this target count.
⚠️ The NEET trap ✗ Rounding 2.745 to 2.75 because 'the digit dropped is 5, so always round up.' ✓ Use the even rule: 2.745 has an even digit (4) before the 5, so drop the 5 and keep 2.74. Only round up when the preceding digit is odd (2.735 to 2.74). 🧠 The exact-5 case is where NTA catches students.
Real NEET questions
NEET 2022
The area of a rectangular field (in m^2) of length 55.3 m and breadth 25 m, after rounding off the value for correct significant digits, is:
A · 138 x 10^1
B · 1382
C · 1382.5
D · 14 x 10^2 ✓
Solution: Step 1: Multiply. Area = 55.3 x 25 = 1382.5 m^2.
Step 2: Find the fewest significant figures. 55.3 has 3, but 25 has only 2. So the answer must have 2 significant figures.
Step 3: Round 1382.5 to 2 significant figures. The first two significant digits are 1 and 3, so we write it as 13.825 x 10^2. The digit to drop is 8 (more than 5), so raise 3 to 4. Result: 14 x 10^2 m^2. Answer: D.
NEET 2023
The diameter of a spherical bob, measured with vernier callipers, yielded the values 3.33 cm, 3.32 cm, 3.34 cm, 3.33 cm and 3.32 cm. The mean diameter to appropriate significant figures is:
A · 3.33 cm ✓
B · 3.32 cm
C · 3.328 cm
D · 3.3 cm
Solution: Step 1: Find the mean. Sum = 3.33 + 3.32 + 3.34 + 3.33 + 3.32 = 16.64 cm. Mean = 16.64 / 5 = 3.328 cm.
Step 2: The raw data each has 3 significant figures, so the mean must be rounded to 3 significant figures.
Step 3: Round 3.328 to 3 significant figures. The digit to drop is 8 (more than 5), so raise the 2 to 3. Result: 3.33 cm. Answer: A.
NEET 2026
Each side of a metallic cube of mass 5.580 kg is measured to be 9.0 cm. Keeping the significant figures in view, the density of the material of the cube can be best expressed as X x 10^3 kg m^-3, where the value of X is:
A · 7.654
B · 7.6
C · 7.65
D · 7.7 ✓
Solution: Step 1: Convert side to metres. 9.0 cm = 0.090 m. Volume = (0.090)^3 = 7.29 x 10^-4 m^3.
Step 2: Density = mass / volume = 5.580 / (7.29 x 10^-4) = 7654 kg m^-3 = 7.654 x 10^3.
Step 3: The side 9.0 cm has only 2 significant figures, so the answer keeps 2 significant figures. Round 7.654 to 2 significant figures: the digit to drop is 5, followed by more digits (654...), so treat it as more than 5 and raise 6 to 7. Result: 7.7 x 10^3. X = 7.7. Answer: D.
Solved Units And Measurements NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Does the even rule apply if there are digits after the 5?
No. The even rule (round-half-to-even) applies only when the digit to be dropped is exactly 5 with nothing after it, like 2.745. If there is anything after the 5, such as 2.7451 or 7.654, the total dropped part is more than 5, so you simply round up the preceding digit.
Is 2.735 rounded to 2.74 or 2.73?
It is 2.74. The digit to drop is 5, and the digit before it is 3, which is odd. By the even rule, an odd preceding digit is raised by 1, so 3 becomes 4, giving 2.74.
Why does NCERT use the even rule instead of always rounding up?
Always rounding 5 upward would push averages slightly too high over many measurements, adding a small bias. The even rule rounds up half the time and down half the time, so errors cancel out. This keeps experimental results honest, which is why NEET follows it.
How do I decide the number of significant figures before rounding?
First apply the correct operation rule. For multiply or divide, the answer keeps the least significant figures among the inputs. For add or subtract, the answer keeps the least decimal places. Only after fixing this target do you round the number to that many digits.
Should intermediate steps be rounded in a long numerical?
No. Keep one or two guard digits through all intermediate steps and round only the final answer. Rounding too early can shift the last significant figure and cost you the mark in NEET.
