Lesson 30 of 100 | Quantitative Aptitude / Percentages / प्रतिशत
Finding the original quantity from a percentage change
Learning outcome
Recover an original positive quantity from a final amount and a stated percentage change, including successive changes and exact recurring percentages.
Undo the multiplier
In a forward calculation, original × change factor = final. Therefore the reverse calculation is original = final/change factor. Division undoes multiplication; subtracting a percentage of the final amount generally does not.
After a p% increase, the final amount is (100 + p)% of the original, so divide by 1 + p/100. After a p% decrease, it is (100 − p)% of the original, so divide by 1 − p/100. Decreases here are below 100%, making the divisor positive.
Why is subtraction wrong after an increase? The stated increase was calculated from the original, whereas subtracting that percentage of the final uses a larger base. The reverse operation must remove the effect of the multiplier, not reuse the percentage on a different reference.
After an increase, the original is smaller than the final; after a decrease, it is larger. Use these comparisons to check your answer.
For successive changes, divide by the product of all factors. Alternatively, undo the steps in reverse chronological order to recover intermediate quantities. Keep fractions exact and verify by applying the stated changes forward.
Worked example 1
A fee becomes ₹2,070 after a 15% increase. Find the original fee.
The final fee is 115% of the original. Therefore original = 2070/1.15 = ₹1800. Check: 1800 × 0.15 = ₹270, and 1800 + 270 = ₹2070.
Worked example 2
After 15% of a tank's water is removed, 51 L remains. Find the original volume.
The remaining fraction is 85% = 0.85. Thus original = 51/0.85 = 60 L. Checking, 15% of 60 L is 9 L, and 60 − 9 = 51 L. Removal is assumed to be the only change.
Worked example 3
A price becomes ₹2,376 after a 10% increase followed by a 10% decrease. Find the original.
The combined factor is 1.10 × 0.90 = 0.99. Original = 2376/0.99 = ₹2400. Forward checking gives 2400 × 1.10 = ₹2640, then 2640 × 0.90 = ₹2376. Treating the two changes as cancellation would miss the original price.
Common traps
Do not subtract the original increase percentage from the final amount. Identify whether the final represents an increased or retained fraction. For multiple changes, reverse every factor, not just the last one.
Practice questions
- A balance becomes ₹2,916 after an 8% increase, with no other change. Find the original balance.
- After 12.5% of a rope is cut off, 49 m remains. Find its original length.
- A price rises by 20%, then falls by 25%, ending at ₹1,620. Find the original price.
- Water use becomes 90 L after a reduction of exactly (100/3)%. Find the original use.
Worked answers
- The final represents 108% of the original. Divide: 2916/1.08 = ₹2700. Checking, 2700 × 1.08 = ₹2916.
- The retained fraction is 87.5% = 7/8. Original = 49 ÷ (7/8) = 56 m. Removing 7 m leaves the stated 49 m.
- The combined factor is 1.20 × 0.75 = 0.90. Original = 1620/0.90 = ₹1800. Forward prices are ₹2160 and then ₹1620, confirming both changes.
- The reduction fraction is (100/3)/100 = 1/3, leaving 2/3. Original = 90 ÷ (2/3) = 135 L. A reduction of 45 L leaves usage of 90 L.