The length of a field is increased by 50%, and its width is decreased by 40%. The percentage change in its area is

Question

The length of a field is increased by 50%, and its width is decreased by 40%. The percentage change in its area is

1. A. 5% increase
2. B. 15% increase
3. C. 20% decrease
4. D. 10% decrease (Correct answer)

Correct Answer

Option D — 10% decrease

Detailed Solution & Explanation

The correct answer is 10% decrease.

Key Points

• Let the original length and width be $l$ and $b$, so the original area is $lb$.
• Length increases by 50% → new length $= 1.5l$
• Width decreases by 40% → new width $= 0.6b$
• New area:
  • $A' = 1.5l \times 0.6b = 0.9\,lb$
• The area becomes 0.9 times the original, which is a 10% decrease.

Shortcut Trick

• Use the successive change formula:
  • $\text{Net }\% = x + y + \frac{xy}{100}$
• With $x = +50$ and $y = -40$:
  • $50 + (-40) + \frac{50 \times (-40)}{100} = 10 - 20 = -10$
• The negative sign means a 10% decrease.

Alternate Method

• Assume convenient numbers: let $l = 100$ and $b = 100$, so area $= 10{,}000$.
• New length $= 150$, new width $= 60$, new area $= 9{,}000$.
• Change $= \frac{10{,}000 - 9{,}000}{10{,}000} \times 100 = 10\%$ decrease.

Additional Information

• The multiplying-factor method is fastest for chained percentage changes: convert each change to a factor (+50% → 1.5, −40% → 0.6), multiply them, then compare with 1. A product below 1 is a decrease, above 1 an increase.
• This extends to any number of changes — three successive changes simply mean three factors multiplied together.