Question
A number is increased by 20% and then again by 20%. By what per cent should the increased number be reduced so as to get back the original number ?
Answer: Option D Answer: (d)
Let the number be 100.
After 20% increase, number = 120
After 20% increase of 120, number
=$120 × 120/100$ = 144
Per cent decrease = $44/144 × 100$
= $275/9 = 30{5}/9%$
Using Rule 7 and Rule 8,
Increase % = $(20 + 20 + {20 × 20}/100)% = 44%$
Required % = $(44/{100 + 44}) × 100%$
= $4400/144%$
=$275/9% = 30{5}/9%$
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