The price of an item is increased twice successively. The final price is ₹2,295 and the initial price was ₹1,000. The percentage increment the second time was twice the percentage increment the first time. By what percentage was the price increased the second time?

Question

Accepted Answer

Let the percentage increase the first time be x%.  
Then the percentage increase the second time is 2x%.

Initial price = 1000  
After first increase, price = 1000 * (1 + x/100)  
After second increase, price = 1000 * (1 + x/100) * (1 + 2x/100) = 2295

So, (1 + x/100)(1 + 2x/100) = 2295/1000 = 2.295

Expanding:  
(1 + x/100)(1 + 2x/100) = 1 + 2x/100 + x/100 + 2x^2/10000 = 1 + 3x/100 + 2x^2/10000

Set equal to 2.295:   
1 + 3x/100 + 2x^2/10000 = 2.295

Subtract 1:    
3x/100 + 2x^2/10000 = 1.295

Multiply both sides by 10000:    
300x + 2x^2 = 12950

Rewrite:    
2x^2 + 300x - 12950 = 0

Divide entire equation by 2:    
x^2 + 150x - 6475 = 0

Use quadratic formula:    
x = [-150 ± sqrt(150^2 + 4*6475)] / 2   
= [-150 ± sqrt(22500 + 25900)] / 2   
= [-150 ± sqrt(48400)] / 2    
= [-150 ± 220] / 2

Two roots:   
1) (-150 + 220)/2 = 70/2 = 35   
2) (-150 - 220)/2 = -370/2 = -185 (discard negative)

So, x = 35%

The second increase = 2x = 2 * 35 = 70%

Therefore, the price was increased by 70% the second time.

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