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A 6 cm long cigarette burns up in 15 minutes if no puff is taken.For every puff, it burns three times as fast during the duration of the puff.If the cigarette burns itself in 13 minutes, then how many puffs has the smoker taken if the average puff lasted 3 seconds?

Answer & Solution
Correct Answer: Option C
Let the number of puffs abe,
$$3x \times \left( {3 \times \frac{1}{{150}}} \right) + \left( {13 \times 60 - 3x} \right) \times \frac{1}{{150}} = 6$$
On solving, We get, x = 20 puffs.
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3 Comments
পৰীক্ষিত কাকতি
পৰীক্ষিত কাকতি 7 years ago
Let no of puffs=n
Without puff ,speed = 6/15
With puff, speed=6*3/15
As, avg puff lasts for 3 sec.So, per puff length of ciggerate burnt=6*3*3/(15*60)
For, 'n' puff cigg.burnt be= 6*3*3*n/(15*60)=54n/(15*60) min
Time left after 'n' puffs=13-3n/60 min

By question,
(13-3n)*6/(60*15) + 54n/(15*60)=6
Solving n=20
পৰীক্ষিত কাকতি
পৰীক্ষিত কাকতি 7 years ago
Understood nothing....
Where is 150 comes from?
Laiba Tanveer
Laiba Tanveer 8 years ago
Please give full solution