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?
A. 17
B. 18
C. 20
D. 22
E. None of these
Answer: Option C
Solution (By Examveda Team)
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.This Question Belongs to Arithmetic Ability >> Speed Time And Distance
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
Understood nothing....
Where is 150 comes from?
Please give full solution