A 6 cm long cigarette burns up in 15 minutes if no puff is

Question

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)20
b)30
c)15
d)50
e)10

A. 20
B. 30
C. 15
D. 50
E. 10

Abhishekgmat87 · Answer

1 minutes puff is equivalent to 3 minutes without puff. As cigarette lasts for 13 minutes, it was puffed for 1 min and not puffed for 12 minutes (someone who want equation, x+y=13, x+ 3y = 15)

No. of puffs = 60/3 = 20

Option A

JerryAtDreamScore · Answer

Breaking Down the Info:

The puffs burn the cigarette three times as fast. For example, if you puff for 10 seconds straight, then we effectively used up 30 seconds of the actual cigarette. This means we expended 20 seconds of the cigarette by puffing.

We can start with the fact that it would take 15 min total with no puffs. The actual time taken with the puffs was 13 min,  so 2 minutes were expended by the puffs.

By using the ratio above, this means we spent 1 min puffing in total.   Thus we puffed for 1 whole min, which is 20 times.

Answer: A

Mugdho · Answer

is there any other method?

Mugdho · Answer

is there any other method?

Posted from my mobile device

san10789 · Answer

Normal Scenario:
 15 minutes to burn 6 cm cigarette ; 
 15*60 seconds to burn 6 cm cigarette.
In 1 second 6/(15*60) cm of cigarette is burnt.  Normal rate.

During puff duration: 
 It is 3 times faster i.e 3 times Normal rate.
 in 1 second 3 * 6/(15*60) cm of cigarette is burnt.  Puff rate.

Actual Total time taken = 13 minutes = 13*60 seconds.

Let the number of Puff be p.
Duration of 1 puff = 3 seconds
 Total puff duration = 3p seconds. 
Using  Puff rate  , In 3p seconds: 3p * 3 * 6/(15*60) cm of cigarette is burnt.. .........(1)

If 3p seconds is time for puff duration then 
(13*60 - 3p) is the time for normal burn with  Normal Rate. 
 In (13*60 - 3p) seconds (13*60 - 3p) * 6/(15*60) cm  of cigarette is burnt......(2)

Total 6 cm of cigarette is burnt. 
Using 1 and 2.
3p * 3 * 6/(15*60) + (13*60 - 3p) * 6/(15*60) = 6

Solving for p;  p = 20

GMATGuruNY · Answer

Average rate of burning =  \frac{6-cm}{13-minutes} = \frac{6}{13}= \frac{30}{65} 
Regular rate of burning =  \frac{6-cm}{15-minutes} = \frac{6}{15} = \frac{2}{5} = \frac{26}{65} 
Rate during puffing = 3 times the regular rate =  3 * \frac{26}{65} = \frac{78}{65}

The regular rate and the puffing rate are MIXED to form the average rate.
The approach below is called ALLIGATION: a great method for mixture problems.
Let R = the regular rate and P = rate during puffing.
The fractions above have all been put over the same denominator so that the alligation can be performed using only the numerators.
To determine the ratio of R to P in the mixture, proceed as follows:

Step 1: Plot the 3 numerators on a number line, with the numerators for R and P on the ends and the numerator for the average rate in the middle. 
R 26------------30------------78 P

Step 2: Calculate the distances between the numerators. 
R 26----- 4 -----30----- 48 -----78 P

Step 3: Determine the ratio of R to P. 
The ratio of R to P is equal to the RECIPROCAL of the distances in red.
R : P = 48:4 = 12:1

When alligation is used for a rate problem, the result indicates that TIME RATIO for the two rates.
Since the total time here is 13 minutes, the ratio above implies that R=12 minutes and P=1 minute, yielding a total time of 13 minutes.
Since P=1 minute, the cigarette is puffed for 60 seconds.
Since each puff lasts for 3 seconds, we get:
Number of puffs  = \frac{total-number-of-seconds}{seconds-per-puff} = \frac{60}{3} = 20

A

GMATGuruNY · Answer

An algebraic approach:

Let t = the time spent puffing, implying that 13-t is the time spent not puffing.

Since the non-puffing rate  = \frac{6-cm}{15-minutes} = \frac{6}{15} = \frac{2}{5} , the distance burnt while not puffing for 13-t minutes  = rt = \frac{2}{5}(13-t)

Puffing rate = 3 times the non-puffing rate  = 3*\frac{2}{5} = \frac{6}{5} 
Thus, the distance burnt while puffing for t minutes  = rt = \frac{6}{5}t

Since the total distance = 6 cm, we get:
 \frac{6}{5}t + \frac{2}{5}(13-t) = 6 
 6t + 2(13-t) = 30 
 6t + 26 - 2t = 30 
 4t = 4 
 t = 1

1 minute = 60 seconds
Since the cigarette is puffed for 60 seconds, and each puff lasts for 3 seconds, the number of puffs  = \frac{total-number-of-seconds}{seconds-per-puff} = \frac{60}{3} = 20

A

Dereno · Answer

Interesting question, let’s first focus on the data given.

A  Normal Burn Rate of t  he cigarette length 6 cm is 15 minutes (900 seconds) - no puff.

Each puff triples the burn rate. That’s efficiency of puff : non puff = 3:1.

As   efficiency is inversely proportional to time.   Time ratio of puff : non puff = 1:3.

The cigar has burned   with puff and non puff   together = 13 minutes =13*60 =  780 seconds.

With   No puff   the time taken is 15 minutes =  900 seconds.

The reduction of time is caused due to the puff which caused some extra burning. With puff each lasting for 3 seconds.

Hence, 900 - 780 = 120 seconds.

Lets now take the time ratio we arrived earlier: puff : non puff = 1:3

If  puff is for 3 seconds (ratio 1); then non puff (ratio 3) equals 9 seconds.

Each puff effectively   saves 6 seconds   of normal burn time. (9 seconds - 3 seconds).

Calculate the number of puffs:

Total time saved = 120 seconds (900 -780 seconds).

Each puff saves 6 seconds of usual burn time.

Number of puffs = 120/6 =  20 puffs. Option A

A 6 cm long cigarette burns up in 15 minutes if no puff is

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A 6 cm long cigarette burns up in 15 minutes if no puff is

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