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

is there any other method?

is there any other method?

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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

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 \(= \frac{total-number-of-seconds}{seconds-per-puff} = \frac{60}{3} = 20\)

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 \(= \frac{total-number-of-seconds}{seconds-per-puff} = \frac{60}{3} = 20\)