The Cell Phone

let $x$ be the duration (in hours) that the phone is ON .

$\frac{x}{1.5} + \frac{8-x}{9} = 1$

$x = 0.2$ hours.

Converting into minutes, we get

$x = 0.2(60) = 12$ min . $\square$

To put up a more easy understanding...Here's something

Let $x$ be the no. of hours it is off and $y$ be the no. of hours it is ON .

Then $x+y=8$ $\dots \dots$ [Total phone working hours]

Also the part of battery used in $x$ hours is $x/9$ similarly $y$ hours of on phone takes $y/\frac{3}{2}$ $\Rightarrow$ $2y/3$

Putting everything

$x/9$ $+$ $2y/3$ $=$ $1$ $\dots \dots$ [1 = TOTAL PHONE CHARGE]

$\Rightarrow$ $\frac {x+6y}{9}$ $=$ $1$

$\Rightarrow$ $x+6y = x+y+1$

$\Rightarrow$ $5y=1$

$\Rightarrow$ $y=1/5$ hours

Hence solving further to get the time in minutes,

$y=\boxed{12}$ min.

(1.5*8)/9=1.333, 1.5-1.3=(0.2 * 60) =12

I will use the same technique as I have been using lately which can be used to solve any puzzle including Rubik Cube.

1. Make some calculated moves and see the outcome.
2. Find a relation between input and outcome.
3. Work out the input to get your desired outcome.

Now coming to the problem

It is obvious that when turned on it discharges 6 times faster than when it is off or more precisely it will take one-sixth of the time to discharge when on as compared to when it was off.

We know that if we leave it off it will take 9 hours to discharge.

To make a total battery time of 8 hours we need to reduce it by one hour

Now let us suppose if at the end of 8 hours we switch it on then it will discharge in 10 minutes instead of one hour, which means leaving it on for 10 minutes will reduce the total battery time by 50 minutes.

Simple arithmetic will tell that leaving it on for 12 minutes will reduce the battery time by one hour and down to 8 hours.

Let's check it. To have a total battery time of 8 hours and on-time of 12 minutes means off time of 7 hours and 48 minutes. Now at the end of this time if we decide to leave it off it will run for another 1 hour and 12 minutes or 72 minutes. Switching it on will reduce this time to one-sixth which will be 12 minutes thus reducing the time by one hour to a total time of 8 hours.

FOR THOS WHO WANT TO UNDERSTAND While the phone is off the charge decrease by $\frac{1}{9}$ of the total charge per hour . While the phone is on the charge of the battery decrease by \frac{1}{\(\frac{3}{2} }) of the total charge per hour . Translating that into an equation will be>> $\frac{1}{9}$ X + $\frac{2}{3}$ Y =1 (total charge of the battery) . (X: is the number of hours the battery is being discharged while the phone is off ) . (Y: is the number of hours the battery is being discharged while the phone is on). Making another equation which is very clear from his words >> X+Y=8 By solving the two equations the answer will be X=7.8 hour =468 min.and Y=0.2 hour =12 min(THE ANSWER)

Let x equal the portion of the phones battery life that it is off for.

Thus,

8= 9x+1.5(1-x)

8=9x+1.5-1.5x

6.5=7.5x

x=13/15

Thus it was on for 2/15 of its battery life (Its whole battery life minus the 13/15 it is off). Multiply this by 1.5 hours gets you 1/5 an hour= 12 minutes.

The solution is by solving, x/1.5 + (8-x)/9 =1

Let F be the rate of discharge while mobile is OFF and N, the discharge rate while its ON.

90N is full discharge, mathematically.

540F also is full discharge, mathematically.

90N=540F, as the question suggests.

N=6F

If x is the duration in minutes the mobile was ON in the question,

540F = xN+(480-x)F

540F = x6F+480F-Fx

540F = 6Fx+480F-Fx

60F = 5Fx

x=60F/5F=12 minutes.