For the first x minutes, the first tap works alone at a rate of 1 cistern/20 minutes. We can express the total amount filled in this time as:

$$x\ \min\left(\frac{1\ cistern}{20\ \min}\right)$$

For the remainder of the 15 minutes (in other words, for 15 - x minutes), the two taps work together at their respective rates. We can express the total amount filled in this time as:

$$\left(15\ -x\right)\ \min\left(\frac{1\ cistern}{20\ \min}+\frac{1\ cistern}{30\ \min}\right)$$

If we add the total amount filled in the first x minutes to the total amount filled in the remained of the 15 minutes, we should get one full cistern:

$$x\ \min\left(\frac{1\ cistern}{20\ \min}\right)+\left(15\ -x\right)\ \min\left(\frac{1\ cistern}{20\ \min}+\frac{1\ cistern}{30\ \min}\right)=1\ cistern$$

Or with our units cancelled:

$$x\left(\frac{1}{20}\right)+\left(15\ -x\right)\left(\frac{1}{20}+\frac{1}{30}\right)=1$$

Then we simplify to solve for x:

$$\frac{x}{20}+\frac{15-x}{20}+\frac{15-x}{30}=1$$ $$\frac{3x}{60}+\frac{45-3x}{60}+\frac{30-2x}{60}=1$$ $$3x+45-3x+30-2x=60$$ $$75-2x=60$$ $$2x=15$$ $$x=7.5$$

So the value of x is 7.5.

BTGmoderatorLU wrote:Two taps can fill a cistern in 20 minutes and 30 minutes. The first tap was opened initially for x minutes after which the second tap was opened. If it took a total of 15 minutes for the tank to be filled, what is the value of x?

A. 5.0
B. 7.5
C. 9.0
D. 10.0
E. 12.5

We are given that the rate of the first cistern is 1/20 and the rate of the second cistern is 1/30. The first tap worked for 15 minutes, and the second tap worked for (15 - x) minutes. Thus:

(1/20)(15) + (1/30)(15 - x) = 1

3/4 + (15 - x)/30 = 1

Multiplying the entire equation by 60, we have:

45 + 30 - 2x = 60

2x = 15

x = 7.5

Answer: B