Question:medium

A, B and C are three taps connected to a tank. A and B together can fill the tank in 6 h, B and C together can fill it in 10 h and A and C together can fill it in \(7\frac{1}{2}\) h. In how much time would all three take to fill the tank?

Updated On: May 6, 2026
  • \(6\) h
  • \(5\) h
  • \(10\) h
  • \(12\) h
  • \(8\) h
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The problem asks for the total time required to fill a tank if all three taps (A, B, and C) are opened simultaneously.
The provided text contains a repetitive typo from the OCR, stating "A and C together can fill it in 10 h" and then "A and C together can fill it in 7(1/2)".
Standard logical structure for such problems implies the three unique pairings are A+B, B+C, and A+C.
Thus, the correct intended values are A and B in 6 hours, B and C in 10 hours, and A and C in 7.5 hours ($15/2$ hours).
Step 2: Key Formula or Approach:
To solve this, we will determine the portion of the tank filled by each pair in one hour.
Let the rates of the taps be denoted by $\frac{1}{A}$, $\frac{1}{B}$, and $\frac{1}{C}$.
Adding the three combined equations yields twice the combined rate of all three taps together: $2\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right)$.
Step 3: Detailed Explanation:

First, let us write down the 1-hour work rates for each pair of taps based on the corrected problem statement.

The part filled by A and B in 1 hour is $\frac{1}{A} + \frac{1}{B} = \frac{1}{6}$.

The part filled by B and C in 1 hour is $\frac{1}{B} + \frac{1}{C} = \frac{1}{10}$.

The part filled by A and C in 1 hour is $\frac{1}{A} + \frac{1}{C} = \frac{1}{7.5} = \frac{2}{15}$.

We will now add these three equations together.

\[ \left(\frac{1}{A} + \frac{1}{B}\right) + \left(\frac{1}{B} + \frac{1}{C}\right) + \left(\frac{1}{A} + \frac{1}{C}\right) = \frac{1}{6} + \frac{1}{10} + \frac{2}{15} \]

This simplification results in:

\[ 2\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right) = \frac{1}{6} + \frac{1}{10} + \frac{2}{15} \]

Next, we find a common denominator for the fractions on the right side, which is 30.

\[ \frac{1}{6} = \frac{5}{30} \]

\[ \frac{1}{10} = \frac{3}{30} \]

\[ \frac{2}{15} = \frac{4}{30} \]

Now, substitute these back into the sum.

\[ 2\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right) = \frac{5 + 3 + 4}{30} = \frac{12}{30} \]

We can reduce the fraction $\frac{12}{30}$ to $\frac{2}{5}$.

\[ 2\left(\frac{1}{A} + \frac{1}{B} + \frac{1}{C}\right) = \frac{2}{5} \]

Dividing both sides by 2 gives the combined 1-hour work rate of all three taps working simultaneously.

\[ \frac{1}{A} + \frac{1}{B} + \frac{1}{C} = \frac{1}{5} \]

Since the combined rate is $\frac{1}{5}$ of the tank per hour, all three taps will take exactly 5 hours to fill the tank completely.

Step 4: Final Answer:
The time taken by all three taps to fill the tank is 5 h.
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