Two pipes A and B together can fill a tank in 40 minutes. Pipe A is twice as fast as pipe B. Pipe A alone can fill the tank in :
1 hour
This problem asks us to find the time taken by Pipe A alone to fill a tank, given the time taken by Pipe A and Pipe B together, and the relationship between their speeds. We can solve this by using the concept of work rates.
Understanding Work Rate in Pipe Problems
The work rate of a pipe is the fraction of the tank it can fill in one unit of time (like a minute or an hour). If a pipe fills a tank in \(T\) minutes, its work rate is \(\frac{1}{T}\) tank per minute.
When pipes work together to fill a tank, their individual work rates are added to find the combined work rate.
Setting Up Variables for Pipe A and Pipe B
Let \(t_B\) be the time taken by Pipe B alone to fill the tank in minutes.
We are told that Pipe A is twice as fast as Pipe B. This means Pipe A takes half the time Pipe B takes to do the same work (fill the tank).
So, the time taken by Pipe A alone to fill the tank is \(t_A = \frac{t_B}{2}\) minutes.
Calculating Individual and Combined Work Rates
The work rate of Pipe B is \(\frac{1}{t_B}\) fraction of the tank per minute.
The work rate of Pipe A is \(\frac{1}{t_A} = \frac{1}{t_B/2} = \frac{2}{t_B}\) fraction of the tank per minute.
The combined work rate of Pipe A and Pipe B when working together is the sum of their individual work rates:
Combined Work Rate = Work Rate of A + Work Rate of B
Combined Work Rate = \(\frac{2}{t_B} + \frac{1}{t_B} = \frac{3}{t_B}\) fraction of the tank per minute.
Using the Given Combined Time
We are given that Pipe A and Pipe B together can fill the tank in 40 minutes. This means their combined work rate is \(\frac{1}{40}\) fraction of the tank per minute.
Now we can equate the two expressions for the combined work rate:
\(\frac{3}{t_B} = \frac{1}{40}\)
Solving for the Time Taken by Each Pipe
To find \(t_B\), we can cross-multiply:
\(3 \times 40 = t_B \times 1\)
\(t_B = 120\) minutes.
So, Pipe B alone takes 120 minutes to fill the tank.
Now we find the time taken by Pipe A alone using the relationship \(t_A = \frac{t_B}{2}\):
\(t_A = \frac{120}{2}\)
\(t_A = 60\) minutes.
Pipe A alone can fill the tank in 60 minutes.
Converting to Hours and Checking Options
The time taken by Pipe A is 60 minutes. We know that 1 hour equals 60 minutes.
So, Pipe A alone can fill the tank in 1 hour.
Let's compare this with the given options:
- 1 hour (60 minutes)
- 2 hours (120 minutes)
- 80 minutes
- 20 minutes
Our calculated time for Pipe A (1 hour) matches the first option.
| Pipe | Time (minutes) | Work Rate (fraction/min) |
|---|---|---|
| Pipe B | \(t_B\) | \(\frac{1}{t_B}\) |
| Pipe A | \(t_A = t_B/2\) | \(\frac{1}{t_A} = \frac{2}{t_B}\) |
| A & B Together | 40 | \(\frac{1}{40}\) |
Equation: \(\frac{2}{t_B} + \frac{1}{t_B} = \frac{1}{40} \implies \frac{3}{t_B} = \frac{1}{40}\)
Solving gives \(t_B = 120\) minutes.
Then \(t_A = 120 / 2 = 60\) minutes.
60 minutes = 1 hour.
Revision Table: Key Concepts for Pipe and Tank Problems
| Concept | Definition | How it Applies Here |
|---|---|---|
| Work Rate | Amount of work done per unit of time. For a pipe filling a tank, it's the fraction of the tank filled per minute/hour. | Used to represent the efficiency of Pipe A and Pipe B. Rate = 1 / Time. |
| Relative Speed/Efficiency | Comparison of how fast one entity does work compared to another. If A is twice as fast as B, Rate(A) = 2 × Rate(B). | Used to find the relationship between \(t_A\) and \(t_B\) and their work rates. \(t_A = t_B/2\). |
| Combined Work Rate | The sum of individual work rates when multiple entities work together on the same task (like filling). | Work Rate (A+B) = Work Rate (A) + Work Rate (B). This sum equals the reciprocal of the combined time. |
Additional Information: Solving Work and Time Problems
Work and time problems often involve understanding the inverse relationship between the time taken and the rate of work. If someone works faster, they take less time to complete the same amount of work.
A common approach is to calculate the amount of work done in a unit of time (like 1 day, 1 hour, or 1 minute). This unit work is the rate.
Total Work = Rate \(\times\) Time.
In tank filling problems, the 'total work' is filling the entire tank, which is often represented as 1 unit of work.
If a pipe empties a tank, its work rate is considered negative.
Always ensure consistent units for time throughout the calculation (e.g., stick to minutes or convert everything to hours).
Understanding the relationship between time, rate, and work is crucial for solving these types of quantitative aptitude problems.
Important Questions from Determinants
An even number is the determinant of which of the following matrices?
(A) \(\begin{bmatrix} 1 & -1 \\ -1 & 5 \end{bmatrix}\)
(B) \(\begin{bmatrix} 13 & -1 \\ -1 & 15 \end{bmatrix}\)
(C) \(\begin{bmatrix} 16 & -1 \\ -11 & 15 \end{bmatrix}\)
(D) \(\begin{bmatrix} 6 & -12 \\ 11 & 15 \end{bmatrix}\)
Choose the correct answer from the options given below:
There are 6 cards numbered 1 to 6, one number on one card. Two cards are drawn at random without replacement.
Let X denote the sum of the numbers on the two cards drawn.
Then P(X > 3) is:
A random variable X has the following probability distribution:
X | -2 | -1 | 0 | 1 | 2 -------------------------------------------- P(X) | 0.2 | 0.1 | 0.3 | 0.2 | 0.2
The variance of X will be:
The angle between two lines whose direction ratios are proportional to \( (\sqrt{3} - 1) \), \( (-\sqrt{3} - 1) \), and -4 is:
Given the determinant:
\[ \Delta = \begin{vmatrix} 1 & \cos x & 1 \\ -\cos x & 1 & \cos x \\ -1 & -\cos x & 1 \end{vmatrix} \]
Which of the following statements are correct?
(A) \( \Delta = 2(1 - \cos^2 x) \)
(B) \( \Delta = 2(2 - \sin^2 x) \)
(C) Minimum value of \( \Delta \) is 2
(D) Maximum value of \( \Delta \) is 4
Choose the correct answer from the options given below:
