Pipes and Cisterns: Formulas, Shortcuts and Solved Examples — MCQs with Answers (Set 1)

The inlet’s rate is 1/x tank per hour. The outlet’s rate is 1/y tank per hour. Since x is greater than y, the outlet is the faster pipe. Opposing rates subtract, not add. So the net rate is 1/y minus 1/x, and the tank empties.

The inlet’s rate is 1/x tank per hour. The outlet’s rate is 1/y tank per hour. Since y is greater than x, the inlet is the faster pipe. Opposing rates subtract, not add. So the net rate is 1/x minus 1/y, and the tank fills.

The pipe fills 1 whole tank in x hours. Rate equals total work divided by total time. Total work here is 1 tank, and total time is x hours. So in 1 hour, it fills 1/x of the tank.

An outlet is a pipe that drains water out of a tank. It lowers the water level over time. The opposite kind of pipe is called an inlet. So the correct term here is outlet.

An inlet is a pipe that adds water into a tank. It raises the water level over time. The opposite kind of pipe is called an outlet. So the correct term here is inlet.

The pipe’s rate is 1/27 of the tank per hour. In 5 hours, it fills 5 x 1/27 = 5/27 of the tank. The greatest common factor of 5 and 27 is 1. Dividing numerator and denominator by 1 gives 5/27. So 5/27 of the tank is filled. Check: 5 x 27 = 135, and 27 x 5 = 135, equal as expected.

The pipe’s rate is 1/13 of the tank per hour. In 6 hours, it fills 6 x 1/13 = 6/13 of the tank. The greatest common factor of 6 and 13 is 1. Dividing numerator and denominator by 1 gives 6/13. So 6/13 of the tank is filled. Check: 6 x 13 = 78, and 13 x 6 = 78, equal as expected.

The pipe’s rate is 1/30 of the tank per hour. In 15 hours, it fills 15 x 1/30 = 15/30 of the tank. The greatest common factor of 15 and 30 is 15. Dividing numerator and denominator by 15 gives 1/2. So 1/2 of the tank is filled. Check: 1 x 30 = 30, and 2 x 15 = 30, equal as expected.

The pipe’s rate is 1/8 of the tank per hour. In 6 hours, it fills 6 x 1/8 = 6/8 of the tank. The greatest common factor of 6 and 8 is 2. Dividing numerator and denominator by 2 gives 3/4. So 3/4 of the tank is filled. Check: 3 x 8 = 24, and 4 x 6 = 24, equal as expected.

The pipe’s rate is 1/19 of the tank per hour. In 15 hours, it fills 15 x 1/19 = 15/19 of the tank. The greatest common factor of 15 and 19 is 1. Dividing numerator and denominator by 1 gives 15/19. So 15/19 of the tank is filled. Check: 15 x 19 = 285, and 19 x 15 = 285, equal as expected.

The pipe’s rate is 1/9 of the tank per hour. In 3 hours, it fills 3 x 1/9 = 3/9 of the tank. The greatest common factor of 3 and 9 is 3. Dividing numerator and denominator by 3 gives 1/3. So 1/3 of the tank is filled. Check: 1 x 9 = 9, and 3 x 3 = 9, equal as expected.

The pipe’s rate is 1/17 of the tank per hour. In 10 hours, it fills 10 x 1/17 = 10/17 of the tank. The greatest common factor of 10 and 17 is 1. Dividing numerator and denominator by 1 gives 10/17. So 10/17 of the tank is filled. Check: 10 x 17 = 170, and 17 x 10 = 170, equal as expected.

The pipe’s rate is 1/25 of the tank per hour. In 17 hours, it fills 17 x 1/25 = 17/25 of the tank. The greatest common factor of 17 and 25 is 1. Dividing numerator and denominator by 1 gives 17/25. So 17/25 of the tank is filled. Check: 17 x 25 = 425, and 25 x 17 = 425, equal as expected.

The pipe’s rate is 1/12 of the tank per hour. In 5 hours, it fills 5 x 1/12 = 5/12 of the tank. The greatest common factor of 5 and 12 is 1. Dividing numerator and denominator by 1 gives 5/12. So 5/12 of the tank is filled. Check: 5 x 12 = 60, and 12 x 5 = 60, equal as expected.

The pipe’s rate is 1/21 of the tank per hour. In 13 hours, it fills 13 x 1/21 = 13/21 of the tank. The greatest common factor of 13 and 21 is 1. Dividing numerator and denominator by 1 gives 13/21. So 13/21 of the tank is filled. Check: 13 x 21 = 273, and 21 x 13 = 273, equal as expected.

The pipe’s rate is 1/15 of the tank per hour. In 3 hours, it fills 3 x 1/15 = 3/15 of the tank. The greatest common factor of 3 and 15 is 3. Dividing numerator and denominator by 3 gives 1/5. So 1/5 of the tank is filled. Check: 1 x 15 = 15, and 5 x 3 = 15, equal as expected.

The pipe’s rate is 1/13 of the tank per hour. In 11 hours, it fills 11 x 1/13 = 11/13 of the tank. The greatest common factor of 11 and 13 is 1. Dividing numerator and denominator by 1 gives 11/13. So 11/13 of the tank is filled. Check: 11 x 13 = 143, and 13 x 11 = 143, equal as expected.

Let pipe A fill 1/10 of the tank each hour. Together, A and B fill 1/5 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/5 – 1/10. Using common denominator 50, this is 10/50 – 5/50. That gives 5/50 tank per hour for pipe B. This reduces to 1/10 tank per hour. Time is the reciprocal of the rate: 10 hours. Check: 1/10 + 1/10 = 1/5, which equals 1/5.

Let pipe A fill 1/6 of the tank each hour. Together, A and B fill 1/2 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/2 – 1/6. Using common denominator 12, this is 6/12 – 2/12. That gives 4/12 tank per hour for pipe B. This reduces to 1/3 tank per hour. Time is the reciprocal of the rate: 3 hours. Check: 1/6 + 1/3 = 1/2, which equals 1/2.

Let pipe A fill 1/15 of the tank each hour. Together, A and B fill 1/10 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/10 – 1/15. Using common denominator 150, this is 15/150 – 10/150. That gives 5/150 tank per hour for pipe B. This reduces to 1/30 tank per hour. Time is the reciprocal of the rate: 30 hours. Check: 1/15 + 1/30 = 1/10, which equals 1/10.

Let pipe A fill 1/18 of the tank each hour. Together, A and B fill 1/9 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/9 – 1/18. Using common denominator 162, this is 18/162 – 9/162. That gives 9/162 tank per hour for pipe B. This reduces to 1/18 tank per hour. Time is the reciprocal of the rate: 18 hours. Check: 1/18 + 1/18 = 1/9, which equals 1/9.

Let pipe A fill 1/20 of the tank each hour. Together, A and B fill 1/4 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/4 – 1/20. Using common denominator 80, this is 20/80 – 4/80. That gives 16/80 tank per hour for pipe B. This reduces to 1/5 tank per hour. Time is the reciprocal of the rate: 5 hours. Check: 1/20 + 1/5 = 1/4, which equals 1/4.

Let pipe A fill 1/14 of the tank each hour. Together, A and B fill 1/13 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/13 – 1/14. Using common denominator 182, this is 14/182 – 13/182. That gives 1/182 tank per hour for pipe B. This reduces to 1/182 tank per hour. Time is the reciprocal of the rate: 182 hours. Check: 1/14 + 1/182 = 1/13, which equals 1/13.

Let pipe A fill 1/24 of the tank each hour. Together, A and B fill 1/15 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/15 – 1/24. Using common denominator 360, this is 24/360 – 15/360. That gives 9/360 tank per hour for pipe B. This reduces to 1/40 tank per hour. Time is the reciprocal of the rate: 40 hours. Check: 1/24 + 1/40 = 1/15, which equals 1/15.

Let pipe A fill 1/12 of the tank each hour. Together, A and B fill 1/10 of the tank each hour. Rates add, so pipe B’s rate is the combined rate minus A’s rate. Pipe B’s rate = 1/10 – 1/12. Using common denominator 120, this is 12/120 – 10/120. That gives 2/120 tank per hour for pipe B. This reduces to 1/60 tank per hour. Time is the reciprocal of the rate: 60 hours. Check: 1/12 + 1/60 = 1/10, which equals 1/10.