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### Question 1

Tap M can fill a cistern in 18 mins. And, a tap N can empty it in 11 mins. In how many minutes will the cistern be emptied if both the taps are opened together when the tank is $13/17$^{th} already empty?

**A**

$6{78/119}$ mins.

**B**

$7{85/118}$ mins.

**C**

$5{68/121}$ mins.

**D**

$9{60/121}$ mins.

**Soln.**

**Ans: a**

1 filled cistern can be emptied in ${18 × 11}/{18 - 11}$ mins. So $1 - 13/17$ = $4/17$ filled cistern can be emptied in ${18 × 11}/{18 - 11}$ × $4/17$ = ${792/119}$, which is same as: $6{78/119}$ mins.

### Question 2

A tank is filled in 13 minutes by three taps running together. Tap A is twice as fast as tap B, and tap B is twice as fast as tap C. How much time will tap A take to fill the tank?

### Question 3

Pipe A can fill a cistern in 30 minutes, while the pipe B can fill it in 20 minutes. They are alternately open for 1 minute. How long will it take the cistern to fill completely?

### Question 4

A tank is filled in 6 minutes by three taps running together. Times taken by the three taps independently are in an AP[Arithmetic Progression], whose first term is a and common difference d. Then, a and d satisfy the relation?

**A**

a^{3} - 18a^{2} - ad^{2} + 6d^{2} = 0.

**B**

a^{3} - 12a^{2} + ad^{2} + 6d^{2} = 0.

**C**

a^{3} - 6a^{2} - ad^{2} + 6d^{2} = 0.

**D**

a^{3} - 30a^{2} + ad^{2} + 6d^{2} = 0.

**Soln.**

**Ans: a**

Let the times taken by the three taps be a - d, a and a + d. Then 6 minutes work of all the taps should add to 1. So we have, $6 × 1/{a - d} + 6 × 1/a + 6 × 1/{a + d}$ = 1, which is same as a^{3} - 18a^{2} - ad^{2} + 6d^{2} = 0.

### Question 5

A tank is filled in 3 minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?

**A**

2.

**B**

3.

**C**

1.

**D**

4.

**Soln.**

**Ans: a**

Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then 3 minutes work of all the taps should add to 1. So we have, 3 × $(1/{1 - d} + 1/1 + 1/{1 + d})$ = 1, which is same as $2/{1 - d^2} + 1$ = ${1/3}$. Solving we get d = ±2.

### More Chapters | See All...

Compound Interest | Clocks and Calendars | Course of Action | Missing Numbers | Inequalities | Time and Work | Venn Diagrams | Statements and Conclusions | Percentages | Coding Decoding | More...

This Blog Post/Article "Pipes and Cisterns Quiz Set 019" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Updated on 2019-08-18.