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### Question 1

Two taps A and B can fill a tank in 16 and 32 minutes respectively. Both the taps are turned on at the same time. After how many minutes should B be turned off so that the tank can be filled in 12 minutes?

### Question 2

A tank is filled in 11 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} - 33a^{2} - ad^{2} + 11d^{2} = 0.

**B**

a^{3} - 22a^{2} + ad^{2} + 11d^{2} = 0.

**C**

a^{3} - 11a^{2} - ad^{2} + 11d^{2} = 0.

**D**

a^{3} - 55a^{2} + ad^{2} + 11d^{2} = 0.

**Soln.**

**Ans: a**

Let the times taken by the three taps be a - d, a and a + d. Then 11 minutes work of all the taps should add to 1. So we have, $11 × 1/{a - d} + 11 × 1/a + 11 × 1/{a + d}$ = 1, which is same as a^{3} - 33a^{2} - ad^{2} + 11d^{2} = 0.

### Question 3

A tap can fill a tank in 2 hours. Because of a leak it took $2{3/8}$ hours to fill the tank. When the tank has been completely filled, the tap is closed. How long will the water last in the tank?

### Question 4

Two taps X, Y and Z can fill a tank in 5, 17 and 4 minutes respectively. All the taps are turned on at the same time. After how many minutes is the tank completely filled?

**A**

$1{167/173}$ mins.

**B**

$2{169/172}$ mins.

**C**

${167/175}$ mins.

**D**

$4{159/175}$ mins.

**Soln.**

**Ans: a**

Let the time be x mins. Then sum of works done by X, Y and Z = 1. $x/5 + x/17 + x/4 = 1$. Solving, we get x = $1{167/173}$. Or use the shortcut ${abc}/{ab + bc + ca}$. Another thing, instead of solving the entire calculation, you can keep an eye on the options to find the nearest answer.

### Question 5

A tank is filled in 17 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} - 51a^{2} - ad^{2} + 17d^{2} = 0.

**B**

a^{3} - 34a^{2} + ad^{2} + 17d^{2} = 0.

**C**

a^{3} - 17a^{2} - ad^{2} + 17d^{2} = 0.

**D**

a^{3} - 85a^{2} + ad^{2} + 17d^{2} = 0.

**Soln.**

**Ans: a**

Let the times taken by the three taps be a - d, a and a + d. Then 17 minutes work of all the taps should add to 1. So we have, $17 × 1/{a - d} + 17 × 1/a + 17 × 1/{a + d}$ = 1, which is same as a^{3} - 51a^{2} - ad^{2} + 17d^{2} = 0.

### More Chapters | See All...

Coding Decoding | Hidden Figures | Basic Simplification | Time and Work | abba Series | Statements and Conclusions | Problems on Numbers | Analogies | Pipes and Cisterns | Passwords and Inputs | More...

This Blog Post/Article "Pipes and Cisterns Quiz Set 003" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.

Updated on 2017-05-17.