Aptitude Discussion

Q. |
There are four varieties of pipes Pipe $A$, Pipe $B$, Pipe $C$ and Pipe $D$. Each pipe can be either an inlet pipe or an outlet pipe but cannot be both. there are 5 tanks of equal volume. Tank $P$ is filled by Pipe $A$ and Pipe $B$ Tank $Q$ is filled by Pipe $A$ and Pipe $C$ Tank $R$ is filled by Pipe $A$ and Pipe $D$ Tank $S$ is filled by Pipe $B$ and Pipe $C$ Tank T is filled by Pipe $C$ and Pipe $D$ Time taken for the first 3 tanks ($P$, $Q$ and $R$) to get filled are in the ratio $1:2:4$ and the time taken for the $S$ and $T$ tanks to be filled are in the ratio $7:10$. Find the outlet pipes among the 4 varieties. |

✖ A. |
Pipes $C$ and $D$ |

✖ B. |
Pipe $D$ |

✖ C. |
Pipes $A$ and $B$ |

✔ D. |
Pipe $A$ |

**Solution:**

Option(**D**) is correct

Let $A,B,C$ and $D$ do $a,b,c$ and $d$ with of work in an hour.

Let $A$ and $B$ fill the tank in 1 hour.

Then $$A$ and $C$ would fill the tank in 2 hours while $A$ and $D$ in 4 hours.

\(a+b=\dfrac{1}{2}\)---------- (i)

\(a+c=\dfrac{1}{2}\)---------- (ii)

\(a+d=\dfrac{1}{4}\)---------- (iii)

Let $B$ and $C$ take $7k$ hours while $c$ and $D$ take $10k$ hours to fill the tank

\(b+c=\dfrac{1}{7k}\) --------------- (iv)

\(c+d=\dfrac{1}{10k}\) -------------- (v)

\(a=\dfrac{(i)+(ii)-(iv)}{(ii)}=\dfrac{(ii)+(iii)-(v)}{(ii)}\)

\(\Rightarrow a=\dfrac{1+\dfrac{1}{2}-\dfrac{1}{7}}{2}=\dfrac{\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{10}}{2}\)---------- (vi)

\(\Rightarrow k=\dfrac{4}{70}\)

On substituting value of $k$ in equation (vi) we get $a<0$

$⇒ a<0,b>0,c>0$ and $d>0$

Hence **only A** is the outlet pipe.

**Gaurav Karnani**

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Didn't understand the solution. Following is my sort of solution:

On reading question clearly option A & C are eliminated as if they would have been outlet pipes then the tank won't get filled in any time.

Now let us assume time taken by A & B as 1 hour. now we get the equations as : $ a+b = 2(a+c) $ & $ a+c = 2(a+d) $ which on solving gives $ a=b-2c $ & $ a=c-2d $. also we have third equation as $ a+b =4a+d $ which on solving becomes $ 3a =b-4d $. now we get fourth equation as $ 10b+10c=7c+7d $ which gives $ 3c=7d-10b $. now from the first two equations $ b-2c=c-2d $ which gives $ b+2d=3c$. putting value of $ 3c=7d-10b$ in $b+2d=3c $ and on solving we get $11b = 5d$. now putting value of d in equation $3a=b-4d$ we get $3a =5/11d-4d$ which implies a is clearly negative. hence A is the outlet pipe.