Question

. 6 pipes are required to fill a rank in 1 hour 20 minutes. How long will it take if only 5 pipes of the same type are used?

Answer 1

NO OF PIPES REQUIRED TO FILL THE TANK=6
TIME=1H 20MIN=80MINS
TIME TAKEN BY 6 PIPE =80 MINS
TIME TAKEN BY 5 PIPES= a
since it is inversely proportional therefore
a=80x6/5
a=96mins
ans.)THE TIME TAKEN TO FILL THE TANK WITH 5 PIPES IS 96 MINS OR 1 HOUR 36 MINS

Answer 2

$\large\mathfrak{\pmb{{\underline{Given}}:}}$

• 6 pipes are required to fill a tank in 1 hour 20 minutes.

$\large\mathfrak{\pmb{{\underline{To~Find}}:}}$

• How long will it take if only 5 pipes of the same type are used?

$\large\mathfrak{\pmb{{\underline{Solution}}:}}$

Here

HereConverting hour in minute.

HereConverting hour in minute.As we know

${\bf\clubsuit} \: \underline{\boxed{ \sf{1 \: hour = 60 \: minutes }}}$

So,

$: \implies \sf{ 1 \: hour \: 20 \: minutes }$

$\sf : \implies(1\times 60) \: minutes + 20 \: minutes:$

$\sf : \implies60 \: minutes + 20 \: minutes:$

$\bf \purple{ : \implies{80 \: minutes}}:$

$\begin{gathered}\\ \\\end{gathered}$

Now, According to the Question

${ \because\sf{6 \: pipes \: are \: fill \: a \: rank \: in \: 80 \: minutes.}}$

$\because \sf{1 \: pipe = 80×6}$

$\bf\red {\because{1 \: pipe = 480}}$

${ \because\sf{5\: pipes \: are \: fill \: a \: tank = \dfrac{480}{5} }}$

${ \because\sf{5\: pipes = \cancel \dfrac{480}{5} }}$

$\bf {\red {\therefore{5\: pipes =96 \: minutes }}}$

$\begin{gathered}\\ \\\end{gathered}$

Converting minutes into hour

Converting minutes into hourAs we know that

${\bf\clubsuit} \: \underline{ \boxed{\sf{1 \: minute = \dfrac{1}{60} }}}$

So

$: \implies \sf 96 \: minutes:$

$: \implies \sf \dfrac{96}{60} \: minutes:$

$: \implies { \bf{\purple{1\: hour \: 36\: minutes}}}$

$\large\mathfrak{\pmb{{\underline{Therefore}}:}}$

96 minutes or 1 hour 36 minutes will take if only 5 pipes of the same type are used.