Question

2 taps running a together can feel a
tank in 6 min. If 1 tap takes 5 min more than
the other to fill the tank
alone find times require for each
tap to fill tank separately .
​

Answer 1

Given:

• 2 taps running a together can fill a tank in 6 min

• tap takes 5 min more than the other to fill the tank

To Find:

• time required for each tap to fill tank separately

Solution:

❍ Now let the time taken by,

• Faster tap be : ${\bf{\gray{ X}}}$
• Slower tap be : ${\bf{\gray{ X + 5}}}$

↦ In 1 hour the faster tap fills :

• ${\pink{\leadsto{ \sf \: \frac{1}{x} \: part \: of \: the \: tank }}}$

↦ In 1 hour the faster tap fills :

• ${\pink{\leadsto{ \sf \: \frac{1}{x+5} \: part \: of \: the \: tank }}}$

When , both of them are open they fill,

$\longrightarrow \sf \: \frac{1}{x} + \frac{1}{x + 5} \: \: part \: of \: the \: tank$

hence, It is said that they fill the tank in 6 hours

So, when both of them are running together they fill ⅙th part In 1 hour

After framing an equation:

$\longrightarrow \sf \: \frac{1}{x} + \frac{1}{x + 5} = \frac{1}{6} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf 6(2x + 5) = x(x + 5) \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: {x}^{2} - 7x - 30 = 0 \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \\ \longrightarrow \sf \: {x}^{2} + 10x - 3x - 30 = 0 \\ \\ \\ \longrightarrow \sf (x - 10)(x + 3) = 0 \: \: \: \: \: \: \: \:$

Now ,

$\longrightarrow \sf x \: - 10 = 0 \\ \\ \\\longrightarrow { \orange{ \underline{\boxed{ \mathfrak {x = 10 }}}} \bigstar} \:$

or

$\longrightarrow \sf x \: + 3 = 0 \\ \\ \\\longrightarrow { \orange{ \underline{\boxed{ \mathfrak {x = - 3 }}}} \bigstar}$

As the time can't be negetive so,

• ${\bf{\gray{ X= 10mins}}}$

Hence ,

$\longrightarrow \sf time \: taken \: by \: faster \: tap \: = { \orange{10mins \bigstar}} \\ \\ \\ \longrightarrow \sf time \: taken \: by \: slow \: tap \: = { \orange{15mins \bigstar}} \: \:$