#### Exercise 8.12

**Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish a work in 12 days, find the time taken by B to finish the piece of work.**

**Sol:**

Let us consider B tales x days to complete the piece of work

Bâ€™s 1 day work = \(\frac{1}{x}\)

Now, A takes 10 days less than that of B to finish the same piece of work that is (x-10) days

Aâ€™s 1 day work =Â \(\frac{1}{x-10}\)

Same work in 12 days

(A and B)â€™sÂ 1 dayâ€™s work = \(\frac{1}{12}\)

According to the question

Aâ€™s 1 day work + Bâ€™s 1 day work = \(\frac{1}{x-10}\) +\(\frac{1}{x}\)

= \(\frac{1}{x}+\frac{1}{x-10}=\frac{1}{12}\)

= \(\frac{x-10+x}{x(x-10)}=\frac{1}{12}\)

= 12(2x-10) = x(x-10)

= 24x-120 = x^{2}-10x

= x^{2}-10x-24x+120 =0

= x^{2}-34x+120 =0

= x^{2}-30x-4x+120 =0

= x(x-30)-4(x-30) =0

=(x-30)(x-4) =0

Either x-30 =0 therefore x =30

Or, x-4 =0 therefore x=4

We observe that the value of x cannot be less than 10 so the value of x = 30

Time taken by B to finish the piece of work is 30 days

**Question 2:Â If two pipes function simultaneously, a reservoir will be filled in 12 hours. One pipe fills the reservoir 10 hours faster than the other. How many hours will the second pipe take to fill the reservoir?**

**Soln:**

Let us assume that the faster pipe takes x hours to fill the reservoir

Portion of reservoir filled by faster pipe in one hour = \(\frac{1}{x}\)

Now, slower pipe takes 10 hours more than that of faster pipe to fill the reservoir that is (x+10 ) hours

Portion of reservoir filled by slower pipe = \(\frac{1}{x+10}\)

Given that, if both the pipes function simultaneously, the same reservoir can be filled in 12 hours

Portion of the reservoir filled by both pipes in one hour = \(\frac{1}{12}\)

Now ,

Portion of reservoir filled by slower pipe in one hour + Portion of reservoir filled by faster pipe in one hour = \(\frac{1}{x}\) + \(\frac{1}{x+10}\)

And portion of reservoir filled by both pipes = \(\frac{1}{12}\)

= \(\frac{1}{x}\) + \(\frac{1}{x+10}\) = \(\frac{1}{12}\)

= 12(2x+10) = x(x+10)

= x^{2}-14x-120 =0

= x^{2}-20x+6x-120 =0

= x(x-20)+6(x-20) =0

=(x-20)(x+6) =0

Either x-20 = 0 therefore x =20

Or, x+6 =0 therefore x =-6

Since the value of time cannot be negative so the value of x is 20 hours

Time taken by the slower pipe to fill the reservoir = x+10 = 30 hours

**Question 3:Â Two water taps together can fill a tank in \(9\frac{3}{8}\). The tap of larger diameter takes 10hours less than the smaller one to fill the tank separately. Find the time in which each tap can **be fill** separately the tank.**

**Soln:**

Let the time taken by the tap of smaller diameter to fill the tank be x hours

Portion of tank filled by smaller pipe in one hour = \(\frac{1}{x}\)

Now, larger pipe diameter takes 10 hours less than the smaller diameter pipe in one hour = \(\frac{1}{x-10}\)

Given that,

Two taps together can fill the tank in \(9\frac{3}{8}\).

= \( \frac{75}{8}\).

Now, portion of the tank filled by both the taps together in one hour

= \(\frac{1}{\frac{75}{8}}=\frac{8}{75}\)

We have ,

Portion of tank filled by smaller pipe in one hour + Portion of tank filled by larger pipe in one hour

= \( \frac{8}{75}\)

= \(\frac{1}{x}+\frac{1}{x-10}=\frac{8}{75}\)

= \(\frac{x-10+x}{x(x-10)}=\frac{8}{75}\)

= 75(2x-10) = 8x(x-10)

= 150x-750 = 8x^{2}-80x

= 8x^{2}-230x+750 =0

= 4x^{2}-115x+375 =0

Here a= 4 , b = -115 , c = 375

\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

= \(x=\frac{115\pm \sqrt{(-115)^{2}-4(4)(375)}}{2(4)}\)

= \(x=\frac{115\pm \sqrt{(13225-6000)}}{8}\)

= \(x=\frac{115\pm\sqrt{85}}{8}\)

The value of x can either be 8 or 3.75 hours.

The value of x is 8 hours

**Question 4:Â **Two pipes running together can fill the tank in \(11\frac{1}{9}\) minutes. if one pipe takes 5 minutes more than the other to fill the tank separately. Find the time in which each pipe would fill the tank separately.

**Sol:**

Let us take the time taken by the faster pipe to fill the tank as x minutes

Portion of tank filled by faster pipe in one minute =Â \(\frac{1}{x}\)

Now,

Time taken by the slower pipe to fill the same tank is 5 minutes more than that of faster pipe = x+5 minutes

Portion of the tank filled by the slower pipe = \(\frac{1}{x+5}\)

Given that,

The two pipes together can fill the tank in \(11\frac{1}{9}\) =\(\frac{100}{9}\)

Portion of tank filled by two pipes together in 1 minute = \(\frac{9}{100}\)

Portion of tank filled by faster pipe in one minute + Portion of the tank filled by the slower pipe

= \(\frac{9}{100}\) = \(\frac{1}{x}+\frac{1}{x+5}\)

= \(\frac{x+5+x}{x(x+5)}= \frac{9}{100}\)

= 9x(x+5) = 100(2x+5)

= 9x^{2}+45x = 200x+500

= 9x^{2}-155x-500 =0

= 9x^{2}-180x+25x-500 =0

= 9x(x-20)+25(x-20) =0

= (x-20)(9x+25) =0

Either x-20 therefore x =20

Or, 9x+25 =0 therefore \(\frac{-25}{9}\)

Since time cannot be negative

So the value of x =20 minutes

The required time taken to fill the tank is 20 minutes

Time taken by the slower pipe is x+5 = 20+5 = 25 minutes

Times taken by the slower and faster pipe are 25 minutes and 20 minutes respectively.