Question

9.) Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter takes10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.​

Answer 1

||✪✪ QUESTION ✪✪||

Two water taps together can fill a tank in 75/8 hours. The tap of larger diameter takes10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. ?

|| ✰✰ ANSWER ✰✰ ||

Let us assume that, The TAP with Smaller Diameter can fill the Tank in = x hours .

So, Tap with larger Diameter will take = (x - 10) hours.

So,

→ Per hour Smaller diameter Tap fill = (1/x) litre ,

→ per hour Larger diameter tap fill = 1/(x-10) litre .

Now. , Given that, both Fill the tank in 75/8 hours.

So,

→ 1/x + 1/(x-10) = 8/75

Taking LCM,

→ (x-10 + x) / (x² -10x) = 8/75

Cross - Multiply,

→ 75(2x-10) = 8(x² - 10x)

→ 150x - 750 = 8x² - 80x

→ 8x² - 230x + 750 = 0

Dividing by 2 both sides,

→ 4x² - 115x + 375 = 0

Splitting the Middle Term now,

→ 4x²-100x-15x +375 = 0

→ 4x(x-25)-15(x-25)=0

→ (x-25)(4x-15) = 0

Putting both Equal to zero now,

→ x = 25 or (15/4) .

______________

If x= 25

Than, (x-10) = 25-10 = 15

if x = (15/4)

(x-10) = 15/4 - 10 = (15-40)/4 = -25/4.

As , Time can't be in Negative..

So, x = 25 hours...

Hence, The TAP with Smaller Diameter can fill the Tank in 25 hours and The TAP with Larger Diameter can fill the Tank in 15 hours .

Answer 2

$\huge\boxed{Answer}$

Time taken by smaller tap is 25 hrs and by larger tap is 15 hrs

$\huge\boxed{solution}$

✎ Lets assume that the time taken by smaller tap to fill the tank is x hrs .

✎ It's given that the larger tap will fill the tank 10 hrs early . So larger tap will fill the tank in x-10 hrs .

______________________

✎ Now the volume of water filled in 1hr

By smaller tap = $\frac{1}{x}$ hrs .

By larger tap = $\frac{1}{x-10}$ hrs .

We are given that time taken by both taps is $\frac{75}{8}$ hrs .

✎Now we'll find volume filled by each tap in 75/8 hours .

By smaller tap = $\frac{75}{8}$ × $\frac{1}{x}$ hrs .

= $\frac{75}{8}$ × $\frac{1}{x-10}$ .

= $\frac{75}{8(x-10}$ .

✎ Now we know that Sum of volume filled by both taps will be equivalent to 1 .

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

$\frac{75}{8}$ $\times$ $\frac{1}{x}$ +

$\frac{1}{x-10}$ . = 1

x - 10 + X/ x² -10x = 8/75

(2x-10)75 = (x²-10x)8

150x - 750 = 8x² - 80x

150x + 80x = 8x² + 750

230x = 8x² + 750

8x² - 230x + 750 = 0 .

✎ Now using quadratic formula .

a = 8

b = -230

c = 750

.

.

$\frac{230( + - ) \sqrt{52900 - 24000} }{16}$

$\frac{230( + - ) \sqrt{28900} }{16}$

$\frac{230( + - ) 170}{16}$

✎ Taking positive sign .

$x = \frac{230 + 170}{16} = 25$

✎ Taking negative sign

$x = \frac{230 - 170}{16} = \frac{15}{4}$

✎ Now we have two AnSwers of x . Let's find x-10

If x = 25

x- 10 = 15 hrs.

If x= 15/4

x- 10 = -25/4

As time can't be negative so time taken by larger tap is 15 hrs and by smaller tap is 25 hrs