Question

19. Two pipes X and Y can fill a cis-
tern in 24 minutes and 32 min-
utes respectively. If both the
pipes are opened together, then
after how much time (in minutes)
should Y be closed so that the
tank is full in 18 minutes ?
(1) 10 (2) 8
(3) 6
(4) 5
​

Answer 1

A work for 18mints so amount of workdone in 18mins. 1/24(18)=3/4. work left=1-3/4 =1/4. complete work is done by b in 32mins. So1/4 work will be done in =1/4*32=8mints

Answer 2

Answer:

8 minutes

Among the given option $<b><u>2) 8 minutes</u></b>$ is the correct option.

Step-by-step explanation:

Method ( 1 ) :

Part filled by X in 1 min = $\frac {1}{24}$

Part filled by Y in 1 min = $\frac {1}{32}$

Let Y is closed after x min. Then, [ Part filled by (X + Y) in x min ] +

[ Part filled by X in ( 18 - x )min ] = 1

∴ x ( $\frac {1}{24}$ + $\frac {1}{32}$ ) + ( 18 - x ) × $\frac {1}{24}$ = 1

⇒ x ( $\frac {4 \:+ \:3}{96}$ ) + $\frac {(\: 18\: -\: x\: )}{24}$ = 1

⇒ $\frac {7x}{96}$ + $\frac {(\: 18\: -\: x\: )}{24}$ = 1

⇒ $\frac {7x \: + \:4(\: 18\: -\: x\: )}{96}$ = 1

⇒ 7x + 4 ( 18 - x ) = 96

⇒ 7x + 72 - 4x = 96

⇒ 7x - 4x = 96 - 72

⇒ 3x = 24

⇒ x = 8 min

$<b><u>Hence, Y must be closed after 8 min </u></b>$

Method ( 2 ) [ The Trick ] :

Two pipes X and Y can fill a tank in x min and y min, respectively. If both the pipes are opened together, then the time after which Pipe Y should be closed so that the tank is full in t minutes, is [ y ( 1 - $\frac{t}{x}$ ) ] min.

Here, x = 24, y = 32 and t = 18

Time after which Pipe Y should be closed

= [ y ( 1 - $\frac{t}{x}$ ) ]

= [ 32 ( 1 - $\frac{18}{24}$ ) ]

= [ 32 × ( $\frac{1}{4}$ ) ]

= 8 minutes