Question

a tank can be filled by pipe a in 15 minutes and bi Pi bi in 10 minutes another pipe C can empty the tank in 30 minutes all three pipes in Operation stimulataneously after 3 minutes pipe C is not used in how many minutes will the remaining tank will be filled by pipe a and b
$time and work$
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Answer 1

Answer:3.6 min

Step-by-step explanation:suppose the total capacity of the tank is 30l. By taking lcm of 15,10 and 30.

So a can fill 2L in 1 min

B can fill 3L in 1 min and

C can empty 1L in 1 min, so if they work simultaneously they can fill 4L in one minute.

After 3 min c is closed so the capacity filled is 3*4=12

Now c is closed so 30-12=18

Now a and b will fill in 18/5=3.6 min

Answer 2

A IN ONE MIN: 1/15 OF TANK

B IN ONE MIN : 1/10 OF TANK

C IN ONE MIN : 1/30 OF TANK

AFTER 3 MINS OF THEM TURNED ON:

3(1/15 + 1/10 + 1/30) = 3 × 6/30 = 3/5 OF THE TANK IS NOW FILLED, LEFT OVER 1 - 3/5 = 2/5

NOW FOR A AND B TO FILL THE REMAINING TANK

WE CAN SAY THAT A + B = 2/5

SO 1/15 + 1/10(PIPE FILLING IN EVERY MINUTE) = 2/5 (PIPE TO BE FILLED BY A & B )

=》 5/30 (PIPE FILLING IN EVERY MINUTE) = 2/5 (PIPE TO BE FILLED BY A & B )

=》》 2/5 × 30/5 = 12/5 MINS

===> WILL BE THE TIME TAKEN TO FILL THE REMAINING TANK AFTER 3 MINS OF FILLING BY ALL THE TANKS.

TO FIND THE VALUE OF THE TOTAL TIME TAKEN OF A AND B SINCE THEY STARTED WE SHOULD USE :

A + B = 3/5 AS WE ARE CALCULATING THE TOTAL TIME SINCE THE START.

SO AGAIN YOU HAVE.....

1/15 + 1/10(PIPE FILLING IN EVERY MINUTE) = 3/5

======》 3/5 × 6 = 18/5 MINS NOW CONVERT THIS INTO A MIXED FRACTION FOR CLEARENCE