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Standard VIII

Mathematics

Relation for Direct Proportion

Six taps can ...

Question

Six taps can fill an empty cistern in $8$ hours. How much more time will be taken, if two taps go out of order? Assume all the taps supply water at the same rate.

1

hour

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2

hours

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3

hours

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4

hours

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Solution

The correct option is B $4$ hours
Given six taps fill an empty cistern in $8$ hours
If two taps go out of order then no. of taps $= 6 - 2 = 4$
Then $\frac{6}{4} = \frac{x}{8}$
$⟹ 4 x = 6 \times 8$
$⟹ x = \frac{48}{4} = 12$ hours
Then, more time $= 12 - 8 = 4$ hours.

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Similar questions

Q. A tap A can fill a cistern in 4 hours and the tap B can empty the full cistern in 6 hours. If both the taps are opened together in the empty cistern, in how much time will the cistern be filled up?

A tap can fill a cistern In 8 hours and another tap can empty the full cistern in 16 hours. If both the taps are open. the time taken to fill the cistern Is

(a) $5 \frac{1}{3}$ hours

(b) 10 hours

(d) 20 hours

Q. Tick (✓) the correct answer:
A tap can fill a cistern in 8 hours and another tap can empty the full cistern in 16 hours. If both the taps are open, the time taken to fill the cistern is
(a)

5 \frac{1}{3}

hours
(b) 10 hours
(c) 16 hours
(d) 20 hours

Q. Four taps can individually fill a cistern of water in

1

hour,

2

hours,

3

hours and

6

hours respectively. If all the four taps are opened simultaneously, the cistern can be filled in how many minutes?

Q. A cistern can be filled by a tap in

4

hours while it can be emptied by another tap in

9

hours. If both the taps are opened simultaneously then after how much time will the cistern get filled ?

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Six taps can fill an empty cistern in 8 hours. How much more time will be taken, if two taps go out of order? Assume all the taps supply water at the same rate.

The correct option is B 4 hoursGiven six taps fill an empty cistern in 8 hoursIf two taps go out of order then no. of taps =6−2=4Then 64=x8⟹4x=6×8⟹x=484=12 hoursThen, more time =12−8=4 hours.

Six taps can fill an empty cistern in $8$ hours. How much more time will be taken, if two taps go out of order? Assume all the taps supply water at the same rate.

The correct option is B $4$ hours
Given six taps fill an empty cistern in $8$ hours
If two taps go out of order then no. of taps $= 6 - 2 = 4$
Then $\frac{6}{4} = \frac{x}{8}$
$⟹ 4 x = 6 \times 8$
$⟹ x = \frac{48}{4} = 12$ hours
Then, more time $= 12 - 8 = 4$ hours.