
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.
a. 1 hours.
b. 2 hours.
c. 3 hours.
d. 4 hours
Answer
596.1k+ views
Hint: The key to this question is the element in the problem which you take as a variable. I suggest taking the water flow rate of each tap to be x. So, the water flow rate of 6 taps will be 6x and of 4 taps would be 4x.
Complete step by step answer:
Let the rate of flow of water through each tap be ‘x’ liter per hour.
Total water filled by ‘n’ taps: $=n\times \left( \text{water supplied by each tap per unit time} \right)\times \left( \text{time for which water is supplied} \right)$
Now, try to interpret the statement given in the question in mathematical terms using the above formula, we get:
$\text{Capacity of cistern=}6\times x\times 8=48x.................(i)$ .
If two taps go out of order, then only 4 taps are left working.
Also, the capacity of the cistern is fixed and won’t change due to the taps going out of order.
Let the time taken by four taps to fill the cistern be t.
$\therefore \text{capacity of cistern=4}\times \text{t}\times \text{x}$ .
Substituting the capacity of the cistern from equation (i).
$\text{48x=4}\times \text{t}\times \text{x}$
$\Rightarrow 48=4\times t $
$\Rightarrow \dfrac{48}{4}=t$
$\therefore t=12\text{ hours}\text{.}$
Calculating the excess time taken by 4 taps over the case when all 6 taps were working:
(time taken by 4 taps to fill the cistern)- (time taken by 6 taps to fill the cistern)
$=12-8$
$=4$ hours.
$\therefore $ The excess time taken by 4 taps to fill the cistern over the case when all 6 taps were working is 4 hours.
Hence, our answer is option (d) 4 hours.
Note: Questions including water flow, have two things to be wisely selected. One is the element of the problem that you are treating as a variable, and the other is the unit of water flow. The units of water flow can have different variations like the length of water column coming out per hour of the pipe or the volume of water coming out per hour, as we assumed in our above question.
Complete step by step answer:
Let the rate of flow of water through each tap be ‘x’ liter per hour.
Total water filled by ‘n’ taps: $=n\times \left( \text{water supplied by each tap per unit time} \right)\times \left( \text{time for which water is supplied} \right)$
Now, try to interpret the statement given in the question in mathematical terms using the above formula, we get:
$\text{Capacity of cistern=}6\times x\times 8=48x.................(i)$ .
If two taps go out of order, then only 4 taps are left working.
Also, the capacity of the cistern is fixed and won’t change due to the taps going out of order.
Let the time taken by four taps to fill the cistern be t.
$\therefore \text{capacity of cistern=4}\times \text{t}\times \text{x}$ .
Substituting the capacity of the cistern from equation (i).
$\text{48x=4}\times \text{t}\times \text{x}$
$\Rightarrow 48=4\times t $
$\Rightarrow \dfrac{48}{4}=t$
$\therefore t=12\text{ hours}\text{.}$
Calculating the excess time taken by 4 taps over the case when all 6 taps were working:
(time taken by 4 taps to fill the cistern)- (time taken by 6 taps to fill the cistern)
$=12-8$
$=4$ hours.
$\therefore $ The excess time taken by 4 taps to fill the cistern over the case when all 6 taps were working is 4 hours.
Hence, our answer is option (d) 4 hours.
Note: Questions including water flow, have two things to be wisely selected. One is the element of the problem that you are treating as a variable, and the other is the unit of water flow. The units of water flow can have different variations like the length of water column coming out per hour of the pipe or the volume of water coming out per hour, as we assumed in our above question.
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