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Three hoses, x, y, and z, each pump water at a constant rate. How long will it take the hoses to fill up a 30,000-gallon tank?
(1) Hose x pumps water at 3 times the speed of z.
(2) Hose y pumps water at half the rate of x.
(1) Hose x pumps water at 3 times the speed of z.
(2) Hose y pumps water at half the rate of x.
18 Apr 2017, 08:42
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Three hoses, x, y, and z, each pump water at a constant rate. How long will it take the hoses to fill up a 30,000-gallon tank?
We would need either the combined rate for hoses x,y & z to pump water or the individual rates for x, y & z
(1) Hose x pumps water at 3 times the speed of z.
This statement just gives the ratio of speeds of x and z. moreover it says nothing about y. Hence this statement is not sufficient.
(2) Hose y pumps water at half the rate of x.
This statement only talks about y and x and nothing about z. Also this statement just gives the ratio of speeds of x and y.Hence this statement is not sufficient.
Combining both statements also is not sufficient, as we only know the ratios but no actual value is given.
Answer is E
We would need either the combined rate for hoses x,y & z to pump water or the individual rates for x, y & z
(1) Hose x pumps water at 3 times the speed of z.
This statement just gives the ratio of speeds of x and z. moreover it says nothing about y. Hence this statement is not sufficient.
(2) Hose y pumps water at half the rate of x.
This statement only talks about y and x and nothing about z. Also this statement just gives the ratio of speeds of x and y.Hence this statement is not sufficient.
Combining both statements also is not sufficient, as we only know the ratios but no actual value is given.
Answer is E
24 Jul 2017, 03:05
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Bunuel
Three hoses, x, y, and z, each pump water at a constant rate. How long will it take the hoses to fill up a 30,000-gallon tank?
(1) Hose x pumps water at 3 times the speed of z.
(2) Hose y pumps water at half the rate of x.
(1) Hose x pumps water at 3 times the speed of z.
(2) Hose y pumps water at half the rate of x.
If we make the equation
1/6x = x
1/2x = z
1/x = y
1/6x + 1/2x + 1/x = 1/30,000
Why cant be C be correct as we can solve to get X
Please help
