DS Q

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DS Q

by Inspired » Tue Jan 19, 2010 2:27 am
Two water tanks, X and Y, were drained simultaneously. If X contained 30 more gallons of water than Y, and both tanks became empty at the same time, how long did it take the tanks to empty?

(1) For every gallon drained from tank Y, 2 gallons were drained from tank X.
(2)Tank Y was drained at a constant rate of 20 gallons per hour.

pls ans
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by ajith » Tue Jan 19, 2010 3:06 am
Inspired wrote:Two water tanks, X and Y, were drained simultaneously. If X contained 30 more gallons of water than Y, and both tanks became empty at the same time, how long did it take the tanks to empty?

(1) For every gallon drained from tank Y, 2 gallons were drained from tank X.
(2)Tank Y was drained at a constant rate of 20 gallons per hour.

pls ans

Say tank X contained x gallons of water and Tank Y contained y gallons of water
Now the question says x = y+30

The rate of drainage for Tank X is say a and rate of Drainage for tank Y is say b

now the time taken for draining X = Time Taken for draining Y
=> x/a = y/b


=> y+30/a = y/b

Now we want the time required to drain the tanks
that is x/a or y/b

Now let us look at the options

option 1 says a = 2b

=> y+30/2b = y/b
y+30 = 2y
y =30

but we want y/b
so 1 alone is not sufficient

Option 2 says
b= 20 gallons per hour

which alone is not sufficient

but if we combine 1&2

Time taken to drain the tank y/b = 30/20 = 1.5 hours

thus the question can be solved combining 1&2
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by swati chaturvedi » Thu Jan 21, 2010 9:20 pm
This question may be solved with only option 2 as below:

X-Y=30 ---- (1)

Rx = X/t and Ry = Y/t (as time is same)

Ry = 20 gln/hr

20 = Y/t or Y = 20 t --- (2)

Y = X- 30 = 20 t

X- Y = 30 = 20 t

t = 30/20 = 1.5 hrs

Option 1 gives only capacity Y as 30 glns but can't be solved further.

Answer = B, only option 2 is sufficient.

Hope to be correct.

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by ajith » Thu Jan 21, 2010 9:43 pm
swati chaturvedi wrote:

Y = X- 30 = 20 t

X- Y = 30 = 20 t

The second statement does not follow from the first statement

Y= X-30 = 20 t

=> -Y = 30 - X = -20t (multiplying by -1)

=> X-Y = 30 = X-20t (adding X everywhere)

which yields X-20t = 30 rather than
20t = 30
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by swati chaturvedi » Thu Jan 21, 2010 11:12 pm
got that, thanks a ton.

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by Concordio » Sat Jan 23, 2010 7:49 pm
IMO answer would be C. (2) is not sufficient alone because the rate of drainage for tank X cannot be found from the information given. (1) is not sufficient because it gives no rate of drainage for either tank. Using both, we know that the rate of drainage for tank X is double that of tank Y.

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by shashank.ism » Fri Feb 19, 2010 10:46 am
Inspired wrote:Two water tanks, X and Y, were drained simultaneously. If X contained 30 more gallons of water than Y, and both tanks became empty at the same time, how long did it take the tanks to empty?

(1) For every gallon drained from tank Y, 2 gallons were drained from tank X.
(2)Tank Y was drained at a constant rate of 20 gallons per hour.

pls ans
x= 30 +Y
both tanks became empty at the same time.
st.1) rate of X drained in terms of y drained is known but not exact rate of any simgle ======== insufficient
st.2) rat e of y drained is known but not the relation bet. X and y ======== insufficient
combined both are known : Y was drained at a constant rate of 20 gallons per hour.
For every gallon drained from tank Y, 2 gallons were drained from tank X. so X was drained at a constant rate of 40 gallons per hour.
so sufficient


======== sufficient
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