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by sunilrawat » Sat Jun 11, 2011 4:28 am
a rainstorm increased the amount of water from 124 bn gallons to 138 bn gallons. If storm increased the amount of water to 82% of the total capacity, approx how many bn gallons of water was the reservoir short of the total capacity prior to the storm.

a 9
b 14
c 25
d 30
e 44
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Source: — Problem Solving |

by Frankenstein » Sat Jun 11, 2011 4:41 am
Hi,
Let V be the total capacity of reservoir.
138 bn gallons is 82% of total capacity of reservoir.
So, (0.82)V = 138 => V =~ 168.
Initially volume of water is 124.
So, unfilled volume of reservoir is 168-124 = 44 bn gallons

Hence, E
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by manpsingh87 » Sat Jun 11, 2011 4:43 am
sunilrawat wrote:a rainstorm increased the amount of water from 124 bn gallons to 138 bn gallons. If storm increased the amount of water to 82% of the total capacity, approx how many bn gallons of water was the reservoir short of the total capacity prior to the storm.

a 9
b 14
c 25
d 30
e 44
let the capacity of reservoir be x. Therefore,
138=82%of x;
x=168 approx.
so before storm reservoir was 168-124=44 bn gallons of water short of its capacity.

hence E
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by sunilrawat » Sat Jun 11, 2011 4:55 am
i solved it like this,

if 124 if x% of capacity (V)

0.82V = 138
0.01xV = 124

solving, we get x ~ 73%, i.e 27% short of V

then instead of calculating 27%of V, I marked 25 as the option...silly mistake
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by cans » Tue Jun 14, 2011 3:28 am
82% of total=138
total=138*100/82 = 6900/41 = 168. something
thus remaining when 124 was filled = 168-124 = 44
IMO E
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