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When n liters of fuel was added to

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by ska7945 » Sat Aug 23, 2008 6:55 pm
When n liters of fuel was added to a tank that was already 1/3 full, the tank was filled to 7/9 of its capacity. In terms of n, what is the capacity of the tank, in liters?

1.10n/9
2.4n/3
3.3n/2
4.9n/4
5.7n/3



oa D
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by vishubn » Sat Aug 23, 2008 8:36 pm
let the original fuel capacity be x
and tank is already 1/3rd full 1/3x

when n litres of fuel is added
Now new capacity =(1/3)x +n=(7/9)x
n=(7/9)x-(1/3)x
n=x(7/9-1/3)
n=x(7-3/9)
n=x(4/9)
x=9n/4 is the answer hence D is the option

hope it helps

Vishu
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by dnairo1981 » Sun Aug 24, 2008 7:13 pm
I also found picking numbers easy

Let capacity = 27

The tank was filled to 7/9 capacity

7/9 * 27 = 21

tank was already 1/3 full

1/3 * 27 = 9

n = 12 ( 12 + 9 = 21)

Use n = 12 in anwer choices

D. 9n/4

9 * 12/ 4 = 9 * 3 = 27
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Re: When n liters of fuel was added to

by sudi760mba » Sun Mar 08, 2009 3:17 pm
ska7945 wrote:When n liters of fuel was added to a tank that was already 1/3 full, the tank was filled to 7/9 of its capacity. In terms of n, what is the capacity of the tank, in liters?

1.10n/9
2.4n/3
3.3n/2
4.9n/4
5.7n/3



oa D
1. n+ 1/3 = 7/9
2. n= (7/9)- (1/3)
3. n=(7/9) - (3/9)
4. n=(4/9)
5. 9n=4 (multiply either side by 9)
6.9n/4 (divide both sides by 4)
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Re: When n liters of fuel was added to

by logitech » Sun Mar 08, 2009 3:30 pm
1/3 = 3/9

So 3/9 + n = 7/9

n = 4/9 ---> 9n/4 = 1
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Re: When n liters of fuel was added to

by Scott@TargetTestPrep » Sat May 01, 2021 6:16 am
ska7945 wrote: ↑
Sat Aug 23, 2008 6:55 pm
 When n liters of fuel was added to a tank that was already 1/3 full, the tank was filled to 7/9 of its capacity. In terms of n, what is the capacity of the tank, in liters?

1.10n/9
2.4n/3
3.3n/2
4.9n/4
5.7n/3



oa D
Solution:

We can let the full capacity of the tank be C liters and create the equation:

1/3 C + n = 7/9 C

3/9 C + n = 7/9 C

n = 4/9 C

9/4 n = C

Answer: D

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