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
When n liters of fuel was added to
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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
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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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
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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1. n+ 1/3 = 7/9ska7945 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
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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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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