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

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

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

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)

1/3 = 3/9 So 3/9 + n = 7/9 n = 4/9 ---> 9n/4 = 1

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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