Question

a tank has 120 gallons of water and is being drained at a rate of \\(\frac{1}{2}\\) gallon each second. another tank has 100 gallons of water and is being drained at a rate of \\(\frac{1}{4}\\) gallon each second. which equation could be used to determine \\(x\\), the number of seconds, when the two tanks have the same amount of water?\
\\(\boldsymbol{\text{o a) } 120x + \frac{1}{2} = 100x + \frac{1}{4}}\\)\
\\(\boldsymbol{\text{o b) } 120 + \frac{1}{2}x = 100 + \frac{1}{4}x}\\)\
\\(\boldsymbol{\text{o c) } 120x - \frac{1}{2} = 100x - \frac{1}{4}}\\)\
\\(\boldsymbol{\text{o d) } 120 - \frac{1}{2}x = 100 - \frac{1}{4}x}\\)

Explanation:

Step1: Define first tank's water amount

The first tank starts with 120 gallons and loses $\frac{1}{2}$ gallon per second. After $x$ seconds, its water is $120 - \frac{1}{2}x$.

Step2: Define second tank's water amount

The second tank starts with 100 gallons and loses $\frac{1}{4}$ gallon per second. After $x$ seconds, its water is $100 - \frac{1}{4}x$.

Step3: Set amounts equal

When the tanks have the same water, equate the two expressions: $120 - \frac{1}{2}x = 100 - \frac{1}{4}x$.

Answer:

D) $120 - \frac{1}{2}x = 100 - \frac{1}{4}x$