A commercial building has two water reservoirs. Reservoir X contains 1,200 gallons of wate… — ACT

A commercial building has two water reservoirs. Reservoir $X$ contains $1,200$ gallons of water and is draining at a constant rate of $18$ gallons per minute. Reservoir $Y$ contains $360$ gallons of water and is being filled at a constant rate of $22$ gallons per minute. After how many minutes will both reservoirs contain the exact same amount of water?

Answer: 21 minutes

Answer

The two reservoirs will contain the same amount of water after 21 minutes.

The correct answer is 21 minutes. By setting the expressions for the volume of both reservoirs equal (

1,200 - 18t = 360 + 22t

) and isolating the variable, we find

40t = 840

, which simplifies to

t = 21

Step-by-Step Solution

Translate the physical scenario for Reservoir

X

into an algebraic expression.

1,200 - 18t

Reservoir

X

starts with

1,200

gallons and loses

18

gallons per minute over

t

minutes.

Translate the physical scenario for Reservoir

Y

into an algebraic expression.

360 + 22t

Reservoir

Y

starts with

360

gallons and gains

22

gallons per minute over

t

minutes.

Set the two expressions equal to each other and solve for

t

1,200 - 18t = 360 + 22t \implies 840 = 40t \implies t = 21

Equating the two volume expressions allows us to find the time

t

at which the volumes are equal.

Key Concept

Translating and Solving Algebraic Word Problems

Alternative Method

Instead of solving algebraically, one can check the rates of change relative to each other. The distance between the initial volumes is

1,200 - 360 = 840

gallons. Since they are moving toward each other (one draining, one filling), their relative rate of convergence is

18 + 22 = 40

gallons per minute. Dividing the total volume difference by the rate of convergence gives

840 / 40 = 21

minutes.

Estimated Time:1m 30s

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Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Alternative Method

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