The total amount of water in 2 containers is 35quarts. If 7 quarts gets poured from the container with less water into the continer with more water, the container with more water will have 4times as much water as the other container. How many quarts are in the container with less water? Also is it asking for the quarts before or after the water was taken away?
if gets poured ... will have
not --- was poured ... has
so, before pouring
L + M = 35 ... M = 35 - L
4 (L - 7) = M + 7 ... 4 L = M + 35
substituting ... 4 L = 35 - L + 35
... solve for L
Initially:
X qts. in first container.
35 - x qts. in the 2nd container.
(35-x)+7 = 4(x-7).
42-x = 4x - 28,
5x = 70,
X = 14.
x-7 = 14 - 7 = 7 qts. in container with less water.
To solve this problem, let's set up equations based on the given information.
Let's assume that the container with less water initially has x quarts of water.
Therefore, the container with more water initially has (35 - x) quarts of water since the total amount of water in both containers is 35 quarts.
According to the problem, when 7 quarts are poured from the container with less water into the container with more water, the container with more water will then have 4 times as much water as the other container.
After pouring the water, the container with less water will have (x - 7) quarts.
The container with more water will have (35 - x + 7) quarts, which simplifies to (42 - x) quarts.
The problem states that the container with more water will have 4 times as much water as the container with less water. So, we can set up the equation:
(42 - x) = 4(x - 7)
Now, solve the equation for x:
42 - x = 4x - 28
Combine like terms:
3x = 70
Divide by 3:
x = 23.33 (rounded to two decimal places)
Therefore, the container with the less water initially had approximately 23.33 quarts of water.
It is important to note that this answer represents the amount of water in the container with less water BEFORE 7 quarts were poured out.