Question 272631: How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion? (Round to the nearest tenth.)
Found 5 solutions by mananth, ikleyn, n2, josgarithmetic, greenestamps:
Answer by mananth(16949)   (Show Source):
You can put this solution on YOUR website! How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion? (Round to the nearest tenth.)
Let the quantity of pure antifreeze added be x
pure(100%)--------------- 10%------------------------- 20 % mix
x-------------------------6-------------------------- x+6 quarts
1x+0.6x=0.2(x+6)
1.6x= 0.2x + 1.2
1.6x-0.2x= 1.2
1.4x= 1.2
x= 1.2/1.4
0.85 quarts
Answer by ikleyn(53742)  (Show Source):
You can put this solution on YOUR website! .
How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion?
(Round to the nearest tenth.)
~~~~~~~~~~~~~~~~~~~~~~~
The solution in the post by @mananth is INCORRECT: its governing equation
is written incorrectly.
I came to provide a correct solution.
Let x be the volume of pure antifreeze to add, in quarts.
Then the governing equation is
x + 0.1*6 = 0.2(x+6) <<<---=== the amount of the pure antifreeze in ingredients (left side)
and in the mixture (right side)
Simplify and find x
x + 0.6 = 0.2x + 1.2,
x - 0.2x = 1.2 - 0.6,
0.8x = 0.6,
x = 0.6/0.8 = 6/8 = 3/4 = 0.75.
ANSWER. 0.75 quarts of pure antifreeze should be added.
Solved correctly.
Notice that the instruction for rounding is not appropriate to the problem.
It tells me that the problem's composer was not focused on the problem when created it.
Or simply does not understand what he/she composes.
Answer by n2(78)  (Show Source):
You can put this solution on YOUR website! .
How many quarts of pure antifreeze must be added to 6 quarts of a 10% solution to obtain a 20% antifreeze soltuion?
~~~~~~~~~~~~~~~~~~~~~~~
Let x be the volume of pure antifreeze to add, in quarts.
Then the governing equation is
x + 0.1*6 = 0.2(x+6) <<<---=== the amount of the pure antifreeze in ingredients (left side)
and in the mixture (right side)
Simplify and find x
x + 0.6 = 0.2x + 1.2,
x - 0.2x = 1.2 - 0.6,
0.8x = 0.6,
x = 0.6/0.8 = 6/8 = 3/4 = 0.75.
ANSWER. 0.75 quarts of pure antifreeze should be added.
Solved correctly.
Answer by josgarithmetic(39790) (Show Source):
Answer by greenestamps(13326)  (Show Source):
You can put this solution on YOUR website!
While the problem was probably intended as one to be solved by formal algebra, here is quick and easy method for solving any 2-part mixture problem like this. The method is based on the fact that the ratio in which the two ingredients need to be mixed is exactly determined by where the target percentage lies between the two given percentages.
Without any more words than necessary, here is the solution to this problem using this method.
(1) The percentages of the two ingredients are 10% and 100%.
(2) The target percentage, 20%, is "8 times as close" to 10% as it is to 100% (the difference between 10% and 20% is 10%; the difference between 20% and 100% is 80%; 80% is 8 times as much as 10%).
(3) That means the amount of the 10% ingredient must be 8 times the amount of the 100% ingredient.
(4) The given amount of the 10% ingredient is 6 quarts, so the needed amount of the 100% ingredient is 1/8 of 6 quarts, which is 6/8 = 3/4 quarts.
ANSWER 3/4 of a quart, or 0.75 quarts
|
|
|
