SOLUTION: Audrey, Beth, Corrine, Dione had some stickers. The
number of Audrey’s stickers was 70% fewer than the total
number of stickers of Beth, Corrine and Dione. The ratio
of
Algebra ->
Percentage-and-ratio-word-problems
-> SOLUTION: Audrey, Beth, Corrine, Dione had some stickers. The
number of Audrey’s stickers was 70% fewer than the total
number of stickers of Beth, Corrine and Dione. The ratio
of
Log On
Question 1184378: Audrey, Beth, Corrine, Dione had some stickers. The
number of Audrey’s stickers was 70% fewer than the total
number of stickers of Beth, Corrine and Dione. The ratio
of the number of Beth’s stickers to the total number of
stickers of Audrey, Corrine and Dione was 4:9. The ratio
of the number of Corrine’s stickers to the total number of
stickers of Audrey, Beth and Dione was 3:23. After Audrey
lost 57 stickers and Dione lost 81 stickers, the ratio of the
number of Audrey’s stickers to the number of Dione’s
stickers became 1:3. Find the difference in number of
stickers between Beth and Corrine at first. Answer by greenestamps(13203) (Show Source):
Often you want to try to set up solving a problem using a single variable, since it will almost always make the solution easier. But the way the information is given in this problem makes that difficult if not impossible.
So use four variables and convert each piece of given information into an equation in those variables.
(1) The number of Audrey’s stickers was 70% fewer than the total number of stickers of Beth, Corrine and Dione
Read "70% fewer" as "30% as much":
(2) The ratio of the number of Beth’s stickers to the total number of stickers of Audrey, Corrine and Dione was 4:9
Convert the ratio x:y=4:9 to 9x=4y:
(3) The ratio of the number of Corrine’s stickers to the total number of stickers of Audrey, Beth and Dione was 3:23
Do the same as with (2):
(4) After Audrey lost 57 stickers and Dione lost 81 stickers, the ratio of the number of Audrey’s stickers to the number of Dione’s stickers became 1:3
Again the same thing:
There are our four equations in A, B, C, and D. Because of the kinds of equations we have, a purely algebraic solution is going to be rather messy. Probably the easiest way to solve the system is to convert each equation to the form pA+qB+rC+sD=t and solve using matrices.
