SOLUTION: Jon had red and blue pencils. He put them into Boxes X and Y. Box X had 2 times as many pencils as Box Y. All the pencils in Box X were red. In Box Y, the ratio of the number of re
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-> SOLUTION: Jon had red and blue pencils. He put them into Boxes X and Y. Box X had 2 times as many pencils as Box Y. All the pencils in Box X were red. In Box Y, the ratio of the number of re
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Question 1202861: Jon had red and blue pencils. He put them into Boxes X and Y. Box X had 2 times as many pencils as Box Y. All the pencils in Box X were red. In Box Y, the ratio of the number of red pencils to the number of blue pencils was 8:1.
(a) What fraction of the total number of pencils in both boxes were blue? Give your answer in the simplest form.
(b) There were 90 more red pencils in Box X than in Box Y. How many blue pencils were there altogether? Found 2 solutions by greenestamps, ikleyn:Answer by greenestamps(13200) (Show Source):
Very likely, another tutor will supply a response showing a different setup for solving the problem than the one I use below....
When the given information for a problem includes a ratio, the work required to get to the answer is often (usually?) less if you set up the problem using that ratio. So....
Let 8n = # of red pencils in box Y
Let n = # of blue pencils in box Y
Now use that start to work the problem.
Box X had 2 times as many pencils as box Y:
Box Y has a total of 8n+n = 9n pencils, so the number of pencils in box X is 18n.
(a) What fraction of the total number of pencils in both boxes were blue? Give your answer in the simplest form.
All the pencils in box X were red, so the number of red pencils in box X was 18n. So the total number of red pencils in the two boxes was 18n+8n = 26n. The number of blue pencils was n, so the total number of pencils was 26n+n = 27n. The fraction of the pencils that were blue is then .
ANSWER: 1/27
(b) There were 90 more red pencils in Box X than in Box Y. How many blue pencils were there altogether?
(this part revised to correct a small error....)
The difference between the numbers of red pencils in the two boxes was 18n-8n = 10n. Since that difference was 90...
The number of blue pencils altogether (in box Y only) was n=9.
You can put this solution on YOUR website! .
Jon had red and blue pencils. He put them into Boxes X and Y.
Box X had 2 times as many pencils as Box Y.
All the pencils in Box X were red.
In Box Y, the ratio of the number of red pencils to the number of blue pencils was 8:1.
(a) What fraction of the total number of pencils in both boxes were blue?
Give your answer in the simplest form.
(b) There were 90 more red pencils in Box X than in Box Y.
How many blue pencils were there altogether?
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Let "b" be the number of blue pencils in box Y.
Then the number of red pencils in box Y is 8b,
the number of all pencils in box Y is b + 8b = 9b,
and the number of red pencils in box X is 2*(9b) = 18b.
Now the answer to question (a) is = = .
For (b), we have 8b red pencils in box Y and 18b red pencils in box X.
From the problem, we have this equation
18b - 8b = 90
10b = 90
b = 90/10 = 9.
So, there are 9 blue pencils in box Y, which means that there are 9 blue pencils in both boxes, altogether.
