<PRE><B>Question 1202861</B>
&lt;br&gt;
Very likely, another tutor will supply a response showing a different setup for solving the problem than the one I use below....&lt;br&gt;
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....&lt;br&gt;
Let 8n = # of red pencils in box Y
Let n = # of blue pencils in box Y&lt;br&gt;
Now use that start to work the problem.&lt;br&gt;
Box X had 2 times as many pencils as box Y:&lt;br&gt;
Box Y has a total of 8n+n = 9n pencils, so the number of pencils in box X is 18n.&lt;br&gt;
(a) What fraction of the total number of pencils in both boxes were blue? Give your answer in the simplest form.&lt;br&gt;
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 {{{n/(27n)=1/27}}}.&lt;br&gt;
ANSWER: 1/27&lt;br&gt;
(b) There were 90 more red pencils in Box X than in Box Y. How many blue pencils were there altogether?&lt;br&gt;
(this part revised to correct a small error....)&lt;br&gt;
The difference between the numbers of red pencils in the two boxes was 18n-8n = 10n.  Since that difference was 90...&lt;br&gt;
{{{10n=90}}}
{{{n=9}}}&lt;br&gt;
The number of blue pencils altogether (in box Y only) was n=9.&lt;br&gt;
ANSWER: 9&lt;br&gt;
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