SOLUTION: Hi A B C D had some stickers.5/11 of the total belonged to B. The ratio of the number of stickers A had to the total number owned by B C D was 1;7 respectively. C had 1/3 as many

Algebra ->  Customizable Word Problem Solvers  -> Misc -> SOLUTION: Hi A B C D had some stickers.5/11 of the total belonged to B. The ratio of the number of stickers A had to the total number owned by B C D was 1;7 respectively. C had 1/3 as many      Log On

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Question 1177096: Hi
A B C D had some stickers.5/11 of the total belonged to B. The ratio of the number of stickers A had to the total number owned by B C D was 1;7 respectively. C had 1/3 as many as what A B and D had in total. If C had 72 fewer than B how many did they have altogether.
Thank you

Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!


B has 5/11 of the stickers; that is given directly.

"The ratio of the number of stickers A had to the total number owned by B C D was 1:7." That means A has 1/8 of the stickers.

"C had 1/3 as many as what A B and D had in total." That means C had 1/4 of the stickers.

C had 72 fewer stickers than B:

%285%2F11%29x-%281%2F4%29x+=+72
%2820%2F44%29x-%2811%2F44%29x+=+72
%289%2F44%29x+=+72
x+=+72%2844%2F9%29+=+44%2A8+=+352

ANSWER: Together they had 352 stickers.