SOLUTION: A box contains some red ,yellow and blue counters. The ratio of the number of red counters to that of yellow counters is 2:3. The ratio of the number of blue counters to that of re

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Question 1079756: A box contains some red ,yellow and blue counters. The ratio of the number of red counters to that of yellow counters is 2:3. The ratio of the number of blue counters to that of red counters is 4:7. What fraction of the counters are yellow
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
A box contains some red ,yellow and blue counters. The ratio of the number of red counters to that of yellow counters is 2:3. The ratio of the number of blue counters to that of red counters is 4:7. What fraction of the counters are yellow
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Fraction of counters YELLOW, highlight%2821%2F43%29.

red     yellow     blue
-----------------------------
 2        3
-----------------------------
 7                  4
-----------------------------

Find the slot for yellow in the second row using %287%2F2%29%2A3=21%2F2.
red     yellow     blue
-----------------------------
 2        3
-----------------------------
 7       21%2F2     4
-----------------------------


  RED     YELLOW    BLUE
--------------------------
  14        21       8         total 43 parts
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Answer by ikleyn(52786) About Me  (Show Source):
You can put this solution on YOUR website!
.
You are given 

r%2Fy = 2%2F3   and   b%2Fr = 4%2F7,   and they want you calculate y%2F%28y+%2B+b+%2B+r%29.


From the condition, 

y%2Fr = 3%2F2   and b%2Fr = 4%2F7.


It implies

%28y%2Bb%29%2Fr = 3%2F2+%2B+4%2F7 = 21%2F14+%2B+8%2F14 = %2821%2B8%29%2F14 = 29%2F14.


Then %28y%2Bb%2Br%29%2Fr = %28y%2Bb%29%2Fr+%2B+1 = 29%2F14%2B1 = %2829%2B14%29%2F14 = 43%2F14.


Thus we have {{r/(y+b+r)}}} = 14%2F43,

and the last step is

y%2F%28y%2Bb%2Br%29 = %28r%2F%28y%2Bb%2Br%29%29%2A%28y%2Fr%29 = %2814%2F43%29%2A%283%2F2%29 = 21%2F43.


Answer.   y%2F%28y+%2B+b+%2B+r%29 = 21%2F43.

Solved.