SOLUTION: Hi paul put some beads in jars A B C.the ratio of the number of beads in A to B 2 to 3. The ratio of beads inB to C is2 to 1. If paul transfers an equal number of beads from B

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Question 1105696: Hi
paul put some beads in jars A B C.the ratio of the number of beads in
A to B 2 to 3. The ratio of beads inB to C is2 to 1. If paul transfers
an equal number of beads from B to A and C he will have an equal number
of beads in A and B and the total of beads in C will increase to 297.
find the number of beads in all 3 jars.
thanks

Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20063) About Me  (Show Source):
You can put this solution on YOUR website!
It's solved below, under my pseudonym, AnlytcPhil :)

Edwin

Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
paul put some beads in jars A B C.
Suppose in the beginning,
 
jar A contains "a" beads,
jar B contains "b" beads
jar C contains "c" beads

the ratio of the number of beads in A to B 2 to 3.
a%2Fb=2%2F3
3a=2b

The ratio of beads in B to C is 2 to 1.
b%2Fc=2%2F1
b=2c

If paul transfers an equal number of beads from B to A
and C he will have an equal number of beads in A and B
Let's do this in steps so that we don't get mixed up:

1. He takes n beads out of B, so B now contains only b-n beads.
2. He then puts those n beads in A, so A now contains a+n beads.
3. He then takes n more beads out of B, so B now contains only 
   b-2n beads. (He took another n beads out of B).
4. He puts those n beads in C, so C now contains c+n beads.

Now A and B have the same number of beads, so we set them
equal:

a%2Bn=b-2n
a%2B3n=b

and the total of beads in C will increase to 297.
c%2Bn=297

So we have 4 equations in 4 unknown

system%283a=2b%2Cb=2c%2Ca%2B3n=b%2Cc%2Bn=297%29

find the number of beads in all 3 jars.
b appears in 3 of the equations, so we substitute
2c for b in the 1st, 3rd and 4th:

system%283a=2%282c%29%2Ca%2B3n=2c%2Cc%2Bn=297%29

system%283a=4c%2Ca%2B3n=2c%2Cc%2Bn=297%29

c occurs in all three equations, so we solve the second 
for c = 297-n and substitute 297-n for c in the first two
equations:

system%283a=4%28297-n%29%2Ca%2B3n=2%28297-n%29%29

system%283a=1188-4n%2Ca%2B3n=594-2n%29%29

system%283a=1188-4n%2Ca%2B5n=594%29%29

Solve the second equation for a = 594-5n and
substitute 594-5n for a in the first equation

3%28594-5n%29=1188-4n%29

1782-15n=1188-4n%29

-11n=-594%29

n=54

Substitute 54 for n in

a%2B5n=594%29%29
a%2B5%2854%29=594%29%29
a%2B270=594
a=324

Also substitute 54 for n in 

c%2Bn=297
c%2B54=297
c=243

Substitute 243 for c in

b=2c

b=2%28243%29
b=486

So A contained a=324 beads\
B contained b=486 beads
C contained c=243 beads

Checking:
 
1. When he took n=54 beads from B, B then contained only 
   486-54=432 beads.
2. When he put those 54 beads in A, A then contained 
   324+54=378 beads. 378.
3. When he took n=54 more beads out of B, B then contained 
   only 432-54=378 beads.

Aha!  That checks for A and B then both contained the same
number of beads,  

To finish,

4. When he put those n=54 beads in C, C then contained 
   243+54=297 beads.

That checks.  So we're right.  

Edwin