Question 1205779: There were 5c 10c and 20c coins in Joel's wallet in the ratio of 2:6:7 respectively. If he takes out half the number of 5c and half the number of 10c coins they would make up $4.20
How many 20c coins in his wallet.
How much money does he have altogether
Found 3 solutions by josgarithmetic, Edwin McCravy, ikleyn:
Answer by josgarithmetic(39618)   (Show Source):
Answer by Edwin McCravy(20056)  (Show Source):
You can put this solution on YOUR website!
We use the fact that if 3 quantities are in the ratio
of a:b:c, then the fraction of the total which the first,
second and third quantities are, are respectively,
 ..
There were 5c 10c and 20c coins in Joel's wallet in the ratio
of 2:6:7 respectively. If he takes out half the number of 5c
and half the number of10c coins they would make up $4.20.
How many 20c coins in his wallet?let x = number of 5c coins.
let y = number of 10c coins.
let z = number of 20c coins.
The fraction of 5c coins is  .
The fraction of 10c coins is  .
the fraction of 20c coins is  .
Since he had x 5c coins, then the value of them would be 5x cents.
When he takes out half of them, only 2.5x cents.
Suppose he had y 10c coins, then the value of them would be 10y cents.
When he takes out half of them, that's only 5y cents.
2.5x + 5y = 420
Multiply through by 2
5x + 10y = 840
Divide through by 5
x + 2y = 168
x = 168 - 2y
Substitute for x
Divide the first through by 2 and the second through by 3
Solve the first for z
z = 1092-14y
Substitute in the second equation
-336+7y-2(1092-14y) = 0
-336+7y-2184+28y = 0
35y-2520 = 0
35y = 2520
y = 72
Substitute in
z = 1092-14y
z = 1092-1008
z = 84
x = 168 - 2y
x = 168 - 2(72)
x = 168 - 144
x = 24
Substitute in
So there are 24 5c coins, 72 10c coins, 84 20c coins
Edwin
Answer by ikleyn(52794)  (Show Source):
You can put this solution on YOUR website! .
There are 5c, 10c, and 20c coins in Joel's wallet in the ratio of 2:6:7 respectively.
If he takes out half the number of 5c and half the number of 10c coins they would make up $4.20
(a) How many 20c coins in his wallet.
(b) How much money does he have altogether
~~~~~~~~~~~~~~~~~~~~~~~
I changed the first line in your post, in order for
the time forms be consistent in all parts of the post.
According to the problem, there are 2x of 5c coins; 6x of 10c coins and 7x of 20c coins,
where x is the common measure of these different quantities.
Half of 2x 5c coins is x 5c coins; half of 6x 10c coins is 3x 10c coins.
Their total is $4.20, or 420 cents
5x + 10(3x) = 420 cents.
Simplify and find x
5x + 30x = 420
35x = 420
x = 420/35 = 12.
Now we are in position to answer the problem's questions.
(a) The number of 20c coins in his wallet is 7x = 7*12 = 84.
(b) The total money in his wallet is 5*(2x) + 10*(6x) + 20*(7x) = 5*(2*12) + 10*(6*12) + 20*(7*12) = 2520 cents or $25.20.
Solved.
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