Found 4 solutions by Theo, josgarithmetic, ikleyn, MathTherapy:
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the problem is in the total value being in dollars rather than being in cents.
to correct for that, you either have to make all denominations in cents or all denominations in dollars.
converting all denominations to cents, you get:
50x + 20y + 10z = 1540
converting all denominations to dollars, you get:
.5x + .2y + .1z = 15.40
then everything will come out ok.
when everything is converted to cents, the equation is 50x + 20y + 10z = 1540
you are given that x = 2y and y = 5x.
since x = 2y, then y = x/2.
you get 50x + 20y + 10z = 1540 becoming 50x + 20 * x/2 + 10 * 5x = 1540.
combine like terms to get 110x = 1540.
solve for x to get x = 14.
this makes y = 7 and z = 70
50 * 14 + 7 * 20 + 70 * 10 = 1540.
this is good.
when everything is converted to dollars, the equation is .5x + .2y + .1z = 15.40
you are given that x = 2y and y = 5x.
since x = 2y, then y = x/2.
you get .5x + .2y + .1z = 15.40 becoming .5x + .2 * x/2 + .1 * 5x = 15.40
combine like terms to get 1.1 * x = 15.40
solve for x to get x = 14.
this makes y = 7 and z = 70
.5 * 14 + .2 * 7 + .1 * 70 = 15.40.
this is also good.
Answer by josgarithmetic(39630) (Show Source): You can put this solution on YOUR website!
COIN HOW MANY VALUE
50c 2x 0.50*2x
20c x 0.20x
10c 5*2x=10x 0.10*10x
SUM $15.40
the sum of the values, first summarized,
-
--------------of the 20 c coins.
This means of the 50 c coins.
Answer by ikleyn(52879) (Show Source): You can put this solution on YOUR website!
.
Gopal had twice as many 50c coins as 20c coins. He also had 5 times as many 10cent coins as 50c coins.
If he had $15.40 how many 50c coins did he have,
My son found the solution to have fractions.
~~~~~~~~~~~~~~~~~~~~~~~~
I will solve it MENTALLY and will explain it to you
and to your son on how to do it.
As the problem says, for every one 20c coin, there are two 50c coins.
(In particular, the number of 50c coins is even).
Next, the problem says that for every 50c coin, there are five 10c coins.
So, you can group the coins in sets, placing one 20c coin, two 50c coins
and ten 10c coins into each set.
Thus, every such set of coins is worth 20c + 2*50c + 10*10c = 220 cents.
The number of such sets (groups) is = 7.
Since each set contains two 50c coins, the number of the 50c coins is 2*7 = 14. ANSWER
Solved.
Answer by MathTherapy(10556) (Show Source): You can put this solution on YOUR website!
Hi
Gopal had twice as many 50c coins as 20c coins. He also had 5 times as many 10cent coins as 50c coins. If he had $15.40 how many 50c coins did he have,
My son found the solution to have fractions.
Thanks
Let the number of 50c coins be F
Then number of 20c coins =
Also, number of 10c coins = 5F
Values of the 50c, 20c, and 10c coins = .5F, , and .1(5F), or .5F, respectively
With TOTAL amount being $15.40, we get: .5F + .1F + .5F = 15.4
1.1F = 15.4
Number of 50c coins, or
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