Jack has twice as many dimes as quarters. If the total value of the coins is $6.30, how many dimes does he have?
I don't know how to write in LaTeX in web2.0calc.com, so I'm going to try without it.
Set variables for the number of dimes and the number of quarters. The number of dimes can be d, and the number of quarters can be q. We already know there are TWICE the amount of dimes than quarters. So, we can say d=2q. since they have cost, we must multiply d or 2q by 10, and q by 25. My work is below:
20q+25q=630(in cents!!!!!) 45q=630(I used a calculator to do 630/45) q=14. 2q=28.
Check: 20(14)+25(14)=630(I used a calculator to do 25•14) 280+350=630. 630=630.
So, Jack has 28 dimes. :)
Jack has twice as many dimes as quaters.
Total value = $6.30
Total value = $6.30 = 630 cents
Now Let the total number of dimes be x, and total number of quaters be y.
Hence it is given,
\(x=2y\)
Further we know that the total amount is 630 cents.
Therefore we can write it as,
\(10x + 25y=630\)
Now we shall solve the two equations using substitution, put \(x = 2y\) in equation \(10x + 25y = 630\)
we get,
10 (2y) + 25y = 630
20y + 25y = 630
45y = 630
y = 14
Now we know \(x = 2y,\) Therfore \(x = 2(14)=28\)
The total number of dimes he has are 28.