Pablo has $3.15in dimes and guaters. How many guaters and dimes does he have?
Pablo has $3.15 in quarters and dimes. he has more quarters than dimes. How many quarters and dimes does he have
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number of "guaters" --- x
number of dimes ------ y
25x + 10y = 315
5x + 2y = 63
y = (63 - 5x)/2 , both x and y must be whole numbers
so 63-5x must be even to be divisible by 2
then x can only be an odd number
x= 1 , y = 29
x = 3 , y = 24
x = 5 , y = 19
x = 7 , y = 14
x = 9 , y = 9
x = 11 , y = 4
no unique solution, I have 6 different ones, each one works
IT SAYS MORE "Guaters" THAN DIMES!!!!!!!!!!!!!
To find out how many quarters and dimes Pablo has, we need to set up an algebraic equation based on the information provided.
Let's assume the number of dimes Pablo has as "x" and the number of quarters as "y".
We know that the total value of dimes and quarters is $3.15. Since there are 10 cents in a dime and 25 cents in a quarter, we can set up the equation:
10x + 25y = 315
Now, we have one equation with two variables. It is impossible to find a unique solution with just one equation. However, we can try to find different pairs of values (x, y) that satisfy the equation.
Let's substitute some values for y and solve for x:
- If y = 0, the equation becomes 10x + 25(0) = 315, which simplifies to 10x = 315. Dividing both sides by 10, we get x = 31.5, but since we can't have a fraction of a dime, this solution is not valid.
- If y = 1, the equation becomes 10x + 25(1) = 315, which simplifies to 10x + 25 = 315. Subtracting 25 from both sides gives 10x = 290, then dividing by 10, we get x = 29. This means Pablo has 29 dimes and 1 quarter.
Therefore, Pablo has 29 dimes and 1 quarter.