Simple Ratios problem from old SAT book

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The problem involves finding the total cent value of 20 coins composed of nickels and dimes, with a ratio of 2:3. The equations derived from the ratio and total coin count lead to the conclusion that there are 12 nickels and 8 dimes. The confusion arises in calculating the number of dimes, which should be 8, not 6. The total value can be calculated by determining the value of the nickels and dimes based on their quantities. The final solution involves confirming the correct counts and calculating the total cent value from these amounts.
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Homework Statement



Aggregate # of nickels and dimes = 20 coins. If the ratio between nickels and dimes is 2:3, what is cent value of 20-coin set?

Homework Equations



2x = 3y


The Attempt at a Solution



20 = nickels(x)+dimes(y); ratio = 2x:3y; 2x = 3y; y = 2x/3

20 = x + 2x/3; x = 12 nickels; 6 dimes? ...confused from there.
 
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Ognerok said:

Homework Statement



Aggregate # of nickels and dimes = 20 coins. If the ratio between nickels and dimes is 2:3, what is cent value of 20-coin set?

Homework Equations



2x = 3y


The Attempt at a Solution



20 = nickels(x)+dimes(y); ratio = 2x:3y; 2x = 3y; y = 2x/3

20 = x + 2x/3; x = 12 nickels; 6 dimes? ...confused from there.

2x=3y => 24=3y => y=8, not 6!
 
N:D = 2:3
total ratio =5

thus, number of N = 2/5 * 20. I think you can do the rest.
 

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