Question 1114664: Justin had only nickels and dimes in his pocket.When he counted his change,he had $2.80.If he had 3 times more dimes then nickels,how many nickels did he have?
Answer by greenestamps(13203)   (Show Source):
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With the problem as worded, there is no solution to the problem, because the answers are not whole numbers.
The trouble is with the phrase "3 times more dimes than nickels". Grammatically, that means the number of dimes is the number of nickels, plus three more times that number -- i.e., four times AS MANY dimes as nickels. When you try to solve the problem with that correct interpretation of the words, you get numbers of nickels and dimes that are not whole numbers.
Apparently the author of the problem intended to say Justin had "3 times AS MANY dimes as nickels"; with those words, there is a reasonable solution to the problem.
I would solve the problem informally, rather than with formal algebra. If he has 3 times as many dimes as nickels, then it is possible to group all the coins into groups of 3 dimes and 1 nickel, each group with a value of 35 cents. Then the number of those groups that he has is found by dividing the total of $2.80 (280 cents) by the value of each group (35 cents).
So Justin has 8 groups of coins, each group consisting of 3 dimes and 1 nickel; therefore he has 24 dimes and 8 nickels.
Answer (after correcting the statement of the problem): Justin has 8 nickels.
Algebraically....
Let x be the number of nickels; then 3x is the number of dimes.
The value of the coins is 280 cents; 5 cents for each of the x nickels and 10 cents for each of the 3x dimes:
The number of nickels is 8.
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