Question 1017726: A boy has a total of $.93 in pennies nickels dimes and quarters he has two more quarters and dimes than he has nickels and if he has as many pennies as quarters how many of each coin does he have
Found 3 solutions by CubeyThePenguin, greenestamps, josgarithmetic: Answer by CubeyThePenguin(3113) (Show Source):
You can put this solution on YOUR website! p = number of pennies
n = number of nickels
d = number of dimes
q = number of quarters
p + 5n + 10d + 25q = 93
q = n + 2
d = n + 2
p = q
Write everything in terms of n.
(n + 2) + 5n + 10(n + 2) + 25(n + 2) = 93
41n + 72 = 93
The solution to this equation is not an integer, so this situation is not possible.
Answer by greenestamps(13200) (Show Source):
You can put this solution on YOUR website!
"... he has two more quarters and dimes than he has nickels..."
That could mean that the number of quarters is two more than the number of nickels AND the number of dimes is ALSO two more than the number of nickels.
Or it could mean that the TOTAL number of quarters and dimes is two more than the number of nickels.
So the statement of the problem is ambiguous, making it impossible to know how to set up the problem.
An attempt to solve the problem as stated is most easily done informally, using logical reasoning and simple mental arithmetic -- not with formal algebra.
(1) The total value of the nickels, dimes, and quarters is a multiple of 5 cents. Since the total value of all the coins is 93 cents, the number of pennies must be 3, or 8, or 13, or ....
(2) Then, since the number of quarters is the same as the number of pennies, there must be 3 pennies and 3 quarters.
(3) That makes 78 cents, leaving 15 cents to be made with the nickels and dimes. There are only two possibilities -- 3 nickels, or 1 dime and 1 nickel.
But neither of those possibilities satisfies the conditions of the problem with ANY interpretation of the given information.
So the problem is doubly bad -- there are two possible interpretations of the given information; and NEITHER of them leads to a solution that satisfies all the conditions.
Re-post, using unambiguous language.
And make sure the problem as stated has a solution....
Answer by josgarithmetic(39620) (Show Source):
|
|
|