Math word problem help!
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Math word problem help!

[From: ] [author: ] [Date: 11-11-02] [Hit: ]
-Let D be the number of dimes and N be the number of nickels. We know that D + N =67, because that is the total number of coins. Let this be your first equation.The total amount of money is $4.20=420 cents.......
A box contains 67 coins, only dimes and nickels. The amount of money in the box is $4.20. How many dimes and how many nickels are in the box?

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Let D be the number of dimes and N be the number of nickels. We know that D + N =67, because that is the total number of coins. Let this be your first equation.

The total amount of money is $4.20=420 cents. If a dime is worth 10 cents, then 10D represents the total value of dimes. A nickel is worth 5 cents so 5N represents the total value in nickels. Now we have a second equation:
10D + 5N = 420

From equation 1, we solve for N:
D + N=67
N=67-D

Now plug this value into the second equation:
10D + 5N = 420
10D + 5(67-D)=420
10D +335 - 5D= 420
5D= 85
D=85/5 =17

Therefore there are 17 dimes. To find the number of nickels, plug the value of D into the first equation:
D + N = 67
17 + N = 67
N= 67- 17= 50

You have 17 dimes and 50 nickels.

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17 dimes and 50 nickels
man $$%% i solved it and when i submitted it 2 other people already answered it.

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10D + 5 (67 -- D) = 420 giving D = (420 -- 335) / (10 -- 5) = 17
17 dimes and 50 nickels
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