Question 148577: I"m Trying to solve a word problem where I have to make three equations and solve. But after reading the problem I only see two equations...
Mike has 19 coins (nickels, dimes, and quarters) in his piggy bank with a total value of $2.35. There are two more dimes than nickels. How many of each type of coins does mike have.
So far I have n+d+q=235 and N= d+2 (I think?) and would the first equation answer the last sentence.
Please help me solve
Answer by aswathytony(47)   (Show Source):
You can put this solution on YOUR website! hello, you know 1 nickel = 5 cent, 1 dime = 10 cent, 1 quarter = 25 cent.
given there are 19 coins. so let us take the number of nickels are N, dimes are D, and quarters are Q .
i.e N + D + Q = 19 .........(1)
also it is given there are two more dimes than nickels,
i.e. D = N + 2 ........(2)
It is given the total value of all the coins is $2.35 = 235 cents
i.e N nickels + D dimes + Q quarters = 235 cents
N * 5 + D * 10 + Q * 25 = 235 .........(3) ( Substituting the
values of nickel, dime & quarter in cents)
substituting (2) in (1)we get,
N + N+ 2 + Q = 19
2N + 2 + Q = 19
2N + Q = 19 - 2 = 17
Q = 17 - 2N ............(4)
SUBSTITUTING (2) & (4) in (3)
N * 5 + (N+2) * 10 + (17 - 2N) * 25 = 235
5N + 10N + 2*10 + 17*25 - 2N *25 = 235
15N + 20 + 425 - 50N = 235
445 - 35N = 235
445 - 235 = 35 N
210 = 35 N
N = 210/ 35 = 6
put N= 6 in (2)
D = N + 2 = 6 + 2 = 8 .
put N=6 in (4)
Q = 17 - 2N = 17 - 2* 6 = 17- 12 = 5 .
i.e. mike have 6 coins of nickel, 8 coins of dimes and 5 quarter coins .
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