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lana66690 [7]
3 years ago
9

A collection of coins consists of nickels, dimes, and quarters. There are three fewer quarters than nickels and six more dimes t

han quarters. How many of each kind of coin is in the collection if the total value of the collection is $4.35?
PLEASE HELP ME THIS IS DUE TOMORROW!!!
Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

You have 15 dimes. You have 9 quarters. You have 12 nickels

Step-by-step explanation:

lets set some variables:

let "n" = the number of nickels

let "d" = the number of dimes

let "q" = the number of quarters

So, the total amount of money you have should be: $4.35 = 0.25q + 0.10d + 0.05n

Now let's look at the relationships between the coins:

"There are three fewer quarters than nickels": n - 3 = q

"six more dimes than quarters": q + 6 = d

So now you have three equations with three variables, all you need to do is solve.

\left \{ {{4.35 = 0.25q + 0.10d + 0.05n} \atop {n - 3 = q}} \atop {q + 6 = d}}\right.

first, you can substitute "n-3" for "q" (according to the 2nd equation) in the 1st and 3rd equation, you get:

\left \{ {{4.35=0.25(n-3)+0.10d+0.05n} \atop {(n-3)+6=d}} \right.

You now only have two equations and two variables.

Simplify:

\left \{ {{4.35=0.25n-0.75+0.10d+0.05n} \atop {n+3=d}} \right.

\left \{ {{4.35=0.30n-0.75+0.10d} \atop {n+3=d}} \right.

Now substitute "n+3" for "d" (according to the 2nd equation) in the 1st equation:

4.35=0.30n-0.75+0.10(n+3)

simplify:

4.35=0.30n-0.75+0.10n+0.30

4.35=0.40n-0.45

4.35+0.45=0.40n

4.80=0.40n

n=12

You have 12 nickels. Now sub "n" back into your equations to find the number of dimes and quarters:

n - 3 = q

12 - 3 = q

q = 9

You have 9 quarters.

q + 6 = d

9 + 6 = d

d = 15

You have 15 dimes.

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