Question

Plz help! Easy question. Medal and fan :)

A collection of nickels and dimes is worth $2.95. There are 47 coins in all. How many nickels are there?

A. 12

B. 18

C. 29

D. 35

Answer 1

@RainbowBrony555

Answer 2

@Gokuporter

Answer 3

Help plzzzzzzz

Answer 4

Okay, there are a bunch of ways you could solve this.

Here's the seat of the pants method: assume all of the coins (all 47) are nickels. How much would 47 nickels be worth? Compare that value with the target of $2.95. If you have less than $2.95, you need to make up the difference by switching some nickels to be dimes. For each nickel you switch, your total goes up by 5 cents. So, if your 47 nickels came out to be worth $2.70 (it doesn't, of course), you would say "oh, I need $2.95-$2.70 = 25 cents more" and switch 5 of the nickels for dimes, which would increase your total by 5*5 =25 cents.

The way your teacher probably wants you to do it involves writing a system of equations and solving.

Let's say we call \(n\) the number of nickels, and \(d\) the number of dimes.

We know there are 47 coins, only dimes and nickels, so one equation we can write is

\[n+d = 47\]

We know that nickels are worth 5 cents and dimes are worth 10 cents, and that together, our coins are worth $2.95 = 295 cents:

\[5n + 10d = 295\]

Our system of equations is thus
\[n+d=47\]\[5n+10d=295\]

this can be easily solved by taking the first equation and solving for one of the variables in terms of the other. Say we solve it for \(n\). We'll get an equation like \(n = 23-d\) (not the real equation). Then we substitute \((23-d)\) wherever we see \(n\) in the other equation, which gives us an equation only in terms of one variable (\(d\)) instead of two. You solve that to get the value of that variable. Then plug that value into the substitution equation to find the value of the other variable.

As an example:

13 coins, dimes and quarters only, totaling $2.05

\[d+q = 13\]\[10d+25q=205\]\[d=13-q\]\[10(13-q)+25q=205\]\[130-10q+25q=205\]\[130+15q=205\]\[15q=205-130\]\[15q=75\]\[q=5\]\[d=13-q = 13-5 = 8\]

checking: 8 dimes + 5 quarters = 8*10 + 5*25 = 80+125 = 205 cents

Answer 5

^^