A collection of 29 coins, all nickels and quarters, is worth $4.65. How many of each type of coin is there?

Question

A collection of 29 coins, all nickels and quarters, is worth $4.65. How many of each type of coin is there? Show the steps with system of equations. 

Answer 1

Sam,

 

x = number of nickels

y = number of quarters

x + y = 29

5x + 25y = 465   //easier to work with than .50x+.25y=4.65

multiply (x + y = 29) by 5, and get 5x + 5y = 145

Subtract (5x + 5y = 145) from (5x + 25y = 465) and get 

0x + 20y = 320, => y = 16, => x + 16 = 29 => x = 13

 

13(5) + 16(25) = 465

Answer 2

First set up variables:  Let N = the number of nickels and Q = the number of quarters

 

Since there are 29 coins, that means that N + Q = 29

 

(Let's do the rest using cents instead of dollars to avoid decimals.)

 

The value of all the nickels is 5N, and the value of all the quarters is 25Q.  So the total value is 5N + 25Q.  But we know that that is 465 (cents).

 

So we have two equations in two unknowns:

 

N + Q = 29

5N + 25Q = 465

 

Multiply the first equation by 5:

5N + 5Q = 29*5 = 145

 

Subtract this new equation from the second equation:

5N + 25Q = 465

5N + 5Q = 145

--------------------

20Q = 320

 

So Q = 16

 

Since there are a total of 29 coins, N must be 13

 

See if that works...

 

13 nickels = 13*5 = 65 cents

16 quarters = 16*25 = 400 cents

 

That adds up to 465 cents, or $4.65.  Correct!