algebra help

Question

Bob has 20 coins totaling $1.35. If he has only dimes and nickels, how many of each coin does he have?

nickels=

dimes=

Answer 1

Hi Corina,

 

Bob has 20 coins totaling $1.35. If he has only dimes and nickels, how many of each coin does he have?

nickels=

dimes=

 

 

Let n=the number of nickels

     d=the number of dimes

 

Since Bob has 20 coins

 

n+d=20

 

Since Bob has $1.35 we have to put a value in money for each variable

 

.05n+.10d=1.35

 

Since we have two equations and two unknowns, we have a system of equations

 

n+d=20

.05n+.10d=1.35  

 

To make it easier to solve, multiply the 2nd equation by 100 to eliminate the decimals

 

100[.05n+.10d=1.35]  

5n+10d=135

 

 

Put the two equations together

5n+10d=135

n+d=20            

 

multiply the 2nd equation by -5

 

5n+10d=135        5n+10d=135     5n+10d=135
n+d=20             -5[n+d=20]        -5n+-5d=-100

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

                                                          5d=35

                                                            d=7

 

Substitute value into 2nd equation

 

n+d=20

n+7=20

n=13

 

Check

.05n+.10d=1.35

.05(13)+.10(7)=1.35

.65+.70=1.35

1.35=1.35  it checks

 

ANSWER:There are 13 nickels and 7 dimes

 

Hope this helps you understand the process.

 

 

Answer 2

Hi Corina,

 

Let's see what we know here.

 

We have only nickels (N) and dimes (D).

We have a total of 20 coins N+D=20

Nickels are worth 0.05 and dimes are worth 0.10

 

The total value of the coins is 1.35

 

Our two equations are:

.05N + .10D = 1.35

N + D = 20

 

We can rewrite the 2nd equation, solving for N as

N = 20-D 

and then substitute this into the first equation.

 

.05(20-D) + .10D = 1.35

 

Let's multiply by 100 to get rid of the decimal points.

5(20-D) + 10D = 135

100-5D + 10D = 135

100 + 5D = 135

5D = 35

D = 7

 

And since N = 20-D

N = 13

 

We have 13 nickels and 7 dimes.