Chapter 3.1, Problem 8P

a.

To determine

To choose: Two variables that represent the number asked in the given sentence

Answer to Problem 8P

Final Answer: The number of nickels = 2x, the number pennies= x

The number of Dimes = y +4, the number of quarters= y

Explanation of Solution

Given Information: It is given that a bag contains twice as many pennies as nickels and four more dimes than quarters. We need to choose two variables to represent these numbers

Calculation: Since it is given that the bag contains twice as many pennies as nickels

Let the number of nickels = x

Then the number of pennies = 2x

And as it is given that the bag contains four more dimes than quarters.

Let the number of quarters = y

Then the number of dimes= y +4

b.

To determine

To write: An open sentence relating the variables x and y in the given condition

Answer to Problem 8P

Final Answer: $x+5y=23$

Explanation of Solution

Given Information: It is given that the total value of all pennies, nickels, dimes and quarters taken together is $2.01

Formula Used: 1 nickel = 5 cents, 1 penny = 1 cent, 1dime = 10 cents, 1 quarter = 25 cents,

1 dollar = 100 cents.

Calculation: Since we know that the number of nickels = x

Number of pennies = 2x, number of quarters = y, number of dimes = y+ 4

Since all coins taken together gives the value of $2.01 that is, 201 cents

Therefore,

According to the question,

  $5x+2x+25y+10(y+4)=201$

Or  $7x+35y=161$

Or  $x+5y=23$

c.

To determine

To find: All the possibilities for the number of each coin satisfying the given condition

Answer to Problem 8P

Final answer: All the possible number of nickels are {3,8,13,18,23}

The possible number of pennies are {6,16,26,36,46}

The possible number of quarters are {0,1,2,3,4}

The possible number of dimes are {4,5,6,7,8}

Explanation of Solution

Given Information: It is given that we have x nickels, 2x pennies, y quarters, ( y + 4) dimes

And the condition that we have obtained is

  $x+5y=23$

Calculation: Since we have obtained

  $x+5y=23$

Or  $5y=23-x$ ...........(1)

From (1), we can say $23-x$ is a multiple of 5.

So, x can take the value of 3, 8, 13, 18, 23

When x = 3, y = 4 x = 13, y = 2

x = 8, y = 3 x = 18, y = 1

x = 23, y = 0

Now since the number of nickels = x

So possible number of nickels are {3,8,13,18,23}

Since the number of pennies = 2x

So possible number of pennies are {6,16,26,36,46}

Since the number of quarters = y

So possible number of quarters are {0,1,2,3,4}

Since the number of dimes = y + 4

So possible number of dimes are {4,5,6,7,8}