Pre-Algebra Student Edition
Pre-Algebra Student Edition
1st Edition
ISBN: 9780078957734
Author: Glencoe/McGraw-Hill
Publisher: MCG
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Question
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Chapter 0, Problem 5P
To determine

To find: The total possible ways to change a dollar

Expert Solution & Answer
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Answer to Problem 5P

Thetotal possible ways are 29

Explanation of Solution

Given:

Consider the values of dollar, quarter, dime and nickel in terms of cents as follows:

1 dollar =100 ȼ

1 quarter =25 ȼ

1 dime =10 ȼ

1 nickel =5 ȼ

Calculation:

Consider the values of dollar, quarter, dime and nickel in terms of cents as follows:

1 dollar =100 ȼ

1 quarter =25 ȼ

1 dime =10 ȼ

1 nickel =5 ȼ

Now, change for dollar could be in terms of single coin or in mix of them.

Comment

Step 2 of 3

Considering single coins,

1. 4 quarters make $11 way

2. 10 dimes make $11 way

3. 20 nickels make $11 way

Now, consider mix of 2 types of coins

4. Quarters and dimes: since dime involves factor of 10 quarters must have value which is factor of 10. Only combination of quarters between 25 ȼ and 100 ȼ (not including 100,as it has been included above ) 50 ȼ.

Thus, it will have 2 quarters and 5 dimes -1 way.

5. Quarters and nickels: Here, we can have 1 quarter, 2 quarters or 3 quarters in the mix

Thus − 3 ways.

6. Dimes and Nickels: Here we can have mix ranging from 1 dime to 9 dimes (as 0 and 10 arealready covered under single coins)

Thus - 9 ways.

Now, considering mix of all three types of coins as,

Various combinations are possible here,

7. with 1 quarter. With 25 cents gone, there can be 1 to 7 dimes (as 0 dimes is already considered under 2 type mix and 8 will exceed limit of $1)

Thus - 7 ways.

8. with 2 quarters: With 50 cents gone, there can be 1 to 4 dimes (as 0 and 5 dimes is alreadyconsidered under 2 type mix

Thus - 4 ways.

9. with 3 quarters: With 75 cents gone, there can be 1 to 2 dimes (as 0 dimes is alreadyconsidered under 2 type mix and 3 will exceed limit of $1 )

Thus there are 2 ways.

Counting all ways of all three scenarios,

The total possible ways are,

  =1+1+1+1+3+9+7+4+2

=29

Conclusion:

Hence, the total possible ways are 29

Chapter 0 Solutions

Pre-Algebra Student Edition

Ch. 0.2 - Prob. 7ECh. 0.2 - Prob. 8ECh. 0.2 - Prob. 9ECh. 0.2 - Prob. 10ECh. 0.2 - Prob. 11ECh. 0.2 - Prob. 12ECh. 0.2 - Prob. 13ECh. 0.2 - Prob. 14ECh. 0.2 - Prob. 15ECh. 0.2 - Prob. 16ECh. 0.2 - Prob. 17ECh. 0.2 - Prob. 18ECh. 0.2 - Prob. 19ECh. 0.3 - Prob. 1ECh. 0.3 - Prob. 2ECh. 0.3 - Prob. 3ECh. 0.3 - Prob. 4ECh. 0.3 - Prob. 5ECh. 0.3 - Prob. 6ECh. 0.3 - Prob. 7ECh. 0.3 - Prob. 8ECh. 0.3 - Prob. 9ECh. 0.3 - Prob. 10ECh. 0.3 - Prob. 11ECh. 0.3 - Prob. 12ECh. 0.3 - Prob. 13ECh. 0.3 - Prob. 14ECh. 0.3 - Prob. 15ECh. 0.3 - Prob. 16ECh. 0.3 - Prob. 17ECh. 0.3 - Prob. 18ECh. 0.3 - Prob. 19ECh. 0.3 - Prob. 20ECh. 0.3 - Prob. 21ECh. 0.3 - Prob. 22ECh. 0.3 - Prob. 23ECh. 0.3 - Prob. 24ECh. 0.3 - Prob. 25ECh. 0.3 - Prob. 26ECh. 0.4 - Prob. 1ECh. 0.4 - Prob. 2ECh. 0.4 - Prob. 3ECh. 0.4 - Prob. 4ECh. 0.4 - Prob. 5ECh. 0.4 - Prob. 6ECh. 0.4 - Prob. 7ECh. 0.5 - Prob. 1ECh. 0.5 - Prob. 2ECh. 0.5 - Prob. 3ECh. 0.5 - Prob. 4ECh. 0.5 - Prob. 5ECh. 0.5 - Prob. 6ECh. 0.5 - Prob. 7ECh. 0.5 - Prob. 8ECh. 0.5 - Prob. 9ECh. 0.5 - Prob. 10ECh. 0.5 - Prob. 11ECh. 0.5 - Prob. 12ECh. 0.5 - Prob. 13ECh. 0.6 - Prob. 1ECh. 0.6 - Prob. 2ECh. 0.6 - Prob. 3ECh. 0.6 - Prob. 4ECh. 0.6 - Prob. 5ECh. 0.6 - Prob. 6ECh. 0.6 - Prob. 7ECh. 0.6 - Prob. 8ECh. 0.6 - Prob. 9ECh. 0.6 - Prob. 10ECh. 0.6 - Prob. 11ECh. 0.6 - Prob. 12ECh. 0.6 - Prob. 13ECh. 0.6 - Prob. 14ECh. 0.6 - Prob. 15ECh. 0.6 - Prob. 16ECh. 0.6 - Prob. 17ECh. 0.6 - Prob. 18ECh. 0.6 - Prob. 19ECh. 0.6 - Prob. 20ECh. 0.6 - Prob. 21ECh. 0.6 - Prob. 22ECh. 0.7 - Prob. 1ECh. 0.7 - Prob. 2ECh. 0.7 - Prob. 3ECh. 0.7 - Prob. 4ECh. 0.7 - Prob. 5ECh. 0.7 - Prob. 6ECh. 0.7 - Prob. 7ECh. 0.7 - Prob. 8ECh. 0.7 - Prob. 9ECh. 0 - Prob. 1PRCh. 0 - Prob. 2PRCh. 0 - Prob. 3PRCh. 0 - Prob. 4PRCh. 0 - Prob. 5PRCh. 0 - Prob. 6PRCh. 0 - Prob. 7PRCh. 0 - Prob. 8PRCh. 0 - Prob. 9PRCh. 0 - Prob. 10PRCh. 0 - Prob. 11PRCh. 0 - Prob. 12PRCh. 0 - Prob. 13PRCh. 0 - Prob. 14PRCh. 0 - Prob. 15PRCh. 0 - Prob. 16PRCh. 0 - Prob. 17PRCh. 0 - Prob. 18PRCh. 0 - Prob. 19PRCh. 0 - Prob. 20PRCh. 0 - Prob. 21PRCh. 0 - Prob. 22PRCh. 0 - Prob. 23PRCh. 0 - Prob. 24PRCh. 0 - Prob. 25PRCh. 0 - Prob. 26PRCh. 0 - Prob. 27PRCh. 0 - Prob. 28PRCh. 0 - Prob. 29PRCh. 0 - Prob. 30PRCh. 0 - Prob. 31PRCh. 0 - Prob. 32PRCh. 0 - Prob. 33PRCh. 0 - Prob. 34PRCh. 0 - Prob. 35PRCh. 0 - Prob. 36PRCh. 0 - Prob. 37PRCh. 0 - Prob. 38PRCh. 0 - Prob. 39PRCh. 0 - Prob. 40PRCh. 0 - Prob. 41PRCh. 0 - Prob. 42PRCh. 0 - Prob. 43PRCh. 0 - Prob. 44PRCh. 0 - Prob. 45PRCh. 0 - Prob. 46PRCh. 0 - Prob. 47PRCh. 0 - Prob. 48PRCh. 0 - Prob. 49PRCh. 0 - Prob. 50PRCh. 0 - Prob. 51PRCh. 0 - Prob. 52PRCh. 0 - Prob. 53PRCh. 0 - Prob. 1PCh. 0 - Prob. 2PCh. 0 - Prob. 3PCh. 0 - Prob. 4PCh. 0 - Prob. 5PCh. 0 - Prob. 6PCh. 0 - Prob. 7PCh. 0 - Prob. 8PCh. 0 - Prob. 9PCh. 0 - Prob. 10PCh. 0 - Prob. 11PCh. 0 - Prob. 12PCh. 0 - Prob. 13PCh. 0 - Prob. 14PCh. 0 - Prob. 15PCh. 0 - Prob. 16PCh. 0 - Prob. 17PCh. 0 - Prob. 18PCh. 0 - Prob. 19PCh. 0 - Prob. 20PCh. 0 - Prob. 21PCh. 0 - Prob. 22PCh. 0 - Prob. 23PCh. 0 - Prob. 24PCh. 0 - Prob. 25PCh. 0 - Prob. 26PCh. 0 - Prob. 27PCh. 0 - Prob. 28PCh. 0 - Prob. 29PCh. 0 - Prob. 30PCh. 0 - Prob. 31PCh. 0 - Prob. 32PCh. 0 - Prob. 33PCh. 0 - Prob. 34PCh. 0 - Prob. 35PCh. 0 - Prob. 36PCh. 0 - Prob. 37PCh. 0 - Prob. 38PCh. 0 - Prob. 39PCh. 0 - Prob. 40PCh. 0 - Prob. 41PCh. 0 - Prob. 42PCh. 0 - Prob. 43PCh. 0 - Prob. 44PCh. 0 - Prob. 45PCh. 0 - Prob. 46PCh. 0 - Prob. 47PCh. 0 - Prob. 48PCh. 0 - Prob. 49PCh. 0 - Prob. 50PCh. 0 - Prob. 51PCh. 0 - Prob. 52PCh. 0 - Prob. 53PCh. 0 - Prob. 54PCh. 0 - Prob. 55PCh. 0 - Prob. 56PCh. 0 - Prob. 57PCh. 0 - Prob. 58P
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