Algebra: Structure And Method, Book 1
Algebra: Structure And Method, Book 1
(REV)00 Edition
ISBN: 9780395977224
Author: Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole
Publisher: McDougal Littell
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Concept explainers

Unitary Method
The word “unitary” comes from the word “unit”, which means a single and complete entity. In this method, we find the value of a unit product from the given number of products, and then we solve for the other number of products.
Speed, Time, and Distance
Imagine you and 3 of your friends are planning to go to the playground at 6 in the evening. Your house is one mile away from the playground and one of your friends named Jim must start at 5 pm to reach the playground by walk. The other two friends are 3 …
Profit and Loss
The amount earned or lost on the sale of one or more items is referred to as the profit or loss on that item.
Units and Measurements
Measurements and comparisons are the foundation of science and engineering. We, therefore, need rules that tell us how things are measured and compared. For these measurements and comparisons, we perform certain experiments, and we will need the experime…
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Question
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Chapter 3.7, Problem 18P
To determine

To determine the number of coins of each person.

Expert Solution & Answer
Check Mark

Answer to Problem 18P

Natalie had 13 nickels

Dirk had 9 dimes

Quincy had 4 quarters

Explanation of Solution

Given:

Natalie has some nickels; Dirk has some dimes and Quincy has some quarters.

Dirk has five more dimes than Quincy has quarters.

If Natalie gives Dirk a nickel, Dirk gives Quincy a dime and Quincy gives Natalie a quarter, they will have the same amount of money.

Calculation:

Let,

Quincy has x quarters.

Then, Dirk has (x+5) dimes.

And let, Natalie has y nickels.

If Natalie gives Dirk a nickel, Dirk gives Quincy a dime and Quincy gives Natalie a quarter, they will have the same amount of money.

So, after giveaway,

Quincy has x - 1 quarters and 1 dime; These worth [10+25(x−1)] cents.

Dirk has 1 nickel and (x+4) dimes; These worth [5+10(x+4)] cents.

Natalie has nickels and 1 quarters; These worth [25+5(y−1)] cents.

According to the problem,

  10+25(x−1)=5+10(x+4)=25+5(y−1)

Taking first two parts:

  ⇒10+25(x−1)=5+10(x+4)⇒10+25x−25=5+10x+40⇒25x−10x=45+15⇒15x=60⇒x=4

Now taking the last two parts:

  ⇒5+10(x+4)=25+5(y−1)⇒5+10(4+4)=25+5y−5⇒5+80=20+5y⇒5y=85−20⇒5y=65⇒y=13

Conclusion:

Therefore, Natalie had 13 nickels, Dirk had 9 dimes and Quincy had 4 quarters.

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Chapter 3.7, Problem 17P
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Chapter 3.7, Problem 1MRE

Chapter 3 Solutions

Algebra: Structure And Method, Book 1

Ch. 3.1 - Prob. 1OECh. 3.1 - Prob. 2OECh. 3.1 - Prob. 3OECh. 3.1 - Prob. 4OECh. 3.1 - Prob. 5OECh. 3.1 - Prob. 6OECh. 3.1 - Prob. 7OECh. 3.1 - Prob. 8OECh. 3.1 - Prob. 9OECh. 3.1 - Prob. 10OE
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Prob. 2OECh. 3.8 - Prob. 3OECh. 3.8 - Prob. 1WECh. 3.8 - Prob. 2WECh. 3.8 - Prob. 3WECh. 3.8 - Prob. 4WECh. 3.8 - Prob. 5WECh. 3.8 - Prob. 6WECh. 3.8 - Prob. 7WECh. 3.8 - Prob. 8WECh. 3.8 - Prob. 9WECh. 3.8 - Prob. 10WECh. 3.8 - Prob. 11WECh. 3.8 - Prob. 1MRECh. 3.8 - Prob. 2MRECh. 3.8 - Prob. 3MRECh. 3.8 - Prob. 4MRECh. 3.8 - Prob. 5MRECh. 3.8 - Prob. 6MRECh. 3.8 - Prob. 7MRECh. 3.8 - Prob. 8MRECh. 3.8 - Prob. 9MRECh. 3.8 - Prob. 10MRECh. 3.8 - Prob. 11MRECh. 3.8 - Prob. 12MRECh. 3.8 - Prob. 1STCh. 3.8 - Prob. 2STCh. 3.8 - Prob. 3STCh. 3 - Prob. 1CRCh. 3 - Prob. 2CRCh. 3 - Prob. 3CRCh. 3 - Prob. 4CRCh. 3 - Prob. 5CRCh. 3 - Prob. 6CRCh. 3 - Prob. 7CRCh. 3 - Prob. 8CRCh. 3 - Prob. 9CRCh. 3 - Prob. 10CRCh. 3 - Prob. 11CRCh. 3 - Prob. 12CRCh. 3 - Prob. 1CTCh. 3 - Prob. 2CTCh. 3 - Prob. 3CTCh. 3 - Prob. 4CTCh. 3 - Prob. 5CTCh. 3 - Prob. 6CTCh. 3 - Prob. 7CTCh. 3 - Prob. 8CTCh. 3 - Prob. 9CTCh. 3 - Prob. 10CTCh. 3 - Prob. 11CTCh. 3 - Prob. 12CTCh. 3 - Prob. 13CTCh. 3 - Prob. 14CTCh. 3 - Prob. 15CTCh. 3 - Prob. 16CTCh. 3 - Prob. 17CTCh. 3 - Prob. 18CTCh. 3 - Prob. 19CTCh. 3 - Prob. 1CLRCh. 3 - Prob. 2CLRCh. 3 - Prob. 3CLRCh. 3 - Prob. 4CLRCh. 3 - Prob. 5CLRCh. 3 - Prob. 6CLRCh. 3 - Prob. 7CLRCh. 3 - Prob. 8CLRCh. 3 - Prob. 9CLRCh. 3 - Prob. 10CLRCh. 3 - Prob. 11CLRCh. 3 - Prob. 12CLRCh. 3 - Prob. 13CLRCh. 3 - Prob. 14CLRCh. 3 - Prob. 15CLRCh. 3 - Prob. 16CLRCh. 3 - Prob. 17CLRCh. 3 - Prob. 18CLRCh. 3 - Prob. 19CLRCh. 3 - Prob. 20CLRCh. 3 - Prob. 21CLRCh. 3 - Prob. 22CLRCh. 3 - Prob. 23CLRCh. 3 - Prob. 24CLRCh. 3 - Prob. 25CLRCh. 3 - Prob. 26CLRCh. 3 - Prob. 27CLRCh. 3 - Prob. 28CLRCh. 3 - Prob. 29CLRCh. 3 - Prob. 30CLRCh. 3 - Prob. 31CLRCh. 3 - Prob. 32CLRCh. 3 - Prob. 33CLRCh. 3 - Prob. 34CLRCh. 3 - Prob. 35CLRCh. 3 - Prob. 36CLRCh. 3 - Prob. 37CLRCh. 3 - Prob. 38CLRCh. 3 - Prob. 39CLRCh. 3 - Prob. 40CLRCh. 3 - Prob. 1MSCh. 3 - Prob. 2MSCh. 3 - Prob. 3MSCh. 3 - Prob. 4MSCh. 3 - Prob. 5MSCh. 3 - Prob. 6MSCh. 3 - Prob. 7MSCh. 3 - Prob. 8MSCh. 3 - Prob. 9MSCh. 3 - Prob. 10MSCh. 3 - Prob. 11MSCh. 3 - Prob. 12MSCh. 3 - Prob. 13MSCh. 3 - Prob. 14MSCh. 3 - Prob. 15MSCh. 3 - Prob. 16MSCh. 3 - Prob. 17MSCh. 3 - Prob. 18MSCh. 3 - Prob. 19MSCh. 3 - Prob. 20MSCh. 3 - Prob. 21MSCh. 3 - Prob. 22MSCh. 3 - Prob. 23MSCh. 3 - Prob. 24MSCh. 3 - Prob. 25MSCh. 3 - Prob. 26MSCh. 3 - Prob. 27MSCh. 3 - Prob. 28MSCh. 3 - Prob. 29MSCh. 3 - Prob. 30MSCh. 3 - Prob. 31MSCh. 3 - Prob. 32MSCh. 3 - Prob. 33MSCh. 3 - Prob. 34MSCh. 3 - Prob. 35MSCh. 3 - Prob. 36MSCh. 3 - Prob. 1MPSCh. 3 - Prob. 2MPSCh. 3 - Prob. 3MPSCh. 3 - Prob. 4MPSCh. 3 - Prob. 5MPSCh. 3 - Prob. 6MPSCh. 3 - Prob. 7MPSCh. 3 - Prob. 8MPSCh. 3 - Prob. 9MPSCh. 3 - Prob. 10MPSCh. 3 - Prob. 11MPSCh. 3 - Prob. 12MPSCh. 3 - Prob. 13MPSCh. 3 - Prob. 14MPSCh. 3 - Prob. 15MPSCh. 3 - Prob. 16MPSCh. 3 - Prob. 17MPS

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Publisher:McGraw-Hill Education
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