Lesson Percentage word problems (Type 1 problems, Finding the Part)

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<H2>Percentage word problems (Type 1 problems: Finding the Part)</H2>

This lesson presents solution examples of word problems on percentage.
It is is a continuation of the lesson <A HREF= http://www.algebra.com/algebra/homework/percentage/lessons/Percentage-problems.lesson> Percentage problems</A> in this site.

Let me remind you that 1% of some number is one hundredth (1/100) part of the number.
10% of some number is one tenth (10/100 = 1/10) part of the number.
20% of some number is one fifth (20/100 = 1/5) part of the number.
25% of some number is one fourth (25/100 = 1/4) part of the number.

The percentage problems include three numbers.
One number is the <B>base</B> <B>B</B>. It represents the total amount of something or the measure of something.
Second number is the <B>rate</B> <B>R</B>. It is a measure of the part relatively to the whole thing, expressed in percents, like 3%, 7.5%, 12.75% (percentage).
Third number is the <B>part</B> <B>P</B>. It is the amount or the measure of the part.

<BLOCKQUOTE>The base, the part and the percentage are connected by the formula
{{{P = (R/100)*B}}}. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(*)
</BLOCKQUOTE>
For example, if the base <B>B</B> is equal to 80 and the percentage <B>R</B> is 5%, then the part <B>P</B> is equal to {{{P = (5/100)*80 = 4}}}.

Problems considered in this lesson are all of the same type: you are given two of three numbers, namely the base <B>B</B> and the percentage <B>R</B>. The third number, the part <B>P</B>, is unknown, and you should find it. These problems are of the <B>Type 1 problems on percentage</B>, as defined in the lesson <A HREF= http://www.algebra.com/algebra/homework/percentage/lessons/Percentage-problems.lesson> Percentage problems</A>.

Below are examples of the Type 1 word problems on percentage.

<H3>Problem 1. Test results</H3>

The student answered 80 questions on the mathematics test.
95% of answers were correct.
How many answers were correct?

<B>Solution</B>

Since 95% of 80 answers were correct, the number of correct answers was equal to {{{(95/100)*80 = 76}}}.

<B>Answer</B>. There were 76 correct answers.

<H3>Problem 2. Tax calculation</H3>

Calculator is advertised at $24.99 plus 7.5% tax at the store.
What is the total cost including tax?

<B>Solution</B>

Tax rate is 7.5%, which is equal to {{{7.5/100 = 0.075}}}.
So, the tax amount is {{{0.075*24.99 = 1.87}}} dollars.
Hence, the total cost including tax is {{{24.99 + 1.87 = 26.86}}} dollars.

<B>Answer</B>. The total cost including tax is $26.86.

<H3>Problem 3. Discount </H3>

Brian bought a printer on sale at the store. The original price was $90.00, the discount was 15%.
What was the sale price of the printer? How much money Brian saved?

<B>Solution</B>

First, find the amount of the discount.
Since the discount was 15% of $90.00, it is equal to 15/100 of $90.00, that is {{{(15/100)*90.00 = 13.50}}}.
So, Brian saved $13.50 with the discount.
Now, find how much Brian paid for the printer.
Since the original price for the printer was $90.00 and the discount was $13.50, Brian paid
{{{90.00 - 13.50 = 76.50}}}.

<B>Answer</B>. Brian paid $76.50 for the printer on sale. Brian saved $13.50.

<H3>Problem 4. Population growth</H3>

The population of Georgetown was 50000 in 2009. The population had increased by 3 percents in 2010.
What was the population at the end of 2010?

<B>Solution</B>

First, find the amount of the population increase.
Since the population had increased by 3 percents in 2010, the amount of the population increase is 3/100 of 50000, that is 
{{{(3/100)*50000 = 1500}}}.

So, the population had increased by 1500 in 2010.
Now, find the population at the end of 2010.
Since the population was 50000 in 2009 and it had increased by 1500 in 2010, the population at the end of 2010 was
{{{50000 + 1500 = 51500}}}.

<B>Answer</B>. The population of Georgetown was 51500 at the end of 2010.

<H3>Problem 5. Investment</H3>

Kathy invested $25,000.00 to the bank account for 2% per year.
What was the interest Kathy earned after one year?

<B>Solution</B>
The interest Kathy earned after one year is 2% of $25,000.00, that is 2/100 of $25,000.00.
It wass equal to
{{{(2/100)*25000 = 500}}} dollars.

<B>Answer</B>. The interest Kathy earned after one year was $500.00.

<H3>Problem 6. Alloy (mixture)</H3>

An alloy contains 70% of silver.
How much silver is contained in 2.5 pounds of the alloy?

<B>Solution</B>
2.5 pounds of the alloy contain 70% of silver, that is 70/100 of 2.5 pounds.
It is equal to
{{{(70/100)*2.5 = 1.75}}} pounds.

<B>Answer</B>. 2.5 pounds of the alloy contain 1.75 pounds of silver.

Examples of percentage word problems of the <B>Type 2</B> and the <B>Type 3</B> are presented in lessons
<A HREF = http://www.algebra.com/algebra/homework/Percentage-and-ratio-word-problems/Percentage-word-problems-%28Type-2-problems-Finding-the-Rate%29.lesson> Percentage word problems (Type 2 problems: Finding the Rate)</A> and
<A HREF = http://www.algebra.com/algebra/homework/Percentage-and-ratio-word-problems/Percentage-word-problems-%28Type-3-problems-Finding-the-Base%29.lesson> Percentage word problems (Type 3 problems: Finding the Base)</A>
in the section <B>Word problems</B> of this site.