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Algebraic Word Problems

In mathematics we have different and varied topics. It covers from basic operations of addition and subtraction to advanced level calculations like wave length of a spectrum. In simple cases we do such operations with the known numerical values and arrive at the result. But many times we may have to find the unknowns by working backwards. In such cases the concepts of algebra is used.

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Definition 

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Algebra is a topic where some assigned letters are used to represent the unknown quantity or quantities. They are called as variable because its actual value is unknown and can result to any value depending on the context. For example, we can easily what $3$ added to $2$ is by just doing $3 + 2$ = $5$, which is an arithmetic. But if the question was what added to $2$ gives $5$, we assign a variable, say '$x$', and write as $x + 2$. Since this sum is $5$, we can equate $x + 2$ = $5$ and solve for '$x$'. Now it becomes algebra. In this context, the statement is true only for the value of $x$ equal to $3$.

In many real life situations we do come across occasions where a desired result is to be achieved with unknown inputs. For example a chemist may require only a $20 \%$ concentration acid but he may have only $40 \%$ and $10 \%$ concentrated acids in his stock. The concept of algebra helps him to identify the ratio of the acids of available concentrations to be mixed. The following word problems show the applications of algebra in different situations.

Word Problems

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Examples 1: 

Jason pays $ \$12.48$ for a T-shirt which includes a sales tax at the rate of $4 \%$. What is the marked price of the shirt and how much of sales tax is paid?

Solution:

We only know the final price and we do not know what the original price is. Hence let us assume '$x$' dollars for that. Its value can be anything but will be true for only one unique value in this situation.

If the sales tax rate is $4 \%$, then the amount of sales tax is $4 \%$ of '$x$', means $0.04x$ dollars. Hence the total bill value is $ \$$ $(x + 0.04x)$ = $ \$1.04x$. But the actual amount paid is $ \$12.48$. Since both refers to the same situation,

$1.04x$ = $12.48$ or $x$ = $\frac{12.48}{1.04}$ = $12$.  

Therefore, the marked price of the shirt is $ \$12$.

The amount sales tax paid = Final price paid – Marked price = $ \$12.28 - \$ 12.00$ = $ \$0.48$.
Example 2: 

Tom has in his piggy box a total of $48$ coins. Some of the coins are in quarters and the remaining coins are in dimes.  If the value of his collections in that box is $ \$9.30$, how many coins of each type are there?

Solution: 

We do not the number of coins in either of the denominations. Let us choose the variable '$x$' to represent the number of quarters. The immediate thought that comes into mind is, the number of dimes has to be $(48 - x)$, since the total is $48$.

Now, we can evaluate the vales of '$x$' quarters and $(48 - x)$ dimes and add those two values because the total value of all the coins is given as $ \$9.30$.

The value of each quarter is $ \$0.25$ and hence the value of '$x$' quarters is $0.25x$ dollars. Similarly, the value of each dime is $ \$0.10$ and hence the value of $(48 - x)$ dimes is $(0.10)(48 - x)$ dollars. Therefore, the algebraic expression for the total value is is $0.25x + (0.10)(48 - x)$ which can be simplified as $0.25x + 4.80 - 0.10x$ or $0.15x + 4.80$. But the actual value is given as $ \$9.30$. Therefore we can set up an algebraic equation as,

$0.15x + 4.80$ = $9.30$ or $0.15x$ = $4.50$ or $x$ = $\frac{4.50}{0.15}$ = $30$. 

So, the number if quarters = $30$ and the number of dimes = $48 - 30$ = $18$.
Example 3: 

Alicia is $3$ times the age of her daughter Sara. $5$ years back Alicia's age was exactly $4$ times the age of Sara. What are their present ages?   

Solution:

In this situation, it is better to assume the age of Sara as '$x$' years. Then, the present age of Alicia is $3 \times x$ = $3x$ years.

Now let us analyze the situation $5$ years back. At that time Sara's age was $(x - 5)$ years and Alicia's age must have been $(3x - 5)$.

As per the problem statement, $5$ years back, 

$(3x - 5)$ = $4(x - 5)$

$3x - 5$ = $4x - 20$

$-x$ = $-15$ or $x$ = $15$

So, the present age of Sara is $15$ years and that of her mother is $3 \times 15$ = $45$ years.
Example 4: 

A chemist has received an order for $18$ gallons of acid with $20 \%$ concentration. But he finds he has stocks of acids of only $40 \%$ concentration and $10 \%$ concentration. How many gallons of each quality of the available acids to be mixed to execute the order?

Solution:

In the field of chemistry, 'concentration' means the ratio of the volume of a particular substance in a mixture to the total volume of the mixture. In case of acids, pure acid is mixed with water to get a diluted strength of acid.

Now, let '$x$' be the gallons of $40 \%$ concentrate acid is used for the given purpose and obviously $(18 - x)$ gallons of $10 \%$ concentrate are required to make the total volume as $48$ gallons.

The amount of pure acid in $x$ gallons of $40 \%$ acid is $x \times$ $(\frac{40}{100})$ gallons and that in $(18 - x)$ gallons of $10 \%$ acid is $(18 - x) \times$ $(\frac{10}{100})$ gallons. When both are added, the total acid content is $[x \times$ $(\frac{40}{100})$ + $(18 - x) \times$ $(\frac{10}{100})$] gallons. This must be equal to $18$ $(\frac{20}{100})$ gallons in the final product. Therefore,

$[x \times$ $(\frac{40}{100})$ + $(18 - x) \times$ $(\frac{10}{100})$]  = $18$ $(\frac{20}{100})$
  
Or, $40x + (18 - x)(10)$ = $18(20)$, removing the denominator $100$ from all the terms

Or, $40x + 180 - 10x$ = $360$

Or, $30x$ = $180$

Or $x$ = $6$ and $18 - x$ = $12$

Therefore, the chemist must mix $6$ gallons of $40 \%$ acid and $12$ gallons of $10 \%$ acid to finally get $18$ gallons of $20 \%$ acid.
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