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Pre Algebra Word Problems

Pre-algebra is the precursor to algebra. The concepts, basics and topics that we need to know before we venture into hard core algebra are covered under pre-algebra. For example, number system that would include natural numbers, whole numbers, integers, fractions, decimals, negative numbers; in general all rational numbers; operations on rational numbers such as addition, subtraction, multiplication, division and exponentiation; factorization of whole numbers; properties of whole numbers, integers and rational numbers such as associative, closure, distributive, etc; square roots and cubed roots of simple whole numbers that are perfect squares or perfect cubes; etc.

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Important Formulas

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1) $\frac{Area\ of\ square}{Rectangle}$ = $Base \times Height$

2) Area of Triangle = $(Base \times Height) ÷ 2$

3) Area of Trapezoid = $(Height \times (base_1 + base_2)) ÷ 2$

4) Area of Circle = $p \times r \times r$

5) Circumference of Circle = $p \times d$

6) Pythagorean Theorem: $a^2 + b^2$ = $c^2$

7) Order of Operations: Parenthesis, Exponents, $\frac{Multiplication}{Division}$ (left to right), $\frac{Addition}{Subtraction}$ (left to right).

8) Commutative Property: $a + b$ = $b + a; a \times b$ = $b \times a$

9) Associative Property: $(a + b) + c$ = $a + (b + c); a \times (b \times c)$ = $(a \times b) \times c$

10) Distributive Property: $a \times (b + c)$ = $ab + ac$

11) Identity Property: $a + 0$ = $a;\ a \times 1$ = $a$

12) Inverse Property: $a+(-a)$ = $0;\ a \times$ $\frac{1}{a}$ = $1,\ a \neq 0$

13) Zero Property: $a \times 0$ = $0$

14) Distance = $Speed \times Time$

15) Metric Conversion Table:

Metric Conversion Table

Word Problems

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Example 1:

Vennica was out to buy fruits. She bought $4kg\ 200gms$ of apples, $2kg\ 350 gms$ of bananas and $5kg\ 650gms$ of oranges. What is the total weight of the fruits that she bought? 

Solution:

Quantity of apples that Vennica bought = $4kg\ 200 gms$

Quantity of bananas = $2 kg\ 350 gms$

Quantity of oranges =$5 kg\ 650 gms$

Therefore total weight of all the fruits put together

= $4kg\ 200 gms + 2 kg\ 350 gms + 5 kg\ 650 gms$

= $(4kg + 2kg + 5kg)\ +\ (200gms + 350gms + 650gms)$

= $(11 kg)\ +\ (1200 gms)$

= $11 kg + (1kg\ 200 gms)$

= $11kg + 1kg + 200 gms$

= $12kg\ 200 gms$

Thus the total weight of the fruits that Vennica bought was $12 kg\ 200 gms$.
Example 2: 

If in the above question, the apples cost $ \$ 23.00\ per\ kg$, the bananas cost $ \$ 14.00\ per\ kg$ and the oranges cost $ \$ 18.00\ per\ kg$, then find the total amount that she spent on all those fruits.

Solution: 

Quantity of apples that Vennica bought = $4 kg\ 200 gms$

= $4$ $\frac{200}{1000}$ $kg$

= $4$ $\frac{2}{10}$ $kg$

= $4$ $\frac{1}{5}$ $kg$

Price of the apples = $ \$23.00\ per\ kg$

Therefore the cost of $4$ $\frac{1}{5}$ $Kg$ apples = $4$ $\frac{1}{5}$ $\times\ 23$

= $\frac{21}{5}$ $\times\ 23$

= $\frac{21 \times 23}{5}$

= $\frac{483}{5}$

= $ \$96$ $\frac{3}{5}$

= $  \$96$ $\frac{60}{100}$

= $ \$96$  cents $60$

Similarly with the bananas.

Quantity of bananas that Vennica bought = $2kg\ 350 gms$

= $2kg\ +$ $\frac{350}{1000}$ $kg$

= $2$ $\frac{35}{100}$ $kg$

= $2$ $\frac{7}{20}$ $kg$

Price of the bananas = $ \$14.00\ kg$ 

Therefore the cost of bananas would be = $2$ $\frac{7}{20}$ $kg\ \times \$14.00/kg$

= $2$ $\frac{7}{20}$ $\times 14$

= $\frac{47}{20}$ $\times 14$

= $\frac{47 \times 14}{20}$

= $\frac{47 \times 7}{10}$

= $\frac{329}{10}$

= $32$ $\frac{9}{10}$

= $ \$32$ and cents $90$

Lastly with the oranges.

Quantity of oranges that Vennica bought = $5 kg\ 650 gms$

= $5 kg +$ $\frac{650}{1000}$ $kg$

= $5$ $\frac{65}{100}$ $kg$

= $5$ $\frac{13}{20}$ $kg$

Price of the oranges = $ \$18.00/kg$

Therefore the cost of oranges = $5$ $\frac{13}{20}$ $kg\ \times\ \$18/kg$

= $5$ $\frac{13}{20}$ $\times\ 18$

= $\frac{113}{20}$ $\times\ 18$

= $\frac{113 \times 18}{20}$

= $\frac{113 \times 9}{10}$

= $\frac{1017}{10}$

= $101$ $\frac{7}{10}$

= $101$ $\frac{7}{10}$

= $ \$101$ and cents $70$

Now let’s add them up. 

The total amount that she spent would be:

$ \$96$ cents $60 + \$32$ cents $90 + \$101$ cents $70$

= $( \$96 + \$32 + \$101)\ +\ (cents\ 60 + cents\ 90 + cents\ 70)$

= $( \$229)\ +\ (cents\ 220)$

= $ \$229 + \$2\ cents\ 20$

= $ \$231\ cents\ 20$

Thus the answer for the total amount spent by Vennica on fruits is $ \$231$ cents $20$.
Example 3: 

Myra works part time at a departmental store. She works for $5$ hours a day. She however works only $5$ days a week. If she earns  $ \$12$ per hour, then how many dollars will she make in the two weeks of her summer break?

Solution:

Myra works $5$ hours in a day and works $5$ days in a week. So the number of hours that she works in a week would be:

$5 \times 5$ = $25$ hours

Her summer break is of $2$ weeks. So the number of hours that she would work in two weeks would be:

$25 \times 2$ = $50$ hours

Now, her income per hour is given to be $ \$12$. So that is the amount that she earns in one hour. Then the amount that she would earn in $50$ hours would be:

$12 \times 50$

=$ \$600$

Thus, answer for the amount that she would earn in two weeks of her summer break would be: $ \$600/-$.
Example 4: 

A rectangular swimming pool with length $100 ft$ and width $40 ft$ has a triangular lounging area attached to its width. The distance from the side of the pool to the apex of the lounging area is $30 ft$. Find the total area of the pool along with its lounging area.

Solution:

To begin with let us make a picture of the situation described in the question.
 
Pre Algebra Word Problem

From our list of the formulas we know that the area of a rectangle is given by length times width. So,

Area of rectangle = $length \times width$

Area of pool = $100 \times 40$ = $4000 sq ft$

Now let us find the area of the triangular lounge area. From our list of formulas above we know that:

Area of a triangle = $\frac{1}{2}$ $\times\ base \times altitude$

So for this triangle, the base is $40$ and altitude is given to be $30$. Thus,

Area of triangle = $\frac{1}{2}$ $\times\ 40 \times 30$ = $600 sq ft$

Thus the total area of the pool and the lounge would be:

Total area = Area of rectangle+area of triangle

Total area = $4000 + 600$

Total area = $4600 sq ft\ \leftarrow Answer!$

Therefore the total area is $4600 ft^2$.
Example 5: 

Vicky flies in an aircraft traveling at a speed of $550 mph$ from London to New York. The journey takes about $7$ hours. Then from New York he takes a train to Boston which reaches Boston in about $6$ hours when the train is traveling at a speed of about $200$ miles per hour. From Boston he boards a bus to Portland, ME that travels at $100$ miles per hour and takes $3$ hours to reach Portland. Find the total distance covered by Vicky from London to Portland, ME.

Solution:

From our list of formulas we know that:

Distance = speed $\times$ time

Therefore, 

Distance from London to NewYork = $550 \times 7$ = $3850$ miles

Distance from New York to Boston = $200 \times 6$ = $1200$ miles

Distance from Boston to Portland, ME = $100 \times 3$ = $300$ miles

Therefore total distance = $3850 + 1200 + 300$

Total distance = $5350$ miles $\leftarrow Answer!$

Thus the total distance travelled by Vicky from London to Portland, ME is $5350$ miles.
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