Algebra Word problems

Example of Word Algebra problems are many. The objective of this unit is to give you the skills you need to solve a variety of algebra Word problems

Example # 1:

A football team has lost 5 yards and then won 9. What is the progress of the team?

Solution

For loss, use negative. To get gain, use positive.

Progress =-5 + 9 = 4 yards

Example # 2:

Use the distributive property to solve the problem below:

Maria bought 10 laptops and 5 pens costing $ 2 each.How the Maria did pay?

Solution

2 x (10 + 5) = 2 × 2 + 10 × 10 = 20 + 10 = 30 $

Example # 3:

A client pays $ 50 for a coffee after a $ 20 discount

What is the origin of the coffee manufacturer price?

Solution

Let x be the initial price.

x – 20 = 50

x 20 + 20 = 50 + 20

x + 0 = 70

x = 70

Example # 4:

Half a number more 5 is 11 what is the number?

Solution

Let x is the number. Always replace “is” by an equal sign

x (1/2) + 5 = 11.

(1/2) x + 5-5 = 11-5.

x (1/2) = 6

2 × (1/2) x = 6 × 2

x = 12

Example # 5:

The sum of two consecutive even integers is 24. What are the two numbers?

Solution

Let be the first even integer 2n and let 2n + 2 be the second integer

2n + 2n + 2 = 24

4n + 2 = 24

4n + 2-2 = 24-2.

4N = 22

n = 5.5

Therefore the first even integer is 2n = 2 × 5.5 = 10 and the second is 10 + 2 = 12

(1/2) x + 5-5 = 11-5 x (1/2) = 6

2 × (1/2) x = 6 × 2

x = 12

Below you will find more complicated algebra Word problems

Example # 6:

The ratio of two numbers is 5 to 1. The sum is 18 years. What are the two numbers?

Solution

Let x is the number of the first. Is the second number y

x / y = 5 / 1

x + y = 18

Using x / y = 5 / 1, we get x = 5y after the multiplication of the cross

Replaces x = 5y in x + y = 18, we get 5y + y = 18

6y = 18

y = 3

x = 5y = 5 × 3 = 15

As you can see, 15/3 = 5, this ratio is correct and 3 + 15 = 18, so the sum is correct.

Example # 7: Algebra Word problems can be as complicated as example # 7. Carefully consider!

Peter has six times more dimes as quarters in his piggy bank. It has 21 pieces in his piggy bank totalling $2.55

How many of each type of room has?

Solution

Let x is the number of quarters. 6 X the number of Dimes

A quarter is equal to 25 cents, x quarters equals x × 25 cents, or 25 cents x

Since a dime is equal to 10 cents, 6 dimes x is equal to 6 x × 10 cents and 60 cents x

Since a 1 dollar equals 100 cents, $ 2.55 amounted to 2.55 × 100 = 255 cents

Putting it all together, 25 cents x + 60 cents x = 255 cents

85 cents x = 255 cents

85 cents x / 85 cents = 255 cents / 85 cents

x = 3

x 6 = 6 × 3 = 18

Peter therefore has 3 18 dimes and quarters

Example # 8:

The area of a rectangle is x 2 + 4 x-12. What are the dimensions of the rectangle (length and width)?

Solution

The main idea is to factor x 2 + 4 x – 12

Since -12 = -2 × 6 and -2 + 6 = 4

x-2 + 4 x 12 =-(x + 2) × (x + 6)

Since the length is generally longer, length = x + 6 and width = x +-2

Example # 9: A must know how in solving algebra Word problems

The area of a rectangle is 24 cm2. The width is two less than the length. What is the length and width of the rectangle?

Solution

Let x be length and let x – 2 is width

Area = length × width = × x (x – 2) = 24

× x (x – 2) = 24

x 2 + 2 x = 24

x 2 + – 2 x – 24 = 0.

Since – 24 = 4 × – 6 and 4 + – 6 = – 2, we get:

(x + 4) ×-(x + 6) = 0

This leads to two equations to solve:

x + 4 = 0 and x – 6 = 0.

x + 4 = 0 gives x =-4. Reject this value as a dimension cannot be negative

-x + 6 = 0 gives x = 6

Therefore, length = 6 and width = x – 2 = 6-2 = 4.

Example # 10:

The sum of two numbers is 16 years. The difference is 4. What are the two numbers?

Let x is the number of the first. Ley is be the second number

x + y = 16

x – y = 4

Solution

Let x is the number of the first. Ley is be the second number

x + y = 16

x – y = 4

Solve the system of equations by elimination

Add the left side and right side gives:

x + x + y + -y = 16 + 4.

x 2 = 20

x = 10

Since x + y = 16, 10 + y = 16

10 + y = 16

10 – 10 + y = 16-10.

y = 6

The figures are 10-6

The word Algebra problems that I solved above are typical questions. You will meet them much in algebra. Hope you had fun to solve these word Algebra problems.

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