8th grade math can throw a lot of surprises at students, not the least of which is the introduction of algebra into the curriculum. While the lucky few who actually enjoy math may find this a welcome addition, the majority will not. 8th grade math is the gateway to the rest of high school math so if students fall back here, they could be struggling for the next four years. There is no reason to let a few numbers and equations stand in your way to learning math and scoring good grades. We don’t live in the Internet age for nothing. Like it does for many of our other needs, the internet gives you viable options to learn math easily and thoroughly. Learn 8th grade math using online tools which are at your disposal any time of the day. Use 8th grade math problems online and watch your grades take an upward turn.

## 8th Grade Math Word Problems

### Solved Examples

**Question 1:**If the scale on a road map is 1 inch is 200 miles, how many inches would represent 1600 miles ?

**Solution:**

Given,

1 inch = 200 miles

To determine how many inches on the map represent 1600 miles.

Let x inches would represent 1600 miles.

The problem can be expressed as two ratios set equal to each other.

Solve for x,

$\frac{1}{200} = \frac{x}{1600}$

=> 1 * 1600 = 200 * x

=> 1600 = 200x

=> x = $\frac{1600}{200}$

=> 8

Hence 8 inches would represent 1600 miles.

1 inch = 200 miles

To determine how many inches on the map represent 1600 miles.

Let x inches would represent 1600 miles.

The problem can be expressed as two ratios set equal to each other.

Solve for x,

$\frac{1}{200} = \frac{x}{1600}$

=> 1 * 1600 = 200 * x

=> 1600 = 200x

=> x = $\frac{1600}{200}$

=> 8

Hence 8 inches would represent 1600 miles.

**Question 2:**The difference between the square of two consecutive numbers is 57. What are the numbers.

**Solution:**

First number = x

Second number = x + 1

The problems states:

(x + 1)

=> x

=> 1 + 2x = 57

=> 2x = 57 - 1 = 56

=> x = $\frac{56}{2}$

=> x = 28

=> First number = 28

Second number = 28 + 1 = 29.

Second number = x + 1

The problems states:

(x + 1)

^{2}- x^{2}= 57=> x

^{2}+1 + 2x - x^{2}= 57=> 1 + 2x = 57

=> 2x = 57 - 1 = 56

=> x = $\frac{56}{2}$

=> x = 28

=> First number = 28

Second number = 28 + 1 = 29.

**Question 3:**Two year ago a man was six times as old as his son. In 18 years he will be twice as old as his son. Determine their present ages.

**Solution:**

Let Father's present age = x

and his son's present age = y

Two year ago:

Father's age = x - 2

and his son's age = y - 2

Man was six times as old as his son.

=> x - 2 = 6( y - 2)

=> x - 2 = 6y - 12

=> x - 6y = -10

=> x = - 10 + 6y .....................................(1)

In 18 years:

Father's age = x + 18

and his son's age = y + 18

Father will be twice as old as his son:

=> x + 18 = 2(y + 18)

=> x + 18 = 2y + 36

=> x - 2y = 36 - 18

=> x - 2y = 18 .........................................(2)

Put equation (1) in equation (2)

=> - 10 + 6y - 2y = 18

=> - 10 + 4y = 18

=> 4y = 28

=> y = 7, put in (1)

=> x = - 10 + 6 * 7

=> x = - 10 + 42 = 32

Hence, the age of father = 32 years

and age of his son = 7 years.

and his son's present age = y

Two year ago:

Father's age = x - 2

and his son's age = y - 2

Man was six times as old as his son.

=> x - 2 = 6( y - 2)

=> x - 2 = 6y - 12

=> x - 6y = -10

=> x = - 10 + 6y .....................................(1)

In 18 years:

Father's age = x + 18

and his son's age = y + 18

Father will be twice as old as his son:

=> x + 18 = 2(y + 18)

=> x + 18 = 2y + 36

=> x - 2y = 36 - 18

=> x - 2y = 18 .........................................(2)

Put equation (1) in equation (2)

=> - 10 + 6y - 2y = 18

=> - 10 + 4y = 18

=> 4y = 28

=> y = 7, put in (1)

=> x = - 10 + 6 * 7

=> x = - 10 + 42 = 32

Hence, the age of father = 32 years

and age of his son = 7 years.