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## Solve My Math Word Problem

### Solved Examples

**Question 1:**42 oranges are to be distributed among some boys and girls. If each boy is given 3 oranges, then each girl gets 6 oranges and if each boy gets 5 oranges, then each girl gets 3 oranges. Find the number of boys and girls.

**Solution:**

Let the number of boys be x and the number of girls be y.

Number of oranges = 42

The problem states:

Each boy gets 3 oranges + each girl gets 6 oranges = 42

=> 3x + 6y = 42 ................................(1)

Each boy gets 5 oranges + each girl gets 3 oranges = 50

=> 5x + 3y = 42 ..................................(2)

Solve (1) and (2)

Multiply equation (2) by 2 and subtract from equation (1)

=> 3x + 6y - 2(5x + 3y) = 42 - 2 * 42

=> 3x + 6y - 10x - 6y = 42 - 84

=> - 7x = - 42

Divide each side by - 7

=> y = 6

Put y = 6 in (1)

=> 3x + 6 * 6 = 42

=> 3x + 36 = 42

=> 3x = 42 - 36

=> 3x = 6

Divide each side by 3

=> x = 2

Hence, the number of boys is 2 and the number of girls is 6.

Number of oranges = 42

Step 1:Step 1:

The problem states:

Each boy gets 3 oranges + each girl gets 6 oranges = 42

=> 3x + 6y = 42 ................................(1)

Each boy gets 5 oranges + each girl gets 3 oranges = 50

=> 5x + 3y = 42 ..................................(2)

**Step 2**:Solve (1) and (2)

Multiply equation (2) by 2 and subtract from equation (1)

=> 3x + 6y - 2(5x + 3y) = 42 - 2 * 42

=> 3x + 6y - 10x - 6y = 42 - 84

=> - 7x = - 42

Divide each side by - 7

=> y = 6

Step 3:Step 3:

Put y = 6 in (1)

=> 3x + 6 * 6 = 42

=> 3x + 36 = 42

=> 3x = 42 - 36

=> 3x = 6

Divide each side by 3

=> x = 2

Hence, the number of boys is 2 and the number of girls is 6.

**Question 2:**Age of two boys are in the ratio 5:7. Eight years ago their ages were in the ratio 7:13. Find their present ages.

**Solution:**

Let the present ages of the two boys be x and y years

Step 1:

Age of two boys are in the ratio 5:7

=> $\frac{x}{y} = \frac{5}{7}$

=> 7x = 5y

or 7x - 5y = 0 .................................(1)

Step 2:

Ages of the two boys be x + 8 and y + 8 years

Age of two boys are in the ratio 7:13

=> $\frac{x - 8}{y - 8} = \frac{7}{13}$

=> 13(x - 8) = 7(y - 8)

=> 13x - 104 = 7y - 56

=> 13x - 7y = - 56 + 104

=> 13x - 7y = 48 ....................................(2)

Step 3:

Solve (1) and (2)

Multiply (1) by 7 and (2) by 5

=> 7( 7x - 5y = 0 )

=> 49x - 35y = 0 .................................... (3)

and 5(13x - 7y = 48)

=> 65x - 35y = 240 .....................................(4)

Subtract (3) from (4)

=> 65x - 35y - (49x - 35y) = 240 - 0

=> 65x - 35y - 49x + 35y = 240

=> 16x = 240

=> x = 15

Step 4:

Put x = 15 in(3)

=> 49 * 15 - 35y = 0

=> 735 - 35y = 0

=> 35y = 735

=> y = 21

Hence, the ages of two boys are, 15 years and 21 years.

Step 1:

Age of two boys are in the ratio 5:7

=> $\frac{x}{y} = \frac{5}{7}$

=> 7x = 5y

or 7x - 5y = 0 .................................(1)

Step 2:

Eight years ago :Eight years ago :

Ages of the two boys be x + 8 and y + 8 years

Age of two boys are in the ratio 7:13

=> $\frac{x - 8}{y - 8} = \frac{7}{13}$

=> 13(x - 8) = 7(y - 8)

=> 13x - 104 = 7y - 56

=> 13x - 7y = - 56 + 104

=> 13x - 7y = 48 ....................................(2)

Step 3:

Solve (1) and (2)

Multiply (1) by 7 and (2) by 5

=> 7( 7x - 5y = 0 )

=> 49x - 35y = 0 .................................... (3)

and 5(13x - 7y = 48)

=> 65x - 35y = 240 .....................................(4)

Subtract (3) from (4)

=> 65x - 35y - (49x - 35y) = 240 - 0

=> 65x - 35y - 49x + 35y = 240

=> 16x = 240

=> x = 15

Step 4:

Put x = 15 in(3)

=> 49 * 15 - 35y = 0

=> 735 - 35y = 0

=> 35y = 735

=> y = 21

Hence, the ages of two boys are, 15 years and 21 years.