We help you to reach the solutions of different types of word  problems step by step. We help you to understand how to deal with different types of  word problems in algebra.
The following are the steps to reach a solution of any word problem

  • Read the problem twice or thrice.
  • Identify the variables involved
  • Identify the equation in the problem
  • Solve the equations using various mathematical methods.
 

Math Word Problems and Answers

Math word problems with answers will help you to get help with all types of word problem.  You can enter the word problems based on any topic in math. Most of the math word problems can be solved with the help of algebra.
Here are a few examples of word problems with answers
Example 1: -
Three consecutive even integers are such that the square of the third is 100 more than the square of the second.  Find the integers.
Solution: -
Let the even integers be x - 2, x, x + 2
Given that the square of the third is 100 more than the square of the second.
That is (x + 2)2 = x2 + 100
x2 + 4x + 4 = x2 + 100
cancelling x2 on both sides, we get
4x + 4 = 100
Subtracting 4 on both sides, we get
4x = 96
Dividing both sides by 4, we get
x = 96/4 = 24
When x = 24, the integers are 22, 24 and 26
Therefore the required integers are 22, 24 and 26.

Example 2: -

Jack is 15 years older than Johnny. The sum of their ages is 85. Find their ages.
Solution: -
Let x be the age of Jack and y be the age of Johnny.
Given Jack is 15 years older than Johnny.
That is x = y + 15...(1)
Also given that sum of their ages is 85.
That is x + y = 85...(2)
From (1), we get
x - y = 15...(3)
Adding (2) and (3), we get
2x = 100
Dividing both sides by 2
x = 100/2 = 50.
Substituting in (2), we get
50 + y = 85
y = 85 - 50 = 35
Jack's age is 50 years
Johnny's age is 35 years

Example 3: -
The product of two numbers is 27 and their sum is 12. Find the two numbers.
Solution: -

Let the two numbers be x and y.
Given that the product of two numbers is 27.  That is xy = 27...(1)
Now given that the sum of the numbers is 12.  That is x + y =12...(2)
Now lets solve these equations to find the two numbers.
From equation (2), we get y = 12 - x
Substituting this in equation (1), we get
x(12 - x) = 27
⇒ 12x - x2 = 27
⇒ x2 -12x + 27 = 0
Now we split the middle term
⇒x2 - 9x - 3x + 27 = 0
Now we take x common from first two terms and -3 common from last two terms
⇒x(x - 9) - 3(x - 9) = 0
Taking x - 9 common we get
(x - 9)(x - 3) = 0
Equating each factor to 0, we get
x = 9 and x = 3
When x = 9, y = 12 - 9 = 3
When x = 3, y = 12 - 3 = 9
So the required numbers are 9 and 3.
Example 4 : -
Mary sells 10 pens and 4 books for $ 200. Again she sells 9 pens and a book for $ 87.  Find the cost of book and cost of pen.
 Solution: -
 Let x be the cost of pen and y be the cost of book she sold.Given that she sells 10 pens and 4 books for $ 200
That is 10x + 4y = 200…(1)
Also she sells 9 pens and a book for $ 87.
That is 9x + y = 87 ...(2)
Solving the two equations we get the required answer.
For this first we multiply the second equation by 4, we get
36x + 4y = 348…(3)
Subtracting equation (2) from (3) we get
26y = 148
y = 148/26 ≈ 6
Substituting in equation (2), we get
9x  + 6 = 87
9x = 87 – 6
9x = 81
Dividing by 9, we get
x = 81/9 = 9
Therefore Cost of the pen she sold =$ 9
cost of the book she sold  = $6.