Solving Math Word Problems in Numbers Using Algebra Part 2

This is the fourth part of the Word Problem Solving on Number Problems and this is the continuation of the Solving Word Problems in Numbers in Algebra.  In this post, we discuss more examples on how to solve number problems. We start with the fourth example.

Example 4

The sum of three consecutive numbers is 78. What are the three numbers?


In the previous post in this series, we have already discussed how to solve problems about two consecutive odd integers. In this example, there are three consecutive integers, not odd and not even. As we can see, 11, 12, and 13 are consecutive integers and we only add 1 each time to get the next number. This means that if x is the smallest number, then x + 1 and x + 2 are the next two integers.

The sum of the three consecutive numbers is 78. So, if we add the numbers, x, x + 1 and x + 2, the sum will be equal to 78. Therefore, we can have the following equation.

x + (x + 1) + (x + 2) = 78.

Solving the equation, we have

3x + 3 = 78

3x = 75

x = 25.

Therefore, the consecutive numbers are 25, 26 and 27. Now, check if their sum is really 78.


Example 5

The length of the second side of a triangle is 2 more than twice the length of the first side. The length of the third side is thrice that of the first side. If the perimeter of the triangle is 62 cm, what are the dimensions of the triangle.


If you can observe, the problem describes the length of the second and the third side. It does not mention the length of the first side.

1st side: ?

2nd side: 2 more than twice the length of the first side

3rd side: thrice the length of the first side

In problems such as this, the length or the number that is not talked about is usually the unknown. So, if we let the first side be equal to x we will have the following representations.

1st side: x

2nd side: 2x + 2

3rd side: 3x

Now, the perimeter  of a triangle is the total length of all the sides, so it means that we have to add the length of the three sides which add up to 62

x + (2x + 2) + 3x = 62

6x + 2 = 62

6x = 60

x = 10

This means that the length of the second side is 2(10) + 2 = 22 and the length of the third side is 3(10) = 30.

Checking, we have 10 + 22 + 30 = 62 which confirms that we have solved the problem correctly.


Example 6 

Separate 81 into two parts such that one part is 5 more than the other. What are the numbers?


In this example, we can use the solution in Example 1 in the previous part of this series, but we can also show the fact that if one number is x, then the other number is 81 - x. For instance, if one number is 10, then we are sure that the other number is 81 - 10.

Now, if we let x be the larger number and 81 - x be the smaller, then if we subtract the two numbers, their difference will be 5. In equation form, we have

x - (81 - x) = 5

Solving the equation,

x - 81 + x = 5

2x = 86

x = 43.

Therefore, the other number is 81 - 43 = 38

As you can see, if we check 43 – 38$, the result is 5

In the next part of this series, we will discuss more problems.

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