# How to Solve Number Word Problems Part 2

This is the second part of the the Solving Number  Word Problems Series. In this part, we will discuss how to solve various number problems.  Note that some of these problems are not really number problems per se, but the strategy in solving them is technically the same. You could say that they are really “number problems in disguise.”

We already had three problems in the first part of this series, so let’s solve the fourth problem.

Problem 4

If $8$ is subtracted from three times a number, then the result is $34$. What is the number?

Scratch work

In the How to Solve Number Word Problems Part 1, I mentioned that sometimes, if it is hard to convert the words in the problem to equations, it is helpful to think of a particular number. For example, in this problem, the phrase is “three subtracted from three times a number.” So, if we choose a number, say for example, $5$, we want to subtract $8$ from three times $5$. In numerical expression, that is $3(5) - 8$. So, if a number is $x$, the expression is $3x - 8$. Now, this $3x - 8$ results to $42$ as stated above. Intuitively, it is saying that $3x - 8$ is equal to $42$. That’s our equation!

Solution

Let $x$ be the number.

$3x - 8$ is 8 less than 3 times the number.

Now, $3x - 8 = 34$.

$3x - 8 + 8 = 34 + 8$

$3x = 42$

$x = 14$

Check: The problem says that if $8$ is subtracted three times the number, the result is $34$. Now, three times $14$ is $42$. Now, if we subtract $8$ from $42$ the result is $34$ and we are correct.

Problem 5

Separate $60$ into two parts such that the one exceeds the other by $24$. What are the numbers?

Scratch Work

If we separate $60$ into two parts, and one part is, for example, $10$, then the other part is $60 - 10$ which is $50$. This means that if one part of $60$ is $x$, then the other part is $60 - x$.

Now it says that the larger number exceeds the smaller number by $24$. This means that

larger number – smaller number = 24.

The only part left now is to choose which is larger, $x$ or $60 - x$. It won’t really matter.

Solution

Let $60 - x$ be the larger and $x$ be the smaller number.

Then,

$(60 - x) - x = 24$

$60 - 2x = 24$.

Subtracting $60$ from both sides results to

$-2x = -36$.

Dividing both sides by $-2$, we get

$x = 18$.

So, the smaller number is $18$ and the larger number is $60 - 18 = 42$.

Check: Is the sum of the two numbers is $60$? Does $42$ exceed $18$ by $24$? If both answers are yes, then we are correct.

Problem 6

The sum of the ages of Abby, Bernice, and Cherry is $76$. Bernice is twice as old as Abby, while Cherry is 4 years older than Abby. What are the ages of the three ladies?

Scratch Work

As I have mentioned above, some problems are really number problems in disguise. This problem is one of them.

From the problem, it is easy to see that the youngest in the group is Abby. Let us say, Abby is $20$. So, Bernice is twice as old or $2(20) = 40$ years old. Then, Cherry is four years older than Abby or $20 + 4 = 24$.

So, from this analysis, if Abby is $x$ years old, then, Bernice is $2x$. Since Cherry is four years older than Abby, then here age is $x + 4$.

In the first sentence, it says that the sum of the ages of the three ladies is $76$. Therefore, we must add their ages ($x$, $2x$ and $x + 4$) and equate it to $76$. That is our equation.

Solution

Let $x$ be Abby’s age, $2x$ be Bernice’s age and $x + 4$ be Cherry’s age.

$x + 2x + (x + 4) = 76$

$4x + 4 = 76$

$4x + 4 - 4 = 76 - 4$

$4x = 72$

Dividing both sides by $4$, we have $x = 18$.

So, Abby is $18$, Bernice is $2(18)$ and Cherry is $18 + 4 = 22$ years old.

Check: Is the sum of their ages $76$?

In the next post in this series, we are going to discuss how to solve problems involving consecutive integers.

### 2 Responses

1. March 19, 2014

[…] introduction to this series, we have learned How to Solve Number Problems Mentally. In Part 1 and Part 2, we have discussed the basic number word problems, and in Part 3, we have learned how to solve word […]

2. August 20, 2015

[…] (3) How to Solve Number Problems Part 2 […]