# 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 is subtracted from three times a number, then the result is . 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, , we want to subtract from three times . In numerical expression, that is . So, if a number is , the expression is . Now, this results to as stated above. Intuitively, it is saying that is equal to . That’s our equation!

**Solution**

Let be the number.

is 8 less than 3 times the number.

Now, .

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

**Problem 5**

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

**Scratch Work**

If we separate into two parts, and one part is, for example, , then the other part is which is . This means that if one part of is , then the other part is .

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

**larger number – smaller number = 24**.

The only part left now is to choose which is larger, or . *It won’t really matter*.

**Solution**

Let be the larger and be the smaller number.

Then,

.

Subtracting from both sides results to

.

Dividing both sides by , we get

.

So, the smaller number is and the larger number is .

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

**Problem 6**

The sum of the ages of Abby, Bernice, and Cherry is . 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 . So, Bernice is twice as old or years old. Then, Cherry is four years older than Abby or .

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

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

**Solution**

Let be Abby’s age, be Bernice’s age and be Cherry’s age.

Dividing both sides by , we have .

So, Abby is , Bernice is and Cherry is years old.

Check: Is the sum of their ages ?

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

Correction for “Problem 6”

if your instructions told us that 76 divided by 4

76 / 4 is 19 not 18

A = 19

C = 19 + 4 = 23

B = 19 * 2 = 38

19 + 23 + 38 = 80

Sum of their ages is 80?(wrong answer)

maybe you’ve missed something on making your example,

also i know i missed something.. where did i go wrong?

Thank You.

Thank you. Note that you are dividing 72 by 4 and not 76 by 4 since we already subtracted 4 from both sides. A = 18, b = 36, C = 22. A + B + c = 18 + 36 + 22 = 76. 🙂