Solving Word Problems in Numbers using Algebra Part 3

This is the last part of the Solving Number Problems Series. In this post, we are going to solve number problems in disguise, or numbers problems with different contexts. The previous two parts in this series you may want to read are Solving Word Problems in Numbers using Algebra are Part 1 and Part 2.

Example 7

Jack is twice as old as Rose. Two years from now, the sum of their ages is 40. How old are they now?


This problem is an age problem but it is very similar to number problems.  As stated, there are two points in time: now and 2 years from now.

Now, Jack is twice as old as Rose. That means that if Rose is 15, then Jack is 30. That means that if the age of Jack is 2x, then Rose’s age is x. Therefore, we have the following representation. 


Rose: x

Jack’s Age: 2x


Two years from now, both of them will be 2 years older. This means that we add 2 years to their ages. So, we represent their ages as follows.

Two Years from Now

Rose’ age: x + 2

Jack’s age: 2x+ 2


The Sum of Their Ages (2 years from now)

(x + 2) + (2x + 2) = 40

3x + 4 = 40

x = 12

So, the age of Rose now is 12 and Jack is 24. Two years from now, Rose will be 14 and Jack will be 26. Checking, we have 14 + 26 = 40. This means that our answer is correct.


Example 8

The sum of the three digit number is 15. The one’s digit is thrice the hundred’s digit. The ten’s digit is one more than the one’s digit. What is the number?


Consider the following descriptions.

one’s digit: thrice the hundred’s digit

ten’s digit: one more than the one’s digit

hundred’s digit: ?

In the previous post, we have discussed that the given which has no description is usually the unknown. Therefore, if we represent algebraically,  we have

one’s digit: 3x

ten’s digit: 3x + 1

hundred’s digit: x

Since the sum of the 3-digit number is 15, we can add the three digits and equate to 15. That is,

3x + (3x + 1) + x = 7x + 1 = 15.

Solving, we have x = 2 (hundred’s digit), 7 (ten’s digit) and 6 (one’s digit). So the number is 276. Now, check each condition in the problem to verify if the answer is correct.


Example 9

Anna has twice as much money as Ria. When she gives $70 to Ria, they will have the same amount of money. How much money does Maria have?


If we let x be Ria’s money, then Anna ha 2x.

Initial Money

Ria’s money: x

Anna’s money: 2x

After Anna gave Ria 70 dollars.

Ria’s money: x + 70

Anna’s money: 2x - 70

Now, after giving Ria the money, the two girls have the same amount of money. Therefore, we can equate both expressions. That is,

2x - 70 = x + 70.

2x - x = 70 + 70

x = 140.

So, Ria has $140 and Anna has $280.

This is the end of this series. The next series will be on Solving Word Problems involving age.

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6 thoughts on “Solving Word Problems in Numbers using Algebra Part 3

  1. Fellow math collegues- let me know what you think:
    I do word problems as extension only now for students so inclined. Although interesting to think of the three time frames and express ages there in (past,present , future) I don’t see what purpose they serve. In my attempt to make all lessons pertinent to real life, I have dropped this concept as a requirement.

  2. Hi sir!

    in example 8, i think you have a typo. and shouldn’t the ten’s digit be 3x + 1 since the one’s digit is 3x?

    nice blog very informative. keep it up

  3. @John

    Personally, I don’t really like giving this kind of problem if I were to teach in high school, but I believe that a good problem solver should try any problem he encounters


    Thanks. I have changed it already. Please inform me if you say more errors.

  4. sir you forgot to change the values after the equation “Solving, we have x = 3 (hundred’s digit), 4 (ten’s digit) and 9 (one’s digit)” 🙂

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