## PCSR REVIEW SERIES WEEK 11: Work Problems

This is the 11th week in our review series. After learning about motion problems which discuss distance, rate and time, let’s now learn how to solve work problems. Below are the articles and videos that you can read and watch. Later, we will have practice problems.

ARTICLES

VIDEOS

Enjoy learning!

## Week 8 Review: Practice Exercises and Problems

After learning how to solve number problems, let’s have some practice exercises.

Week 8 Review: Practice Exercises and Problems

1.) One number is 3 more than the other. Their sum is 27. What are the numbers?

2.) One number is 5 less than the other. Their sum is 51. What are the numbers?

3.) One number is 3 times the other number. Their sum is 48. What are the numbers?

4.) One number is 5 times the other number. Their difference is 52. What are the numbers?

5.) The sum of three numbers is 36. The second number is 5 more than the first number and the third number is 8 less than the first number. What are the three numbers?

6.) The sum of three numbers is 98. The second number is twice the first number and the third number twice the second number. What are the three numbers?
7.) One number is two more than thrice the other. Their sum is 26. What are the two numbers?

8.) One number is thrice the other. When 3 is added to the larger and 7 is subtracted from the smaller, their sum becomes 32. What are the two numbers?

9.) The sum of two positive consecutive numbers is 91. What are the two numbers?

10.) The sum of two positive consecutive EVEN integers is 66. What are the two numbers?

11. ) The sum of two positive consecutive ODD integers is 36. What are the two numbers?

12.) The sum of three positive consecutive ODD integers is 81. What are the three integers?

13.) The sum of the smallest and the largest of five positive consecutive integers is 108. What is the third integer?

14.) The average of four positive consecutive EVEN integers is 19. What is the largest integer?

15.) The average of seven positive consecutive integers is 31. What is the smallest integer?

Enjoy solving!

## PCSR REVIEW SERIES WEEK 8: Number Problems

This is what everybody has been waiting for. After learning all the basic math from Week 1 to 7, we are now ready to solve some problems. Let’s start with some Number Word Problems. Read the articles below and watch the videos and later, we are going to have some practice exercises.

ARTICLES

How to Solve Consecutive Number Problems

VIDEOS

How to Solve Number Problems Mentally

How to Solve Number Problems

Consecutive Number Problems

Enjoy learning!

## How to Solve Word Problems by Working Backwards Part 3

In part 1 and part 2 of this series, we have learned how to solve number age problems by working backward. In this post, we are going to learn how to solve backward using inverse operations. Recall that multiplication and division are inverse operations and addition and subtraction are inverse operations.

Example 5

A number is multiplied by 4 and then, 3 is added to the product. The result is 31. What is the number?

Solution

The key phrases in this problem are (1) multiplied by 4 and (2) added to (3) the result is 31. Since we are working backward, we start with 31, and then find the inverse of “added to 3” which is “subtract 3.” So, 31 – 3 = 28.

Next, we find the inverse of “multiplied by 4,” which is “divided by 4.” So, 28/4 = 7.

So, the answer to this problem is 7.

Check: 7(4) + 3 = 31

Example 6

Think of a number. Divide it by 8. Then subtract 4 from the quotient. The result is 5. What is the number?

Solution

The key phrases in this problem are (1) divided by 8 (2) subtract 4 and (3) the result is (3) the result is 5.

We start with the result which is 5 and find the inverse of “subtract 4” which is “add 4.” So, 5 + 4 = 9. Next, we find the inverse of “divide by 8” which is “multiply by 8.” So, 9(8) = 72.

So, the correct answer is 72.

Check: 72/8 – 4 = 9 – 4 = 5.

In the next post, we will discuss more about solving math word problems by working backward.

## How to Solve Quadratic Problems Part 2

In the previous post, we have used quadratic equations to solve a word problem involving consecutive numbers. In this post, we discuss more quadratic problems. This is the second problem in the series.

Problem 2

Miel is 12 years older than Nina. The product of their ages is 540.

Solution

Let x = age of Nina
x + 12 = age of Miel

The product of their ages is 540, so we can multiply the expressions above and equate the product to 540. That is,

x(x + 12) = 540.

Multiplying the expressions, we have $x^2 + 12x = 540$.

Subtracting 540 from both sides, we obtain $x^2 + 12x - 540 = 0$.

We want to find two numbers whose product is -540 and whose sum is 12. Those numbers are -18 and 30.

This means that the factors are

(x – 18)(x + 30) = 0.

Equating each expression to 0, we have

x – 18 = 0, x = 18
x + 30 = 0, x = – 30.

Since we are talking about age, we take the positive answer x = 18.

This means that Nina is 18 years old. Therefore, Miel is 18 + 12 = 30 years old.

## How to Solve Word Problems Involving Ratio Part 4

This is the fourth and the last part of the solving problems involving ratio series. In this post, we are going to solve another ratio word problem.

Problem

The ratio of two numbers 1:3. Their difference is 36. What is the larger number?

Solution and Explanation

Let x be the smaller number and 3x be the larger number.

3x – x = 36
2x = 36
x = 18

So, the smaller number is 18 and the larger number is 3(18) = 54.

Check:

The ratio of 18:54 is 1:3? Yes, 3 times 18 equals 54.
Is their difference 36? Yes, 54 – 18 = 36.

Therefore, we are correct.

## How to Solve Digit Problems Part I

Digit Problems is one of the word problems in Algebra. To be able to solve this problem, you must understand how our number system works. Our number system is called the decimal number system because the numbers in each place value is multiplied by powers of 10 (deci means 10). For instance, the number 284 has digits 2, 8, and 4 but has a value of 200 + 80 + 4. That is, $(100 times 2) + (10 times 8) + (4 times 1) = 284$.

As you can observe, when our number system is expanded, the hundreds digit is multiplied by 100, the tens digit is multiplied by 10, and the units digit (or the ones digit) is multiplied by 1. Then, all those numbers are added. The numbers 100, 10, and 1 are powers of 10: $10^2 = 100$, $10^1 = 10$, and $10^0 = 1$. So, numbers with $h$, $t$, and $u$ as hundreds, tens, units digits respectively has value $100h + 10t + u$.

It is clear that this is also true for higher number of digits such as thousands, ten thousands, hundred thousands, and so on.

Many of the given numbers in this type of problem have their digits reversed. As we can see, if 10t + u is reversed, then it becomes $10u + t$. For instance, $32 = 10(3) + 1(2)$ when reversed is $23 = 10(2)+ 1(3)$. Now, that we have already learned the basics, we proceed to our sample problem.

Worked Example

The tens digit of a number is twice the units digit. If the digits are reversed, the new number is 18 less than the original. What are the numbers?

Solution and Explanation

The tens digit of a number is twice the unit digit. This means that if we let the units digit be $x$, then the tens digit is $2x$. As we have mentioned above, we multiply the tens digit with 10 and the units digit with 1. So, the number is $(10)(2x) + x$.

Now, when the digits are reversed, then x becomes the tens digit and $2x$ becomes the ones digit. So, the value of the number is $(10)(x) + 2x$.

From the problem above, the number with reversed digit is 18 less than the original number. That means, that if we subtract 18 from original number, it will equal the new number. That is, $(10)(2x) + x - 18 = 10(x) + 2x$ $20x + x - 18 = 12x$ $21x - 18 = 12x$ $9x = 18$ $x = 2$ $2x = 4$

So, the number is 42 and the reversed number is 24.

Check: 42 – 24 = 18. 