Browse Tag: math word problem

How to Solve Problems by Working Backwards Part 2

In the previous post, we have learned how to solve number problems by working backward.  In this post, we discuss age problems. We already had 2 examples in the previous post in this series, so we start with the third example.

Example 3

Arvin is 5 years older than Michael.  The sum of their ages is 37.  What are their ages?

Solution

This is very similar to the problems in the previous post in this series.  Arvin is 5 years older than Michael, so if we subtract 5 from Arvin’s age, their ages will be equal.  But if we subtract 5 from Arvin’s age, we also have to subtract 5 from the sum of their ages. That is,

37 – 5 = 32.

Now, their ages are equal, so we can divide the sum by 2.  That is 32 ÷ 2 = 16.  This means that the younger person is 16 since we subtracted 5 from Arvin’s age. Therefore, Michael is 16 and Arvin is 16 + 5 = 21.

Check

16 + 21 = 37.

Example 4

Mia is 3 years older than Pia.  In 4 years, the sum of their ages is 35. What are their ages?

Solution

There are two of them and we added 4 to both ages, so subtracting 8 from 35 (the sum) will determine the sum of their present age.  That is 35 – 8 = 27 is the sum of their present age.

Next, Mia is 3 years older than Pia, so if we subtract 3 from the sum of their present age, their ages will be equal. So, 27 – 3 = 24.

We can now divide the sum of their ages by 2.  That is 24/2 = 12.

This means that 12 is the age of the younger person because we subtracted 3 from the age of the older person.

So, Pia is 12 and Mia is 15.

Check 12 + 15 = 22.

In the next post, we will discuss more problems that can be solved by working backward.

How to Solve Quadratic Word Problems Part 1

In the previous posts, we have learned how to solve quadratic equations by getting the extracting the square root, by factoring, and by quadratic formula. We continue this series by learning how to solve math word problems using quadratic equations. Most of the time, we need to rewrite the equation to the general form which is ax^2 + bx + c = 0.

Problem 1

The product of two consecutive positive even numbers is 48. What are the numbers?

Solution and Explanation

This problem can be solved mentally and by simple guess and check; however, we will solve it algebraically in order to illustrate the method of using quadratic equations.

Let
x = smaller number
x + 2 = larger number

From the given above, we can form the following equation.

Smaller number times larger number = 48

x(x + 2) = 48

By the distributive property, this results to

x^2 + 2x = 48

Now, we need to make this equation in general form ax^2 + bx + c = 0 so we can factor easily. To do this, we subtract 48 from both sides resulting to

x^2 + 2x - 48 = 0

By factoring, we need two numbers whose sum is 2 and product is -48 where the absolute value of the larger number is greater than that of the smaller. With this restriction in mind, we have the following pairs of factors whose product is -48.

{48, -1}, {24, -2}, {12, -4}, {8,-6}

From these pairs, 8 and -6 has a sum of 2. Therefore, the factors are

(x + 8)(x - 6) = 0

Equating to 0, we have

x + 8 = 0, x = -8
x - 6 = 0, x = 6

Since we are looking for positive integers, we will take x = 6 and x + 2 = 6 + 2 = 8.

Therefore, the two consecutive numbers are 6 and 8.

Check: The numbers 6 and 8 are consecutive numbers and their product is 48. Therefore, we are correct.

Area of Rectangle: Worked Examples

In the previous post, we have learned some problems on how to solve area of squares. In this post, we are going to solve problems involving area of rectangles. The area of rectangle is one of the most common questions in the Geometry part of the Civil Service Examination. In this post, we are going to learn how to solve problems involving area of rectangles. The formula for the Area (A) can be calculated by multiplying the length (l) and the width (w). That is,

A = l \times w.

Problem 1

Find the area of a rectangle with length 8cm and width 5 cm.

Solution

A = l \times w

A = 8 \times 5

A = 40

So, the area of the rectangle is 40 cm^2

Problem 2

The area of a rectangle is 12 cm^2 and its width is 3 cm. What is its length?

Solution

A = l \times w

Substituting the given, we have

12 = l \times 3.

To get l, we have to divide both sides by 3.

\dfrac{12}{3} = \dfrac{l \times 3}{3}

4 = l

Therefore, the length is equal to 3.

Problem 3

The length of a rectangle is twice its width. Its area is 32 square units. What are the dimensions of the rectangle?

Solution

Let x = width of the rectangle

Since the length is twice, it is 2x.

So,

w = x
l = 2x

A = l \times w

A = 2w \times w

32 = 2w^2

Dividing both sides by 2, we have

16 = w^2.

Getting the square root of both sides, we have

4 = w

So, l = 2w = 2(4) = 8.

So, the dimension of the rectangle is 4 by 8.

Problem 4

The length of a rectangle is two more than its width. Its area is 48 square units. What are the dimensions of the rectangle?

Solution

Let length be equals x and width be equals x + 2. That is

w = x
l = x + 2

A = lw
48 = x(x + 2)
48 = x^2 + 2x

Subtracting 48 from both sides, we have
x^2 + 2x - 48 = 0.

Factoring, we have

(x + 8)(x - 6) = 0

x = -8, x = 6

So, the width is 6 units and the length is 6 + 2 = 8 units.

Check:
Is the length two more than 6?
Is 6 \times 8 equals 48?

Introduction to Venn Diagrams

According to some examinees, there are several items in the October 2014 Civil Service Exam that requires the use of Venn diagrams, so I will discuss in details how to use Venn diagrams in a series of posts.  In this post, I am going to introduce its concept and its use as layman as possible. I will limit the discussion to the preparation of solving word problems involving Venn diagrams.

Venn diagrams are used to represent logical relations between sets. In word problems involving diagrams, elements of sets are usually people who choose or prefer a particular thing (e.g. color, food, hobbies). For instance, we have two available desserts, biscuits and cookies, and then Abby ate a biscuit and Bella ate a cake. Chubby, being extremely hungry, ate a biscuit and a cake. So, the situation can be represented as follows using a Venn diagram.  Continue Reading