The Ratio Word Problems Tutorial Series

This is a series of tutorials regarding ratio word problems. Ratio is defined as the relationship between two numbers where the second number is how many times the first number is contained. In this series of problems, we will learn about the different types of ratio word problems.

How to Solve Word Problems Involving Ratio Part 1 details the intuitive meaning of ratio.  It uses arithmetic calculations in order to explain its meaning. After the explanation, the algebraic solution to the problem is also discussed.

How to Solve Word Problems Involving Ratio Part 2 is a continuation of the first part. In this part, the ratio of three quantities are described. Algebraic methods is used as a solution to solve the problem.

How to Solve Word Problems Involving Ratio Part 3 in this post, the ratio of two quantities are given. Then, both quantities are increased resulting to another ratio.

How to Solve Word Problems Involving Ratio Part 4 involves the difference of two numbers whose ratio is given.

If you have more math word problems involving ratio that are different from the ones mention above, feel free to comment below and let us see if we can solve them.

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 Word Problems Involving Ratio Part 3

In the previous two posts, we have learned how to solve word problems involving ratio with two and three quantities. In posts, we are going to learn how to solve a slightly different problem where both numbers are increased.

Problem

The ratio of two numbers is 3:5 and their sum is 48. What must be added to both numbers so that the ratio becomes 3:4?

Solution and Explanation

First, let us solve the first sentence. We need to find the two numbers whose ratio is 3:5 and whose sum is 48.

Now, let x be the number of sets of 3 and 5.

3x + 5x = 48
8x = 48
x = 6

Now, this means that the numbers are 3(6) = 18 and 5(6) = 30.

Now if the same number is added to both numbers, then the ratio becomes 3:4.

Recall that in the previous posts, we have discussed that ratio can also be represented by fraction. So, we can represent 18:30 as \frac{18}{30}. Now, if we add the same number to both numbers (the numerator and the denominator), we get \frac{3}{4}. If we let that number y, then

\dfrac{18 + y}{30 + y} = \dfrac{3}{4}.

Cross multiplying, we have

4(18 + y) = 3(30 + y).

By the distributive property,

72 + 4y = 90 + 3y

4y - 3y = 90 - 72

y = 18.

So, we add 18 to both the numerator and denominator of \frac{18}{30}. That is,

\dfrac{18 + 18}{30 + 18} = \dfrac{36}{48}.

Now, to check, is \dfrac{36}{48} = \frac{3}{4}? Yes, it is. Divide both the numerator and the denominator by 12 to reduce the fraction to lowest terms.

How to Solve Word Problems Involving Ratio Part 2

This is the second part of a series of post on Solving Ratio Problems. In the first part, we have learned how to solve intuitively and algebraically problems involving ratio of two quantities. In this post, we are going to learn how to solve a ratio problem involving 3 quantities.

Problem 2

The ratio of the red, green, and blue balls in a box is 2:3:1. If there are 36 balls in the box, how many green balls are there?

Solution and Explanation

From the previous, post we have already learned the algebraic solutions of problems like the one shown above. So, we can have the following:

Let x be the number of grous of balls per color.

 

2x + 3x + x = 36

6x = 36

x = 6

So, there are 6 groups. Now, since we are looking for the number of green balls, we multiply x by 3.

So, there are 6 groups (3 green balls per group) = 18 green balls.

Check:

From above, x = 6(1) is the number of blue balls. The expression 2x represent the number of red balls, so we have 2x = 2(6) = 12 balls. Therefore, we have 12 red balls, 18 green balls, and 6 blue balls.

 

We can check by adding them: 12 + 18 + 6 = 36.

 

This satisfies the condition above that there are 36 balls in all. Therefore, we are correct.

 

How to Solve Word Problems Involving Ratio Part 1

In a dance school, 18 girls and 8 boys are enrolled. We can say that the ratio of girls to boys is 18:8 (read as 18 is to 8). Ratio can also be expressed as fraction so we can say that the ratio is 18/8. Since we can reduce fractions to lowest terms, we can also say that the ratio is 9/4 or 9:4. So, ratio can be a relationship between two quantities. It can also be ratio between two numbers like 4:3 which is the ratio of the width and height of a television screen.

Problem 1

The ratio of boys and girls in a dance club is 4:5. The total number of students is 63. How many girls and boys are there in the club?

Solution and Explanation

The ratio of boys is 4:5 means that for every 4 boys, there are 5 girls. That means that if there are 2 groups of 4 boys, there are also 2 groups of 5 girls. So by calculating them and adding, we have

4 + 5 = 9
4(2) +5(2) =18
4(3) +5(3) =27
4(4) +5(4) = 36
4(5) +5(5) = 45
4(6) +5(6) =54
4(7) +5(7) =63

As we can see, we are looking for the number of groups of 4 and, and the answer is 7 groups of each. So there are 4(7) = 28 boys and 5(7) = 35 girls.

As you can observe, the number of groups of 4 is the same as the number of groups of 5. Therefore, the question above is equivalent to finding the number of groups (of 4 and 5), whose total number of persons add up to 63.

Algebraically, if we let x be the number of groups of 4, then it is also the number of groups of 5. So, we can make the following equation.

4 x number of groups + 5 x number of groups of 5 = 63

Or

4x + 5x = 63.

Simplifying, we have

9x = 63
x = 7.

So there are 4(7) = 28 boys and 5(7) = 35 girls. As we can see, we confirmed the answer above using algebraic methods.

How to Solve Investment Word Problems in Algebra

Investment word problems in Algebra is one of the types of problems that usually come out in the Civil Service Exam. In solving investment word problems, you should know the basic terms used. Some of these terms are principal (P) or the money invested, the rate (R) or the percent of interest, the interest (I) or the return of investment (profit), and the time or how long the money is invested. Investment is the product of the principal, the rate, and the time, and therefore, we have the formula

I = PRT.

This tutorial series discusses the different types of problems in investment and discussed the method and strategies used in solving them.

How to Solve Investment Problems Part 1 discusses the common terminology used in investment problems. It also discusses an investment problem where the principal is invested at two different interest rates.

How to Solve Investment Problems Part 2 is a discussion of another investment problem just like in part 1. In the problem, the principal is invested at two different interest rates and the interest in one investment is larger than the other.

How to Solve Investment Problems Part 3 is very similar to part 2, only that the smaller interest amount is described.

How to Solve Investment Problems Part 4 discusses an investment problem with a given interest in one investment and an unknown amount of investment at another rate to satisfy a percentage of interest for the entire investment.

How to Solve Investment Problems Part 4

This is the fourth part of the Solving Investment Problems Series. In this part, we discuss a problem which is very similar to the third part. We discuss an investment at two different interest rates.

Problem

Mr. Garett invested a part of $20 000 at a bank at 4% yearly interest. How much does he have to invest at another bank at a 8% yearly interest so that the total interest of the money is 7%.

Solution and Explanation

Let x be the money invested at 8%

(1) We know that the interest of 20,000 invested at 4% yearly interest is

20,000(0.04)

(2) We also know that the interest of the money invested at 8% is

(0.08)(x)

(3) The interest of total amount of money invested is 7%. So,

(20,000 + x)(0.07)

Now, the interest in (1) added to the interest in (2) is equal to the interest in (3). Therefore,

20,000(0.04) + (0.08)(x) = (20,000 + x)(0.07)

Simplifying, we have

800 + 0.08x = 1400 + 0.07x

To eliminate the decimal point, we multiply both sides by 100. That is

80000 + 8x = 140000 + 7x
8x – 7x = 140000 – 80000
x = 60000

This means that he has to invest $60,000 at 8% interest in order for the total to be 7% of the entire investment.

Check:
$20,000 x 0.04 = $800
$60,000 x 0.08 = 4800

Adding the two interest, we have $5600. We check if this is really 7% of the total investment.
Our total investment is $80,000.

Now, $80,000 x 0.07 = $5600.

How to Solve Investment Problems Part 3

This is the third part of the Solving Investment Problems Series. In this part, we discuss solve invest problem which is very similar to the second part. We discuss an investment at two different interests.

Problem

A government employee invested a part of Php60000 in bonds at 6% yearly interest and the remaining part in stocks at a 5% yearly interest. The annual interest in stocks is Php850 less than the annual interest in bonds. How much was invested at each rate?

Solution and Explanation

Let x = amount invested on bonds (6% yearly interest)
60,000 – x = amount invested on stocks (5% yearly interest).

The interest in bonds is 6% per year times the amount invested or

(0.06)(x)

when the percentage is converted to decimals.

In addition, the interest in stocks is 5% per year times the amount invested or

(0.05)(60000 – x)

when the percentage is converted to decimals.

Now, the interest in stocks is 850 less than the interest in bonds which means that if we subtract 850 from the interest in bonds they will be equal. That is

interest in bonds – 850 = interest in stocks.

Substituting the expressions of each, we have

(0.06)(x) – 850 = (0.05)(60000 – x).

Simplifying, we have

0.06x – 850 = 3000 – 0.05x.

To eliminate the decimal numbers, we multiply everything by 100. The equation becomes

6x – 85000 = 300000 – 5x

We simplify by adding 85000 and 5x to both sides. The equation becomes.

11x = 385000.

Dividing both sides by 11, we have
x = 35000.

So, 35000 was invested in bonds and 60000-35000 = 25000 was invested in stocks.

Check:

35000 × 0.06 = 2100
25000 × 0.05 = 1250

Indeed, the amount invested in bonds is 2100 – 1200 = 850 less than the interest in stocks.

Related Posts Plugin for WordPress, Blogger...
1 2 3 4 5 6 10