Category: Math Word Problems

Week 3 Review Answers and Solutions

Math Word Problems, Practice Tests, Review Series

These are the answers and solutions to the Week 3 Practice Exercises and Problems.

Solutions to Practice Exercise 1

a.) 2 1/5 + 3 2/5

We can add the whole numbers first, 2 + 1 = 3. Then, add the fractions: 1/5 + 2/5 = 3/5.
We then combine the whole number and the fraction, so the answer is 3 3/5.

b.) 8 1/4 + 2 3/4

We can add the whole numbers first, 8 + 2 = 10. Then, add the fractions: 1/4 + 3/4 = 4/4 = 1
We then add 10 + 1 = 11.

c.) 5 + 2 1/4

We can just add the whole numbers: 5 + 2 = 7. Then, we append the fraction. So the correct answer is 7 ¼.

d.) 5 1/2 + 1/5

We just add the fractions and combine the sum with the whole number 5 later. To add dissimilar fractions, we get the LCM of the denominators. The LCM of 2 and 5 is 10.

The equivalent fraction of ½ = 5/10.
The equivalent fraction of 1/5 = 2/10.
5/10 + 2/10 = 7/10

We now append 5. So, the correct answer is 5 7/10.

e.) 3 1/3 + 4 1/4 + 5 1/5

Just like in (d), we can separately add the whole numbers and then add the fractions.

Whole numbers: 3 + 4 + 5 = 12

To add dissimilar fractions, we get the LCM of the denominators. The LCM of 3, 4, and 5 is 60.

The equivalent fraction of 1/3 = 20/60.
The equivalent fraction of 1/4 = 15/60.
The equivalent fraction of 1/5 = 12/60.

20/60 + 15/60 +12/60 = 47/60

Appending the whole number, the final answer is 12 47/60.

Solutions to Practice Exercises 2

a.) 4 6/7 – 3/7

Solution

We just subtract the fractions and append the whole number. 6/7 – 3/7 = 3/7. So, the final answer is 4 3/7.

b.) 8 – 3/4

Solution

One strategy here is to borrow 1 from 8 and make the fraction 4/4. This means that 8 becomes 7 4/4.
So, 7 4/4 – ¾ = 7 ¼.

c.) 12 – 5 2/9

Solution

Our minuend is a whole number, so we can make a fraction out of it. To do this, we can borrow 1 from 12 and make the fraction 9/9. This means that 12 becomes 11 9/9.
So, 11 9/9 – 5 2/9 = 6 7/9.

d.) 7 3/10 – 7/10

We cannot subtract 3/10 – 7/10, so we borrow 1 from 7 and make the fraction 6 10/10. But since we already have 3/10, we add it to 6 10/10 making it 6 13/10.
So, 6 13/10 – 7/10 = 6 6/10 = 6 3/5.

e.) 6 1/5 – 3/4

Another strategy in subtracting fractions is to convert mixed fractions to improper fractions. The improper fraction equivalent of 6 1/5 is 31/5. Then, we find the LCM of 5 and 4 which is 20.

Now, the equivalent fraction of 31/5 is 124/20.
The equivalent fraction of 3/4 = 15/20.
124/20 – 15/20 = 109/20

Converting 109/20 to mixed fraction, we have 5 9/20.

f.) 9 3/8 – 4 5/7

9 3/8 – 4 5/7 = 8 3/8+8/8 – 4 5/7 = 8 11/8 – 4 5/7

The LCM of 8 and 7 is 56, so

4 77-40/56 = 4 37/56.

Solutions to Practice Problems

1.) 1 3/5 + 4/5 = 1 7/5 = 2 2/5

2.) Converting the improper fractions, we have
2 5/8= 21/8
1 5/6 = 11/6.

This means that we need to perform.
21/8-11/6.

Since they are dissimilar fractions, we get their LCM which is 48.
(126-88)/48= 38/48 reduce lowest term by dividing the numerator and denominator by 2, we get 19/24

3.) 2 5/6 – 17/8 = 17/6 – 17/8

LCD: 24
68/24 – 51/24 = 17/24

4.) 3/8 + 1/4
LCD: 8
3/8 + 2/8 = 5/8

Whole pizza – 5/8
8/8 – 5/8
= 3/8

5.) d = 3 4/15 + 5/8
d= 49/15 + 5/8
d= (49(8)+5(15))/120
d= (392+75)/120
d= 467/120
d=3 107/120

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Week 3 Review: Practice Exercises and Problems

Math Word Problems, Mathematics

In the last post, we learned about addition and subtraction of mixed fractions.  To test your knowledge about these topics, answer the exercises and word problems below. The answer answers and solutions will be posted soon.

Practice Exercises 1

a.) 2 1/5 + 3 2/5
b.) 8 1/4 + 2 3/4
c.) 5 + 2 1/4
d.) 5 1/2 + 1/5
e.) 3 1/3 + 4 1/4 + 5 1/5

Practice Exercises 2

a.) 4 6/7 – 3/7
b.) 8 – 3/4
c.) 12 – 5 2/9
d.) 7 3/10 – 7/10
e.) 6 1/5 – 3/4
f.) 9 3/8 – 4 5/7

Practice Problems

1.) Leo’s family drank 1 3/5 liters of juice yesterday morning and 4/5 liters of juice yesterday afternoon. How much juice did Leo’s family drank in all yesterday?

2.) A train station is between a school and a clinic. The distance between the school and the clinic is 2 5/8 kilometers and the distance between the train station and the clinic is 1 5/6 kilometers. What is the distance between the school and the train station?

3.) A piece of iron rod weighs 2 5/6 kg and another piece weighs 17/8 kilograms. Which is heavier and by how much?

4.) Gina bought a pizza. She gave 3/8 of it to her kids and 1/4 to her neighbor. What part of the pizza was left?

5.) Jaime’s house is two rides away from school. The jeepney ride is 3 4/15 kilometers and the tricycle ride is 5/8 kilometers. How far is Jaime’s school from his house?

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Solving Ratio and Proportion Problems Part 2

Math Word Problems, Mathematics

In the previous post, we have learned the meaning and notation of ratio. We can write the ratio of 8 girls and 12 boys as 8:12 or as 8/12. However, if we represent this as fraction, we can also reduce the fraction to its lowest terms which is equal to 2/3. Converting to lowest term is dividing the numerator and denominator by the largest possible integer known as the greatest common factor or greatest common divisor. In the example above, the greatest common divisor of 8 and 12 is 4, and 8 divided by 4 is 2, and 12 divided by 4 is 3, so, 8:12 can also be represented as 2:3.

The Meaning of Direct Proportion

Consider the following problem.

A car is traveling at an average speed of 60 kilometers per hour. What is the total distance it traveled after 5 hours?

Solution

We can solve the problem above by simply multiplying 5 hours by 60 kilometers per hour giving us 300 kilometers. We can also answer the problem by simply constructing the table below.

proportion

Notice from the table that if the number of hours is multiplied by 2, then the distance is also multiplied by 2. For example, from 1 hour to 2 hours, the number of hours is multiplied by 2, and the distance is also multiplied by 2, that is 60 × 2 = 120 hours. From 2 hours to 4 hours, the number of hours is also multiplied by 2 and the distance is also multiplied by 2, that is 120 × 2 = 240 hours. If the number of hours is multiplied by 3, the distance is also multiplied by 3. From 1 hour to 3 hours, the number of hours is multiplied by 3 and the distance is also multiplied by 3, that is 60 × 3 = 180 hours.

Suppose we have two quantities and if we multiply one quantity by a number, then the other quantity is also multiplied the same number, then we say that the two quantities are directly proportional. In the example above, time and distance are the two quantities that are directly proportional.

Representing Direct Proportions

We can represent the problem above in ratio. The first ratio is 60 kilometers and 1 hour. The second ratio is 5 hours and an unknown number of kilometers. If we let the unknown number of hours be n, then the ratios are

1 hr :60 km and 5 hrs :n km

Notice that the number of hours is multiplied by 5 (1 hr to 5 hrs), so the distance should also be multiplied by 5. That is, 60 × 5 = 300.

Now that we found the answer to the problem above, let us represent them in ratios as shown.

1:60 and 5:300

Observe from the representation that the product of the outer terms (1 and 300) is equal to the product of the inner terms (60 and 5). The product are both 300. This property is always true in directly proportional quantities: the product of the outside terms (extremes) is equal to the product of the inside terms (means). In the original representation, we had

1:60 = 5:n

Using the relationship between the means and extremes, we can solve for n algebraically. That is,

1 × n = 60 × 5. So, n = 300.

This can also be represented in fraction as 1/60 = 5/n. Cross multiplying, we have n = (60)(5) = 300.

Summary

In a directly proportional relationship, if the ratios are a:b and c:d, then

a: b = c: d

and a × d = b × c.

Practice Problem

Three cubes of sugar is needed to make 1 cups of coffee. How many cubes of sugar is needed to make 20 cups of coffee?

Solution

Let x = number of cubes of sugar needed to make 20 cups of coffee.

3:1 = x:20

The product of the extremes is equal to the product of the means, so solving algebraically, we have

1(x) = (3)(20)
x = 60.

Therefore, we need 60 cubes of sugar for 20 cups of coffee.

In the next post, we are going to have some practice problems on how to solve direct proportion problems.

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Practice Exercises on Subtraction of Integers

Math Word Problems, Mathematics

To answer the exercises below, it is assumed that you have already finished reading addition of integers and subtraction of integers.

In subtraction of integers, we have learned two rules:

(1) a – b = a + (-b)
(2) a – (-b) = a + b

We will use these rules in answering the exersises below.

Exercises

1. 2 – 5
2. 18 – ( – 2)
3. 16 – 7
4. -17 – 3
5. -9 – (-3)
6. 0 – (-11)
7. -18 – (-25)
8. -10 – 9
9. 12 – (-9)
10. -6 – 3

Solutions/Answers

1. 2 – 5

Solution 1: 5 is greater than 2. If you subtract two numbers, if the subtrahend is larger than the minuend, the answer will be negative. So, the answer is -3.

Solution 2: From rule 1, a – b = a + (-b), so 2 + 5 = 2 + (-5) = -3
Answer:

2. 18 – ( – 2)

Solution: From rule 2, a – (-b) = a + b, so 18 + 2 = 20.

Answer: 20

3. 16 – 7

Answer: 9

4. -17 – 3

Solution: From rule 1, -17 – 3 = -17 + (- 3) = -20. Recall that in adding two negative numbers, we just add the numbers and then the answer will be negative.

Answer: -20

5. -9 – (-3)

Solution: From rule 2, -9 – (-3) = -9 + 3 = -6.

6. 0 – (-11)

Solution: From rule 2, 0 – (-11) = 0 + 11 = 11.

7. -18 – (-25)

Solution: From rule 2, a –(-b) = a + b. So, -18 + 25 = 7.

8. -10 – 9

Solution: From rule 1, a – b = a + (-b), so -10 + (- 9) = – 19.

9. 12 – (-9)

Solution: From rule 2, a –(-b) = a + b, so 12 + 9 = 21.

10.- 6 – 3

Solution: From rule 1, a – b = a + (-b) = -6 + -3 = -9.

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Solving Word Problems by Working Backwards: Summary

Math Word Problems, Mathematics

In the previous posts, I have shown to you a series on how to solve word problems by working backward.  Although I recommend that you learn this method, it is also very important to learn algebraic methods and others since some problems are difficult to solve by working backward. Below are the posts in this series.

How to Solve Word Problems by Working Backward Part 1 discusses solving number problems by working backward.  Two examples are given.

How to Solve Word Problems by Working Backward Part 2 discusses solving age problems by working backward. Two sample problems are solved in this post.

How to Solve Word Problems by Working Backward Part 3 discusses more complicated number problems. Two sample problems are discussed in this post.

If you are interested to solve more word problems, please visit the Word Problems page. It contains word problems about number, age, motion, work, discount and many more.

 

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How to Solve Word Problems by Working Backwards Part 3

Math Word Problems

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.

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How to Solve Problems by Working Backwards Part 2

Math Word Problems, Mathematics

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.

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How to Solve Word Problems by Working Backwards Part 1

Math Word Problems, Mathematics

Most of us would always take a pen and solve for x if we see word problems. But did you know that you can solve them by working backward or even mentally? In this post, I am going to teach you some techniques on solving problems by working backward.

Example 1: One number is three more than the other. Their sum is 45. What are the numbers?

Solution

In the given, one number is 3 more than the other. This means that if you subtract 3 from the larger number they will be equal. Note that if we subtract 3 from one of the numbers, then we should also subtract 3 from their sum. Therefore, their sum will be 45 – 3 = 42. Since the numbers are equal, we now divide the sum by 2. That is, 42/2 = 21.

So, the smaller number is 21 and the larger is 21 + 3 = 24.

Check: 21 + 24 = 45

Example 2: One number is 5 less than the other. Their sum is 43. What is the smaller number?

Solution

This is very similar to Example 1. Here, one number is 5 less than the other; so, if we add 5 to the smaller number, they will be equal. If we add 5 to the smaller number, we should also add 5 to their sum. Therefore, their sum will be 43 + 5 = 48. Since the two numbers are equal, we can divide the sum by 2. That is 48/2 = 24. Since we added 5, it means that 24 is the larger number. So, the smaller number is 24 – 5 = 19.

Check: 19 + 24 = 43

In the next post, we are going to discuss more examples.

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