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!

Week 4 Review Answers and Solutions

Below are the solutions and answers to the Week 4 Practice Problems and Solutions.

Practice Exercises 1

Note: In multiplying fractions, we multiply the numerator by the numerator of the other fraction, and then multiply the denominator by the denominator of the other fraction. For whole numbers, we can put 1 as the denominator. All fractions must be in lowest terms.

A. 1/2 × 1/3 = 1/6
B. 2/3 × 4/5 = 8/15
C. 8/1 × 5/6 = 40/6 = 6 4/6 or 6 2/3
D. 2 5/8 × 3 = 21/8 × 3/1 = 63/8 = 7 7/8
E. 3 1/8 × 4/5 = 25/8 × 4/5 = 100/40 = 2 20/40 or 2 1/2
F. 1 2/3 × 2 3/4 = 5/3 × 11/4 = 55/12 = 4 7/12

Practice Exercises 2

A. When dividing fractions you get the reciprocal of the divisor, and then multiply. In 1/5÷ 3/10, the divisor 3/10 and the reciprocal of 3/10 is 10/3. So, 1/5 × 10/3 = 10/15 or 2/3

B. 1/2 ÷ 3/8 = 1/2 × 8/3 = 8/6 = 1 2/6 or 1 1/3

C. 9 ÷ 3/7 = 9 × 7/3 = 63/3 or 21

D. 2 5/8 ÷ 2

First, we convert 2 5/8 to improper fraction as follows. That is 2\frac{5}{8} = \frac{8 \times 2 + 5}{8} = \frac{21}{8}. Don’t forget that the denominator of the mixed fraction is the same as the denominator of the improper fractions.

Here, the reciprocal of 2 is 1/2. So, 21/8 × 1/2 = 21/16 = 1 5/16

E. 3 1/8 ÷ 3/5 = 25/8 × 5/3 = 125/24 = 5 5/24

F. 2 3/4 ÷ 1 1/8 = 11/4 ÷ 9/8 = 11/4 × 8/9 = 88/36 = 2 16/36 or 2 4/9

Practice Problems

Note: In multiplication and division of fractions, all mixed fractions must be converted to improper fractions (see Practice Problem 2D above).

1.) 2/3 × 1/4 = 2/12 or 1/6

2.) 3/5 × 35/1=105/5 = 21 (women)
2/5 × 35/1 = 70/5 = 14 (men)

3.) 2 3/4 × 7 = (11/4) ×  7 = 77/4 = 19 1/4

4.) A = L × W
A = (35 1/4) × (20 1/2) = (141/4) × (41/2)
= 5781/8= 722 5/8

5.) 2 4/5 × 5/1= 14/5 × 5/1 = 14

6.) 1 1/2 L juice is to be shared equally by 6 friends, so 1 1/2 ÷ 6.

The mixed fraction  1 1/2 is 3/2 in improper form.

Dividing by 6 is the same as multiplying by 1/6, so 3/2 × 1/6 = 3/12

3/12 is not yet in its lowest term. So get its lowest term, divide the numerator and denominator by their GCF which is 3.  So, 3/12 will become 1/4.

The final answer is 1/4 L.

7.) 8 ÷ 2/5 = 8 × 5/2 = 8/1 × 5/2 = 40/2 = 20.

Answer: 20 cakes

8.) Bookshelf length divided by book’s thickness = number of books that will fit in the bookshelf

5 1/4 feet ÷ 1 1/2 inches = ?

Notice that the units are in feet and inches. We cannot proceed until the units are the same, so we need  to convert feet into inches. (1 ft = 12 in) So 5 ft × 12 in per ft = 60 in. We still have 1/4 ft, so 1/4 of 12 which is 3 in. All in all, we have 63 in. Now our equation is

63  ÷ 3/2 = 63/1 × 2/3

= 126/3 = 42.

Answer: 42 books

9.) 5 pumpkin pies are to be shared equally among 12 persons, equals 5/12.

10.) We have 8 3/4 hectares ÷ 4 children

35/4 ÷ 4 =35/4 × 1/4 = 35/16

Converting 35/16 to mixed fraction, we have 2 3/16.

Week 4 Review: Practice Exercises and Problems

In the previous post, you have learned about multiplication and division of fractions. Now, let’s solve some exercises and problems.

Practice Exercises 1

a.) 1/2 × 1/3
b.) 2/3 × 4/5
c.) 8 × 5/6
d.) 2 5/8 × 3
e.) 3 1/8 × 4/5
f.) 1 2/3 × 2 3/4

Practice Exercises 2
a.) 1/5 ÷ 3/10
b.) 1/2 ÷ 3/8
c.) 9 ÷ 3/7
d.) 2 5/8 ÷ 2
e.) 3 1/8 ÷ 3/5
f.) 2 3/4 ÷ 1 1/8

Practice Problems

1.) What is 2/3 of 1/4?

2.) In a dance studio, 3/5 are women and 2/5 are men. If there are 35 persons in the dance studio, how many are men? How many are women?

3.) 2 3/4 liters of water is needed to water a flower bed. How many liters is needed to water 7 flower beds?

4.) A rectangular fish pond is 35 1/4 feet long and 20 1/2 wide. What is its area?

5.) 2 4/5 deciliters or soda is needed to make a punch. How many deciliters of soda is needed to make 5 punches?

6.) A 1 1/2 L juice is to be shared equally by 6 friends. How many L of soda is the share of each person?

7.) Two-fifth cup of oil is needed to make a birthday cake. How many birthday cakes can be made using 8 cups?

8.) The length of a bookshelf is 5 1/4 feet long. Each book on the shelf is 1 1/2 inches thick. How many books will fit on the shelf?

9.) Five pumpkin pies are to be shared equally among 12 persons. How much pumpkin pie does each person get?

10.) Jessie has 8 3/4 hectares of land. He decided to divide it equally among his four children. How many hectares of land will each receive?

Week 3 Review Answers and Solutions

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

Week 3 Review: Practice Exercises and Problems

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?

Solving Ratio and Proportion Problems Part 2

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.

Practice Exercises on Subtraction of Integers

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.

Solving Word Problems by Working Backwards: Summary

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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