Go through the Math in Focus Grade 7 Workbook Answer Key Chapter 2 Lesson 2.5 Operations with Rational Numbers to finish your assignments.

## Math in Focus Grade 7 Course 2 A Chapter 2 Lesson 2.5 Answer Key Operations with Rational Numbers

### Math in Focus Grade 7 Chapter 2 Lesson 2.5 Guided Practice Answer Key

**Copy and Complete**

Question 1.

Answer:

–\(\frac{1}{9}\) + \(\frac{2}{-5}\)

Write equivalent fractions using the LCD

\(\frac{(-1)(-5)}{(9)(-5)}\) + \(\frac{2(9)}{(-5)(9)}\)

= \(\frac{5}{-45}\) + \(\frac{18}{-45}\)

= \(\frac{5+18}{-45}\)

= \(\frac{23}{-45}\)

Question 2.

Answer:

-1 \(\frac{1}{6}\) + 3\(\frac{4}{9}\)

= -1 – \(\frac{1}{6}\) + 3 + \(\frac{4}{9}\)

= -1 + 3 = 2

= –\(\frac{1}{6}\) + \(\frac{4}{9}\)

LCD of \(\frac{1}{6}\) and \(\frac{4}{9}\) is 18

= –\(\frac{3}{18}\) + \(\frac{8}{18}\) = \(\frac{5}{18}\)

2 + \(\frac{5}{18}\) = 2\(\frac{5}{18}\)

Question 3.

\(\frac{1}{6}\) + \(\left(\frac{-5}{9}\right)\) + \(\left(\frac{-1}{3}\right)\)

Add two rational numbers at a time, working from left to right.

Answer:

Method 2

Use a common denominator for all three fractions.

Answer:

\(\frac{1}{6}\) + \(\left(\frac{-5}{9}\right)\) + \(\left(\frac{-1}{3}\right)\)

= \(\frac{3}{18}\) + \(\left(\frac{-10}{18}\right)\) + \(\left(\frac{-6}{18}\right)\)

= (3 -10 – 6)/18

= -13/18

**Evaluate each expression.**

Question 4.

\(\frac{1}{4}\) – \(\frac{3}{10}\)

Answer:

The fractions have unlike denominators.

So, find the LCD for 1/4 and 3/10 that is 20.

\(\frac{1}{4}\) × \(\frac{5}{5}\) – \(\frac{3}{10}\) × \(\frac{2}{2}\)

\(\frac{5}{20}\) – \(\frac{6}{20}\) = – \(\frac{1}{20}\)

Question 5.

\(\frac{7}{8}\) – \(\frac{9}{10}\)

Answer:

Given,

\(\frac{7}{8}\) – \(\frac{9}{10}\)

The fractions have unlike denominators.

LCD of 8 and 10 is 40

\(\frac{7}{8}\) × \(\frac{5}{5}\) – \(\frac{9}{10}\) × \(\frac{4}{4}\)

\(\frac{35}{40}\) – \(\frac{36}{40}\) = – \(\frac{1}{40}\)

Question 6.

3\(\frac{1}{4}\) – 7\(\frac{5}{6}\)

Answer:

Given

3\(\frac{1}{4}\) – 7\(\frac{5}{6}\)

Rewriting the equation

3 + \(\frac{1}{4}\) – 7 – \(\frac{5}{6}\)

3 – 7 = -4

Now subtract the fractions

\(\frac{1}{4}\) – \(\frac{5}{6}\)

LCM of 4 and 6 is 12

\(\frac{1}{4}\) × \(\frac{3}{3}\) – \(\frac{5}{6}\) × \(\frac{2}{2}\)

\(\frac{3}{12}\) – \(\frac{10}{12}\) = – \(\frac{7}{12}\)

-4 – \(\frac{7}{12}\) = -4\(\frac{7}{12}\)

Question 7.

\(\frac{3}{7}\) – \(\frac{27}{28}\) – \(\frac{3}{14}\)

Answer:

Given

\(\frac{3}{7}\) – \(\frac{27}{28}\) – \(\frac{3}{14}\)

LCM of 7, 14, 28 is 28.

\(\frac{3}{7}\) × \(\frac{4}{4}\) – \(\frac{27}{28}\) × \(\frac{1}{1}\) – \(\frac{3}{14}\) × \(\frac{2}{2}\)

\(\frac{12}{28}\) – \(\frac{27}{28}\) – \(\frac{6}{28}\)

= (12 – 27 – 6)/28

= -21/28

= -3/4

**Solve.**

Question 8.

Philadelphia suffered a severe snowstorm in 1996 that left 30\(\frac{7}{10}\) inches of snow on the ground. Another severe snowstorm occured in 2010, when 28\(\frac{1}{2}\) inches of snow fell.

a) Write a subtraction expression for the difference in depth of these two record snowfalls.

Answer:

30\(\frac{7}{10}\) – 28\(\frac{1}{2}\)

b) Rewrite the expression as an addition expression.

Answer: – 28\(\frac{1}{2}\) + 30\(\frac{7}{10}\)

c) Find the difference in these two record snowfalls.

Answer:

30\(\frac{7}{10}\) – 28\(\frac{1}{2}\)

Rewriting the equation

30 + \(\frac{7}{10}\) – 28 – \(\frac{1}{2}\)

30 – 28 = 2

\(\frac{7}{10}\) – \(\frac{1}{2}\)

LCD = 10

\(\frac{7}{10}\) – \(\frac{5}{10}\) = \(\frac{2}{10}\) = \(\frac{1}{5}\)

2 + \(\frac{1}{5}\) = 2 \(\frac{1}{5}\)

**Copy and Complete.**

Question 9.

Answer:

–\(\frac{4}{5}\) × \(\frac{20}{21}\)

Multiply the numerators and multiply the denominators

= (-4 × 20)/(5 × 21)

= –\(\frac{80}{105}\)

Question 10.

Answer:

-3\(\frac{1}{4}\) × -2\(\frac{2}{3}\)

Write as improper fractions

–\(\frac{13}{4}\) × –\(\frac{8}{3}\)

= \(\frac{104}{12}\)

= \(\frac{52}{6}\)

= \(\frac{26}{3}\)

= 8\(\frac{2}{3}\)

**Evaluate each quotient.**

Question 11.

\(\frac{3}{20}\) ÷ \(\left(-\frac{6}{35}\right)\)

Answer:

Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.

\(\frac{3}{20}\) × –\(\frac{35}{6}\)

= (3 × -35)/(20 × 6)

= -105/120

= -7/8

Question 12.

-3\(\frac{1}{3}\) ÷ \(\left(-1 \frac{1}{4}\right)\)

Answer:

Given,

-3\(\frac{1}{3}\) ÷ \(\left(-1 \frac{1}{4}\right)\)

\(\frac{10}{3}\) ÷ \(\frac{5}{4}\)

LCD of 3 and 4 is 12.

\(\frac{10}{3}\) × \(\frac{4}{4}\) ÷ \(\frac{5}{4}\) × \(\frac{3}{3}\)

\(\frac{40}{12}\) ÷ \(\frac{15}{12}\)

\(\frac{40}{12}\) = 2 \(\frac{2}{3}\)

Question 13.

Answer:

Dividing two fractions is the same as multiplying the first fraction by the reciprocal of the second fraction.

\(\frac{1}{4}\) ÷ \(\frac{-3}{8}\) = \(\frac{-2}{3}\)

**Solve. Show your work.**

Question 14.

A pancake recipe requires 1\(\frac{2}{3}\) cups of flour to make 20 pancakes and you have 9 cups of flour.

a) How many pancakes can you make with 1 cup of flour?

Answer:

Given,

A pancake recipe requires 1\(\frac{2}{3}\) cups of flour to make 20 pancakes

Let 1 cup of flour is x

1 \(\frac{2}{3}\) x = 20

x = 12

b) How many pancakes can you make with 9 cups of flour?

Answer:

1 cup = 12 pancakes

9 cups = 9 × 12 = 108

Thus you can make 81 pancakes with 9 cups of flour.

c) Do you have enough to make 100 pancakes? Explain your reasoning.

Answer:

1 cup of flour = 12 pancakes

9 cups = 9 × 12 = 108

Thus you can make 81 pancakes with 9 cups of flour.

As I can make 108 pancakes with 9 cups of flour, I can conclude that I have enough flour to make 100 pancakes

### Math in Focus Course 2A Practice 2.5 Answer Key

**Evaluate each expression. Give your answer in simplest form.**

Question 1.

\(\frac{1}{2}\) + \(\left(-\frac{5}{6}\right)\)

Answer:

Given,

\(\frac{1}{2}\) + \(\left(-\frac{5}{6}\right)\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

\(\frac{1}{2}\) – \(\frac{5}{6}\)

LCD of 2 and 6 is 6.

\(\frac{1}{2}\) × \(\frac{3}{3}\) – \(\frac{5}{6}\) × \(\frac{1}{1}\)

\(\frac{3}{6}\) – \(\frac{5}{6}\) = – \(\frac{2}{6}\) or – \(\frac{1}{3}\)

Question 2.

–\(\frac{6}{7}\) + \(\frac{3}{14}\)

Answer:

Given,

–\(\frac{6}{7}\) + \(\frac{3}{14}\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

LCD of 7 and 14 is 14.

–\(\frac{12}{14}\) + \(\frac{3}{14}\) = –\(\frac{9}{14}\)

Question 3.

–\(\frac{1}{7}\) + \(\left(\frac{-3}{5}\right)\)

Answer:

Given,

–\(\frac{1}{7}\) + \(\left(\frac{-3}{5}\right)\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

-(\(\frac{1}{7}\) + \(\frac{3}{5}\))

LCD of 7 and 5 is 35

-(\(\frac{5}{35}\) + \(\frac{21}{35}\))

–\(\frac{26}{35}\)

Question 4.

\(\frac{1}{2}\) + \(\left(-\frac{2}{5}\right)\) + \(\frac{1}{4}\)

Answer:

Given,

\(\frac{1}{2}\) + \(\left(-\frac{2}{5}\right)\) + \(\frac{1}{4}\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

\(\frac{1}{2}\) + \(\frac{1}{4}\) – \(\frac{2}{5}\)

\(\frac{3}{4}\) – \(\frac{2}{5}\)

LCD of 4 and 5 is 20.

\(\frac{15}{20}\) – \(\frac{8}{20}\) = \(\frac{7}{20}\)

Question 5.

–\(\frac{1}{7}\) + \(\left(\frac{-5}{6}\right)\) + \(\left(\frac{-1}{3}\right)\)

Answer:

Given,

–\(\frac{1}{7}\) + \(\left(\frac{-5}{6}\right)\) + \(\left(\frac{-1}{3}\right)\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

–\(\frac{1}{7}\) – \(\frac{5}{6}\) – \(\frac{1}{3}\)

– (\(\frac{1}{7}\) + \(\frac{5}{6}\) + \(\frac{1}{3}\) )

LCD of 7, 6 and 3 is 42.

– (\(\frac{6}{42}\) + \(\frac{35}{42}\) + \(\frac{14}{42}\))

– (6 + 35 + 14)/42

– \(\frac{55}{42}\)

Question 6.

\(\frac{3}{5}\) – \(\frac{2}{3}\)

Answer:

Given,

\(\frac{3}{5}\) – \(\frac{2}{3}\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

LCD of 5 and 3 is 15.

\(\frac{9}{15}\) – \(\frac{10}{15}\) = – \(\frac{1}{15}\)

Question 7.

–\(\frac{1}{7}\) – \(\frac{3}{14}\)

Answer:

Given,

–\(\frac{1}{7}\) – \(\frac{3}{14}\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

LCD of 7 and 14 is 14.

-(\(\frac{1}{7}\) + \(\frac{3}{14}\))

-(\(\frac{2}{14}\) + \(\frac{3}{14}\))

– \(\frac{5}{14}\)

Question 8.

–\(\frac{1}{5}\) – \(\left(\frac{-2}{7}\right)\)

Answer:

Given,

–\(\frac{1}{5}\) – \(\left(\frac{-2}{7}\right)\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

– \(\frac{1}{5}\) + \(\frac{2}{7}\)

LCD of 7 and 5 is 35

-(\(\frac{7}{35}\) + \(\frac{10}{35}\))

–\(\frac{17}{35}\)

Question 9.

\(\frac{1}{3}\) – \(\left(-\frac{2}{5}\right)\) – \(\frac{3}{4}\)

Answer:

Given,

\(\frac{1}{3}\) – \(\left(-\frac{2}{5}\right)\) – \(\frac{3}{4}\)

The fractions of the denominators are not the same. So, we have to find the LCD of the denominators and rewrite the fractions.

\(\frac{1}{3}\) + \(\frac{2}{5}\) – \(\frac{3}{4}\)

LCD of 3, 5, and 7 is 60.

\(\frac{20}{60}\) + \(\frac{24}{60}\) – \(\frac{45}{60}\)

\(\frac{44}{60}\) – \(\frac{45}{60}\) = – \(\frac{1}{60}\)

**Evaluate each product. Give your answer in simplest form.**

Question 10.

–\(\frac{7}{25}\) • \(\frac{5}{14}\)

Answer:

Given,

–\(\frac{7}{25}\) • \(\frac{5}{14}\)

-(7 × 5)/(25 × 14)

= -1/10

Question 11.

\(\frac{5}{8}\) • \(\left(-\frac{4}{15}\right)\)

Answer:

Given,

\(\frac{5}{8}\) • \(\left(-\frac{4}{15}\right)\)

(5 × -4)/(8 × 15)

= -20/120

= -1/6

Question 12.

\(\frac{7}{30}\) • \(\left(-\frac{6}{7}\right)\)

Answer:

Given,

\(\frac{7}{30}\) • \(\left(-\frac{6}{7}\right)\)

(7 × -6)/(30 × 7)

= -1/5

Question 13.

–\(\frac{8}{27}\) • \(\left(-\frac{9}{40}\right)\)

Answer:

Given,

–\(\frac{8}{27}\) • \(\left(-\frac{9}{40}\right)\)

= (-8 × -9)/(27 × 40)

= 1/15

Question 14.

–\(\frac{11}{16}\) • \(\left(-\frac{4}{33}\right)\)

Answer:

Given,

–\(\frac{11}{16}\) • \(\left(-\frac{4}{33}\right)\)

(-11 × -4)/(16 × 33)

1/12

Question 15.

\(\frac{5}{8}\) • \(\left(-2 \frac{4}{5}\right)\)

Answer:

Given,

\(\frac{5}{8}\) • \(\left(-2 \frac{4}{5}\right)\)

\(\frac{5}{8}\) • \(\left(\frac{-6}{5}\right)\)

(5 × -6)/(8 × 5)

-3/4

Question 16.

–\(\frac{3}{22}\) • 1\(\frac{5}{6}\)

Answer:

Given,

–\(\frac{3}{22}\) • 1\(\frac{5}{6}\)

Convert from mixed fraction to the improper fraction to find the product of two fractions.

–\(\frac{3}{22}\) • \(\frac{11}{6}\)

= –\(\frac{3×11}{22×6}\)

= –\(\frac{33}{132}\)

= –\(\frac{1}{4}\)

Question 17.

3\(\frac{1}{8}\) • \(\left(-\frac{3}{10}\right)\)

Answer:

Given,

3\(\frac{1}{8}\) • \(\left(-\frac{3}{10}\right)\)

Convert from mixed fraction to the improper fraction to find the product of two fractions.

\(\frac{25}{8}\) • –\(\frac{3}{10}\)

= –\(\frac{75}{80}\)

= –\(\frac{15}{16}\)

Question 18.

-4\(\frac{1}{2}\) • \(\left(-1 \frac{8}{9}\right)\)

Answer:

Given,

-4\(\frac{1}{2}\) • \(\left(-1 \frac{8}{9}\right)\)

Convert from mixed fraction to the improper fraction to find the product of two fractions.

–\(\frac{9}{2}\) • –\(\frac{17}{9}\)

minus × minus = plus

We get

\(\frac{9}{2}\) • \(\frac{17}{9}\)

= \(\frac{51}{6}\)

= \(\frac{17}{2}\)

**Evaluate each quotient. Give your answer in simplest form.**

Question 19.

-10 ÷ \(\left(-\frac{5}{6}\right)\)

Answer:

Given,

-10 ÷ \(\left(-\frac{5}{6}\right)\)

First cancel the negative signs

We get

10 ÷ \(\frac{5}{6}\)

(10 × 6)/(1 × 5)

= 60/5

= 12

Question 20.

\(\frac{9}{25}\) ÷ (-18)

Answer:

Given,

\(\frac{9}{25}\) ÷ (-18)

(9 × 1)/25 × -28

=\(\frac{9}{-450}\)

= –\(\frac{1}{50}\)

Question 21.

–\(\frac{3}{8}\) ÷ \(\left(-\frac{1}{8}\right)\)

Answer:

Given,

–\(\frac{3}{8}\) ÷ \(\left(-\frac{1}{8}\right)\)

First cancel the negative signs

We get

\(\frac{3}{8}\) ÷ \(\frac{1}{8}\) = 3

Question 22.

–\(\frac{1}{4}\) ÷ \(\frac{3}{8}\)

Answer:

Given,

–\(\frac{1}{4}\) ÷ \(\frac{3}{8}\)

= -(1 × 8)/(4 × 3) = -8/12

= -2/3

Question 23.

\(\frac{5}{12}\) ÷ \(\left(-\frac{1}{6}\right)\)

Answer:

Given,

\(\frac{5}{12}\) ÷ \(\left(-\frac{1}{6}\right)\)

(5 × 6)/(-12 × 1)

= -30/12

= -5/2

= \(\left(-2 \frac{1}{2}\right)\)

Question 24.

-1\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)

Answer:

Given,

-1\(\frac{1}{4}\) ÷ \(\frac{3}{4}\)

Convert from mixed fraction to the improper fraction

–\(\frac{5}{4}\) ÷ \(\frac{3}{4}\)

= -(5 × 4)/(4 × 3)

= –\(\frac{20}{12}\)

= -1\(\frac{2}{3}\)

Question 25.

\(\frac{8}{15}\) ÷ \(\left(-2 \frac{2}{3}\right)\)

Answer:

Given,

\(\frac{8}{15}\) ÷ \(\left(-2 \frac{2}{3}\right)\)

Convert from mixed fraction to the improper fraction

\(\frac{8}{15}\) ÷ –\(\frac{8}{3}\)

= (8 × 3)/(15 × -8)

= –\(\frac{1}{5}\)

Question 26.

3\(\frac{3}{4}\) ÷ \(\left(-\frac{1}{4}\right)\)

Answer:

Given,

3\(\frac{3}{4}\) ÷ \(\left(-\frac{1}{4}\right)\)

Convert from mixed fraction to the improper fraction

\(\frac{15}{4}\) ÷ –\(\frac{1}{4}\)

= (15 × 4)/(4 × -1)

= –\(\frac{60}{4}\)

= -15

Question 27.

2\(\frac{1}{2}\) ÷ \(\left(-1 \frac{2}{3}\right)\)

Answer:

Given,

2\(\frac{1}{2}\) ÷ \(\left(-1 \frac{2}{3}\right)\)

Convert from mixed fraction to the improper fraction

\(\frac{5}{2}\) ÷ –\(\frac{5}{3}\)

= -(5 × 3)/(2 × 5)

= –\(\frac{3}{2}\) or -1\(\frac{1}{2}\)

Question 28.

-2\(\frac{2}{7}\) ÷ \(\left(-1 \frac{3}{7}\right)\)

Answer:

Given,

-2\(\frac{2}{7}\) ÷ \(\left(-1 \frac{3}{7}\right)\)

Convert from mixed fraction to the improper fraction

–\(\frac{16}{7}\) ÷ –\(\frac{10}{7}\)

Cancel the negative signs and then perform division operation

\(\frac{16}{7}\) ÷ \(\frac{10}{7}\)

= (16 × 7)/(7 × 10)

= \(\frac{8}{5}\)

= 1 \(\frac{3}{5}\)

Question 29.

Answer:

Given,

-7 ÷ –\(\frac{7}{3}\)

Cancel the negative signs and then perform division operation

(7 × 3)/(7)

= 21/7

= 3

Question 30.

Answer:

Given,

–\(\frac{2}{3}\) ÷ 4

= -(2 × 1)/(3 × 4)

= -2/12

= –\(\frac{1}{6}\)

Question 31.

Answer:

Given,

Cancel the negative signs and then perform division operation

\(\frac{3}{4}\) ÷ \(\frac{5}{8}\)

= (3 × 8)/(4 × 5)

= 24/20

= 1\(\frac{1}{5}\)

Question 32.

Answer:

Given,

–\(\frac{1}{5}\) ÷ 1\(\frac{2}{15}\)

–\(\frac{1}{5}\) ÷ \(\frac{17}{15}\)

= -(1 × 15)/(5 × 17)

= –\(\frac{3}{17}\)

**Solve. Show your work.**

Question 33.

David hiked 15\(\frac{9}{10}\) miles on Saturday and 6\(\frac{7}{10}\) miles on Sunday. How much farther did David bike on Saturday than on Sunday?

Answer:

David hiked 15\(\frac{9}{10}\) miles on Saturday and 6\(\frac{7}{10}\) miles on Sunday.

15\(\frac{9}{10}\) – 6\(\frac{7}{10}\) = 9 \(\frac{1}{5}\)

Question 34.

A recipe calls for \(\frac{3}{4}\) cup of flour, but Kelli has only \(\frac{1}{4}\) cup of flour. How much more flour does she need?

Answer:

Given,

A recipe calls for \(\frac{3}{4}\) cup of flour, but Kelli has only \(\frac{1}{4}\) cup of flour.

\(\frac{3}{4}\) – \(\frac{1}{4}\) = \(\frac{1}{2}\)

Thus Kelli needs \(\frac{1}{2}\) cup of flour.

Question 35.

A weather report showed that the rainfall in Janesville was 2\(\frac{2}{3}\) inches during the first half of January. At the end of January, the total rainfall was 3\(\frac{1}{4}\) inches. How much did it rain in the second half of January?

Answer:

Given,

A weather report showed that the rainfall in Janesville was 2\(\frac{2}{3}\) inches during the first half of January.

At the end of January, the total rainfall was 3\(\frac{1}{4}\) inches.

3\(\frac{1}{4}\) – 2\(\frac{2}{3}\)

Convert from mixed fraction to the improper fraction

\(\frac{13}{4}\) – \(\frac{8}{3}\)

LCD of 4 and 3 is 12

\(\frac{39}{12}\) – \(\frac{32}{12}\) = \(\frac{7}{12}\)

Question 36.

The sum of two rational numbers is 5\(\frac{1}{2}\). If one of the numbers is 6\(\frac{3}{14}\), find the other number.

Answer:

The sum of two rational numbers is 5\(\frac{1}{2}\). If one of the numbers is 6\(\frac{3}{14}\)

x + 6\(\frac{3}{14}\) = 5\(\frac{1}{2}\)

x = 5\(\frac{1}{2}\) – 6\(\frac{3}{14}\)

x = 5 + \(\frac{1}{2}\) – 6 – \(\frac{3}{14}\)

x = -1 + \(\frac{2}{7}\) = –\(\frac{5}{7}\)

Question 37.

**MathJournal** Peter adds \(\frac{1}{a}\) + \(\left(-\frac{1}{b}\right)\) and says the answer is \(\frac{1}{a-b}\). Give an example to show that Peter is wrong.

Answer:

\(\frac{1}{a}\) + \(\left(-\frac{1}{b}\right)\)

\(\frac{1}{a}\) – \(\frac{1}{b}\)

LCD of a and b is ab

\(\frac{b}{ab}\) – \(\frac{a}{ab}\)

\(\frac{b-a}{ab}\)

By this we can say that Peter is wrong.

Question 38.

**MathJournal** Jo multiplies two mixed numbers, -4\(\frac{3}{5}\) and 1\(\frac{2}{7}\) as follows:

Describe Jo’s mistakes. What is the correct answer?

Answer:

-4\(\frac{3}{5}\) × 1\(\frac{2}{7}\)

Convert from mixed fraction to the improper fraction

–\(\frac{23}{5}\) × \(\frac{9}{7}\)

–\(\frac{207}{35}\)

-5\(\frac{32}{35}\)

Question 39.

A clock’s battery is running low. Every 6 hours, it slows down by \(\frac{1}{2}\) hour. By how much does it slow down in 1 hour?

Answer:

If its running low and every six hours it slows down half an hour, we need to divide 6 from 1/2.

6 divided by 1/2 is 12

It will slow down 12 minutes every hour.

Question 40.

A weighs 5\(\frac{1}{2}\) pounds and package B weighs 1\(\frac{1}{4}\) pounds. Find the average weight of the two packages.

Answer:

Given,

A weighs 5\(\frac{1}{2}\) pounds and package B weighs 1\(\frac{1}{4}\) pounds.

5\(\frac{1}{2}\) + 1\(\frac{1}{4}\) = 6 \(\frac{3}{4}\)

Convert from mixed fraction to the improper fraction

6 \(\frac{3}{4}\) = \(\frac{27}{4}\)

Average of two packets

\(\frac{27}{4}\) ÷ 2 = \(\frac{27}{8}\) = 3 \(\frac{3}{8}\)

Question 41.

A scientist measured the weight of some damp soil. After exposing the soil to the air for 4\(\frac{1}{2}\) weeks, the scientist found that the weight had decreased by 5\(\frac{1}{8}\) ounces. Find the average weight loss per week.

Answer:

Let the weight loss per week is x

A scientist measured the weight of some damp soil. After exposing the soil to the air for 4\(\frac{1}{2}\) weeks, the scientist found that the weight had decreased by 5\(\frac{1}{8}\) ounces.

4\(\frac{1}{2}\) weeks = 5\(\frac{1}{8}\)

1 week = x

x × 4\(\frac{1}{2}\) = 5\(\frac{1}{8}\)

x = 5\(\frac{1}{8}\) ÷ 4\(\frac{1}{2}\)

x = \(\frac{41}{8}\) ÷ \(\frac{9}{2}\)

x = 1 \(\frac{5}{36}\) ounces

Question 42.

A plank measures 4\(\frac{3}{4}\) feet. Elizabeth cuts off \(\frac{2}{5}\) of the plank. How long is the plank now?

Answer:

Given,

A plank measures 4\(\frac{3}{4}\) feet.

Elizabeth cuts off \(\frac{2}{5}\) of the plank.

4\(\frac{3}{4}\) – \(\frac{2}{5}\) = 4 \(\frac{7}{20}\)

Thus the plank is 4 \(\frac{7}{20}\) feet long now.