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Solve the exercises Joint practice page 10 (Chapter 6 Math 7 Connect)

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### Solve lesson 6.11, page 10, Math 7 Textbook Connecting knowledge volume 2

Formulate possible ratios from the equation 3x = 4y (x,y \( \ne \)0)

**Solution method**

If ad= bc (a,b,c,d \( \ne \) 0), we have the following proportions:

\(\dfrac{a}{b} = \dfrac{c}{d};\dfrac{a}{c} = \dfrac{b}{d};\dfrac{d}{b} = \dfrac{ c}{a};\dfrac{d}{c} = \dfrac{b}{a}\)

**Detailed explanation**

The possible ratios are:

\(\dfrac{3}{4} = \dfrac{y}{x};\dfrac{3}{y} = \dfrac{4}{x};\dfrac{x}{4} = \dfrac{ y}{3};\dfrac{x}{y} = \dfrac{4}{3}\)\(\dfrac{3}{4} = \dfrac{y}{x};\dfrac{3} {y} = \dfrac{4}{x};\dfrac{x}{4} = \dfrac{y}{3};\dfrac{x}{y} = \dfrac{4}{3}\)

### Solve lesson 6.12, page 10, Math 7 textbook Connecting knowledge volume 2

Make all possible ratios from 4 numbers: 5; ten; 25; 50.

**Solution method**

Step 1: Find the equality obtained from the above 4 numbers.

Step 2: With ad= bc (a,b,c,d \( \ne \) 0), we have the following proportions:

\(\dfrac{a}{b} = \dfrac{c}{d};\dfrac{a}{c} = \dfrac{b}{d};\dfrac{d}{b} = \dfrac{ c}{a};\dfrac{d}{c} = \dfrac{b}{a}\)

**Detailed explanation**

We have: 5.50 = 10.25

The possible ratios are:

\(\dfrac{5}{{10}} = \dfrac{{25}}{{50}};\dfrac{5}{{25}} = \dfrac{{10}}{{50}}; \dfrac{{50}}{{10}} = \dfrac{{25}}{5};\dfrac{{50}}{{25}} = \dfrac{{10}}{5}\)

### Solve lesson 6.13, page 10, Math 7 Textbook, Connecting knowledge volume 2

Find x and y, knowing:

a) \(\dfrac{x}{y} = \dfrac{5}{3}\) and x+y = 16;

b) \(\dfrac{x}{y} = \dfrac{9}{4}\) and x – y = -15.

**Solution method**

Step 1: Using the property of proportions, deduce 2 equal ratios with 2 numerators x and y

Step 2: Use the property of the series of equal ratios:

a) \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a + c}}{{b + d}}\)

b) \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a – c}}{{b – d}}\)

**Detailed explanation**

a) Because \(\dfrac{x}{y} = \dfrac{5}{3} \Rightarrow \dfrac{x}{5} = \dfrac{y}{3}\)

Applying the property of the series of equal ratios, we have:

\(\begin{array}{l}\dfrac{x}{5} = \dfrac{y}{3} = \dfrac{{x + y}}{{5 + 3}} = \dfrac{{16 }}{8} = 2\\ \Rightarrow x = 2.5 = 10\\y = 2.3 = 6\end{array}\)

So x=10, y=6

b) Because \(\dfrac{x}{y} = \dfrac{9}{4} \Rightarrow \dfrac{x}{9} = \dfrac{y}{4}\)

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Applying the property of the series of equal ratios, we have:

\(\begin{array}{l}\dfrac{x}{9} = \dfrac{y}{4} = \dfrac{{x – y}}{{9 – 4}} = \dfrac{{ – 15}}{5} = – 3\\ \Rightarrow x = ( – 3).9 = – 27\\y = ( – 3).4 = – 12\end{array}\)

So x = -27, y = -12.

### Solve lesson 6.14 page 10 Math 7 textbook Connecting knowledge volume 2

The ratio of the number of students in classes 7A and 7B is 0.95. Ask how many students are in each class, knowing that the number of students in one class is 2 more than the other.

**Solution method**

Let the number of students in 2 classes be x, y (em) respectively (x,y > 0)

Use the property of the series of equal ratios: \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a – c}}{{b – d}}\)

**Detailed explanation**

Let the number of students in 2 classes be x, y (em) respectively (x,y > 0)

Since the ratio of the number of students in classes 7A and 7B is 0.95, \(\dfrac{x}{y} = 0.95\)\( \Rightarrow \dfrac{x}{{0.95}} = \dfrac{y}{1}\) and x < y

And the number of students in one class is 2 more than the other, so y – x = 2

Applying the property of the series of equal ratios, we have:

\(\begin{array}{l}\dfrac{y}{1} = \dfrac{x}{{0.95}} = \dfrac{{y – x}}{{1 – 0.95}} = \dfrac{2}{{0.05}} = 40\\ \Rightarrow y = 40.1 = 40\\x = 40.0.95 = 38\end{array}\)

So the number of students in two classes 7A and 7B are 38 and 40 students respectively.

### Solve lesson 6.15, page 10, Math 7 Textbook Connecting knowledge volume 2

It is planned to make a road in 15 days. A team of 45 workers worked 10 days to get half the job. How many more people must be added to complete the work on time (knowing the labor productivity of each person is the same)?

**Solution method**

The product of the number of people and the time to complete is constant

**Detailed explanation**

Let the number of people who need to complete the work on time is x (person) (x \( \in \)N*)

Since a team of 45 workers worked 10 days to get half of the job, it took a team of 45 workers 20 days to complete the job.

Since the product of the number of people and the time to complete is constant,

15.x=45.20

\( \Rightarrow x = \dfrac{{45.20}}{{15}} = 60\)

So need to add: 60 – 45 = 15 more people to be able to complete the work on time.

### Solve lesson 6.16, page 10, Math 7 Textbook, Connecting knowledge volume 2

Find three numbers x,y,z knowing that: \(\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}\) and x+2y – 3z = -12

**Solution method**

Apply the property of the series of equal ratios:\(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{e}{f} = \dfrac{{a + 2c – 3e} }{{b + 2d – 3f}}\)

**Detailed explanation**

Applying the property of the series of equal ratios, we have:

\(\begin{array}{l}\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4} = \dfrac{{x + 2y – 3z}}{{ 2 + 2.3 – 3.4}} = \dfrac{{ – 12}}{{ – 4}} = 3\\ \Rightarrow x = 3.2 = 6\\y = 3.3 = 9\\z = 3.4 = 12\end{ array}\)

So x = 6, y = 9, z = 12.