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Solve the exercise Joint exercise page 19 (Chapter 6 Math 7 Connect)
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Solve lesson 6.27, page 20, Math 7 Textbook Connecting knowledge volume 2

Solution method

* Check if the ratio of their respective values ​​is always equal.

+ If equal, then the two quantities are proportional

+ If not equal, then the two quantities are not two proportional quantities

* Check the product of 2 of their respective values ​​is always equal or not.

If equal, then the two quantities are inversely proportional

+ If not equal, then the two quantities are not two quantities that are inversely proportional

Detailed explanation

We see: \(\dfrac{{0,5}}{{2,5}} = \dfrac{1}{5} = \dfrac{{1,5}}{{7,5}} = \dfrac {2}{{10}} = \dfrac{{2,5}}{{12,5}}\) so x and y are two proportional quantities.

The related formula: \(x = \dfrac{1}{5}.y\) (or y = 5.x)

Solve problem 6.28 page 20 Math 7 textbook Connecting knowledge volume 2

Solution method

+ Use the definition of two quantities that are directly proportional and inversely proportional:

If y = ax (a is a non-zero constant), then y is proportional to x by the scaling factor a.

If \(y = \dfrac{a}{x}\)(a is a non-zero constant), then y is inversely proportional to x by the scaling factor a

+ Represent the quantity y in terms of z.

If y = k. z (k is a constant) then y and z are proportional quantities.

If \(y = \dfrac{k}{z}\) ( k is a constant) then y and z are two quantities in inverse proportion.

Detailed explanation

a) Assuming y is proportional to x by the scaling factor a so y = ax so \(x = \dfrac{y}{a}\)

y is proportional to z by the scaling factor b so y = bz

Thus, \(x = \dfrac{y}{a} = \dfrac{{bz}}{a} = \dfrac{b}{a}.z\)( \(\dfrac{b}{a} \) is a constant because a and b are constants)

So x is proportional to z and the scaling factor is \(\dfrac{b}{a}\)

b) Assuming y is proportional to x by the scaling factor a so y = ax so \(x = \dfrac{y}{a}\)

y is inversely proportional to z by the scaling factor b so y = \(\dfrac{b}{z}\)

Therefore, \(x = \dfrac{y}{a} = \dfrac{{\dfrac{b}{z}}}{a} = \dfrac{b}{z}:a = \dfrac{b} {z}.\dfrac{1}{a} = \dfrac{{\dfrac{b}{a}}}{z}\)( \(\dfrac{b}{a}\) is constant because a ,b are constants)

So x is inversely proportional to z and the scaling factor is \(\dfrac{b}{a}\)

c) Assuming y is inversely proportional to x by the scaling factor a so y = \(\dfrac{a}{x}\) so x = \(\dfrac{a}{y}\)

y is inversely proportional to z by the scaling factor b so y = \(\dfrac{b}{z}\)

Therefore, \(x = \dfrac{a}{y} = \dfrac{a}{{\dfrac{b}{z}}} = a:\dfrac{b}{z} = a.\dfrac{ z}{b} = \dfrac{a}{b}.z\)( \(\dfrac{a}{b}\) is constant because a,b are constants)

So x is proportional to z and the scaling factor is \(\dfrac{a}{b}\)

Solve lesson 6.29 page 20 Math 7 Textbook Connecting knowledge volume 2

Solution method

Let the mass of copper and zinc to prepare 150 kg of brass be x, y (kg) (x,y > 0) respectively.

Apply the property of the sequence of equal ratios: \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a + c}}{{b + d}}\)

Detailed explanation

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Let the mass of copper and zinc to prepare 150 kg of brass are x, y (kg) respectively (x,y > 0) so x + y = 150

Since pure copper and zinc are in a 6:4 ratio, \(\dfrac{x}{6} = \dfrac{y}{4}\)

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

\(\begin{array}{l}\dfrac{x}{6} = \dfrac{y}{4} = \dfrac{{x + y}}{{6 + 4}} = \dfrac{{150 }}{{10}} = 15\\ \Rightarrow x = 15.6 = 90\\y = 15.4 = 60\end{array}\)

So the mass of copper and zinc to prepare 150 kg of brass is 90 kg and 60 kg respectively.

Solve lesson 6.30 page 20 Math 7 textbook Connecting knowledge volume 2

Solution method

For the same job, time and number of products done are inversely proportional

Detailed explanation

Let the time it takes the apprentice to complete the amount of work the skilled craftsman does in 48 hours x (hours) (x > 0)

Because for the same job, time and number of products done are two quantities that are inversely proportional.

By the property of two quantities in inverse proportion, we have:

12.48 = 8. x \( \Rightarrow x = \dfrac{{12.48}}{8} = 72\)

So the time it takes the apprentice is 72 hours.

Solution 6.31 page 20 Math 7 Textbook Connecting knowledge volume 2

Solution method

Let the number of books 4 classes 7A, 7B, 7C, 7D collect, respectively, be x,y,z,t (book) (x,y,z,t \( \in \)N*)

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

Detailed explanation

Let the number of books 4 classes 7A, 7B, 7C, 7D collect, respectively, be x,y,z,t (book) (x,y,z,t \( \in \)N*)

Since class 7D contributed 4 books more than class 7A, t – x = 4

Since the number of books donated is proportional to the number of students in the class, \(\dfrac{x}{{38}} = \dfrac{y}{{39}} = \dfrac{z}{{40}} = \dfrac{t}{{40}}\)

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

\(\begin{array}{l}\dfrac{x}{{38}} = \dfrac{y}{{39}} = \dfrac{z}{{40}} = \dfrac{t}{{ 40}} = \dfrac{{t – x}}{{40 – 38}} = \dfrac{4}{2} = 2\\ \Rightarrow x = 2.38 = 76\\y = 2.39 = 78\\z = 2.40 = 80\\t = 2.40 = 80\end{array}\)

So the number of books 4 classes 7A, 7B, 7C, 7D collected are 76, 78, 80, 80 books respectively.

Solution 6.32 page 20 Math 7 textbook Connecting knowledge volume 2

Solution method

Let the number of reference books for Grade 6, 7 and 8 Math that the library buys be x, y, z (x, y, z, respectively) \( \in \)\(\mathbb{N}\) )

The number of books and the price of a book are respectively inversely proportional.

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

Detailed explanation

Let the number of reference books for Grade 6, 7 and 8 Math that the library buys be x, y, z (x, y, z, respectively) \( \in \)\(\mathbb{N}\) )

Since there are 121 books in total, we have \(x + y + z = 121\)

Since the amount of money used to buy each type of book is the same, the number of books and the price of a book are respectively inversely proportional.

By the property of two quantities in inverse proportion, we have:

40.x=45.y=50.z

\(\begin{array}{l} \Rightarrow \dfrac{x}{{\dfrac{1}{{40}}}} = \dfrac{y}{{\dfrac{1}{{45}}} } = \dfrac{z}{{\dfrac{1}{{50}}}}\\ = \dfrac{{x + y + z}}{{\dfrac{1}{{40}} + \dfrac {1}{{45}} + \dfrac{1}{{50}}}} = \dfrac{{121}}{{\dfrac{{121}}{{1800}}}} = 121.\dfrac {{1800}}{{121}} = 1800\\ \Rightarrow x = 1800.\dfrac{1}{{40}} = 45\\y = 1800.\dfrac{1}{{45}} = 40 \\z = 1800.\dfrac{1}{{50}} = 36\end{array}\)

So, the number of reference books for grade 6, grade 7, and grade 8 math that the library bought is 45 books, 40 books and 36 books, respectively.