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### Solution 1 page 14 Math textbook 7 Creative horizon volume 2

Let two quantities a and b be proportional to each other. Know that when a = 2 then b = 18.

a) Find the scaling factor k of a with respect to b.

b) Calculate the value of b when a = 5.

**Solution method**

a) Express a in terms of b

b) Substitute a = 5 into the relation between a and b, find b

**Detailed explanation**

a) Since a and b are proportional quantities, a = kb

When a = 2 then b = 18 so 2 = k . 18 \(\Rightarrow k = \dfrac{2}{18}=\dfrac{1}{9}\)

So the scaling factor of a with respect to b is \(\dfrac{1}{9}\)

b) From the formula: \(a = \dfrac{1}{9}b\)

Substituting a = 5 into the formula will get:

\(5 = \dfrac{1}{9}b \Rightarrow 5:\dfrac{1}{9} = b \Rightarrow b = 45\)

So b = 45 at a = 5.

### Solution 2 page 14 Math textbook 7 Creative horizon volume 2

Let two quantities x and y be proportional to each other. Know that when x = 7 then y = 21.

a) Find the scaling factor of y with respect to x and express y in terms of x

b) Find the scaling factor of x with respect to y and express x in terms of y

**Solution method**

If x is proportional to y by the scaling factor k then y is proportional to x by the scaling factor \(\dfrac{1}{k}\)

**Detailed explanation**

a) According to the problem, we have x proportional to y, but at x = 7, y = 21 we have the following ratio:

\( \Rightarrow \dfrac{x}{y} = \dfrac{7}{{21}} = \dfrac{1}{3}\)

\( \Rightarrow \dfrac{x}{y} = \dfrac{1}{3} \Rightarrow 3x = y\)

So the scaling factor of y with respect to x is 3 and y = 3x

b) We have x = \(\dfrac{1}{3}y\) so the scaling factor of x to y is : \(\dfrac{1}{3}\)

Because 3x = y \( \Rightarrow x = \dfrac{1}{3}y\)

### Solution 3 page 14 Math textbook 7 Creative horizon volume 2

Let m and n be two quantities that are directly proportional to each other. Write the formula for m over n and find the unknown values in the following table:

n |
-2 |
-first |
0 |
first |
2 |

m |
? |
? |
? |
-5 |
? |

**Solution method**

Express m over n.

Substitute the value of n into the formula to find the corresponding m.

**Detailed explanation**

We have : \(\dfrac{n}{m} = \dfrac{{ – 2}}{?} = \dfrac{{ – 1}}{?} = \dfrac{0}{?} = \dfrac{ 1}{{ – 5}} = \dfrac{2}{?}\) \( \Rightarrow \dfrac{n}{m} = \dfrac{1}{{ – 5}}\) \( \Rightarrow m = – 5n\)

adsense

Replace \(n = – 2 \Rightarrow m = ( – 2).( – 5) = 10\) \( \Rightarrow ? = 10\)

Replace \(n = – 1 \Rightarrow m = ( – 1).( – 5) \Rightarrow ? = 5\)

Replace \(n = 0 \Rightarrow m = 0.( – 5) \Rightarrow ? = 0\) but ? is the denominator so \(? \ne 0\) \( \Rightarrow ? \in \emptyset \)

Replace \(n = 2 \Rightarrow m = 2.( – 5) \Rightarrow ? = – 10\)

### Solve problems 4 pages 14 Math textbook 7 Creative horizon volume 2

Given that two quantities S and t are proportional to each other:

a) Calculate the unknown values in the above table

b) Write the formula for t in terms of S

**Solution method**

Apply the property of proportional quantities:\(\dfrac{S_1}{t_1}=\dfrac{S_2}{t_2}=\dfrac{S_3}{t_3}=…\)

**Detailed explanation**

a) Since S and t are proportional quantities \( \dfrac{1}{{ – 3}} = \dfrac{2}{?} = \dfrac{3}{?} = \dfrac{4 }{?} = \dfrac{5}{?}\) (proportional quantity property)

\(\Rightarrow t= – 3S\)

Substituting S = 2 we have: t= -3.2 = -6

Substituting S = 3 we have: t= -3.3 = -9

Substituting S = 4 we have: t= -3.4 = -12

Substituting S = 5 we have: t= -3.5 = -15

b) From sentence a, we have the formula for calculating t in terms of S: \(t = – 3S\)

### Solve problems 5 pages 14 Math textbook 7 Creative horizon volume 2

In the following cases, check whether the quantity x is proportional to the quantity y:

a)

x |
2 |
4 |
6 |
-8 |

y |
1.2 |
2.4 |
3.6 |
– 4.8 |

b)

x |
first |
2 |
3 |
4 |
5 |

y |
3 |
6 |
9 |
twelfth |
25 |

**Solution method**

Check the x and y ratios respectively.

+ If the ratios are equal, then the two quantities x and y are proportional.

+ If there are unequal proportions, then the two quantities x and y are not proportional.

**Detailed explanation**

a) We have : \( \dfrac{2}{{1,2}} = \dfrac{4}{{2,4}} = \dfrac{6}{{3,6}} = \dfrac{{ – 8}}{{ – 4.8}}\) so x is proportional to y

b)

We see: \(\dfrac{1}{3} = \dfrac{2}{6} = \dfrac{3}{9} = \dfrac{4}{{12}} \ne \dfrac{5}{ {25}}\)so x is not proportional to y

### Solve lesson 6 page 15 Math textbook 7 Creative horizon volume 2

Two uniform metal rings have volumes \(3c{m^3}\) and \(2c{m^3}\). How many grams does each ring weigh, knowing that the two rings weigh 96.5 g? (Show that mass and volume are two quantities that are directly proportional to each other.)

**Solution method**

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

**Detailed explanation**

Let the weight of the ring \(3c{m^3}\) be A (g) and the other ring B (g) ( A,B > 0)

According to the problem, we have A proportional to B by volume, so we have A : B = 3 : 2 \( \Rightarrow \dfrac{A}{B} = \dfrac{3}{2} \Rightarrow \dfrac{ A}{3} = \dfrac{B}{2}\)

According to the problem 2 rings weigh 96.5g, so A+B = 96.5

Applying the property of the sequence of equal ratios, we have: \( \Rightarrow \dfrac{A}{3} = \dfrac{B}{2} = \dfrac{{A + B}}{5}= \ dfrac{{96.5}}{5}\)

\( \Rightarrow 5A = 3.96.5 \Rightarrow A = 57.9\)

\( \Rightarrow B = 96.5 – 57.9 = 38.6\)

So the ring of volume \(3c{m^3}\) has a mass of 57.9 g and the other has a mass of 38.6 g

### Solve problems 7 page 15 Math textbook 7 Creative horizon volume 2

Four electric coils of the same type have a total mass of 26 kg.

a) Calculate the mass of each roll, knowing that the first roll weighs as much as \(\dfrac{1}{2}\) the second, \(\dfrac{1}{4}\)the third roll and \( \dfrac{1}{6}\) fourth reel.

b) Knowing the first coil is 100m long, calculate how many grams of wire weighs in one meter.

**Solution method**

Let the mass of roll 1 be x and express the mass of the remaining rolls in terms of x

**Detailed explanation**

a) Let the mass of the first roll be x kg

Since the mass of the first roll is equal to \(\dfrac{1}{2}\) of the second roll, we have the mass of the second roll = 2x kg

Since the mass of the first roll is equal to \(\dfrac{1}{4}\)the third roll, we have the mass of the third roll = 4x kg

Since the mass of the first roll is equal to \(\dfrac{1}{6}\) of the 4th roll, we have a mass of the 4th roll of 6x kg

According to the problem, the mass of 4 rolls is 26kg, so we have: \(x + 2x + 4x + 6x = 26\) \( \Rightarrow 13x = 26\)

\( \Rightarrow x = 2\)kg

So the weights of the coils are: 2kg, 4kg, 8kg, 12kg, respectively

b) According to the problem, we have roll 1 100m long and in question a we calculate that roll 1 weighs 2kg

So we have 1 meter of heavy wire: \(\dfrac{2}{{100}}\)= \(0.02\) kg

### Solve problem 8 page 15 Math textbook 7 Creative horizon volume 2

A triangle has three side lengths proportional to 3; 4; 5 and has a circumference of 60 cm. Calculate the lengths of the sides of the triangle.

**Solution method**

Use 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 3 sides of the triangle be a, b, c (cm) respectively (a,b,c > 0)

According to the problem, the 3 sides of a triangle are proportional to 3, 4, 5, so we have the ratio a : b : c = 3 : 4 : 5

And the perimeter of the triangle is 60cm, so we have: a + b + c = 60

\( \Rightarrow \dfrac{a}{3} = \dfrac{b}{4} = \dfrac{c}{5} = \dfrac{{a + b + c}}{{12}} = \dfrac {{60}}{{12}} = 5\)

\( \Rightarrow \) a = 15 ; b = 20 ; c = 25

So the 3 sides of the triangle are 15cm, 20cm, and 25cm .

### Solve problem 9 page 15 Math textbook 7 Creative horizon volume 2

Tien, Hung and Manh go fishing together in the summer. He fished 12 fish, Hung caught 8 fish and Manh caught 10 fish. The total proceeds from the sale of fish are 180,000 VND. Ask if the above amount is divided among friends in proportion to the number of fish each person catches, how much money will each of you receive?

**Solution method**

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

**Detailed explanation**

Let’s call the amount of money 3 friends Tien, Hung, and Manh got, respectively, T,H,M (thousand dong) (T,H,M > 0)

According to problem 3, you sold a total of 180 thousand, so we have:

T + H + M = 180

Divide the amount between you guys proportional to the number of fish each person caught, we will have: \(\dfrac{T}{{12}} = \dfrac{H}{8} = \dfrac{M}{{ ten}}\)

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

\(\dfrac{T}{{12}}=\dfrac{H}{8} = \dfrac{M}{{10}} = \dfrac{{T + H + M}}{{12 + 8 + 10}} = \dfrac{{180}}{{30}}= 6\)

\( \Rightarrow T = 6.12=72; H=6.8=48;M=6.10=60\)

So the amount of money Tien, Hung and Manh can sell are: 72 thousand, 48 thousand and 60 thousand dong, respectively.