## Solve Exercises Lesson 20. Percentage (Chapter 6 Math 7 Connect) – Math Book

Solve Exercises Lesson 20. Percentage (Chapter 5 Math 7 Connect)
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### Solve problem 6.1 page 7 Math textbook 7 Connecting knowledge volume 2 – KNTT

Replace the following ratio with the ratio of integers:

$$a)\dfrac{{10}}{{16}}:\dfrac{4}{{21}};b)1,3:2,75;c)\dfrac{{ – 2}}{5 }:0.25$$

Solution method

Step 1: Calculate the ratio

Step 3: Write the ratio below and put the ratio between the integers

Detailed explanation

$$\begin{array}{l}a)\dfrac{{10}}{{16}}:\dfrac{4}{{21}} = \dfrac{{10}}{{16}}.\ dfrac{{21}}{4} = \dfrac{{105}}{{32}} = 105:32;\\b)1,3:2.75 = \dfrac{{1,3}}{{ 2.75}} = \dfrac{{130}}{{275}} = \dfrac{{26}}{{55}} = 26:55;\\c)\dfrac{{ – 2}}{5 }:0.25 = \dfrac{{ – 2}}{5}:\dfrac{1}{4} = \dfrac{{ – 2}}{5}.\dfrac{4}{1} = \dfrac {{ – 8}}{5} = ( – 8):5\end{array}$$

### Solve lesson 6.2 on page 7 Math 7 textbook Connecting knowledge volume 2

Find the equal ratios of the following ratios and then make the ratio:

$$12:30;\dfrac{3}{7}:\dfrac{{18}}{{24}};2,5:6,25$$$$12:30;\dfrac{3}{7) }:\dfrac{{18}}{{24}};2,5:6,25$$

Solution method

Step 1: Calculate the ratios.

Step 2: Find 2 equal proportions

Step 3: Make a ratio

Detailed explanation

$$\begin{array}{l}12:30 = \dfrac{{12}}{{30}} = \dfrac{2}{5};\\\dfrac{3}{7}:\dfrac{ {18}}{{24}} = \dfrac{3}{7}.\dfrac{{24}}{{18}} = \dfrac{9}{{14}};\\2,5:6 ,25 = \dfrac{{2,5}}{{6,25}} = \dfrac{{250}}{{625}} = \dfrac{2}{5}\end{array}$$

Thus, the equal ratios are: 12:30 and 2.5 : 6.25.

We get the ratio: 12:30 = 2.5 : 6.25

### Solve lesson 6.3, page 7 Math textbook 7 Connecting knowledge volume 2

Find x in the following proportions:

$$a)\dfrac{x}{6} = \dfrac{{ – 3}}{4};b)\dfrac{5}{x} = \dfrac{{15}}{{ – 20}}\ ) Solution method Using the proportionality property: If \(\dfrac{a}{b} = \dfrac{c}{d}$$ then ad =bc

Detailed explanation

$$\begin{array}{l}a)\dfrac{x}{6} = \dfrac{{ – 3}}{4}\\x = \dfrac{{( – 3).6}}{4 }\\x = \dfrac{{ – 9}}{2}\end{array}$$

So $$x = \dfrac{{ – 9}}{2}$$

$$\begin{array}{l}b)\dfrac{5}{x} = \dfrac{{15}}{{ – 20}}\\x = \dfrac{{5.( – 20)}} {{15}}\\x = \dfrac{{ – 20}}{3}\end{array}$$

So $$x = \dfrac{{ – 20}}{3}$$

### Solve problem 6.4, page 7 Math 7 Textbook Connecting knowledge volume 2

Make all possible proportions from Equation 14.(-15)= (-10).21

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:

$$\dfrac{{14}}{{ – 10}} = \dfrac{{21}}{{ – 15}};\dfrac{{14}}{{21}} = \dfrac{{ – 10} }{{ – 15}};\dfrac{{ – 15}}{{ – 10}} = \dfrac{{21}}{{14}};\dfrac{{ – 15}}{{21}} = \dfrac{{ – 10}}{{14}}$$

### Solution 6.5 page 7 Math textbook 7 Connecting knowledge volume 2

To mix physiological saline, people need to mix in the right ratio. Know that for every 3 liters of pure water mixed with 27 g of salt. If there is 45 g of salt, how many liters of pure water should be mixed to get physiological saline?

Solution method

The ratio of the volume of pure water to the mass of salt to be mixed is constant

Detailed explanation

Call the number of liters of pure water to be mixed: x (liter) (x > 0)

We have the final scale: $$\dfrac{3}{{27}} = \dfrac{x}{{45}} \Rightarrow x = \dfrac{{3.45}}{{27}} = 5$$

So need 5 liters of water

### Solve problem 6.6 page 7 Math textbook 7 Connecting knowledge volume 2

To plow a field in 14 days, 18 plows must be used. How many plows should be used to plow that field in 12 days? (Knowing the productivity of the plows is the same)?

Solution method

The product of the number of plows and the completion time remains constant

Detailed explanation

Let the number of plows needed to plow that field in 12 days is: x (machine) (x $$\in$$ N)$$\in$$

Since the product of the number of plows and the time to complete is constant, then:

$$14.18 = 12.x \Rightarrow x = 21$$

So need 21 plows