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Solve Exercises Lesson 21. Properties of equal numbers (Chapter 6 Math 7 Connect)
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Solve problem 6.7, page 9 Math 7 textbook Connecting knowledge volume 2
Find two numbers x and y, knowing: \(\dfrac{x}{9} = \dfrac{y}{{11}}\) and x+y = 40
Solution method
Use the property of the sequence of equal ratios: \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a + c}}{{b + d}}\)
Detailed explanation
Applying the property of the series of equal ratios, we have:
\(\begin{array}{l}\dfrac{x}{9} = \dfrac{y}{{11}} = \dfrac{{x + y}}{{9 + 11}} = \dfrac{ {40}}{{20}} = 2\\ \Rightarrow x = 2.9 = 18\\y = 2.11 = 22\end{array}\)
So x= 18, y = 22.
Solve lesson 6.8, page 9 Math 7 Textbook Connecting knowledge volume 2
Find two numbers x and y, knowing: \(\dfrac{x}{{17}} = \dfrac{y}{{21}}\) and x – y= 8
Solution method
Use the property of the series of equal ratios: \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a – c}}{{b – d}}\)
Detailed explanation
\(\begin{array}{l}\dfrac{x}{{17}} = \dfrac{y}{{21}} = \dfrac{{x – y}}{{17 – 21}} = \ dfrac{8}{{ – 4}} = – 2\\ \Rightarrow x = ( – 2).17 = – 34\\y = ( – 2).21 = – 42\end{array}\)
So x= -34; y = -42
Solve lesson 6.9, page 9 Math 7 textbook Connecting knowledge volume 2
The ratio of products made by the two workers is 0.95. How many products does each person make, knowing that one person makes 10 more products than the other?
Solution method
Use the property of the series of equal ratios: \(\dfrac{a}{b} = \dfrac{c}{d} = \dfrac{{a – c}}{{b – d}}\)
Detailed explanation
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Let the number of products that 2 people can make, respectively, x, y (product) (x, y > 0)
Since one person makes 10 more products than the other, x – y = 10
Since the ratio of product made by the two workers is 0.95, \(\dfrac{y}{x} = 0.95 \Rightarrow \dfrac{y}{{0.95}} = \dfrac{x} {first}\)
Applying the property of the series of equal ratios, we have:
\(\begin{array}{l}\dfrac{x}{1} = \dfrac{y}{{0.95}} = \dfrac{{x – y}}{{1 – 0.95}} = \dfrac{{10}}{{0.05}} = 200\\ \Rightarrow x = 200.1 = 200\\y = 200.0.95 = 190\end{array}\)
So 2 people can make 200 and 190 products respectively.
Solve lesson 6.10, page 9 Math textbook 7 Connecting knowledge volume 2
Three classes 7A, 7B, 7C were assigned the task of planting 120 trees to green the barren hills. Calculate the number of plants that can be planted in each layer, knowing that the number of plants planted in the three classes 7A, 7B, 7C is proportional to 7; 8; 9.
Solution method
Let the number of trees of 3 classes 7A, 7B, 7C be planted as x, y, z respectively (x,y,z > 0)
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 number of trees of 3 classes 7A, 7B, 7C be planted as x, y, z respectively (x,y,z > 0)
Since the total number of plants in 3 classes is 120 trees, x+y+z = 120
Since the number of crops grown in the three classes 7A, 7B, 7C is proportional to 7;8;9, \(\dfrac{x}{7} = \dfrac{y}{8} = \dfrac{z}{9} \)
Applying the property of the series of equal ratios, we have:
\(\begin{array}{l}\dfrac{x}{7} = \dfrac{y}{8} = \dfrac{z}{9} = \dfrac{{x + y + z}}{{ 7 + 8 + 9}} = \dfrac{{120}}{{24}} = 5\\ \Rightarrow x = 5.7 = 35\\y = 5.8 = 40\\z = 5.9 = 45\end{array} \)
So the number of trees of 3 classes 7A, 7B, 7C can be planted, respectively, 35; 40; 45 trees.
