Percentage and ratio word problems require understanding of the relationship between variables from which the question is formed
The options that give the correct values of the duration of the work are;
- [tex]41. \ c) \ \dfrac{4}{x}[/tex]
- [tex]42. \ d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}[/tex]
Reasons:
41. Number of days it takes a man to complete the work alone = x days
Therefore;
[tex]The \ work \ done \ by \ the \ man \ in \ one \ day = \dfrac{1}{x}[/tex]
[tex]The \ work \ done \ in \ four \ days \ by\ the \ man = 4 \times \dfrac{1}{x} = \dfrac{4}{x}[/tex]
The correct option is [tex]c) \ \dfrac{4}{x}[/tex]
42. Number of days it takes a man to complete the work alone = x days
[tex]Work \ done \ by \ a\ man \ in \ one \ day = \dfrac{1}{x}[/tex]
[tex]Work \ done \ by \ four \ men \ in \ one \ day = \dfrac{4}{x}[/tex]
Number of days it takes a boy to complete the work alone = y days
[tex]Work \ done \ by \ a \ boy \ in \ one \ day = \dfrac{1}{x}[/tex]
[tex]Work \ done \ by \ six \ boys \ in \ one \ day = \dfrac{6}{y}[/tex]
4 men and 6 boys work for 5 days to complete the work
Therefore, work done by 4 men and 6 boys in 1 day is therefore;
[tex]\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}[/tex]
The correct option is therefore;
[tex]d) \ \dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}[/tex]
43. As per the case study, we have;
Case 1
[tex]\dfrac{4}{x} + \dfrac{6}{y} = \dfrac{1}{5}[/tex]
Which gives;
[tex]\dfrac{6\cdot x + 4\cdot y}{y \cdot x} = \dfrac{1}{5}[/tex]
30·x + 20·y = y·x
Case 2
[tex]\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}[/tex]
Which gives;
[tex]\dfrac{4\cdot x + 3\cdot y}{y \cdot x} = \dfrac{1}{7}[/tex]
28·x + 21·y = y·x
Therefore;
30·x + 20·y = 28·x + 21·y
∴ 2·x = y
Plugging in the value of y = 2·x, in Case 1 gives;
[tex]\dfrac{4}{x} + \dfrac{6}{2 \cdot x} = \dfrac{1}{5}[/tex]
[tex]\dfrac{2 \times 4 + 6}{2 \times x} = \dfrac{14}{2 \times x} =\dfrac{7}{x} = \dfrac{1}{5}[/tex]
7 × 5 = x
x = 7 × 5 = 35
The number of days, x, it takes a man to complete the work alone, is given by option; a) 35 days
44. For the equation [tex]\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}[/tex], if [tex]a = \dfrac{1}{x}[/tex], and [tex]b = \dfrac{1}{y}[/tex], we have;
[tex]3 \cdot a+ 4\cdot y = \dfrac{1}{7}[/tex]
21·a + 28·y = 1
The correct option is option C. 21·a + 28·b = 1
45. A solution to the equation [tex]\dfrac{3}{x} + \dfrac{4}{y} = \dfrac{1}{7}[/tex], is given by the values of x, and y, that gives;
[tex]\dfrac{1}{14} + \dfrac{1}{14} = \dfrac{1}{7}[/tex]
We have;
3 × 14 = 42
4 × 14 = 56
Therefore, a solution to the equation is (42, 56)
The correct option is [tex]c) \ \underline{ (42, \ 56)}[/tex]
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