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41. Assuming that a man can complete the work alone in x days, his work in four days would be: a) b) X X C d) 4x x 42. If a man alone takes x days to complete the work while a boy takes y days to complete the work. The equation representing case 1 is given by: a)- + = 5 b) 6 = 4 + = у 5 + 5 d) = X 43. As per the case study, In how many days can a man alone complete the work. a) 35 days b) 70 days c) 50 days d) 100 days 44. For the equation + * = — if is taken as “a' and + is taken as 'b', the equation у​

41 Assuming That A Man Can Complete The Work Alone In X Days His Work In Four Days Would Be A B X X C D 4x X 42 If A Man Alone Takes X Days To Complete The Work class=

Answer :

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]
  • 43. a) 35 days
  • 44. c) 21·a + 28·b = 1
  • 45. c) (42, 56)

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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