Answer:
The experienced painter takes 3 hours to paint the room
The inexperienced painter takes 6 hours to paint the room
Step-by-step explanation:
* Lets explain how to solve the problem
- Two painters can paint a room in 2 hours if they work together
- Assume that the experienced painter can paint the room in a hours
∴ Its rate is 1/a
- Assume that the inexperienced painter can paint the room in b hours
∴ Its rate is 1/b
∵ When they working together they will finish it in two hours
∴ Their rate together is 1/2
- Equate the sum of the rate of each one and the their rate together
∴ [tex]\frac{1}{a}+\frac{1}{b}=1/2[/tex]
-To add two fraction with different denominators we multiply the 2
denominators and multiply each numerator by the opposite
denominator
∴ [tex]\frac{b+a}{ab}=\frac{1}{2}[/tex]
- By using the cross multiplication
∴ 2(b + a) = ab
∴ 2b + 2a = ab ⇒ (1)
- The inexperienced painter takes 3 hours more than the experienced
painter to finish the job
∵ The experienced painter can finish the room in a hours
∵ The inexperienced painter can finish the room in b hours
∵ The inexperienced painter takes 3 hours more than the experienced
painter to finish the job
∴ b = a + 3 ⇒ (2)
- Substitute equation (2) in equation (1)
∴ 2(a + 3) + 2a = a(a + 3)
∴ 2a + 6 + 2a = a² + 3a ⇒ add like terms
∴ 4a + 6 = a² + 3a ⇒ subtract 4a from both sides
∴ 6 = a² - a ⇒ subtract 6 from both sides
∴ a² - a - 6 = 0 ⇒ factorize it
∵ a² = (a)(a)
∵ -6 = -3 × 2
∵ -3(a) + 2(a) = -3a + 2a = -a ⇒ the middle term in the equation
∴ a² - a - 6 = (a - 3)(a + 2)
∵ a² - a - 6 = 0
∴ (a - 3)(a + 2) = 0
∴ a - 3 = 0 ⇒ add 3 to both sides
∴ a = 3
- OR
∴ a + 2 = 0 ⇒ subtract 2 from both sides
∴ a = -2 ⇒ rejected because there is no negative value for the time
- Substitute the value of a in equation (2) to find b
∵ b = 3 + 3 = 6
∴ The experienced painter takes 3 hours to paint the room
∴ The inexperienced painter takes 6 hours to paint the room
Experienced painter needs 3 hours to paint the room individually.
Inexperienced painter needs 6 hours to paint the room individually.
This problem is related to the speed of completing the work.
To solve this problem, we must state the formula for the speed.
[tex]\large {\boxed {v = \frac{x}{t}} }[/tex]
where:
v = speed of completing the work( m³ / s )
x = work ( m³ )
t = time taken ( s )
Let's tackle the problem!
Painter A can complete work by herself in t_a hours.
[tex]\text{Painter A's Speed} = v_a = x \div t_a[/tex]
[tex]v_a = x \div t_a[/tex]
Painter B can complete work by herself in t_b hours.
[tex]\text{Painter B's Speed} = v_b = x \div t_b[/tex]
[tex]v_b = x \div t_b[/tex]
The inexperienced painter takes 3 hours more than the experienced painter to finish the job
[tex]\text{Painter B's Time} = 3 + \text{Painter A's Time}[/tex]
[tex]t_b = 3 + t_a[/tex]
Two painters can paint a room in 2 hours if they work together
[tex]\text{Total Speed} = v = v_a + v_b[/tex]
[tex]\frac{x}{t} = \frac{x}{t_a} + \frac{x}{t_b}[/tex]
[tex]\frac{1}{t} = \frac{1}{t_a} + \frac{1}{t_b}[/tex]
[tex]\frac{1}{2} = \frac{1}{t_a} + \frac{1}{3 + t_a}[/tex]
[tex]\frac{1}{2} = \frac{3 + t_a + t_a}{t_a(3 + t_a)}[/tex]
[tex]\frac{1}{2} = \frac{3 + 2t_a}{t_a(3 + t_a)}[/tex]
[tex]t_a(3 + t_a) = 2(3 + 2t_a)[/tex]
[tex]t_a^2 + 3t_a = 6 + 4t_a[/tex]
[tex]t_a^2 + 3t_a - 4t_a - 6 = 0[/tex]
[tex]t_a^2 - t_a - 6 = 0[/tex]
[tex](t_a -3)(t_a + 2) = 0[/tex]
[tex](t_a -3) = 0[/tex]
[tex]t_a = \boxed{3 ~ hours}[/tex]
[tex]t_b = 3 + t_a[/tex]
[tex]t_b = 3 + 3[/tex]
[tex]t_b = \boxed {6 ~ hours}[/tex]
Grade: High School
Subject: Mathematics
Chapter: Linear Equations
Keywords: Linear , Equations , 1 , Variable , Line , Gradient , Point
