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A painter was paid $800 as labor cost for painting the exterior of the

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A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 21 Mar 2018, 23:05
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A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?

(A) 20
(B) 30
(C) 40
(D) 50
(E) 60

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Re: A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 26 Mar 2018, 09:33
1

Solution:



Given:

    • Labor cost for painting the exterior of the front of the house = \($ 800\)

    • Regular rate of the painter = \($R\) per hour

    • Estimated time = \(T\) hours

      o Inference: \(R*T = 800\)

    • Actual time taken to complete the work = \((T + 4)\) hours

    • Reduced Hourly rate = \($(R-10)\) per hour.

Working Out:

We need to find the value of R.

Note: Since the total amount remains the same, the two scenarios: Estimated and Actual, can be equated to find the unknown values.
Translating the above point in the form of an equation we get:

    • R* T = (R-10) * (T+4)

    • Or, R*T = R*T + 4R -10T – 40

    • Or, 10T +40 = 4R

    • Or, R = (10T+40)/ 4

Putting this value of R in the equation \(RT = 800\), we get:

    • \((10T +40)/4 * T = 800\)

    • Or, \(10T^2\) \(+ 40T = 3200\)

    • Or,\(T^2 + 4T = 320\)

    • Or, \(T^2 + 4T-320 = 0\)

    • Or, \((T+20) * (T-16)= 0\)

    • Thus, T can be +16 or -20.

      o Since T is time, and time cannot be negative, we can ignore -20.

      o T = 16 hours.

      o R = $800/16 = $ 50

Answer: Option D

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Re: A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 21 Mar 2018, 23:14
Bunuel wrote:
A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?

(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


R*T = 800
(R-10) * (T+4) = 800
By checking answers
if R = 50 then T = 16
Then R becomes 40 and T becomes 20

Hence D
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A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 22 Mar 2018, 01:21
Bunuel wrote:
A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?

(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


Instead of writing equations, we'll use the answers.
This is an Alternative approach and is especially useful if long word problems tend to confuse you.

Starting with the median - C.
If the regular rate is 40 then the regular time is 800/40 = 20. In this case the new rate was 40-10=30 and the new time was 800/30 = 80/3. Since 20 + 4 does not equal 80/3, this answer is incorrect.
Let's try the next largest answer, (D).
If the regular rate is 50 then the regular time is 800/50 = 16. In this case the new rate was 50-10=40 and the new time was 800/40 = 20. Since 16 + 4 = 20 then all our calculations work out and our answer is (D).

Note that you don't actually need to write down any equations with variables to use this approach.
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Re: A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 22 Mar 2018, 17:09
Bunuel wrote:
A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?

(A) 20
(B) 30
(C) 40
(D) 50
(E) 60

Answer choices

For each answer's rate and decreased rate, find time taken.
The difference in hours should be 4.

Rate * Time = Pay, P
Thus, Time, \(T = \frac{P}{R}\)

Start with C) $40 per hour (estimated). $30 per hour (actual)

Time at $40: \(\frac{$800}{$40per.hr}= 20\) hours
Time at $30: \(\frac{$800}{$30per.hr}=\frac{80}{3}= 26.xx\) hours

800/30 is not an integer. I calculated to find out
whether $40 per hour = too high or too low

6.xx extra hours. Too great. Needs to decrease.

R and T are inversely proportional
(one increases, the other decreases)
T needs to come down. R must go up.
$40 per hour is too low

Try D) $50 per hour (estimated). $40 per hour (actual)

Time at $50: \(\frac{800}{50} = 16\) hours
Time at $40, from above = \(20\) hours

Difference? 4 hours. That's correct.

Answer D

Algebra

(Hourly pay rate) * (# hours worked) = Total pay in dollars
Same basis as \(RT = W\)
\(RT = W\), so \(T = \frac{W}{R}\)

W = 800
Estimated rate = \(r\)
Actual rate = \(r-10\)
Estimated time = \(\frac{800}{r}\)
Actual time = \(\frac{800}{r-10}\)

To find rate, \(r\), use time, which is defined in terms of \(r\)
(Actual time) - (estimated time) = \(4\) hours

\(\frac{800}{r-10} - \frac{800}{r} = 4\)

\(\frac{800r - 800(r-10)}{r(r-10)} = 4\)

\(800r - 800r + 8000 = 4* r(r-10)\)

\(8000 = 4 * r(r-10)\)

\(2000 = r^2 - 10r\)

\(r^2 - 10r - 2000 = 0\)

\((r - 50)(r + 40) = 0\)

\(r = 50\)

Answer D
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Re: A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 23 Mar 2018, 12:41
Bunuel wrote:
A painter was paid $800 as labor cost for painting the exterior of the front of a house. This amount was the painter’s estimated labor cost based on a regular rate of R, in dollars per hour, for an estimated time T, in hours. However, the actual time it took for the painter to do the job was 4 hours longer than the estimated time, thereby reducing the hourly rate for the job by $10 per hour. What is R, the painter’s regular rate, in dollars per hour?

(A) 20
(B) 30
(C) 40
(D) 50
(E) 60


We can create the equation:

RT = 800

T = 800/R

and

(T + 4)(R - 10) = 800

RT + 4R - 10T - 40 = 800

RT + 4R - 10T = 840

Substituting, we have:

R(800/R) + 4R - 10(800/R) = 840

800 + 4R - 8000/R = 840

4R - 8000/R = 40

Multiplying by R, we have:

4R^2 - 8000 = 40R

4R^2 - 40R - 8000 = 0

Dividing by 4, we have:

R^2 - 10R - 2000 = 0

(R - 50)(R + 40) = 0

R = 50 or R = -40

Since R can’t be negative, R = 50.

Answer: D
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Re: A painter was paid $800 as labor cost for painting the exterior of the  [#permalink]

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New post 26 Mar 2018, 11:16
Wasted my time trying to go for an algebraic solution but such questions could best be solved using options.
We know that - R*T=800.
Let C be the correct answer.
Then, R=40; T=20.

Now, T is increased by 4 hours and R is reduced by $10.
R(new) = 30; T(new) = 24. -> RT(New) = 720.

So we know that R cannot be 20 and 30 either because the decrease in number is by 10 while the increase is by 4 only. Thus, the product will get smaller.

Therefore, for easier calculations, let R=50, then T=16.
When R is reduced by $10 and T is increased by 4 hours, then R=40 and T=20.

Thus, both remain the same.

So the correct answer is 50.
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Re: A painter was paid $800 as labor cost for painting the exterior of the   [#permalink] 26 Mar 2018, 11:16
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