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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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Bunuel wrote:
macjas wrote:
At his regular hourly rate, Don had estimated the labour cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?

(A) 28
(B) 24
(C) 16
(D) 14
(E) 12


Say the regular hourly rate was \(r\)$ and estimated time was \(t\) hours, then we would have:

\(rt=336\) and \((r-2)(t+4)=336\);

So, \((r-2)(t+4)=rt\) --> \(rt+4r-2t-8=rt\) --> \(t=2r-4\).

Now, plug answer choices for \(t\) and get \(r\). The pair which will give the product of 336 will be the correct answer.

Answer B fits: if \(t=24\) then \(r=14\) --> \(rt=14*24=336\).

Answer: B.

Hope it's clear.


My confusion is: Will not the estimated repair cost be changed?
I solved the question with the tougher way:
(336/T)- (336/T+4) = 2
It needs lots of calculations.
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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Hi Baten80,

The first sentence of the prompt tell us that Don estimated the cost as $336 and was PAID that amount. The additional information tells us that Don had to work MORE hours and was thus paid LESS per HOUR overall - meaning that the cost did NOT change.

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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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The quadratic algebra can get a little complicated here, and the factor tree method I suggested 5 years ago is not always ideal, so I would also suggest straightforward backsolving (trying the answers and adjusting accordingly) as an alternative, easier way to solve this problem.

What is backsolving? Essentially, it involves trying your answers to see whether the rest of the numerical details in the question stem fall into place, then adjusting in the correct direction (higher or lower) when necessary. If you adjust in the correct direction, then you should never have to try more than 2 answers.

Let's start in the middle with Choice C and test. If Don had estimated 16 hours, then he would be making $336/16 = $21/hour. However, the question tells us that he actually worked 4 hours more than the estimate, at $2/less, which in this case would have been 20 hours x $19/hour = $380. That's too much (the actual total was $336), so he must have estimated more hours at a lower rate. Try a higher number for the estimate, then: 24 hours.

If he estimated 24 hours, then he would be making $336/24 = $14/hour. The question tells us that he actually worked 4 hours more than the estimate, at $2/less, which in this case would have been 28 hours x $12/hr = $336.

This is the correct total, so Choice B is correct.

-Brian

Originally posted by mcelroytutoring on 30 May 2021, 14:55.
Last edited by mcelroytutoring on 31 May 2021, 07:14, edited 4 times in total.
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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Video solution from Quant Reasoning starts at 0:26:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
avigutman wrote:
Video solution from Quant Reasoning starts at 0:26:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1


avigutman

This is so helpful, thank you!!!
Initially, I got stuck using this method because I said wait the prime factorization of 28 has an extra 2, so when 24 and 28 are combined, the prime factors of those two numbers added does not exactly equal the prime factors of 336... so that cannot be my answer. Why aren't we looking for an exact match here in prime factors?
Thank you again.
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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woohoo921 wrote:
Initially, I got stuck using this method because I said wait the prime factorization of 28 has an extra 2, so when 24 and 28 are combined, the prime factors of those two numbers added does not exactly equal the prime factors of 336... so that cannot be my answer. Why aren't we looking for an exact match here in prime factors?


woohoo921 if I understood you correctly, you multiplied 24*28 (24 is 2^3 * 3^1, and 28 is 2^2 * 7^1, so their product is 2^5 * 3^1 * 7^1).
Is that what you did? If so, that was your mistake. 24 and 28 independently need to be factors of 336, but not simultaneously.
As an analogy, 6 is a factor of 12 and 4 is a factor of 12, but 6*4=24 is not a factor of 12.
Multiplying 24 by 28 in this problem isn't necessary. Rather, this problem plays on the equivalence of 24*14 and 28*12.
Does that help?
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
avigutman wrote:
woohoo921 wrote:
Initially, I got stuck using this method because I said wait the prime factorization of 28 has an extra 2, so when 24 and 28 are combined, the prime factors of those two numbers added does not exactly equal the prime factors of 336... so that cannot be my answer. Why aren't we looking for an exact match here in prime factors?


woohoo921 if I understood you correctly, you multiplied 24*28 (24 is 2^3 * 3^1, and 28 is 2^2 * 7^1, so their product is 2^5 * 3^1 * 7^1).
Is that what you did? If so, that was your mistake. 24 and 28 independently need to be factors of 336, but not simultaneously.
As an analogy, 6 is a factor of 12 and 4 is a factor of 12, but 6*4=24 is not a factor of 12.
Multiplying 24 by 28 in this problem isn't necessary. Rather, this problem plays on the equivalence of 24*14 and 28*12.
Does that help?


To clarify, from my understanding your method was that the prime factorization of 336 is= (2^4)(3^1)(7^1). Then, you found the prime factorization of 24 which is (2^3)(3^1). You then said that you are missing (2^2)(2^7) and that the missing primes come from the prime box of 28. The prime box of 28 is then (2^2)(7). Correct? So basically it all evens out?
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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woohoo921 wrote:
Then, you found the prime factorization of 24 which is (2^3)(3^1). You then said that you are missing (2^2)(2^7) and that the missing primes come from the prime box of 28. The prime box of 28 is then (2^2)(7). Correct? So basically it all evens out?


No, woohoo921, at 8:30 I say that once we've accounted for 24, we still need a (2^1)(7^1) to get to 336. That's 14, not 28.
Then, separately I find the prime box of 28 (2^2)(7^1) and discover that I'd need a (2^2)(3^1) to get to 336. That's 12.
In conclusion, answer choice B has us getting to 336 with 24*14 vs with 28*12, and these two products are a match for the description in the question stem (28 is 4 more than 24, and 12 is 2 less than 14).
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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macjas wrote:
At his regular hourly rate, Don had estimated the labor cost of a repair job as $336 and he was paid that amount. However, the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate. What was the time Don had estimated for the job, in hours?

(A) 28
(B) 24
(C) 16
(D) 14
(E) 12


We can Plug In The Answers (PITA). I like trying B and D.

B: If it had taken the time expected, he'd have made 336/24 = $14 per hour. It took 28 hours, so he made 336/28 = $12 per hour. Is that what we wanted? Yep!

Answer choice B.


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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
Can someone please explain the question? How did he receive the same amount when the hours are increased?
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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aksh123456123456 wrote:
Can someone please explain the question? How did he receive the same amount when the hours are increased?


Hi aksh123456123456,

The first sentence of the prompt tell us that Don estimated the cost as $336 and was PAID that amount. However, the additional information tells us that Don had to work MORE hours than he originally estimated the job would take - but since he was paid the estimated cost ($336), he was thus paid LESS per HOUR overall.

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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
Is it just me or does this not feel like a 2min question ? Let’s assume you made it to the quadratic and then started substitution of answers , still takes a lot of time .

Posted from my mobile device
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Re: At his regular hourly rate, Don had estimated the labor cos [#permalink]
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SignUp wrote:
Is it just me or does this not feel like a 2min question ? Let’s assume you made it to the quadratic and then started substitution of answers , still takes a lot of time .

Posted from my mobile device


Hi SignUp,

Most GMAT questions are written so that they can be approached in more than one way - and in this prompt, working through a math-heavy approach using Quadratics would likely take you far more than 2 minutes (you can actually see from the 'stats' for this question that lots of people were not able to solve it in 2 minutes - or even close to it). There's actually a much faster way to answer this question that does not require a Quadratic at all - but you still have to do some basic multiplication and division to find the correct answer. As an aside, the 'goal' to answer every Quant question in 2 minutes or less is a terrible idea - and if you are performing at a high level, then you will absolutely see some Quant questions on the Official GMAT that might take you upwards of 3 minutes to solve.

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At his regular hourly rate, Don had estimated the labor cos [#permalink]
$/hr = $336/hours worked.

Regular hours worked- 336/x

current hours worked - (regular hours worked + 4) -----336/x + 4

$x-2 = 336 / (336/x + 4)

FOIL:
336 + 4X- 672/X - 8 = 336

subtract 336 from both sides and multiply by x

4x^2- 672 - 8x = 0

divide through by 4

x^2 - 2x - 168 = 0

Look through the options for two numbers that add to that multiply to -168 and see if you can manipulate them to also add to -2.

Your two numbers are 14 and 12.

(x-14) (x+12) = 0.

x cant be negative so x =14.

Now the question is for the number of hours. remember that x is dollar per hour.

336/14 = 24 hours worked.
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