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# At his regular hourly rate, Don had estimated the labour cos

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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08 Oct 2015, 10:27
Hi paidlukkha,

You can certainly treat this prompt as a 'system' question (2 variables and 2 unique equations.

336 = (X)(H)

However, your second equation is NOT. Since the number of hours increases by 4 and the difference in hourly pay is 2, the equation should be...

336 = (X - 2)(H + 4)

From here, you can proceed with the Algebra and you'll get to the solution.

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From the information in the question

n Hours (estimated) * x(Don charge for job per hour) = $336 Equation 1: $$nx=336$$ Also, from the information in the question n+4(Hours actualltaken to complete the job)*x-2(Don actual payment per hour)=$336
Equation 2: $$(n+4)(x-2)=336$$

Solving equations 1 and 2 for n and x
Putting nx from equation 1 in equation 2

$$336+4x-2n-8=336$$
$$4x-2n=8$$

Putting value of x in terms of n from Equation 1

$$4(336/n)-2n=8$$
$$672/n-n=4$$
$$n^2+4n-672=0$$
$$n^2+28n-24n-672=0$$
$$(n+28)(n-24)=0$$
$$n=24$$(Ignoring n=-28 as number of hours cannot be negative)

So time Don had estimated to finish the job is 24 hours.

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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25 Feb 2016, 20:06
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Reverse Plug in!

1) Start with the middle number and determine in you should go up or down:
336/16= $21/hr 16 is choice C Next add 4 hours -> 336/20 = 16.8. hour Difference is not 2. Move up in hours 2) 336/24 = 14 24 is choice B Next add 4 hours -> 336/28=12 Difference is 2 so this is the answer Kudos [?]: 6 [1], given: 28 BSchool Forum Moderator Joined: 12 Aug 2015 Posts: 2219 Kudos [?]: 806 [0], given: 595 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 09 Mar 2016, 21:39 Hey Anyone Looking for a more algebraic approach? here is what i actually did let it took x hours to complete the Job So wage per hour = 336/x now as per question => it took x+4 hours to do the job and he was paid => 336/x - 2 /hour now total wage must remain constant hence 336 = (336/x - 2 ) * (x+4) => x^2 +8x - 672 = 0 x=-8+52/2 (neglecting the negative value as hours are non negative ) x= 28 Hence A is sufficient ... Would Love your Thoughts on this approach .. MathRevolution _________________ Give me a hell yeah ...!!!!! Kudos [?]: 806 [0], given: 595 Senior Manager Status: Professional GMAT Tutor Affiliations: AB, cum laude, Harvard University (Class of '02) Joined: 10 Jul 2015 Posts: 394 Kudos [?]: 460 [0], given: 53 Location: United States (CA) Age: 37 GMAT 1: 770 Q47 V48 GMAT 2: 730 Q44 V47 GMAT 3: 750 Q50 V42 GRE 1: 337 Q168 V169 WE: Education (Education) At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 26 May 2016, 17:33 This is one of the toughest questions in the OG: 85% incorrect on GMAT Club! I have found that this one is much easier if you just draw a factor tree and use a little bit of trial and error with the various factors (and combinations thereof). Look for one set of numbers that is 4 apart, and the other 2 apart, and you have your answer. You know that the rate and the time are going to be relatively close to one another, because a difference of$2 in Don's hourly rate results in a difference of 4 hours in the time spent on the job. These numbers are not exactly the same, but they are close, suggesting that r and t are relatively close to one another in value.

Attached is a visual that should help.
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Screen Shot 2016-05-26 at 5.33.02 PM.png [ 108.82 KiB | Viewed 1023 times ]

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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27 May 2016, 02:47
Suppose Don estimated that the job will be completed in t hours.

Then his hourly rate becomes $$\frac{336}{t}$$ $per hour. Now when works for 4 hours longer, his hourly rate is$2 less.
so

$$\frac{336}{t+4}$$ will be the hourly rate

$$\frac{336}{t}$$ - 2 = $$\frac{336}{t+4}$$

t(t+4) = 2 *21*16 = 24 * 28

We get t = 24

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At his regular hourly rate, Don had estimated the labour cos [#permalink]

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27 May 2016, 14:35
algebraic approach with one variable
336/t*(t+4)-2(t+4)=336
t^2+4t-672=0
t=24 hours

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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16 Jun 2016, 06:02
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

To solve this problem we can translate the problem with the given information into an equation. Since we don’t know Don's hourly rate nor the time he had estimated for the job, we use two variables:

w = Don’s hourly rate

t = number of hours he estimated for the job

We are told that the job took 4 hours longer than he had estimated and, consequently, he earned $2 per hour less than his regular hourly rate, thus $$(r-2)(t+4)=336$$. Complete solution is here: https://gmatclub.com/forum/at-his-regul ... l#p1109300 _________________ Kudos [?]: 124810 [0], given: 12079 Senior Manager Joined: 13 Mar 2017 Posts: 458 Kudos [?]: 94 [0], given: 53 Location: India Concentration: General Management, Entrepreneurship GPA: 3.8 WE: Engineering (Energy and Utilities) At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 11 May 2017, 03:31 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 Let the regular hourly rate of Don is x. And the estimated time is y. -> xy =336 Since he took 4 hours longer than the estimated time & was paid the$2 per hour less than his regular hourly rate.
(x-2)(y+4) = 336
-> xy - 2y + 4x - 8 = 336
-> 336 + 4x - 2y - 8 = 336
-> 4x -2y = 8
-> 2x - y = 4
-> 2x - 336/x = 4
-> x - 168/x = 2
-> x^2 - 168 - 2x = 0
-> x^2 - 14x + 12x - 168 = 0
-> x(x-14) + 12(x-14) = 0
-> (x+12)(x-14) = 0
-> x = 14

Estimated time y = 336/14 = 24.

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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11 May 2017, 08:37
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

$$x$$ is the hours that Don had estimated, so $$\frac{336}{x}$$ is the money per hour he earned.

We have $$(x+4)\times (\frac{336}{x}-2)=336$$
$$\implies (x+4)(336-2x)=336x \\ \implies 336x - 2x^2 + 4\times 336 - 8x = 336x \\ \implies 2x^2 +8x = 4 \times 336 \\ \implies x^2 +4x = 2\times 336 = 672 \\ \implies x^2 + 4x + 4 = 676 \\ \implies (x+2)^2 = 676$$

Note that $$676 = 2\times 338 =2 \times 2 \times 169 = 2^2 \times 13^2 = 26^2$$
Hence $$(x+2)^2=26^2 \implies x+2=26$$ since $$x > 0$$

$$\implies x=24$$. The answer is B.
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At his regular hourly rate, Don had estimated the labour cos [#permalink]

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22 May 2017, 13:30
Hi guys,

I have two questions:

1) I understand every part of the solution, but I need a lot of time for factoring (last step):

(t^2)+4t-672=0 --> factor
(t-24)(t+28)=0

How do you do that? What is your approach?
What I would do is, split 672 up into its factors, which are 2^5*3*7 ... and then I try every single calculation to find the right figures.

2) How can questions like this be done within 2 minutes?
Even after I knew this question by heart, it took me 5 1/2 minutes to get it done, by writing all the important steps down, without taking a break to think.
There are worse questions than this one, but still ...

Thanks,
Tom

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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07 Jun 2017, 18:30
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. I am still a confused . Could you please show how you arrived at T=2R-4 ? Kudos [?]: 0 [0], given: 113 VP Joined: 09 Jun 2010 Posts: 1411 Kudos [?]: 153 [0], given: 916 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 08 Jun 2017, 03:34 BrainLab wrote: BrainLab wrote: t(t+4)=672 we can estimate here 20*30=600 --> 22*30=660 So we see that only A or B can be the answer here... 24*28 suits better (22*30) A bit too large numbers to calculate, not really a GMAT Style, though it's an official GMAT Question....... Update We can complete a square here: $$t^2+4t=672$$-> $$t^2+4t+4=672+4$$ --> $$(t+2)^2=26^2$$ T=24, -28 Answer (B) pick the answer choices are most easy for this problem . _________________ visit my facebook to help me. on facebook, my name is: thang thang thang Kudos [?]: 153 [0], given: 916 SVP Joined: 12 Sep 2015 Posts: 1755 Kudos [?]: 2302 [0], given: 355 Location: Canada At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 28 Aug 2017, 14:26 Top Contributor 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 Here's an algebraic solution: Let h = # of hours that Don ESTIMATED for the job. So, h + 4 = ACTUAL # of hours it took Don to complete the job. So, IF Don, had completed the job in h hours, his RATE would have =$336/h
However, since Don completed the job in h+4 hours, his RATE was actually = $336/(h + 4) ...consequently, he earned 2$ per hour less than his regular hourly rate.
In other words, (John's estimated rate) - 2 = (John's actual rate)
So, $336/h - 2 =$336/(h + 4)

ASIDE: since the above equation is a bit of a pain to solve, you might consider plugging in the answer choices to see which one works.

Okay, let's solve this: $336/h - 2 =$336/(h + 4)
To eliminate the fractions, multiply both sides by (h)(h+4) to get: 336(h+4) - 2(h)(h+4) = 336h
Expand: 336h + 1344 - 2h² - 8h = 336h
Simplify: -2h² - 8h + 1344 = 0
Multiply both sides by -1 to get: 2h² + 8h - 1344 = 0
Divide both sides by 2 to get: h² + 4h - 672 = 0
Factor (yeeesh!) to get: (h - 24)(h + 28) = 0
Solve to get: h = 24 or h = -28
Since h cannot be negative (in the real world), h must equal 24.

[Reveal] Spoiler:
B

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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01 Sep 2017, 00:29
Plug the Answer choice always take from middle (C)
here 336/16 = 21, so 19*20 not equal to 336 and also it is 380 so go lesser value
(B)
336 /24 = 14, S0 12* 26 = 336 fits so answer is
B
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Re: At his regular hourly rate, Don had estimated the labour cos   [#permalink] 01 Sep 2017, 00:29

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