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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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22 Oct 2015, 12:44
So you have two equations from the question stem. Let x be the hourly rate and y the time in hours:

$$x*y=336$$

and

$$(y+4)*(x-2) = 336$$ which expands to: $$xy-2y+4x-8=336$$

plug in $$x*y=336$$ in the above formula:

$$336-2y+4x-8=336$$, simplify to:

$$2x-y-2=0$$

Now pick numbers which fit, start with C, then go up to B: 2x-24-4=0, if y=24, then x=14 and 2x=28. So this fits.
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At his regular hourly rate, Don had estimated the labour cos  [#permalink]

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29 Nov 2015, 12:43
Given Info: The total hours estimated by Don differs from the actual number of hours required to finish the job. He had to work 4 more hours to finish the job. For this reason, he received $2 less than was estimated to be given per hour for his job. He received a total amount of$336 to complete his job.

Interpreting the Problem: We have to find the time Don had initially estimated to complete his job. This can be worked out by forming two equations from 2 different conditions given to us. One can be worked out on the number of hours initially estimated and his hourly rate to complete the job, and the other could be worked out on the actual hours worked nd actual hourly rate he received for the job. After that, we will equate both the equations to the total payment received and find the hours estimated for the job.

Solution:
Let us assume the time estimated by Don for the job be n hours and let the cost Don charges per hour for the job be $x per hour. 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. Hence, the Answer is B Current Student Joined: 08 Feb 2016 Posts: 3 Location: United States GMAT 1: 700 Q46 V41 GPA: 3.61 WE: Brand Management (Health Care) Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 25 Feb 2016, 20:06 2 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
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Re: At his regular hourly rate, Don had estimated the labour cos  [#permalink]

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

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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.

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 Answer is B VP Joined: 07 Dec 2014 Posts: 1064 At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 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 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 3161 Location: United States (CA) Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 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 given that Don was paid$336, based on his original estimate, so we can say:

w x t = 336

Next we are given that the job took 4 hours longer and that, as a result, he earned 2 dollars less than his regular rate. This leads us to say:

(w – 2)(t + 4) = 336

We rewrite the equation w x t = 336 as w = 336/t. Now we substitute 336/t for w in the equation (w – 2)(t + 4) = 336. Thus, we have:

[(336/t) – 2](t + 4) = 336

After FOILing we have:

336 + (4x336)/t – 2t – 8 = 336

(4x336)/t – 2t – 8 = 0

Multiplying the entire equation by t, we get:

4 x 336 – 2t^2 – 8t = 0

Dividing the entire equation by 2, we get:

2 x 336 – t^2 – 4t = 0 or 672 – t^2 – 4t = 0

We can also rewrite this as: t^2 + 4t – 672 = 0

Now this is where we should be strategic with our answer choices. To solve this quadratic we are looking for two numbers that sum to a positive 4 and multiply to a negative 672. Our answer choices are:

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

There are only two pairs of answer choices that are 4 units apart: 16 and 12, and 28 and 24. Since 24 multiplied by 28 is 672, we know that the numbers that are needed for the factoring are 24 and 28. Thus, we can say:

(t – 24)(t + 28) = 0

We can see that t = 24 or t = -28. However, since we can’t have a negative number of hours, only t = 24 is the correct answer.

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

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29 Oct 2016, 17:06
336/(t+4) = 336/t -2

(336+2t+8)*t = 336t +1344

2t^2+8t-1344=0

t^2+4t-672=0

(t+28)(t-24)=0

t = 24 or -28

Time cannot be negative.

So t=24.

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

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10 May 2017, 23:11
Hi Bunuel ,

Could you please explain , where it given in the question that
w – 2)(t + 4) = 336

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

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11 May 2017, 02:48
abhisheknandy08 wrote:
Hi Bunuel ,

Could you please explain , where it given in the question that
w – 2)(t + 4) = 336

Regards

Say the regular hourly rate was $$r$$$and estimated time was $$t$$ hours. 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
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At his regular hourly rate, Don had estimated the labour cos  [#permalink]

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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. Answer.... B _________________ CAT 99th percentiler : VA 97.27 | DI-LR 96.84 | QA 98.04 | OA 98.95 UPSC Aspirants : Get my app UPSC Important News Reader from Play store. MBA Social Network : WebMaggu Appreciate by Clicking +1 Kudos ( Lets be more generous friends.) What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish". Senior CR Moderator Status: Long way to go! Joined: 10 Oct 2016 Posts: 1394 Location: Viet Nam Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 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. _________________ Intern Joined: 20 May 2017 Posts: 4 At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 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 CEO Joined: 12 Sep 2015 Posts: 2702 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.

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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]

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04 Oct 2017, 02:42
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'm a bit confused here. I see that once I get to t=2r-4 that I can plug in the answers to see which one fits, but If I try answer a, 28, I get 28=2r-4, r=12, 12*28 = 336 which is the same as I get when I plug in answer choice B. I'm sure I'm missing a step here but I can't figure it out. Thanks! Math Expert Joined: 02 Sep 2009 Posts: 47918 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 04 Oct 2017, 03:00 jboog wrote: 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 $$r=14$$ --> $$rt=14*24=336$$.

Hope it's clear.

I'm a bit confused here. I see that once I get to t=2r-4 that I can plug in the answers to see which one fits, but If I try answer a, 28, I get 28=2r-4, r=12, 12*28 = 336 which is the same as I get when I plug in answer choice B.

I'm sure I'm missing a step here but I can't figure it out.

Thanks!

If t = 28, then $$28=2r-4$$ --> $$32=2r$$and $$r = 16$$, not 12.
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Re: At his regular hourly rate, Don had estimated the labour cos  [#permalink]

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26 Nov 2017, 12:25
Can anyone please comment on the following strategy, given that I was able to write down the proper equations (i.e. rt= 336 and (r-2)(t+4) = 336).

I almost immediately realized that I wouldn't be able to solve that in under 4-5 minutes with my current ability in algebra, so I went for pairing...

Considering one of the answers will give me the time "t", and considering the GMAT loves traps, I looked for a combination of r*t that would give me 336, which wasn't too hard looking at the unit digits...

I fully understand I should practice on getting this solved in about 2 minutes, but this is not where I'm at currently. So on test day I would be left with the option of tanking the test by spending 5-6 minutes or randomly guessing. Please let me know your thoughts.

Thanks!
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Re: At his regular hourly rate, Don had estimated the labour cos &nbs [#permalink] 26 Nov 2017, 12:25

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