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

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At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 31 Jul 2012, 09:01
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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
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 31 Jul 2012, 09:21
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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.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 01 Aug 2012, 09:39
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[336][/X] - [336][/(X+4)]= 2

Solve for X.

Ans= 24 since -28 is not a valid answer.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 07 Dec 2012, 02:03
I have just worked on OG Math practice questions and hardly have I solved this question. That's why I have used Google and found you guys :)
sayak636 wrote:
[336][/X] - [336][/(X+4)]= 2
I have composed the same equation, however its solving has taken me for ages.

I like Bunuel's solution, but I has not guessed to do the same. I'd only slightly change the course of solving. When we get to \(t = 2r - 4\), \(r\) easily seems to be replaced by \(336/t\). Now we have \(t = (2*336/t) - 4\) and can plug answer choices to find out the correct option.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 15 Jun 2013, 19:41
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While substitution does tend to take long for this problem, before substitution you could
factorize 336 to its primes = 2*2*2*2*3*7


Now you can begin to substitute : Ans
Choice A = 28*12 (2*2*7*2*2*3) not equal to 32*10 (clearly its 320 and not 336)
Choice B = 24*14 (2*2*2*3*2*7) equals 28*12 (from prev choice)

thx
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 16 Jun 2013, 03:50
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Shiv636 wrote:
[336][/X] - [336][/(X+4)]= 2

Solve for X.

Ans= 24 since -28 is not a valid answer.


Infact, one doesn't need to solve after this step too:

\(\frac{336}{x} - \frac{336}{(x+4)} = 2\)

336[(x+4)-x] = 2*x(x+4)

x(x+4) = 672

From the given options, we can straightaway eliminate A and C, as because the units digit after multiplication of 28*(28+4) and 16*(16+4) will never be 2.

We also know that 14*20 = 280 and 12*20 = 240. Thus, 14*18(D) or 12*16(E) can never equal 672.

By eliminaion, the answer is B.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 07 Jan 2014, 07:41
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.


On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.

I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 07 Jan 2014, 21:38
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Rdotyung 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


On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.

I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?


You utilize the answer choices whenever you CAN. Here I would keep an eye on the choices right from the start. I would say
R*T = 336 (his regular hourly rate * time he estimated)
The options give us the value of T which is an integer.

\(336 = 2^4*3*7\)

So R*T = 336
(R-2)*(T + 4) = 336
So T as well as T+4 should be factors of 336.
If T is 28, T+4 is 32 which is not a factor of 336 so ignore it.
If T is 24, T+4 is 28. Both are factors of 336. Keep it. If T is 24, R is 14. So (R - 2) is 12. 12*28 does gives us 336 so T = 24 must be the correct answer.

But note that if you want to reduce your mechanical work, you need to be fast in your calculations. You cannot spend a minute working on every option or making calculation mistakes.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 23 Apr 2014, 07:06
How do you go from

> 336[(x+4)-x] = 2*x(x+4)
to > x(x+4) = 672?
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 23 Apr 2014, 19:02
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gciftci wrote:
How do you go from

> 336[(x+4)-x] = 2*x(x+4)
to > x(x+4) = 672?



\(336 * [(x+4)-x] = 2 * x * (x+4)\)

\(336 * [x+4 -x] = 2 * x * (x+4)\)

x and -x get cancelled to give:

\(336 * [4] = 2 * x * (x+4)\)

Divide both sides by 2.
\(336 * 2 = x * (x+4)\)

\(672 = x * (x+4)\)
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 27 May 2014, 09:12
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VeritasPrepKarishma wrote:

On my own I got to the step where I need to utilize the answer choices. I didn't know what to do at that point because it never crosses my mind to use the answer choices and backwards solve like this.

I've only ever seen this kind of method recommended when the problem involves second degree equations. Is that a fair statement? You only backwards solve like this when dealing with second degree equations?
[/quote]

You utilize the answer choices whenever you CAN. Here I would keep an eye on the choices right from the start. I would say
R*T = 336 (his regular hourly rate * time he estimated)
The options give us the value of T which is an integer.

\(336 = 2^4*3*7\)

So R*T = 336
(R-2)*(T + 4) = 336
So T as well as T+4 should be factors of 336.
If T is 28, T+4 is 32 which is not a factor of 336 so ignore it.
If T is 24, T+4 is 28. Both are factors of 336. Keep it. If T is 24, R is 14. So (R - 2) is 12. 12*28 does gives us 336 so T = 24 must be the correct answer.

But note that if you want to reduce your mechanical work, you need to be fast in your calculations. You cannot spend a minute working on every option or making calculation mistakes.[/quote]

Hi Karishma, Why do T and T+4 have to be factors of 336? Why cannot rate be a fraction and difference of two fractions can yield an integer in this case 2? What am I missing here? Thanks!
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 27 May 2014, 20:56
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MensaNumber wrote:

Hi Karishma, Why do T and T+4 have to be factors of 336? Why cannot rate be a fraction and difference of two fractions can yield an integer in this case 2? What am I missing here? Thanks!


All the options are integers so value of T must be an integer. So T+4 must be an integer too. Therefore, T and T+4 must be factors of 336. Also, in GMAT, usually numbers are easy since you do not get calculators. So very rarely will you find that rate or time is a fraction. Even if it will be, it will be a simple fraction such as 1/2 etc.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 28 May 2014, 01:22
Karishma, Thanks for your reply. Yeah that makes sense.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 12 Aug 2014, 16:15
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.


Hi Bunuel,

Can you recommend some similar problems?

Thanks!
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At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 07 Jan 2015, 12:51
Let t be the hourly rate and p the price: t * p = $336

Additional 4 hours and $2 less per hour would yield: (t+4) * (p-2) = $336

Since both equations are equal:

336 : p = (336 : (p-2))-4

Solving for p yields 14 (the other solution is negative, so we do not consider it)

At this point probably we are very pressed on time, so the shortcut is to find the answer the last digit of which multiplied by 4 yields 6. 14 squared is not 336 so by elimination it is 24

Answer: B
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At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 18 Jan 2015, 09:05
I back solved this one like this:

I started with option A, but I will only show the correct option, which is B:

I said that r*t=336, so the amount of hours he worked times the money he got for each hour should be his final salary.
Then I substituted the proposed times for t:
r*24=336
r=14 --> This is how much he should have got per hour worked.

But he worked 4 hours more, so 24+2 = 28. Then he actually got 336/28 = 12, per hour.

12 is 2 less than 14, as it is supposed to, so the correct answer is ANS B.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 20 Apr 2015, 11:06
It is very challanging to solve this question in 2 minutes. Are there any strategies on time saving for this types of questions?
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 20 Apr 2015, 17:48
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Hi RussianDude,

You're not expected to answer every question in the Quant section in under 2 minutes, so if you took a little longer than that on this question, then that's fine (as long as you were doing work and not staring at the screen). If you took more than 3 minutes to answer this question, then chances are that YOUR approach is the "long" approach and that you have to practice other tactics.

Here, since the answer choices ARE numbers, we're really looking for an answer that divides into 336 AND when you add 4 to that answer, that sum ALSO divides evenly into 336. The difference between those two rates should be $2 (as the question states). In that way, you can answer this question with some basic division and note-taking (and likely save time and avoid a long-winded Algebra approach).

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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 20 Apr 2015, 20:41
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RussianDude wrote:
It is very challanging to solve this question in 2 minutes. Are there any strategies on time saving for this types of questions?


Check out this solution: at-his-regular-hourly-rate-don-had-estimated-the-labour-cos-136642.html#p1314465
It will save you some time.
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Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] New post 20 Apr 2015, 21:06
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



\(\frac{336}{X} = \frac{336}{X+4}+2\)

X=24
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Re: At his regular hourly rate, Don had estimated the labour cos   [#permalink] 20 Apr 2015, 21:06
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