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

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

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New post 26 Nov 2017, 11:25
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 labor cos  [#permalink]

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New post 26 Dec 2017, 04:39
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.


Hello Bunuel, here is my solution

let total number of hours be y
let hourly rate be x
Total amount payed x*y= 360 ---> Hourly rate x= y/336
to complete job it took 4 hours longer --> y+4
hourly rate reduced by 2 USD ---> x-2

(y+4)(x-2) = 336
plug in into above equation x= y/336

(y+4)(y/336 - 2)

336y/y - 2y+1344/y - 8 = 336
336 -2y+1344/y-8=336
-2y+1344/y= 336+8-336
-2y+1344/y =8
ok after this I got stuck and confused, where am I going what I am solving :)

can you please explain what have I done wrong ? I decoded the info correctly I mean expressed initially in numbers.

thanks for your help.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 05:57
dave13 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 \(r=14\) --> \(rt=14*24=336\).

Answer: B.

Hope it's clear.


Hello Bunuel, here is my solution

let total number of hours be y
let hourly rate be x
Total amount payed x*y= 360 ---> Hourly rate x= y/336
to complete job it took 4 hours longer --> y+4
hourly rate reduced by 2 USD ---> x-2

(y+4)(x-2) = 336
plug in into above equation x= y/336

(y+4)(y/336 - 2)

336y/y - 2y+1344/y - 8 = 336
336 -2y+1344/y-8=336
-2y+1344/y= 336+8-336
-2y+1344/y =8
ok after this I got stuck and confused, where am I going what I am solving :)

can you please explain what have I done wrong ? I decoded the info correctly I mean expressed initially in numbers.

thanks for your help.


If \(xy = 336\), then \(x = \frac{336}{y}\), not x = y/336.

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

\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

\(y^2 + 4y - 672 = 0\);

\(y(y + 4) = 672\)

Plug options: \(y = 24\) works.

Hope it helps.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 07:50
here is my solution

let total number of hours be y
let hourly rate be x
Total amount payed x*y= 360 ---> Hourly rate x= y/336
to complete job it took 4 hours longer --> y+4
hourly rate reduced by 2 USD ---> x-2

(y+4)(x-2) = 336
plug in into above equation x= y/336

(y+4)(y/336 - 2)

336y/y - 2y+1344/y - 8 = 336
336 -2y+1344/y-8=336
-2y+1344/y= 336+8-336
-2y+1344/y =8
ok after this I got stuck and confused, where am I going what I am solving :)

can you please explain what have I done wrong ? I decoded the info correctly I mean expressed initially in numbers.

thanks for your help.[/quote]

If \(xy = 336\), then \(x = \frac{336}{y}\), not x = y/336.

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

\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

\(y^2 + 4y - 672 = 0\);

\(y(y + 4) = 672\)

Plug options: \(y = 24\) works.

Hope it helps.[/quote]

Bunuel - Many thanks ! Just two more question

could you show in detail how you got this \(y^2 + 4y - 672 = 0\); and another question what is the point of factoring ? \(y(y + 4) = 672\) :? I remember when we see such equation\(y^2 + 4y - 672 = 0\); we need to find discriminant ? like D = B^2-4AC
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 07:58
dave13 wrote:
Bunuel - Many thanks ! Just two more question

could you show in detail how you got this \(y^2 + 4y - 672 = 0\); and another question what is the point of factoring ? \(y(y + 4) = 672\) :? I remember when we see such equation\(y^2 + 4y - 672 = 0\); we need to find discriminant ? like D = B^2-4AC


\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

Cancel 336: \(- 2y + 4*\frac{336}{y} - 8 = 0\);

Reduce by 2: \(- y + 2*\frac{336}{y} - 4 = 0\);

Multiply by y: \(-y^2+672-4y=0\)

Re-arrange: \(y^2 + 4y - 672 = 0\);

Here you can solve for y with discriminant formula but it's easier to factor the way shown and PLUG OPTIONS.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 08:26
Bunuel wrote:
dave13 wrote:
Bunuel - Many thanks ! Just two more question

could you show in detail how you got this \(y^2 + 4y - 672 = 0\); and another question what is the point of factoring ? \(y(y + 4) = 672\) :? I remember when we see such equation\(y^2 + 4y - 672 = 0\); we need to find discriminant ? like D = B^2-4AC


\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

Cancel 336: \(- 2y + 4*\frac{336}{y} - 8 = 0\);

Reduce by 2: \(- y + 2*\frac{336}{y} - 4 = 0\);

Multiply by y: \(-y^2+672-4y=0\)

Re-arrange: \(y^2 + 4y - 672 = 0\);

Here you can solve for y with discriminant formula but it's easier to factor the way shown and PLUG OPTIONS.


Thank you Bunuel. When you reduce by 2, why didn't you reduce fraction by 2 as well 336/y Normally when we reduce and or multiply numbers in equation all numbers are subject to be changed accordingly - No ? :? please correct me if I am wrong.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 08:30
dave13 wrote:
Bunuel wrote:
dave13 wrote:
Bunuel - Many thanks ! Just two more question

could you show in detail how you got this \(y^2 + 4y - 672 = 0\); and another question what is the point of factoring ? \(y(y + 4) = 672\) :? I remember when we see such equation\(y^2 + 4y - 672 = 0\); we need to find discriminant ? like D = B^2-4AC


\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

Cancel 336: \(- 2y + 4*\frac{336}{y} - 8 = 0\);

Reduce by 2: \(- y + 2*\frac{336}{y} - 4 = 0\);

Multiply by y: \(-y^2+672-4y=0\)

Re-arrange: \(y^2 + 4y - 672 = 0\);

Here you can solve for y with discriminant formula but it's easier to factor the way shown and PLUG OPTIONS.


Thank you Bunuel. When you reduce by 2, why didn't you reduce fraction by 2 as well 336/y Normally when we reduce and or multiply numbers in equation all numbers are subject to be changed accordingly - No ? :? please correct me if I am wrong.


No. When you divide \(4*\frac{336}{y}\) you get \(\frac{1}{2}*4*\frac{336}{y}=2*\frac{336}{y}\).
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 08:50
Quote:
Quote:
Thank you Bunuel. When you reduce by 2, why didn't you reduce fraction by 2 as well 336/y Normally when we reduce and or multiply numbers in equation all numbers are subject to be changed accordingly - No ? :? please correct me if I am wrong.


No. When you divide \(4*\frac{336}{y}\) you get \(\frac{1}{2}*4*\frac{336}{y}=2*\frac{336}{y}\).


Bunuel - But isn't 336/y a separate number ? Shouldn't we have done so 1/2 *4 and 1/2 * 336/y ? :?
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 08:55
Quote:
Quote:
dave13 wrote:
Thank you Bunuel. When you reduce by 2, why didn't you reduce fraction by 2 as well 336/y Normally when we reduce and or multiply numbers in equation all numbers are subject to be changed accordingly - No ? :? please correct me if I am wrong.


No. When you divide \(4*\frac{336}{y}\) you get \(\frac{1}{2}*4*\frac{336}{y}=2*\frac{336}{y}\).


Bunuel - But isn't 336/y a separate number ? Shouldn't we have done so 1/2 *4 and 1/2 * 336/y ? :?


No. Let me ask you: what is the value of 4*6 when reduced by 2? Is it 2*6 = 12 or 2*3 = 6? I think you'll benefit if you brush up fundamentals before practicing questions.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 09:06
Quote:
Quote:
Bunuel wrote:

No. When you divide \(4*\frac{336}{y}\) you get \(\frac{1}{2}*4*\frac{336}{y}=2*\frac{336}{y}\).


Bunuel - But isn't 336/y a separate number ? Shouldn't we have done so 1/2 *4 and 1/2 * 336/y ? :?


No. Let me ask you: what is the value of 4*6 when reduced by 2? Is it 2*6 = 12 or 2*3 = 6? I think you'll benefit if you brush up fundamentals before practising questions.


@Bunuel- thanks for asking me this question :) I think if we reduce 4*6 by 2 the answer will be 12 because 4*6 is one number. I have brushed up fundamentals but sometimes in some PS questions i encounter some technical details i have a vague idea about or simply cant use theory in practice. Thanks a lot for explanation :)
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 09:14
dave13 wrote:
@Bunuel- thanks for asking me this question :) I think if we reduce 4*6 by 2 the answer will be 12 because 4*6 is one number. I have brushed up fundamentals but sometimes in some PS questions i encounter some technical details i have a vague idea about or simply cant use theory in practice. Thanks a lot for explanation :)


Generally \(\frac{1}{x}*(a + b) = \frac{a}{x} + \frac{b}{x}\) but \(\frac{1}{x}*(a*b) = \frac{a*b}{x}\)
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Dec 2017, 10:59
Bunuel wrote:
dave13 wrote:
Bunuel - Many thanks ! Just two more question

could you show in detail how you got this \(y^2 + 4y - 672 = 0\); and another question what is the point of factoring ? \(y(y + 4) = 672\) :? I remember when we see such equation\(y^2 + 4y - 672 = 0\); we need to find discriminant ? like D = B^2-4AC


\(336 - 2y + 4*\frac{336}{y} - 8 = 336\);

Cancel 336: \(- 2y + 4*\frac{336}{y} - 8 = 0\);

Reduce by 2: \(- y + 2*\frac{336}{y} - 4 = 0\);

Multiply by y: \(-y^2+672-4y=0\)

Re-arrange: \(y^2 + 4y - 672 = 0\);

Here you can solve for y with discriminant formula but it's easier to factor the way shown and PLUG OPTIONS.


Bunuel could you please tell me in which case should I use factoring and in which case conventional method of finding discriminant ? Thanks a lot answering my questions. Highly appreciated!
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 21 Jul 2018, 02:42
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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.



Solving Quadric equation is taking too much time. So What we take back solving approach.
RT = 336
and, (R-2)(T+4)=336-----(ii)
If we start back solving from C then we get Time 16 and Rate 21, these values don't satisfy equation (ii)
the given answer choices are increasing so we don't have to try D and E options.
Trying B, Time is 24 and Rate is (334/24)= 14
Thus, Puting the values in equation ii, (14-2)(24+4) = 12 x 28 = 336
Ans B.
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Re: At his regular hourly rate, Don had estimated the labor cos  [#permalink]

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New post 26 Sep 2018, 05:00
Well what I did was basically to prime factorize \(336 = \frac{2^4 * 3 * 7}{x}\)

I then tried \(x = 24 = 2^3 * 3\) back in the equation and ended up with \(2 * 7 = 14\)

Now if we add \(x+4 = 24 + 4 = 28\)
\(28 = 2^2 * 7 = 28\) and that would leave \(2^2 * 3 = 12\)
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Re: At his regular hourly rate, Don had estimated the labor cos &nbs [#permalink] 26 Sep 2018, 05:00

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