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

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Director
Joined: 09 Mar 2016
Posts: 608
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 05:39
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. Math Expert Joined: 02 Sep 2009 Posts: 46264 Re: At his regular hourly rate, Don had estimated the labour cos [#permalink] ### Show Tags 26 Dec 2017, 06:57 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 $$r=14$$ --> $$rt=14*24=336$$.

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.
_________________
Director
Joined: 09 Mar 2016
Posts: 608
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 08:50
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
Math Expert
Joined: 02 Sep 2009
Posts: 46264
Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

### Show Tags

26 Dec 2017, 08:58
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.
_________________
Director
Joined: 09 Mar 2016
Posts: 608
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 09:26
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.
Math Expert
Joined: 02 Sep 2009
Posts: 46264
Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 09:30
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 $$\frac{1}{2}*4*\frac{336}{y}=2*\frac{336}{y}$$.
_________________
Director
Joined: 09 Mar 2016
Posts: 608
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 09:50
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 ?
Math Expert
Joined: 02 Sep 2009
Posts: 46264
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 09:55
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.
_________________
Director
Joined: 09 Mar 2016
Posts: 608
At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 10:06
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
Math Expert
Joined: 02 Sep 2009
Posts: 46264
Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 10:14
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 $$\frac{1}{x}*(a*b) = \frac{a*b}{x}$$
_________________
Director
Joined: 09 Mar 2016
Posts: 608
Re: At his regular hourly rate, Don had estimated the labour cos [#permalink]

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26 Dec 2017, 11:59
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!
Re: At his regular hourly rate, Don had estimated the labour cos   [#permalink] 26 Dec 2017, 11:59

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

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