December 14, 2018 December 14, 2018 09:00 AM PST 10:00 AM PST 10 Questions will be posted on the forum and we will post a reply in this Topic with a link to each question. There are prizes for the winners. December 14, 2018 December 14, 2018 10:00 PM PST 11:00 PM PST Carolyn and Brett  nicely explained what is the typical day of a UCLA student. I am posting below recording of the webinar for those who could't attend this session.
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Re: At his regular hourly rate, Don had estimated the labor cos
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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 (r2)(t+4) = 336). I almost immediately realized that I wouldn't be able to solve that in under 45 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 56 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
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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 \((r2)(t+4)=336\); So, \((r2)(t+4)=rt\) > \(rt+4r2t8=rt\) > \(t=2r4\). 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 > x2 (y+4)(x2) = 336 plug in into above equation x= y/336 (y+4)(y/336  2) 336y/y  2y+1344/y  8 = 336 336 2y+1344/y8=336 2y+1344/y= 336+8336 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
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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 \((r2)(t+4)=336\); So, \((r2)(t+4)=rt\) > \(rt+4r2t8=rt\) > \(t=2r4\). 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/336to complete job it took 4 hours longer > y+4 hourly rate reduced by 2 USD > x2 (y+4)(x2) = 336 plug in into above equation x= y/336 (y+4)(y/336  2) 336y/y  2y+1344/y  8 = 336 336 2y+1344/y8=336 2y+1344/y= 336+8336 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
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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/336to complete job it took 4 hours longer > y+4 hourly rate reduced by 2 USD > x2 (y+4)(x2) = 336 plug in into above equation x= y/336 (y+4)(y/336  2) 336y/y  2y+1344/y  8 = 336 336 2y+1344/y8=336 2y+1344/y= 336+8336 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^24AC



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Re: 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
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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^24AC \(336  2y + 4*\frac{336}{y}  8 = 336\); Cancel 336: \( 2y + 4*\frac{336}{y}  8 = 0\); Reduce by 2: \( y + 2*\frac{336}{y}  4 = 0\); Multiply by y: \(y^2+6724y=0\) Rearrange: \(y^2 + 4y  672 = 0\); 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
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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^24AC \(336  2y + 4*\frac{336}{y}  8 = 336\); Cancel 336: \( 2y + 4*\frac{336}{y}  8 = 0\); Reduce by 2: \( y + 2*\frac{336}{y}  4 = 0\); Multiply by y: \(y^2+6724y=0\) Rearrange: \(y^2 + 4y  672 = 0\); 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
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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
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Re: At his regular hourly rate, Don had estimated the labor cos
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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
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26 Dec 2017, 09:14



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Re: At his regular hourly rate, Don had estimated the labor cos
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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^24AC \(336  2y + 4*\frac{336}{y}  8 = 336\); Cancel 336: \( 2y + 4*\frac{336}{y}  8 = 0\); Reduce by 2: \( y + 2*\frac{336}{y}  4 = 0\); Multiply by y: \(y^2+6724y=0\) Rearrange: \(y^2 + 4y  672 = 0\); 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
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21 Jul 2018, 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 \((r2)(t+4)=336\); So, \((r2)(t+4)=rt\) > \(rt+4r2t8=rt\) > \(t=2r4\). 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, (R2)(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, (142)(24+4) = 12 x 28 = 336 Ans B.
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Re: At his regular hourly rate, Don had estimated the labor cos
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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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