Lucas Read 30 of His Book Containing 480 Pages
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Tom read a book containing 480 pages by reading the aforementioned num [#permalink] 
Updated on: 02 April 2013, 14:30
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Question Stats:
68% (02:54) correct 32% (03:22) incorrect based on 242 sessions
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Tom read a book containing 480 pages past reading the same number of pages each day. If he would have finished the book 5 days before by reading 16 pages a 24-hour interval more, how many days did Tom spend reading the book?
A. 10
B. 12
C. 15
D. 16
East. 18
m15 q20
Originally posted by beckee529 on 30 Oct 2007, 06:18.
Final edited by Bunuel on 02 April 2013, xiv:30, edited 1 time in total.
Edited the question.
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Posts: 84583
Re: Tom read a book containing 480 pages past reading the aforementioned num [#permalink] 02 Apr 2013, 14:30
m15 q20
Tom read a book containing 480 pages by reading the aforementioned number of pages each day. If he would have finished the volume 5 days earlier by reading sixteen pages a mean solar day more, how many days did Tom spend reading the book?
A. 10
B. 12
C. 15
D. 16
E. 18
Say the number of days Tom spent reading the book was \(d\). So:
The number of pages he read per twenty-four hours would be \(\frac{480}{d}\);
The number of pages he would read per solar day with an increased speed would exist \(\frac{480}{d-5}\);
Nosotros are told that the number of pages for the second case was xvi pages per day more, so \(\frac{480}{d}=\frac{480}{d-5}-16\). At this indicate it'southward much improve to plug answer choices rather than solve for \(d\). Answer choice C fits: \(\frac{480}{15}=32=\frac{480}{15-5}-16\).
Answer: C.
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Re: quad eqn [#permalink] xxx October 2007, 06:48
beckee529 wrote:
Tom read a book containing 480 pages reading the same number of pages each day. If he had read xvi pages a twenty-four hours more, he would have finished the book 5 days earlier. How many days did Tom spend reading the volume?
10
12
fifteen
16
xviii
so i was able to fix the equation correctly
[480 / x] = [480 / (x+16)] + 5
the working it out
[480 / x] = [480 + v(x + 16)] / (x+16)
480x + 7680 = 5x^2 + 560x
5x^2 + 80x -7680
finally dividing by 5
x^ii + 16x - 1536
my question is without a calculator on the exam what is the easiest manner/ best plan of assail to realize that
x^2 + 16x - 1536 is...
(ten + 48) (ten - 32)
ten = 32
then finally 480/x = 480/32 = 15 which is the answer
there was no fashion i was able to do this within two minutes.. also much factorizing and room for error.. for those larger quadratic equations. would backsolving be a good option here? please advise, thanks.
If I become to a signal where the equation takes too long to solve, I would just kickoff plugging in the respond choices...
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Joined: 30 October 2007
Posts: 5
[#permalink] 30 Oct 2007, 06:53
Actually u can set up 2 equation
P--stands for the pages
D--stands for the days
ane) P*D=480 (we want to observe the Days, and then P=480/D)
2) (P+16)(D-5)=480 => PD-5P+16D-80=480
as the 1) stated u tin put 1) into 2)
=> 480-5P+16D-eighty=480 => 16D-5P=80
put the bold ane into information technology => 16D-5(480/D)=80
the we get the terminal equation 16D^two-2400=80D (divide 16)
=> D^2-5D-150=0
(D-15)(D+10)=0 and so D=15 days
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Joined: 10 Apr 2012
Posts: 21
Concentration: Finance, Economics
Re: Tom read a book containing 480 pages reading the aforementioned number [#permalink] 02 Apr 2013, 14:21
Meet Img for a back up method.
Attachments
Note_20130401_173652_05.jpg [ 111.42 KiB | Viewed 12791 times ]
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Re: Tom read a volume containing 480 pages reading the same number [#permalink] 02 Apr 2013, 14:25
samsonfred76 wrote:
See Img for a back upwardly method.
I like this approach a lot, but on the GMAT information technology is a lilliputian math intensive. Seeing that 480 is divisible by 48 is petty, but seeing that it'southward not divisible by 64 might have a piffling bit longer.
Great approach that works well on these types of questions, and works particularly well on ratio questions. I'd advise using this concept if y'all're quick at math and don't like setting up algebraic formulae. All roads lead to Rome, equally I like to say.
Thank you!
-Ron
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Re: Tom read a volume containing 480 pages by reading the aforementioned num [#permalink] 02 Apr 2013, xx:43
beckee529 wrote:
Tom read a book containing 480 pages by reading the aforementioned number of pages each mean solar day. If he would have finished the book 5 days before by reading 16 pages a twenty-four hours more, how many days did Tom spend reading the volume?
A. 10
B. 12
C. 15
D. 16
E. eighteen
m15 q20
Allow the charge per unit of reading the book(the number of pages being read each day) be p and the total number of days exist d. Thus, r*d = 480. Also, (r+16)*(d-5) = 480. Thus, both d and (d-five) have to be a factor of 480. That eliminates options B,D,Eastward. Left with A and C. Pick up any i and try plugging in. In any case, you plug-in merely once.
C.
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Re: Tom read a book containing 480 pages past reading the same num [#permalink] 13 Nov 2013, 06:10
beckee529 wrote:
Tom read a book containing 480 pages by reading the aforementioned number of pages each day. If he would take finished the volume five days earlier past reading sixteen pages a mean solar day more than, how many days did Tom spend reading the volume?
A. 10
B. 12
C. xv
D. 16
East. eighteen
m15 q20
For these questions Backsolving is definetely the best approach.
If yous start with C the answer should popular in xl seconds tops.
Thank you
J 
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Joined: xiii Dec 2013
Posts: 30
Re: Tom read a book containing 480 pages by reading the same num [#permalink] 18 Jul 2014, 17:25
Started writing the equation and it took besides long.
Then realized it's pretty easy using values:
Started with B:
12 days = 40 pages a day
7 days (i.east. 12 - 5) wouldn't yield 56 pages a twenty-four hours to reach 480 full. It would yield vii*56 = 392 (yet missing 78 pages)
Went to C:
15 days = 32 pages a mean solar day
10 days = 48 pages a day (32 + 16).
Equation holds, so it is the reply.
Hope it helps.
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Re: Tom read a volume containing 480 pages by reading the aforementioned num [#permalink] 18 Sep 2017, 04:09
beckee529 wrote:
Tom read a book containing 480 pages by reading the same number of pages each day. If he would take finished the book 5 days before by reading sixteen pages a day more, how many days did Tom spend reading the book?
A. 10
B. 12
C. xv
D. 16
E. 18
m15 q20
Algebra is i style of doing it. Another is past plugging in values. As usual, we start with the middle value so that we tin can go up or downwardly depending on whether the days are short or in excess.
Say number of days = 15
The Tom reads 480/xv = 32 pages everyday.
In 5 days, he reads 32*5 pages which when split equally across the residue of the days such that each solar day gets 16 pages gives us 32*5/sixteen = ten days
These 10 days are five less than 15 and hence everything fits.
Answer (C)
Of course, we could take needed to do this calculation in one case once more in instance (C) did not fit.
For instance, trying (B), 480/12 = forty pages
And then in five days, the pages read = 40*v. When we try to split them equally amidst the leftover days such that each day gets sixteen pages, we practise non get an integral number of days. Hence this will not be the answer.
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Re: Tom read a book containing 480 pages by reading the aforementioned num [#permalink] 12 December 2021, 06:eleven
beckee529 wrote:
Tom read a book containing 480 pages past reading the same number of pages each day. If he would have finished the book 5 days earlier by reading 16 pages a day more than, how many days did Tom spend reading the book?
A. 10
B. 12
C. xv
D. 16
E. xviii
m15 q20
ATQ,
\(\frac{480}{X}\)-\(\frac{480}{X+sixteen}\) = 5
X= 32 or -48
At present, 480/X = 15
Ans. C
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Re: Tom read a book containing 480 pages by reading the same num [#permalink]
12 December 2021, 06:11
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