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In order to complete a reading assignment on time, Terry

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In order to complete a reading assignment on time, Terry [#permalink]

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In order to complete a reading assignment on time, Terry planned to read 90 pages per day. However, she read only 75 pages per day at first, leaving 690 pages to be read during the last 6 days before the assignment was to be completed. How many days in all did Terry have to complete the assignment on time?

(A) 15
(B) 16
(C) 25
(D) 40
(E) 46

Stuck on this question its from OG 13 - 119
[Reveal] Spoiler: OA

Last edited by Bunuel on 29 May 2014, 01:30, edited 2 times in total.
Renamed the topic and edited the question.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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nelz007 wrote:
In order to complete a reading assignment on time, Terry planned to read 90 pages per day. However, she read only 75 pages per day at first, leaving 690 pages to be read during the last 6 days before the assignment was to be completed. How many days in all did Terry have to complete the assignment on time?

(A) 15
(B) 16
(C) 25
(D) 40
(E) 46

Stuck on this question its from OG 13 - 109


Say Terry read at the rate of 75 pages per day for \(d\) days.

Now, in \(d\) days she read \(75d\) pages, and since after that 690 pages were left, then the number of pages in the book is \(75d+690\).

Next, we also know that she had \(d+6\) days for the assignment. Since Terry planned to read 90 pages per day for this number of days, then the number of pages in book is also equal to \(90(d+6)\).

So, we have that \(75d+690=90(d+6)\) --> \(d=10\). Thus, the total number of days that she has to complete the assignment on time is \(d+6=16\).

Answer: B.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 11 Oct 2012, 20:08
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Here's my attemp..

let say she has x days to complete the task above..

1st Case:

Rate =75
Time = x-6 days
Work = 75(x-6) = 75-450

2nd Case

Rate = ?
Time = 6 days
Work = 690

so rate = 115pages/day

=> total work done => 75x-450+690 =90x

x=16
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Re: Rate/work problem from OG [#permalink]

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New post 11 Oct 2012, 09:43
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Here are my 50 cents
We know that Distance = Time*Speed, thus in our case Distance is the assignment to be completed and T are those days and S is peed, the number of pages per day. From here we get:
D=90*X - according to the plan, assignment to be completed on time reading 90 pages per day for next X days. But, Terry's plans changed so she read as follow:
75 pages for first Y days and 690 pages for last 6 days, we get these equations:
75*Y+690=90*X
X-Y=6 --------->>X planned number of days, Y - actually used reading 75 pages per day and 6 leftover days used to complete a lump 690 pages
From above we find that X=Y+6 and 75Y+690=90Y+540 or 15Y=150 --->>>>> Y=10, hence X=16 (B)
Please, correct me if I went awry.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 25 Oct 2012, 06:35
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690/6 = 115

75 *6 = 450

690-450 =240

240/15 =16 So answer = 16.( just common sense)
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 12 Jun 2015, 03:24
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We can solve this one with weighted average method (see MGMAT)
X*(-15) + 6*(25) = 0
X = 10 --> X+6 = 16 (B)
Let be the first X days at a rate of 75 Pages - It's below planed average of 90 (-15), the last 6 days = 690/6 = 115 --> It's above the average of 90 (+25)
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 23 Dec 2012, 05:32
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I would like to give it a try too.

Initially the planned rate was supposed to be \(90 pages\) per day. So the total number of pages will come out to be \(90*t\) where t is the number of days required to complete the book, when done at the planned rate.

Now the study has been dirtributed. The person reads the book at the rate of \(75 pages\) per day for the \(t-6\) days and for the last \(6\) days, the person reads at \(115\) pages per day.

So the equation becomes \(90t=75(t-6) + 115*6\).
On solving t comes as 16.
Hence the answer.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 25 Oct 2013, 19:20
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Her original goal was to read 90pages per day.
Thus in 6 days she would have read 90*6= 540 pages.
Since she was left with 690 pages means she was 150pages short of target.
Since the difference between 90 pages and 75 pages is 15 pages.
The number of days would be 150/15 = 10 days
Total number of days to complete assignment 10 + 6 days = 16days.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 11 Oct 2012, 19:22
Thanks for the awesome reply javoni and bunuel kudos to both of you.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 14 Nov 2012, 05:21
\(75 pages per day on D number of days + 690 rem. pages\)

\(75 x d + 690 = 90 (d + 6)\)

\(690 - 540 = 90D - 75D\)

\(150 = 15D ==> D = 10 days\)

Answer: \(10 + 6 = 16 days\)
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 23 Dec 2012, 03:35
Well, How about plugging in.

Start off with C=25 days.
We know that for 6 days its 690 pgs. So 25-6=19 days left
Now we have to multiply 75 and 19=1425
Now add 1425 and 690= 2115 pgs
But the planned rate is 90 pgs. Multi ply 90 and 25 we get 2250 pgs.
This is more than 2115. So we need to plugin a lesser number.
Plugin ans choice b and get the answer.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 16 Jun 2013, 17:56
Formula for Avg= Total/no. of units
=> Total= Avg* no. of units

Given,
Planned average reading=90 pages/day
=> Planned total pages for d days=90*d (in pages)

Actual total pages = 75pages/day*(d-6) days +690 pages

equating both since no. of pages are the same :
90d=75(d-6) + 690
d=16

Answer B
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 21 Sep 2013, 02:35
Initial rate :

90 = P/D
Where
P --> total number of pages.
D --> total number of days.

New rate.

75 = P-690 / D-6

Solve both equations.

D=16 days.
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 20 Sep 2014, 07:32
nelz007 wrote:
In order to complete a reading assignment on time, Terry planned to read 90 pages per day. However, she read only 75 pages per day at first, leaving 690 pages to be read during the last 6 days before the assignment was to be completed. How many days in all did Terry have to complete the assignment on time?

(A) 15
(B) 16
(C) 25
(D) 40
(E) 46

Stuck on this question its from OG 13 - 119



Sol:

x= total number of pages

(x-690)/75 + 6 = x/90

x= 240* 6

number of days = x/90 (where x= 240*6) = 16
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Re: In order to complete a reading assignment on time, Terry [#permalink]

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New post 11 Jun 2016, 06:40
nelz007 wrote:
In order to complete a reading assignment on time, Terry planned to read 90 pages per day. However, she read only 75 pages per day at first, leaving 690 pages to be read during the last 6 days before the assignment was to be completed. How many days in all did Terry have to complete the assignment on time?

(A) 15
(B) 16
(C) 25
(D) 40
(E) 46


We can solve this problem in a similar manner to solving a work problem, and we can use a table to organize our information. We can fill in what she was “supposed” to do and what she “did” do.

We are given that her initial plan was to read 90 pages per day for “T” days. We can multiply, getting 90T for the total work that was supposed to be done for the entire reading assignment.

We are next told that she actually started out reading 75 pages per day until she had 6 days left to finish the project. Since T represents the total number of days we can say she read 75 pages per day for (T–6) days. Finally, when we multiply, we see that her work done on the T–6 days was 75T–450.

Image

To finish this problem we need to set up one final equation. We are told that she must read 690 pages in the last 6 days to complete the assignment. From the chart we see that 90T denotes a completed assignment. Thus we can say:

75T – 450 + 690 = 90T

75T + 240 = 90T

15T = 240

T = 16

Answer B.
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Re: In order to complete a reading assignment on time, Terry   [#permalink] 11 Jun 2016, 06:40
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