VeritasKarishma wrote:
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
"In order to complete a reading assignment on time, Terry planned to read 90 pages per day. "Reading 90 pages/day would have completed the assignment.
"However, she read only 75 pages per day at first,"She read 15 fewer pages each day till ...
leaving 690 pages to be read during the last 6 days before the assignment was to be completed.690 pages were left for 6 days. Actually as per plan only 90*6 = 540 pages should have been left for 6 days.
Why were there 690 - 540 = 150 extra pages left to read? Because she read 15 fewer pages each day. To gather 150 extra pages left unread, she must have read 15 fewer pages for 10 days.
"How many days in all did Terry have to complete the assignment on time?"She has already read for 10 days and has 6 more to go so she had total 16 days.
Answer (B)
VeritasKarishma, I came up with a similar approach is it correct.
nick 1816, is this approach logical in your opinion ?
If we visualize on a graph 90 is the median and 75 and 115 opp. spectrum.
So, the excess 25 pages, over and above 90 pages (115 -90) for 6 days compensates for 15 days less read for certain days.
Can we derive a formula 15D=25*6, where D is the first days she read 75 pages.
Hence, D = 10
Thanks
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