PS-Number Properties : PS Archive
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PS-Number Properties

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Senior Manager
Joined: 17 Aug 2005
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Location: Boston, MA
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23 Jan 2006, 13:34
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FYI..i don't have an OA for this as it was a challenge problem of the week, sent from Delta Course. They only keep the answers up for a couple of weeks!! But I would like to know the steps in which to solve this. Pls explain.

When the integer x is divided by the integer y, the remainder is 60. Which of the following is a possible value of the quotient x/y?

I. 15.15
II. 18.16
III. 17.17

(A) I only
(B) II only
(C) III only
(D) I and II only
(E) I and III only
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Joined: 20 Nov 2005
Posts: 2911
Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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23 Jan 2006, 14:39
I think its D.

Here is the explanation:

x = y *q + r where x,y,q, and r are integers.

Case I:
x/y = 15.15 i.e x = 15y + 0.15 y

So 0.15 y = 60 i.e y = 400 (an integer). Its possible.

Case II:
x/y = 18.16 i.e x = 18y + 0.16 y

So 0.16 y = 60 i.e y = 375 (an integer). Its possible.

Case III:

x/y = 17.17 i.e x = 17y + 0.17y

So 0.17y = 60 i.e y = 100/17 (Not integer). This is not possible.
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SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

Senior Manager
Joined: 17 Aug 2005
Posts: 392
Location: Boston, MA
Followers: 2

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23 Jan 2006, 15:44
I attempted to solve it a bit differently, but using the same base equation of x=q*y+r. I also got D. Knowing the remainder is 60 it makes much more sense to set r=60 when solving. Is one way better than the other? Did I get lucky with my method of solving?

Case I:
x=15.15y
x=15y+60
therefore, 15.15y=15y+60
y=120
integer, so it is possible

Case II:
x=18.16y
x=18y+60
therefore, 18.16y=18y+60
y=375
integer, so it is possible

Case III:
x=17.17y
x=17y+60
therefore, 17.17y=17y+60
y=60/.17
not integer, not possible
23 Jan 2006, 15:44
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