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PS 2

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SVP
SVP
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PS 2 [#permalink] New post 01 Jun 2004, 17:42
00:00
A
B
C
D
E

Difficulty:

  5% (low)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
Answer with Solutions.....
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SVP
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 [#permalink] New post 01 Jun 2004, 18:50
assume normally N items are produced at cost C
if additiona n items are produced then actual cost = (N+n) * C

But over production causes savings by x dollars per unit
So savings = (N+n) * x these savings should will be depleted if n itemas are stored for d number of days at y per day

so n*y*d = (N+n)*x
d = x(N+n)/yn = x/y[ 1+N/n ]

none of the answers match this
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 [#permalink] New post 01 Jun 2004, 21:39
anandnk wrote:
assume normally N items are produced at cost C
if additiona n items are produced then actual cost = (N+n) * C

But over production causes savings by x dollars per unit
So savings = (N+n) * x these savings should will be depleted if n itemas are stored for d number of days at y per day

so n*y*d = (N+n)*x
d = x(N+n)/yn = x/y[ 1+N/n ]

none of the answers match this


I guess we have to assume N=0 for this problem in which case the ans is x/y

What's the OA?
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 [#permalink] New post 02 Jun 2004, 06:27
Total savings xn
Total cost of storage yn

OA xn/yn = x/y
Director
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 [#permalink] New post 05 Jun 2004, 20:17
total amount saved for n units XN
total amount spent on storing N units = YND where D is # of days

there fore when YND = XN

D = X/N :)
_________________

Praveen

  [#permalink] 05 Jun 2004, 20:17
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