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# Borja has just returned from abroad. He has been to Mabaland

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Borja has just returned from abroad. He has been to Mabaland [#permalink]  07 Aug 2006, 09:19
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Borja has just returned from abroad. He has been to Mabaland and Turaland and still has 1-lira bills from each company. The Mabalian lira is worth $0.05 and the Turish lira is worth$0.23. If Borja's lira bills are worth $12.57 in all, how many possibilities are there for the number of Turish lira bills he has? (A) 7 (B) 8 (C) 9 (D) 10 (E) more than 10 Last edited by kevincan on 07 Aug 2006, 13:17, edited 1 time in total. VP Joined: 29 Dec 2005 Posts: 1349 Followers: 7 Kudos [?]: 30 [0], given: 0 Re: PS: Lira [#permalink] 07 Aug 2006, 13:09 kevincan wrote: Borja has just returned from abroad. He has been to Mabaland and Turaland and still has 1-lira bills from each company. The Mabalian lira is worth$0.05 and the Turish lira is worth $0.23. If Borja's lira bills are worth$12.57 in all, how many possibilities are there for the number of Turish lira bills he has?

(A) 7 (B) 8 (D) 9 (E) 10 (F) more than 10

Mabaland lira = x
Turaland = y

0.5x + 0.23y = 12.57
the LCM of 0.5 and 0.23 is 1.15.

if x = 1, y = 54.43.
if x = 2, y = 54.22
if x = 3, y = 54.00
so every (12.57 - 0.15) - 23(n)x is divisible by y.
there are (12.57-0.15)/1.15 = 12.42/1.15 = 10.8
so total possibilities = 10.8+1 = 11.8

so should be E (or F).

Last edited by Professor on 07 Aug 2006, 18:27, edited 1 time in total.
VP
Joined: 29 Dec 2005
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kevincan wrote:
Correct! Any easier explanations?

i also seek for any better explanation? laxi, are you arround??????
Intern
Joined: 18 Feb 2006
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we need the ones digit of the 0.23x to equal 2 or 7, so the value of 0.23x can be displayed as 1.15x - 0.23, as it would start the sequence at 0.92.

1.15x - 0.23 < 12.57
1.15x < 12.8
x < 11.13

so 11 possibilities. (E)
CEO
Joined: 20 Nov 2005
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Schools: Completed at SAID BUSINESS SCHOOL, OXFORD - Class of 2008
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E - Total possibilities = 11

5x + 23y = 1257

Only possibilities are when the y is either 4 or 9 because only then last digit of 23y will be 2 or 7.
So all possibilities are 4,14,24,34,44,54 and 9,19,29,39,49.

Because 1257/23 = 54.abcd
So we can't go beyond 54.
_________________

SAID BUSINESS SCHOOL, OXFORD - MBA CLASS OF 2008

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