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# m21 #8

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Director
Status: Final Countdown
Joined: 17 Mar 2010
Posts: 541
Location: India
GPA: 3.82
WE: Account Management (Retail Banking)
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Re: m21 #8 [#permalink]

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14 Sep 2012, 06:44
Initial Ratio:
s/h=9/8
=>s=(9/8)h........(i)

Then,we add more cars to already existing cars so the ratios becomes:
(s+3) / (h+1) =6/5
=> 5s+15=6h+6
=>5{(9/8) * h}=6h+6
=>h=24

so, from (i) s=27

later;
(s+3)/(h+1)=(27+3)/(24+1)=30/25

Hence the difference in no. of the new value of cars;
30-25=5

C wins
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Re: m21 #8 [#permalink]

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17 Sep 2012, 08:06
the original ratio of sedans to hatchbacks was 9:8
3 sedans and 1 hatcback is added, giving a new ratio of 6:5
remember that ratios can also be expressed as fractions

let x be the multiple of the original ratio so that 9x+8x=11x=number of original cars
(9x+3)/(8x+1)=6/5
cross multiply: 5(9x+3)=6(8x+1)
45x+15=48x+6
9=3x
x=3

plugging back in..
what is 9x+3-(8x+1)?
30-25=5
Re: m21 #8   [#permalink] 17 Sep 2012, 08:06

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# m21 #8

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