## Re: Plz sir solve Quant Question of number system

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• Hi, (a+b)/2 = (a-24) [where a b] = a+b = 2a-48 = b = a-48 Also, ab = (b+12)^2 = ab = b*b + 144 + 24b = (b+48)b = b^2 + 24b + 144 = b^2 + 48 b = b^2 +
Message 1 of 4 , Oct 5, 2009
Hi,

(a+b)/2 = (a-24) [where a>b]
=> a+b = 2a-48
=> b = a-48

Also,
ab = (b+12)^2
=> ab = b*b + 144 + 24b
=> (b+48)b = b^2 + 24b + 144
=> b^2 + 48 b = b^2 + 24b + 144
=> b = 144/24

=> b = 6 .....ANS
=> a = b+48
=> a = 54 .....ANS

Regards,
Rahul Jain

--- In ascent4cat@yahoogroups.com, Rajeev kumar <rajeev2mail@...> wrote:
>
> The arithmetic mean of two numbers is smaller by 24 than the larger of the two number and the GM of the same number exceeds by 12 the smaller of the number .Find the number??
>
>
>
> Add whatever you love to the Yahoo! India homepage. Try now! http://in.yahoo.com/trynew
>
• Let the two nos. be a and c For the first condition:AM=(a+c)/2and hence,supposing a c,we havea-(a+c)/2=24= (a-c)/2-24= a-c=48..............eqn1Now with the
Message 2 of 4 , Oct 6, 2009
 Let the two nos. be 'a' and 'c'For the first condition:AM=(a+c)/2and hence,supposing a>c,we havea-(a+c)/2=24=>(a-c)/2-24=>a-c=48..............eqn1Now with the second condition:GM=(ac)^1/2hence,(ac)^1/2=c+12,with the same assumption a being greater than c.now ac=(c+12)^2....eqn2hence from 1,we have a=c+48therefore,putting in eqn 2;c(c+48)=(c+12)^2solving this we have the value of 'c'as 6 and 'a' we can deduce as 54(=c+48).Thanks,Anupam --- On Wed, 30/9/09, Rajeev kumar wrote:From: Rajeev kumar Subject: [2IIM CAT Prep] Plz sir solve Quant Question of number systemTo: "\'ascent4cat@yahoogroups.com\'" Date: Wednesday, 30 September, 2009, 7:15 PM  The arithmetic  mean of  two numbers is smaller by 24 than the larger of the two number  and the GM of  the same number exceeds by 12 the smaller of the number .Find the number?? From cricket scores to your friends. Try the Yahoo! India Homepage!

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