- View SourceHi,

(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

> - View SourceLet 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:*<rajeev2mail@...>*

From: Rajeev kumar <rajeev2mail@...>

Subject: [2IIM CAT Prep] Plz sir solve Quant Question of number system

To: "\'ascent4cat@yahoogroups.com\'" <ascent4cat@yahoogroups.com>

Date: Wednesday, 30 September, 2009, 7:15 PMThe 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??

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