let us consider 5+7+4=36 47 30
first two digits come from ([5+1]*4)+(5+7)=36
then next 2 digits come from [5*(7+1)]+7=47
the last 2 come from 5*(5+1)=30
Similarly if we do for 7+3+5,we get
first 2 digits are ([7+1]*5)+(7+3)=50
next 2 are [7*(3+1)]+3=31
last 2 are 7*(7+1)=56
therefore answer is 503156....
IAS Aptitude Question
- indiainspire
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RE: IAS Aptitude Question
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Rdx
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RE: IAS Aptitude Question
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Varun Galar
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RE: IAS Aptitude Question
1st set: a*c + a+b+c = 5*4+5+7+4 = 36
2nd set: a*b + a+b = 35+12 = 47
3rd set: a*a+a = 9*9+9 = 90
1. 5 + 7 + 4 = 364730.
30 = (5*5)+5
47 = (7*5) + 5+7
36 = (4*5) + 5 + 7 + 4
so 7 + 3 + 5 = 503156
2nd set: a*b + a+b = 35+12 = 47
3rd set: a*a+a = 9*9+9 = 90
1. 5 + 7 + 4 = 364730.
30 = (5*5)+5
47 = (7*5) + 5+7
36 = (4*5) + 5 + 7 + 4
so 7 + 3 + 5 = 503156
Last edited by Varun Galar on Sat Jun 27, 2015 10:42 am, edited 1 time in total.