Latest TCS Aptitude Question SOLUTION: 1. If the equation A+B+C+D+E=FG is two digit number, whose value is 10F+G and the letters A,B,C,D,E,F and G each represents different digits.If FG is as large as possible, what is

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1. If the equation A+B+C+D+E=FG is two digit number, whose value is 10F+G and the letters A,B,C,D,E,F and G each represents different digits.If FG is as large as possible, what is the value of G?
(a)4 (b)5 (c)1 (d)3

Asked In TCS (11 years ago)

Unsolved Read Solution (19)

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G.SOWJANYA  (11 years ago)

9+8+7+6+5=35 is not possible..
so now we can go for 34 which is not possible for any coimbination..
and next for33 is not possible..
and nnow we can go for 32 =9+8+6+5+4 is possible..therfore, 2 is the answer..

Is this solution Helpfull? Yes (22) | No (16)  

prasha  (11 years ago)

9+8+7+5+2=31 then anser is 1

Is this solution Helpfull? Yes (16) | No (5)  

sanjana  (11 years ago)

9+8+7+5+2=31..
ans=1

Is this solution Helpfull? Yes (4) | No (2)  

Archana   (11 years ago)

Given That A,B,C,D,E,F are different digits, so assume the values of them as 4,5,6,7,9 on added them as 4+5+6+7+9=31 which represents 3 for F and 1 for G..and the values are not same of the others....so the ans is 1

Is this solution Helpfull? Yes (3) | No (3)  

thangamari  (11 years ago)

ans is 2
9+8+4+6+5=32

Is this solution Helpfull? Yes (3) | No (3)  

DhineshThyagarajan  (11 years ago)

the ans for g is G=5
for simple just go through option if we consider 4 it will not equal when u sub no. for a=0,b=2,c=3,d=4,e=5 which is not equal to FG. Befor proceeding this fix the values for f=1 and A=0 and if we consider option (b)5 sub in equ 10(1)+5=15, and sub values for a=0,b=2,c=3,d=4,e=6 and the sum is 15 and FG is 15 so ans is 5...

Is this solution Helpfull? Yes (3) | No (6)  

vikas  (11 years ago)

9+8+7+6+5=35
10(f)+g
10(3)+5=35
as,FG should be as large as possible,so 5 ans

Is this solution Helpfull? Yes (2) | No (13)  

vikas  (11 years ago)

guys wer it is mentionened u tak values less than 10.
4+6+7+8+10=10(3)+5=35 so 5 ans

Is this solution Helpfull? Yes (2) | No (1)  

niki  (11 years ago)

a=9,b=8,c=7,d=6,e=5 so fg=35
10f+g=35 so f=3 and g=5.. i guess so

Is this solution Helpfull? Yes (1) | No (16)  

Manisha  (11 years ago)

9+8+1+5+2=31 where FG=31 i.e; f=3,g=1.

Is this solution Helpfull? Yes (1) | No (3)  

snehil  (11 years ago)

9+8+6+5+4=32
ie g=2

Is this solution Helpfull? Yes (1) | No (4)  

Spoorthi M N  (9 years ago)

FG is as large as possible and all the 7 numbers should be different.
By trial and Error method,
9 + 8 + 7 + 6 + 5 = 35…5 is getting repeated twice.
9 + 8 + 7 + 6 + 4 = 34…4 is getting repeated
9 + 8 + 7 + 5 + 4 = 33…3 repeats
9 + 8 + 6 + 5 + 4 = 32
None of the numbers repeat in the above case and 32 is the maximum number FG can have. The value of G is 2.

Is this solution Helpfull? Yes (1) | No   

riddhi soni  (11 years ago)

3+4+5+6+9=27 ans

Is this solution Helpfull? Yes | No (4)  

AAKRITI GUPTA   (11 years ago)

9+8+7+5+4=33 so the ans is 3

Is this solution Helpfull? Yes | No (5)  

mahalakshmi  (11 years ago)

largest possible is 2..
but as per the option 1

Is this solution Helpfull? Yes | No (1)  

amam  (11 years ago)

ans will be 31.bcoz
9+8+7+6+5=35 now we need to take f and g as different digit like (1,2,3,4),but by using it we can not get 35 value.similarly for 32,33,34
but for 31 =9+8+7+5+2=10*3+1. which is true .so ans will be 1 for g.

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siri  (11 years ago)

the sum should b maximum ,n digits r diffrerent.
thr is posibility wth 32 instead of 31,so g=1 is not correct
9+8+6+5+4=32=10(3)+2,
g=2,

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sindhu  (10 years ago)

10+9+11+13+8=51
10*5+1=51

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Priyadharsshini  (8 years ago)

9+8+7+5+4 =33
The ans is 3

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