Latest TCS Aptitude Question SOLUTION: Que. Given A+B+C+D+E= FG. If FG is Maximum as possible and A,B,C,D,E,F,G all are distinct digits(0-9) then what is the value of G. a) 1 b) 2 c) 3 d) 4 e) 5

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Que. Given A+B+C+D+E= FG. If FG is Maximum as possible and A,B,C,D,E,F,G all are distinct digits(0-9) then what is the value of G.
a) 1
b) 2
c) 3
d) 4
e) 5

Asked In TCS Deven Patidar (8 years ago)

Solved milan jena Read Solution (40)

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Read Solution (40) :

Kumari guddu singh  (8 years ago)

G=2
atq, all are distinct digitd(0-9)
9+8+7+6+5=35, not possible coz E=G=5
9+8+7+6+4=34 not possible coz E=G=4
9+8+7+4+5=33 not possible coz F=G=3
9+8+4+6+5=32, possible coz all are distinct
so, ans is 2

Is this solution Helpfull? Yes (23) | No (2)  

milan jena  (8 years ago) Best Solution

A+B+C+D+E=FG
9+8+7+6+5=35 Here E=G, So it is not possible.
again 9+8+7+6+4=34 Here also E=G, so it is also not possible
again 9+8+7+5+4=33 Here F=G
Finally 9+8+6+5+4=32 In this case condition is satisfies, so all are distinct and value of G=2.

Is this solution Helpfull? Yes (12) | No (2)  

Manan Shaw  (8 years ago)

4+6+5+8+9=32 because all distinct and maximum . G=2

Is this solution Helpfull? Yes (3) | No   

viswanadh ilindraparthi  (8 years ago)

9+8+7+6+5+4=35
G=5

Is this solution Helpfull? Yes (1) | No (12)  

kavita  (8 years ago)

e)2
9+8+6+5+4=32

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

Gudivaka Praveen  (8 years ago)

9+8+7+6+5+2 = 31
Since ABCDEFG all are distinct.
G = 1

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

soumita chakraborty  (8 years ago)

A


















































AS FG IS MAXIMUM AS POSSIBLE ,SO VALUE OF A+B+C+D+E=9+9+9+9+9=45=9*5
SO VALUE OF G IS 5 AS 9 IS NOT GIVEN IN OPTION...

Is this solution Helpfull? Yes (1) | No (19)  

Sandeep Patidar  (8 years ago)

0+1+3+4+6=7*2
G=2

Is this solution Helpfull? Yes (1) | No   

tarun  (8 years ago)

3 a=8,b=7,c=1,d=6,e=5 f=9,g=3

Is this solution Helpfull? Yes (1) | No   

Jhilik Sil Sarma  (8 years ago)

according to the question maximum value of A+B+C+D+E=9+9+9+9+9=45
given A+B+C+D+E=FG
now if FG=45 then the value of G will be 5 or 9
but 9 is not given in the option so value of G is 5

Is this solution Helpfull? Yes | No (8)  

himanshu garg  (8 years ago)

for max 5+6+7+8+9=35
ans=5

Is this solution Helpfull? Yes | No (5)  

Deven Patidar  (8 years ago)

JHILIK SIL= A,B,C,D,E,F,G all are distinct so how it can possible 9+9+....

HIMANSHU and VISWANADH : A,B,C,D,E,F,G all are distinct and your solution you have 5 for 2 times
all of three can't be right...

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ravi anand  (8 years ago)

9+8+5+2+0=6*4
so 4 will be the value of g

Is this solution Helpfull? Yes | No (1)  

Supratim Das  (8 years ago)

as all are distinct and we have to maximize so 1st solution should be
9+8+7+6+5=35 (but here 5 exist twice,at E and at G position)
so next target is to find 34(9+8+7+6+4) here also 4 exist twice
so next target is to find 33(9+8+7+6+3) here 3 exist thrice but 33 can be found by summing 9+8+7+4+5
so G is 3

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Deven Patidar  (8 years ago)

KAVITA i your answer is correct.

Supratim Das : 9+8+7+4+5=33 so G=3 but here 3 is 2 times.

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himanshi tayal  (8 years ago)

0+2+3+4+6=15
then g=5

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UDIT GAUTAM  (8 years ago)

0+1+7+8+9=25=fg
g=5

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ashish soni  (8 years ago)

ans is 4

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pranshu singh  (8 years ago)

ans= e)5
A=9
B=8
C=7
D=6
E=5
A+B+C+D+E=35

Is this solution Helpfull? Yes | No (3)  

Gaurav  (8 years ago)

9+8+6+5+4+0=32
2=so,ans

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Kousik Paul  (8 years ago)

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

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soni kumari  (8 years ago)

3bcz root of 9

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rabi  (8 years ago)

9+8+7+4+2=30 ie
6*5 hence ang is 5
here all digits are different

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prabhat ranjan mishra  (8 years ago)

3 is answer

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kishorechandra  (8 years ago)

Answer is b:
Use trail and error method..by giving the last number(G) in the options in the last as one..and giving the nearest possible sum.of their 10th digit place number(assume 3) then by filling the values of E D C B A and adding the by assigning the biggest number to the last one then we get 3 repeated..then we keep 2 in the last position and follow the same procedure..we get the answer by second trail itself..9+8+6+5+4=32.

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BHANU PRAKASH  (8 years ago)

here A+B+C+D+E= 5+ 6+ 4+ 8+ 9=32
NOW G=2

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anthonysudha  (8 years ago)

A,B,C,D,E=9,8,7,6,5soFG=35thenG=5

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Neha  (8 years ago)

9+8+7+6+5=35 wrong
9+8+7+6+4=34 wrong
9+8+7+4+5=33 wrong
9+8+4+6+5=32 right
ans is 2

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shuvrajyoti ghosh  (8 years ago)

5 AS MAXIMUM IS NEEDED SO 9+8+7+6+5=35.... 35=5*7 SO ACCORDING TO THE OPTION ..... 5 IS ANSWER..

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Jyothi Prakash Thaduri  (8 years ago)

9+8+7+4+5=32

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g obulesu  (8 years ago)

max 5+6+7+8+9=35
ans=5

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Mohammad Husain  (8 years ago)

5: 5+6+7+8+9

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Mousuma Roy  (8 years ago)

9+8+6+5+4=32
=> G=2

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Mayank Kumar  (8 years ago)

9+8+7+6+5=35
f=3
g=5

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volam chandan teja  (8 years ago)

distinct and max as possible so 9+8+7+6+5=35 so g=5

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SUBHAM  (8 years ago)

1+5+6+7+8=9*3
ans=3(c)

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sutapa banerjee  (8 years ago)

9+8+7+6+5=35...... E=G
9+8+7+6+4=34.......E=G
9+8+7+6+3=33.......F=G
9+8+7+6+2=32.... G=2 (b)

Is this solution Helpfull? Yes | No (1)  

Saikat Sarangi  (8 years ago)

9+8+6+3+2=28
28=7*4
hence, 9+8+6+3+2=7*4
According to the question all the digits should be distinct and must lie between 0-9
Therefore F=7, G=4
Ans: Option (D)=4.

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swati rathore  (8 years ago)

option (e)- i.e. 5

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Cp Choudhary  (8 years ago)

9+8+7+5+2=31 so according to me answer should be a) 1.

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