Latest TCS Aptitude Question SOLUTION: IF A+B+C=13 C+D+E=13 E+F+G=13 G+H+I=13 Then the value of E ?

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(#M40110201) TCS Logical Reasoning question Keep an EYE

IF
A+B+C=13
C+D+E=13
E+F+G=13
G+H+I=13
Then the value of E ?

Asked In TCS Anand Mishra (9 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

harish  (9 years ago)

can any one explain how have you chosen the values of a,b,c...... ?

Is this solution Helpfull? Yes (14) | No   

chandrakala  (9 years ago)

A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
Value of E = 4

Is this solution Helpfull? Yes (13) | No (5)  

Devendra  (9 years ago)

I got from one site.I just share in detail. i learn you also learn how it solved.

A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
Value of E = 4

13*4 = 52
1+2+3+....+9 = 45

Then C+D+E must be equal to 7
or they must from 1 , 2 or 4.

1 and 2 for E cannot satisfy as because 1 and 2 cannot
come together in C+D+E
hence 1 and 4 and 2 and 4 should be together.

so you get two solutions and putting them any of the either
way you get E = 4. and everynumber gets satisfied with a
unique value.

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

kakali sarkar  (9 years ago)

please explain how are u choosing the values

Is this solution Helpfull? Yes (5) | No   

vishnu murali  (9 years ago)

By given condition, we have
A+B+C = 13...(1)
C+D+E = 13...(2)
E+F+G = 13...(3)
G+H+I = 13...(4)
Add up (1)+(2)+(3)+(4), we get
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45

Hence,C+E+G = 52 - 45 = 7...(5)

By (5), we have E = 7-C-G

Is this solution Helpfull? Yes (4) | No (3)  

Surit Datta  (9 years ago)

A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
Value of E = 4

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

Vasu  (9 years ago)

ans is 4,
Because choose values randomly,
for example
A+B+C=13(A=4,B=5,C=4)
C+D+E=13(C=4,D=5,E=4)
E+F+G=13(E=4,F=5,G=4)
G+H+I=13(G=4,H=5,I=4)

Is this solution Helpfull? Yes (2) | No (6)  

Sakti Prasad Panda  (9 years ago)

By given condition, we have
A+B+C = 13...(1)
C+D+E = 13...(2)
E+F+G = 13...(3)
G+H+I = 13...(4)
Add up (1)+(2)+(3)+(4), we get
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45

Hence,C+E+G = 52 - 45 = 7...(5)
By (5), we have E = 7-C-G

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

prashant shukla  (8 years ago)

e=1
(a=0,b=4,c=9)=13
(c=9,d=3,e=1)=13
(e=1,f=7,g=5)=13
(g=5,h=6,i=2)=13

Is this solution Helpfull? Yes (1) | No   

kiruthiga  (9 years ago)

how u r choose a,b,c,d,.... values

Is this solution Helpfull? Yes | No   

Neeladri Rudra  (9 years ago)

Sol: By given condition, we have
A+B+C = 13...(1)
C+D+E = 13...(2)
E+F+G = 13...(3)
G+H+I = 13...(4)
Add up (1)+(2)+(3)+(4), we get
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45

Hence,C+E+G = 52 - 45 = 7...(5)
By (5), we have E = 7-C-G

Is this solution Helpfull? Yes | No (10)  

ankit  (9 years ago)

this can solve using heat & trail method

Is this solution Helpfull? Yes | No   

BHAVESH GAJWANI  (8 years ago)

A + B + C | 9 + 3 + 1 = 13 = 6 + 5 + 2
C + D + E | 1 + 8 + 4 = 13 = 2 + 7 + 4
E + F + G | 4 + 7 + 2 = 13 = 4 + 8 + 1
G + H + I | 2 + 5 + 6 = 13 = 1 + 3 + 9
Value of E = 4
its all about hit and trial just keep in mind that we dont have to put two big no. together oderwise we would get stuck as sum 13 could not be possible with left over digit..!!

Is this solution Helpfull? Yes | No (3)  

p saha  (10 Months ago)

By given condition, we have
A+B+C = 13...(1)
C+D+E = 13...(2)
E+F+G = 13...(3)
G+H+I = 13...(4)
Add up (1)+(2)+(3)+(4), we get
52=13*4 = A + B + C+ C + D + E+ E + F + G+ G + H + I = C+E+G+(A + B + C+ D +E + F + G + H + I)
= C+E+G+(1+2+ 3+ 4+ 5+ 6+ 7+ 8+ 9) = C+E+G+(1+9)*9/2 = C+E+G + 45

Hence,C+E+G = 52 - 45 = 7...(5)

By (5), we have E = 7-C-G

Then consider
A B C
C TO C is diagonally connected so put C = 1
C D E
As we have to find out E keep E as it is
E F G
G TO G is diagonally connected so put G = 2
G H I

from eq (5)
So E = 7 - C - G = 7 - 1 - 2 = 4

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