GATE2018 CE-2: GA-10

Question

Each of the letters in the figure below represents a unique integer from $1$ to $9$. The letters are positioned in the figure such that each of $(A+B+C), (C+D+E), (E+F+G)$ and $(G+H+K)$ is equal to $13$. Which integer does $E$ represent?

• A. $1$
• B. $4$
• C. $6$
• D. $7$

Comments

A=9 B=3 C=1 D=8 E=4 F=7 G=2 H=6 I=5

We can replace ABC with IHG and D with F but E is constant

Answer 1

We have to get sum $13$ in $4$ ways. 

• $A + B + C = 13$
• $C + D + E = 13$
• $E + F + G = 13$
• $G + H + K= 13$

Adding all $4$ we get

$A + B + D + F + H + K + 2(C + E+ G) = 52$

Since $A \to K$ sum to $45$ we get 

$C + E + G = 52-45 =7$

Only option we can add to $7$ with $3$ distinct positive integers is $1 +2+4.$ So, $C,E,G \in \{1,2,4\}$

We also have $C+D+E = 13.$ If $C$ and $E$ are $1,2$ this is not possible as this requires $D$ to be $10$ which is not allowed. So, at least one of $E$ or $C$ must be $4$ also making $G$ to be either $1$ or $2.$

Since, $E + F + G = 13, E + G $ must also be $\geq4$ meaning $E$ and $G$ cannot be $1,2.$ Since, $G \leq 2,$ it means $E > 2.$

So, only option is $E = 4.$

Correct Option: B.

Comments

@Arjun sir how A to G sum to 45.

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@Vikas As Each of the letters in the figure represents a unique integer from 1 to 9, therefore their sum will also be equal to the sum of integers from 1 to 9 ie 45.

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@Arjun The sum of A to I = 45 and not A to G , Please edit the answer.

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very nice explanation sir!!👌

Answer 2

a+b+c=13 ---- 1

c+d+e=13 ---- 2

e+f+g=13 ---- 3

g+h+k=13 ---- 4

Add all four equations:

(a+b+c+d+e+f+g+h+k)+(c+e+g)=52

$\sum _{i=a} ^k i = 45$

c+e+g= 7 ---- 5

So, Option c and d are eliminated.

Now let's assume e=1, put in eq.2,3 and 5, add eq.2 and 3

c+d+2+f+g=26

c+d+f+g=24 ---- 6

c+1+g=7 from eq5

c+g=6 put in eq. 6

d+f=18, which is not possible

So, option a is eliminated.

Hence, Answer is B.

Answer 3

9     3      1

                8

                4      7      2

                                6

                                5