Latest Infosys Aptitude Question SOLUTION: A + B + C +D = D + E + F + G = G + H + I =17. IF A = 4 WHAT ARE THE VALUES OF D AND G. EACH LETTER TAKEN ONLY ONE OF THE DIGIT FROM 1 TO 9.

Step 1.

a + b + c + d = d + e + f + g = g + h + i = 17

Means, ( a + b + c + d )+ (d +e + f + g )+ (g +h + i ) = 17 +17+17 = 17 x 3 = 51

a + b + c + d + e + f + g + h + i +( d + g ) = 51

Step 2.

But 'a' to ' i ' takes values only from 1 to 9

So a + b + c + d + e + f + g + h + i = sum of the numbers from 1 to 9 = 45

Step 3.

From step 1 and step 2

d + g = 51 - 45 = 6

possible values of d & g are (1, 5), ( 5,1 ), (2,4) & (4, 2 )

( 2, 4 ) & ( 4 , 2 ) are not possible as "a = 4 "

We have to try with other values (1, 5 ) & ( 5 , 1 )

Step 4.

When d = 1 and g = 5

1. a + b + c + d = 4 + b +c + 1 = 17

b + c = 12 ; only possible values of b & c are ( 9, 3 ) or ( 3, 9 )

2. d + e + f + g = 1 + e + f + 5 = 17

e + f = 17 - 6 = 11

possible combinations for 11 are ( 2,9 ) ; ( 3,8 ) ; (4,7 ) & ( 5,6 )

Out of the above four combinations nothing is possible beacuse

b & c takes values 3 & 9 g = 5 (assumed) and a = 4 (given).

So, d = 1 & g = 5 is not the solution.

Case 2.

When d = 5 and g = 1

1. a + b + c + d = 4 + b + c + 5 = 17

b + c = 17 - 9 = 8

only possible combination is 2 & 6

a + b + c + d = 4 + ( 2 + 6 ) + 5 = 17

2. d + e + f + g = 5 + e + f + 1 = 17

e + f = 17 - 6 = 11

The only possible value is 3 & 8

d + e + f + g = 5 + ( 3 + 8 ) + 1 = 17

g + h + i = 1 + h + i = 17

h + i = 17 - 1 = 16

The only possible value for h & i are 7 & 9

g + ( h+ i ) = 1 + ( 7 + 9 ) + 17

So the values of d = 5 & g = 1

D=5 g=1 or g=5 d=1

By trial and error,
a+b+c+d=4+2+6+5=17
d+e+f+g=5+8+3+1=17
g+h+i= 1+9+7=17

therefore, d=5 and g=1