As hard as ABC

as it is given a+c=7 and b+c=11 so, a+2b+3c can also be written as (a+c)+(b+c)+(b+c)=7+11+11 = 29 ANS

First, split the equation into something easier to work with. a+c+b+c+b+c. If you look closely, you see that you can substitute in the numbers 7 and 11 so it looks like this: 7+11+11. Then you add them together to get 29.

if a+c=7 and b+c=11 and the value of a+2b+3c is a=7-c and b=11-c then putting the values of c = 7-c+{11-c}2+3c = 7-c+22-2c+3c = 7+22 = 29

2b+2c = 22 , a + c = 7

=> a+2b+3c = 29

(a+b)+(b+c)+(b+c)

a+2b+c=a+c+2c+2b =(a+c)+2(b+c) =7+2(11) =29

a=7-c b=11-c Substitute: a+2b+3c (7-c) + 2(11-c) + 3c 7-c + 22- 2c + 3c 29

a+2b+3c = (a+c)+(b+c)+(b+c) = 7+11+11 =29

(b+c=11) x 2 2b+2c=22

so, a+c+2b+2c=7+22 a+2b+3c= 29

a+2b+3c= (a+c)+ (b+c) + (b+c) = 7 + 11 + 11 = 7 +22 = 29

Given data, a+c= 7------------------------Equation 1 b+c= 11----------------------Equation 2

Equation 1 + (2*Equation 2)= Answer = 29

B+C=11 if doubled it would be 2B+2C=22. So A+C=7 so if you add 2B+2C=22 to A+C=7 it will give you A+2B+3C which equals 29. This makes your answer 29.

a+c=7 , b+c=11 has many solutions. so we make another equation and eliminate c variable. and we get a-b=-4 . now the possible number (a,b,c) are ..... (1,5,6), (2,6,5),(3,7,4),(4,8,3), (5,9,2)...... we can substitute these solution. Tadaaa.... and the result is same , its 29 ..