Answer: The answer would be B.
I think we must use two numbers which have two digits and one of them must be one digited number.If we try it we can say our numbers are xy,ab and c.
xy + ab + c = 50
10x+y+10a+b+c = 50
10(x+a) + y+b+c = 50
Think about y+b+c. Maximum it can be 5+4+3 = 12 but it can't because there is a number which is multiple of 10.So it can be 10 maximum and x+a must be 4.
We can only use 1+3. 40 + y+b+c = 50 , y+b+c = 10. And other numbers are 2+4+5 = 11. 10 doesn't equal to 11 and it says there isn't any solution for this problem :)
The approximate middle value
The answer for this question is
11/3
Answer:
x = 1
Step-by-step explanation:
