How do you add numbers?
568 + 345 = ???
Last digit ( 8 + 5 = 13 --- 3 is last digit and 1 carry)
tens digit ( 6 + 4 + 1(carry) = 11 --- 1 is tens digit and 1 carry)
hundreds digit = 5 + 3 + 1(carry) = 9
or, 568 + 345 = 913
See here, sum of ten's digit 6 and 4 is 10, but tens digit of the sum is '1'.
Point is if we add, say pqr + stu = xyz (All alphabets are digits)
z = r + u only if r + u < 10.
or, z = last digit of r + u and gives a carry to tens digit, carry1 = 1.
y = q + t only if 'q + t + carry1 < 10' ---------------------(A)
or, y = last digit of 'q + t + carry1' and carry2 = 1 (carry to hundred's digit)
Similarly, do the same for x.
In the given question, in statement 1, it is given that y = q + t.
We are asked if z = p + s.
If y = q + t, means carry1 = 0 and z = r + u <10
Ex: 445 + 133 = 578 satisfies the condition, but 445 + 135 doesn't as 5 + 5 > 9 which will change the tens digit, i.e. 1 carry will be added, 445 + 135 = 580 (In this case sum of tens digit 4 + 3 = 7, not 8)
Also, from (A), q + t must be less than 10, else condition wont be satisfied.
Ex: 335+381 = 716 (3 + 8 = 11, but l=tens digit of answer is 1)
So, sum of ten's digit can't give any carry to hundred's digit.
So, hundreds' digit of number a is equal to sum of that of b and c.
