vipulgoyal wrote:Both a, b, and c are 3-digits integers, where a=b+c. Is the hundreds' digit of number a equal to sum of that of b and c?
1). Tens' digit of a=tens' digit of b+tens' digit of c
2). Units' digit of a=units' digit of b + units' digit of c

my take E

If Tens' digit of a=tens' digit of b+tens' digit of c, it obviously means there is no carry over to the addition of hundred's digit. So, (1) alone is sufficient.
We can't conclude this from statement 2, which is insufficient.
Ans: A

I think an example with numbers would help

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.