enigma123 wrote:
Given that a, b, c, and d are different nonzero digits and that 10d + 11c < 100 – a, which of the following could not be a solution to the addition problem below?
abdc
+ dbca
(A) 3689
(B) 6887
(C) 8581
(D) 9459
(E) 16091
This is the original question and its beyond my head to solve this. can someone please help and try to explain the concept of this?
Also , can you please tell how to solve if the question says COULD BE a solution to the addition problem?
10d + 11c < 100 – a --> 10d+11c+a<100 --> (10d+c)+(10c+a)<100. (10d+c) is the way of writing two-digit integer
dc and (10c+a) is the way of writing two-digit integer
ca. Look at the sum:
ab
dc+db
ca-----
Now, as two-digit integer
dc + two-digit integer
ca is less than 100, then there won't be carried over 1 to the hundreds place and as b+b=2b=even then the hundreds digit of the given sum must be even too. Thus 8
581
could not be the sum of abdc+dbca (for any valid digits of a, b, c, and d).
Answer: C.
As for "could be true" questions:
try our new "Must or Could be True Questions" tag -
http://gmatclub.com/forum/search.php?se ... tag_id=193 to learn more about this type of questions.
Hope it helps.
Best explanation ever. couldn't be more clear.
Sky is the limit. 800 is the limit.