The letters A,B,C,D,E,F, and G, not necessarily in that order, represents 7 consecutive positive integers.
It is known that:
- F is as much less than B as C is greater than D.
- G is greater than F.
- B is the middle term.
- A is precisely 3 greater than D.
Arrange this sequence of letters starting from the lowest magnitude to the highest magnitude.
Provide valid reasoning for your answer.
In (3) we're told B is the middle term. From (1), F is less than B.
From (4), A must be greater than B, since there would not be two numbers whose difference is 3 that are both less than the middle term in a sequence of seven consecutive integers. This also tells us that D is not the first term (since B is 3 more than the first term of the sequence), and that A is not the last term.
Since D is not the first term, then from (1) C must be greater than B, as there would be no way to order F, D and C all below B to make the statement true.
C must be at least 2 greater than D, then. But C cannot be 3 greater than D (4) and it cannot be more than 3 greater than D (1). So C is exactly 2 more than D, which means they must be adjacent to B. Now we can also place F and A as follows:
_FDBCA_
Finally from (2) G is greater than F, so the final sequence must be:
EFDBCAG
|
|
Posted by tomarken
on 2022-03-03 10:55:44 |
Solution
