Help with this one please
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What is r?
(1) The distance between r and 0 is 3 times the distance between m and 0.
(2) 12 is halfway between m and r.
I assume it's clear that neither statement is sufficient on its own. You could do the problem algebraically:
From (2) 12-m = r-12
From (1) |r| = 3|m|
Thus either r = 3m (if they are on the same side of zero) OR r = -3m (if they are on opposite sides of zero)
In each case, you'll get two equations/two unknowns- if you solve, in the first case you'll find m = 6, r = 18; in the second case you'll find m = -12,
r = 36.
Or you could do the problem 'pictorially'. From 1), we don't know if m and r are on the same side of zero, or on opposite sides. Using 1)+2), we know that m and r cannot both be negative -- but they could both be positive, or we could have that r > 0, m < 0. So we should get two different solutions, even using both statements. You can confirm that both cases are possible- you should be able to see that there will be one solution where m is negative and r is a positive number quite far to the right of 12, and another solution where both m and r are positive, and are closer together than in the first case.
In either case, E.
(1) The distance between r and 0 is 3 times the distance between m and 0.
(2) 12 is halfway between m and r.
I assume it's clear that neither statement is sufficient on its own. You could do the problem algebraically:
From (2) 12-m = r-12
From (1) |r| = 3|m|
Thus either r = 3m (if they are on the same side of zero) OR r = -3m (if they are on opposite sides of zero)
In each case, you'll get two equations/two unknowns- if you solve, in the first case you'll find m = 6, r = 18; in the second case you'll find m = -12,
r = 36.
Or you could do the problem 'pictorially'. From 1), we don't know if m and r are on the same side of zero, or on opposite sides. Using 1)+2), we know that m and r cannot both be negative -- but they could both be positive, or we could have that r > 0, m < 0. So we should get two different solutions, even using both statements. You can confirm that both cases are possible- you should be able to see that there will be one solution where m is negative and r is a positive number quite far to the right of 12, and another solution where both m and r are positive, and are closer together than in the first case.
In either case, E.
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