[GMAT math practice question] 7.11
As the figure shows, A = -5 and B = 4. If AC:CD:DB=1:2:3, what are the values of C and D?
A. -3 and -1
B. -3.5 and -0.5
C. -2.5 and 0.5
D. 1 and 3
E. 0 and 1
As the figure shows, A = -5 and B = 4. If AC:CD:DB=1:2:3, wh
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Distance between A to D = 4 - (-5) = 9
Since AC : CD : DB = 1 : 2 : 3
$$1x\ :\ 2x:\ \ 3x=9$$
$$1x+2x+3x=9$$
$$6x=9$$
$$x=\frac{9}{6}=1.5$$
$$Therefore,\ AC=1\cdot1.5=1.5$$ $$CD=2\cdot1.5=3$$ $$DB=3\cdot1.5=4.5$$
$$C=-5+1.5=-3.5$$
$$D=-5+4.5=-0.5$$
Answer = option B
Since AC : CD : DB = 1 : 2 : 3
$$1x\ :\ 2x:\ \ 3x=9$$
$$1x+2x+3x=9$$
$$6x=9$$
$$x=\frac{9}{6}=1.5$$
$$Therefore,\ AC=1\cdot1.5=1.5$$ $$CD=2\cdot1.5=3$$ $$DB=3\cdot1.5=4.5$$
$$C=-5+1.5=-3.5$$
$$D=-5+4.5=-0.5$$
Answer = option B
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Since AC:CD:DB = 1:2:3, we can put AC = k, CD = 2k and DB = 3k.
Then k + 2k + 3k = 6k = 4 - (-5) = 9 and k = 3/2.
Thus, C = A + k = -5 + (3/2) = -(7/2) = -3.5 and D = B - 3k = 4 - 9/2 = -(1/2) = -0.5.
Therefore, B is the answer.
Answer: B
Since AC:CD:DB = 1:2:3, we can put AC = k, CD = 2k and DB = 3k.
Then k + 2k + 3k = 6k = 4 - (-5) = 9 and k = 3/2.
Thus, C = A + k = -5 + (3/2) = -(7/2) = -3.5 and D = B - 3k = 4 - 9/2 = -(1/2) = -0.5.
Therefore, B is the answer.
Answer: B
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