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<x, y> is defined as (2x+y)/(x-2y). What is the value

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<x, y> is defined as (2x+y)/(x-2y). What is the value

by Max@Math Revolution » Tue Sep 10, 2019 11:42 pm

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[GMAT math practice question]

< x, y > is defined as (2x+y)/(x-2y). What is the value of <1/2, 2/3 >- <1/3 , 1/4 >?

A. 5/2
B. 5/3
C. 7/2
D. 3/2
E. 8/3
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by Max@Math Revolution » Fri Sep 13, 2019 6:54 am
=>

< 1/2,2/3 > = (2*(1/2) + (2/3))/((1/2)-2(2/3)) = (1 + (2/3))/((1/2) - (4/3)) = (5/3) / (-(5/6)) = -2.
< 1/3, 1/4 > = (2*(1/3) + (1/4))/((1/3)-2(1/4)) = (2/3+1/4)/(1/3-1/2) = (11/12)/(-(1/6)) = -(11/2)
Thus, < 1/2,2/3 > - < 1/3,1/4 > = (-2)-(-(11/2)) = 7/2

Therefore, C is the answer.
Answer: C
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by deloitte247 » Sat Sep 14, 2019 8:18 am
$$what\ is\ the\ value\ of\ <\frac{1}{2},\frac{2}{3}>-<\frac{1}{3},\frac{1}{4}>?$$
$$for\ <\frac{1}{2},\frac{2}{3}>where\ x=\frac{1}{2}and\ y=\frac{2}{3}$$
$$\frac{\left(2x+y\right)}{\left(x-2y\right)}=\frac{\left(2\left(\frac{1}{\left(2\right)}\right)+\frac{2}{3}\right)}{\frac{1}{2}-2\left(\frac{2}{\left(3\right)}\right)}=\frac{\left(1+\frac{2}{3}\right)}{\frac{1}{2}-\frac{4}{3}}=\frac{\frac{\left(3+2\right)}{3}}{\frac{\left(3-8\right)}{6}}=\frac{\left(\frac{5}{3}\right)}{-\frac{5}{6}}$$
= $$\frac{5}{3}\cdot\frac{-6}{5}=-2----------eqi$$


$$for<\frac{1}{3},\frac{1}{4}>\ where\ x=\frac{1}{3}and\ y=\frac{1}{4}$$ $$for<\frac{1}{3},\frac{1}{4}>\ where\ x=\frac{1}{3}and\ y=\frac{1}{4}$$
$$\frac{\left(2x+y\right)}{\left(x-2y\right)}=\frac{2\left(\frac{1}{3}\right)+\frac{1}{4}}{\frac{1}{2}-2\left(\frac{1}{4}\right)}=\frac{\left(\frac{2}{3}+\frac{1}{4}\right)}{\frac{1}{3}-\frac{1}{4}}=\frac{\frac{\left(8+3\right)}{12}}{\frac{\left(2-3\right)}{6}}=\frac{\frac{11}{12}}{-\frac{1}{6}}$$
$$=\frac{11}{12}\cdot-\frac{6}{1}=-\frac{11}{2}------eqii$$
$$eqni-eqnii=<\frac{1}{2},\frac{2}{3}>-<\frac{1}{3},\frac{1}{4}>$$
$$where<\frac{1}{2},\frac{2}{3}>=-2\ and\ <\frac{1}{3},\frac{1}{4}>=-\frac{11}{2}$$
$$-2-\left(-\frac{11}{2}\right)$$
$$-\frac{2}{1}+\frac{11}{2}=\frac{\left(-4+11\right)}{2}=\frac{7}{2}$

$$Answer\ is\ Option\ C$$
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