Automobile A is traveling at two-thirds the speed that Automobile B is traveling. How fast is Automobile A traveling?
(1) If both automobiles increased their speed by 10 miles per hour, Automobile A would be traveling at three-quarters the speed that Automobile B would be traveling.
(2) If both automobiles decreased their speed by 10 miles per hour, Automobile A would be traveling at half the speed that Automobile B would be traveling
OA D
Source: Princeton Review
Automobile A is traveling at two-thirds the speed that Autom
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$$A=\frac{2}{3}of\ B$$
How fast is Automobile A travelling
Statement 1
If both automobiles increased their speed by 10 miles per hour, Automobile A would be travelling at three - quarter the speed that automobile B will be travelling.
$$A+10=\frac{3}{4}of\ \left(B+10\right)$$
From question
$$A=\frac{2}{3}B$$
$$\frac{2}{3}B+10=\frac{3}{4}B+\frac{80}{4}$$
$$\frac{2}{3}B-\frac{3}{4}B=\frac{30}{4}-10$$
$$\frac{8B-9B}{12}=\frac{30-40}{4}$$
$$-\frac{B}{12}=-\frac{10}{4}$$
$$-\frac{4B}{4}=-\frac{120}{4}$$
$$b=30$$
$$Since\ A=\frac{2}{3}of\ B$$
statement 1 is INSUFFICIENT.
$$\frac{2}{3}\cdot30=20$$
Statement 2
If both automobile reduce their speed by 10 mph
Automobile A will be travelling a 1/2 the speed of that B would be travelling.
Therefore,
$$A-10==\frac{1}{2}\left(B-10\right)where\ A=\frac{2}{3}B$$ $$\frac{2}{3}B-10=\frac{1}{2}B-5$$ $$\frac{2}{3}B-\frac{1}{2}B=-5+10$$ $$\frac{4B-3B}{6}=5$$ $$b=6\cdot5=30mph$$ $$where\ A=\frac{2}{3}of\ B$$ $$where\ A=\frac{2}{3}of\ 30$$ $$where\ A=20mph$$
Statement 2 is also SUFFICIENT.
$$Answer\ is\ Option\ D$$
Both statement alone are SUFFICIENT.
How fast is Automobile A travelling
Statement 1
If both automobiles increased their speed by 10 miles per hour, Automobile A would be travelling at three - quarter the speed that automobile B will be travelling.
$$A+10=\frac{3}{4}of\ \left(B+10\right)$$
From question
$$A=\frac{2}{3}B$$
$$\frac{2}{3}B+10=\frac{3}{4}B+\frac{80}{4}$$
$$\frac{2}{3}B-\frac{3}{4}B=\frac{30}{4}-10$$
$$\frac{8B-9B}{12}=\frac{30-40}{4}$$
$$-\frac{B}{12}=-\frac{10}{4}$$
$$-\frac{4B}{4}=-\frac{120}{4}$$
$$b=30$$
$$Since\ A=\frac{2}{3}of\ B$$
statement 1 is INSUFFICIENT.
$$\frac{2}{3}\cdot30=20$$
Statement 2
If both automobile reduce their speed by 10 mph
Automobile A will be travelling a 1/2 the speed of that B would be travelling.
Therefore,
$$A-10==\frac{1}{2}\left(B-10\right)where\ A=\frac{2}{3}B$$ $$\frac{2}{3}B-10=\frac{1}{2}B-5$$ $$\frac{2}{3}B-\frac{1}{2}B=-5+10$$ $$\frac{4B-3B}{6}=5$$ $$b=6\cdot5=30mph$$ $$where\ A=\frac{2}{3}of\ B$$ $$where\ A=\frac{2}{3}of\ 30$$ $$where\ A=20mph$$
Statement 2 is also SUFFICIENT.
$$Answer\ is\ Option\ D$$
Both statement alone are SUFFICIENT.
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Each Statement gives the same type of relationship between the two speeds, so the answer will be C, D or E. If we let a and b be the two speeds of the cars, we know that a = 2b/3.
Using Statement 1 alone, you might be able to see by inspection that the speeds are 20 and 30. Or we can use algebra:
a + 10 = (3/4)(b + 10) = (3b/4) + 7.5
Substituting for a, we have
(2b/3) + 10 = (3b/4) + 7.5
which is an equation we can clearly solve for b, and thus we can find a. Statement 2 will produce a similar equation, so the answer is D.
Using Statement 1 alone, you might be able to see by inspection that the speeds are 20 and 30. Or we can use algebra:
a + 10 = (3/4)(b + 10) = (3b/4) + 7.5
Substituting for a, we have
(2b/3) + 10 = (3b/4) + 7.5
which is an equation we can clearly solve for b, and thus we can find a. Statement 2 will produce a similar equation, so the answer is D.
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