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22 Oct 2019, 20:01
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What is the value of \(|\frac{a}{b}|\)?
(1) \(\frac{a+b}{a}=1\frac{3}{4}\)
(2) \(\frac{a+b}{a-b}+\frac{a-b}{a+b}=7\frac{1}{7}\)
(1) \(\frac{a+b}{a}=1\frac{3}{4}\)
(2) \(\frac{a+b}{a-b}+\frac{a-b}{a+b}=7\frac{1}{7}\)
Chetan's tricky questions
23 Oct 2019, 12:36
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From Statement 1,
1 + (b/a) = 1 + (3/4)
b/a = 3/4
so |a/b| = 4/3, and Statement 1 is sufficient.
For Statement 2, the two terms in the sum on the left side are reciprocals of each other. If we let x = (a+b)/(a-b), the equation becomes
x + (1/x) = 7 + (1/7)
and we can immediately see that x = 7 and x = 1/7 are two solutions to this equation. Since we'd get a quadratic if we multiply both sides by x, the equation can have at most two solutions, so those are the only two solutions for x. So either
7 = (a + b)/(a - b)
7a - 7b = a + b
6a = 8b
|a/b| = 8/6 = 4/3
or
1/7 = (a + b)/(a-b)
a-b = 7a + 7b
-6a = 8b
|a/b| = |8/(-6)| = 4/3
and since in each case we get the same value for |a/b|, Statement 2 is sufficient and the answer is D.
1 + (b/a) = 1 + (3/4)
b/a = 3/4
so |a/b| = 4/3, and Statement 1 is sufficient.
For Statement 2, the two terms in the sum on the left side are reciprocals of each other. If we let x = (a+b)/(a-b), the equation becomes
x + (1/x) = 7 + (1/7)
and we can immediately see that x = 7 and x = 1/7 are two solutions to this equation. Since we'd get a quadratic if we multiply both sides by x, the equation can have at most two solutions, so those are the only two solutions for x. So either
7 = (a + b)/(a - b)
7a - 7b = a + b
6a = 8b
|a/b| = 8/6 = 4/3
or
1/7 = (a + b)/(a-b)
a-b = 7a + 7b
-6a = 8b
|a/b| = |8/(-6)| = 4/3
and since in each case we get the same value for |a/b|, Statement 2 is sufficient and the answer is D.
25 Oct 2019, 20:07
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chetan2u
What is the value of \(|\frac{a}{b}|\)?
(1) \(\frac{a+b}{a}=1\frac{3}{4}\)
(2) \(\frac{a+b}{a-b}+\frac{a-b}{a+b}=7\frac{1}{7}\)
(1) \(\frac{a+b}{a}=1\frac{3}{4}\)
(2) \(\frac{a+b}{a-b}+\frac{a-b}{a+b}=7\frac{1}{7}\)
Chetan's tricky questions
IanStewart, great solution..
Another way would be...
(1) \(\frac{a+b}{a}=1\frac{3}{4}\)...
\(1+\frac{b}{a}=1+\frac{3}{4}.......\frac{b}{a}=\frac{3}{4}.........\frac{a}{b}=\frac{4}{3}\)...
(2) \(\frac{a+b}{a-b}+\frac{a-b}{a+b}=7\frac{1}{7}\)
Divide by b.....\(\frac{\frac{a}{b}+\frac{b}{b}}{\frac{a}{b}-\frac{b}{b}}+\frac{\frac{a}{b}-\frac{b}{b}}{\frac{a}{b}+\frac{b}{b}}=7\frac{1}{7}\)........\(\frac{\frac{a}{b}+1}{\frac{a}{b}-1}+\frac{\frac{a}{b}-1}{\frac{a}{b}+1}=7\frac{1}{7}\)
Let \(\frac{a}{b}=x\)..
\(\frac{x+1}{x-1}+\frac{x-1}{x+1}=7\frac{1}{7}..........\frac{(x^2+2x+1)+(x^2-2x+1)}{x^2-1}=\frac{50}{7}.....\frac{2x^2+2}{x^2-1}=\frac{50}{7}........14x^2+14=50x^2-50.....36x^2=64....x^2=\frac{64}{36}.....|x|=\frac{8}{6}=\frac{4}{3}\)
D
