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18 May 2023, 04:30
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
64% (02:57) correct
36%
(02:59)
wrong
based on 170
sessions
History
Date
Time
Result
Not Attempted Yet
a, b and c are positive numbers. If \(\frac{a}{2b - c}=\frac{2b}{3a + c}=\frac{a}{b}\), what is the value of \(\frac{a}{b}\)?
A. \(\frac{1}{3}\)
B. \(\frac{1}{2} \)
C. \(\frac{2}{3}\)
D. \(\frac{3}{4} \)
E. \(\frac{4}{5}\)
A. \(\frac{1}{3}\)
B. \(\frac{1}{2} \)
C. \(\frac{2}{3}\)
D. \(\frac{3}{4} \)
E. \(\frac{4}{5}\)
18 May 2023, 13:15
Kudos
Bookmarks
Bunuel
\(\frac{a}{2b - c}=\frac{2b}{3a + c}=\frac{a}{b}\)
\(\frac{a}{2b - c} = \frac{a}{b}\)
Cross-multiplying we get
\(ab = 2ba - ca\) -- (1)
\(\frac{2b}{3a + c} = \frac{a}{b}\)
Cross-multiplying we get
\(2b^2 = 3a^2 + ca\) -- (2)
Adding (1) and (2) we get
\(ab + 2b^2 = 2ba + 3a^2\)
\(2b^2 = ba + 3a^2\)
We can use the options to see which of the options matches the expression \(2b^2 = ba + 3a^2\)
Answer choice elimination
A. \(\frac{1}{3}\)
b = 3; a = 1
\(2b^2 = ba + 3a^2\)
- \(2b^2 = 2(3)^2 = 18\)
- \(ba + 3a^2= (3*1) + 3*(1^2) = 6\)
Therefore \(2b^2 \neq ba + 3a^2\); Eliminate
B. \(\frac{1}{2}\)
b = 2; a = 1
\(2b^2 = ba + 3a^2\)
- \(2b^2 = 2(2)^2 = 8\)
- \(ba + 3a^2= (2*1) + 3*(1^2) = 5\)
Therefore \(2b^2 \neq ba + 3a^2\); Eliminate
C. \(\frac{2}{3}\)
b = 3; a = 2
\(2b^2 = ba + 3a^2\)
- \(2b^2 = 2(3)^2 = 18\)
- \(ba + 3a^2= (2*3) + 3*(2^2) = 18\)
Therefore \(2b^2 = ba + 3a^2\)
Option C
13 Jun 2023, 21:44
Kudos
Bookmarks
Bunuel
\(\frac{a}{2b-c}\) = \(\frac{a}{b}\)
ab = 2ab - ac
ab = ac (a positive) -> b=c
\(\frac{2b}{3a+c}\) = \(\frac{a}{b}\)
3\(a^2\) + ac = 2\(b^2\)
3\(a^2\) + ac - 2\(c^2\) = 0 (b=c, as above)
Divide by \(c^2\)
3\((\frac{a}{c})^2\) + \(\frac{a}{c}\) - 2 = 0
Solve this equation, we have \(\frac{a}{c}\) = \(\frac{2}{3}\) (choose this one) or \(\frac{a}{c}\) = -1 (wrong because a & c are positive)
So, the answer is C.
