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22 May 2021, 18:07
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
19% (02:12) correct
81%
(02:42)
wrong
based on 140
sessions
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If \(x>\frac{14}{13}\) be the solution to the inequality \(bx-(a+b)<0\), where a and b are constants, then what is the solution to the inequality \(ax+2a+b<0\)?
A. x < -15
B. x > -15
C. x < 15
D. x > 15
E. None of the above
A. x < -15
B. x > -15
C. x < 15
D. x > 15
E. None of the above
25 Aug 2021, 07:52
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nick1816
We can solve the inequality in terms of a and b, we have \(bx < (a + b)\).
Knowing the solution has a change in sign, we can confirm b is negative.
Then \(x > \frac{a + b}{b}\) is the solution, and by matching the numbers we have \(\frac{a + b}{b} = \frac{14}{13}\), which is \(\frac{a}{b} = \frac{1}{13}\).
Although there are many combinations of a and b that work, we may simply pick a pair since this is a PS question. b = -13 and a = -1 for example (again b must be negative, which means a is negative as well!).
Then plug in the question: \(-x + -15 < 0\) and \(x > -15\).
Ans: B
20 Feb 2024, 00:24
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ROY13
- \(bx-(a+b)<0\);
\(bx < a+b\).
If b were positive, we'd divide the above by positive b, keep the sign and would get \(x < \frac{a+b}{b}\). However, since we are given that the solution is \(x>\frac{14}{13}\), with > sign instead of the < sign, then b must be negative for the sign of \(bx < a+b\) to flip when we divide by b. Hence, we get \(x > \frac{a+b}{b}\) instead and thus \(\frac{a+b}{b} = \frac{14}{13}\).
Next, \(\frac{a+b}{b} = \frac{14}{13}\) simplifies to \(\frac{a}{b} = \frac{1}{13}\). Since we already established that b is negative, then a must also be negative for their ratio to equal a positive number.
Now, let's work with \(ax+2a+b<0\):
- \(ax+2a+b<0\)
\(ax < -2a-b\)
Divide by a, which is negative and flip the sing:
- \(x > -2-\frac{b}{a}\)
Since \(\frac{a}{b} = \frac{1}{13}\), then \(\frac{b}{a}=13\). Thus, we get:
- \(x > -2-13\)
\(x > -15\)
Answer: B.
