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13 Apr 2008, 07:59
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If a/b=1/2, then the numerical value of which of the following expressions cannot be determined?
A) 2a/b
B) (a+b)/a
C) (a+1)/(b+1)
D) (a+3b)/(a+b)
E) 6a-3b
A) 2a/b
B) (a+b)/a
C) (a+1)/(b+1)
D) (a+3b)/(a+b)
E) 6a-3b
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13 Apr 2008, 08:48
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Hi!
The answer is C.
One of possible ways to solve it is to represent the fractions in terms of a/b. For example, it may be helpful to divide the numerator and denominator by b (the value of fraction will remain the same) and see if you can always get the value. Or, just use that b=2a and see if replacing of b can help.
A. 2a/b – clear.
B. (a+b)/a = 1+b/a – clear.
C. (a+1)/(b+1) = (a/b +1/b)/(1+1/b). You need to know the value of b; knowing a/b is not enough. Checking: a=1, b=2 then (a+1)/(b+1)=2/3. And if a=2, b=3 then (a+1)/(b+1)=3/4. So, here is the answer – C!
D (a+3b)/(a+b) = (a/b+3)/(a/b+1) - clear.
E. 6a-3b. Since a/b=1/2 then b=2a and we have 6a-3*2a=0. We were lucky here! For, say, 7a-3b we would have 7a-6a=a and it wouldn’t be possible to say anything.
The answer is C.
One of possible ways to solve it is to represent the fractions in terms of a/b. For example, it may be helpful to divide the numerator and denominator by b (the value of fraction will remain the same) and see if you can always get the value. Or, just use that b=2a and see if replacing of b can help.
A. 2a/b – clear.
B. (a+b)/a = 1+b/a – clear.
C. (a+1)/(b+1) = (a/b +1/b)/(1+1/b). You need to know the value of b; knowing a/b is not enough. Checking: a=1, b=2 then (a+1)/(b+1)=2/3. And if a=2, b=3 then (a+1)/(b+1)=3/4. So, here is the answer – C!
D (a+3b)/(a+b) = (a/b+3)/(a/b+1) - clear.
E. 6a-3b. Since a/b=1/2 then b=2a and we have 6a-3*2a=0. We were lucky here! For, say, 7a-3b we would have 7a-6a=a and it wouldn’t be possible to say anything.
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
