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16 Sep 2014, 01:23
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Which of the following fractions is the largest?
A. \(\frac{3,252}{3,257}\)
B. \(\frac{3,456}{3,461}\)
C. \(\frac{3,591}{3,596}\)
D. \(\frac{3,346}{3,351}\)
E. \(\frac{3,453}{3,458}\)
A. \(\frac{3,252}{3,257}\)
B. \(\frac{3,456}{3,461}\)
C. \(\frac{3,591}{3,596}\)
D. \(\frac{3,346}{3,351}\)
E. \(\frac{3,453}{3,458}\)
16 Sep 2014, 01:23
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Official Solution:
Which of the following fractions is the largest?
A. \(\frac{3,252}{3,257}\)
B. \(\frac{3,456}{3,461}\)
C. \(\frac{3,591}{3,596}\)
D. \(\frac{3,346}{3,351}\)
E. \(\frac{3,453}{3,458}\)
To solve this problem, notice that the difference between the numerator and the denominator in each fraction is equal to 5. Since all of the numbers are positive, we know that the larger the denominator, the larger the fraction:
\(\frac{1}{4} < \frac{2}{5} < \frac{3}{6} < \frac{4}{7} < \frac{5}{8} < \frac{6}{9} < \frac{7}{10} < ...\)
Using this logic, we can see that the largest fraction is \(\frac{3,591}{3,596}\), which has the largest denominator among all the options.
We can also take an algebraic approach by writing the fractions as \(\frac{n-5}{n}\). Simplifying this expression, we get:
\(\frac{n-5}{n}=\)
\(=\frac{n}{n}-\frac{5}{n}=\)
\(=1-\frac{5}{n}\)
As \(n\) increases, the value of \(\frac{5}{n}\) decreases, making the value of \(1-\frac{5}{n}\) larger. Therefore, we reach the same conclusion as before: the larger the denominator, the larger the fraction.
Answer: C
Which of the following fractions is the largest?
A. \(\frac{3,252}{3,257}\)
B. \(\frac{3,456}{3,461}\)
C. \(\frac{3,591}{3,596}\)
D. \(\frac{3,346}{3,351}\)
E. \(\frac{3,453}{3,458}\)
To solve this problem, notice that the difference between the numerator and the denominator in each fraction is equal to 5. Since all of the numbers are positive, we know that the larger the denominator, the larger the fraction:
\(\frac{1}{4} < \frac{2}{5} < \frac{3}{6} < \frac{4}{7} < \frac{5}{8} < \frac{6}{9} < \frac{7}{10} < ...\)
Using this logic, we can see that the largest fraction is \(\frac{3,591}{3,596}\), which has the largest denominator among all the options.
We can also take an algebraic approach by writing the fractions as \(\frac{n-5}{n}\). Simplifying this expression, we get:
\(\frac{n-5}{n}=\)
\(=\frac{n}{n}-\frac{5}{n}=\)
\(=1-\frac{5}{n}\)
As \(n\) increases, the value of \(\frac{5}{n}\) decreases, making the value of \(1-\frac{5}{n}\) larger. Therefore, we reach the same conclusion as before: the larger the denominator, the larger the fraction.
Answer: C
General Discussion
25 May 2017, 21:48
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Well, if we have a number of proper positive fractions (numerator < denominator), such that difference between the numerator and denominator for each of them be the same:- then, the one with the highest numerator will have the highest value.
Eg, lets consider 5 positive proper fractions:
a/(a+x), b/(b+x), c/(c+x), d/(d+x), e/(e+x)
where x is obviously a positive number.
Now, out of all the numerators, if a<b<c<d<e, Then:
definitely: a/(a+x) < b/(b+x) < c/(c+x) < d/(d+x) < e/(e+x)
Applying the same to our question, we can see that each option is a positive proper fraction where difference between numerator and denominator is 5.
So the one with highest numerator will have the highest value.
Hence C is the answer
Eg, lets consider 5 positive proper fractions:
a/(a+x), b/(b+x), c/(c+x), d/(d+x), e/(e+x)
where x is obviously a positive number.
Now, out of all the numerators, if a<b<c<d<e, Then:
definitely: a/(a+x) < b/(b+x) < c/(c+x) < d/(d+x) < e/(e+x)
Applying the same to our question, we can see that each option is a positive proper fraction where difference between numerator and denominator is 5.
So the one with highest numerator will have the highest value.
Hence C is the answer
