all fractions are close to 1/3
So if we take each
\(I. 47/150<50/150=1/3\)
\(II. 111/330>110/330=1/3\)
\(III. 299/900<300/900=1/3\)
Now between II and III we can easily figure out that 299 is closer to 300 than \(47*3\). So III>I
So the correct order from smallest to largest is : \(I>III>II\)
So shouldn't it be B?
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Yes, you are 100% correct. I changed this in the problem above. I said "smallest to largest", but then when I did it in my head, I put them in order from largest to smallest! Sorry about any confusion.
Mike
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I. 47/150 is almost 50/150 = 1/3. It is (50-47)/150 = 3/150 = 1/50 away from 1/3
II. 111/330 is beyond 110/330 = 1/3
III. 299/900 is almost 300/900 = 1/3. 299/900 is (300-299)/900 = 1/900 away from 1/3

I is the farther to the left of 1/3 than III while II is the largest.

Rank: I,III,II

Try and get all the denominators the same, i.e. in terms of 900:

I: 47x6/150x6 = 282/900

II: 111x3/330x3 =:(approx) 333/990

III: 299/900

From the above, we know that I is the smallest (as it has the smallest numerator) and hence comes first, eliminate options C,D and E. Only left with A & B.

299/900 will definitely be greater than 333/990 as if we find the accurate multiple by which to get 900 in II, it will give us a numerator being less than 299. Hence III<I

Final order: I,III,II

Option B

I. 47/150 = 282/900
II. 333/990 >330/990=1/3= 300/900
III. 299/900

So, II >III>I or as per mikemcgarry I< III < II

Answer B
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It can also be solved by reducing the values.
So, I. 47/150 << 50/150 =1/3
II. 111/330 > 110/330 = 1/3
III. 299/900 just< 300/900 = 1/3

So, II >III>I or as per mikemcgarry I< III < II

Answer B
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I. \(\frac{47}{150}\) =50/150-3/150=1/3-1/50

II. \(\frac{111}{330}\)=110/330+1/330=1/3+1/330

III. \(\frac{299}{900}\)=300/900-1/900=1/3-1/900


Answer (B) I, III, II