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Bunuel
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Bunuel
Rank those three in order from smallest to biggest.

I. 47/150
II. 111/350
III.299/900


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II

Dear Bunuel Please correct the OA..it should be A
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Edited. Thank you for noticing.
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Bunuel
Rank those three in order from smallest to biggest.

I. 47/150
II. 111/350
III.299/900


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II

| vs ||| - 47*6/(150*6)=282/900 vs 299/900 --->||| > |
| vs || - 47*7/(150*7) vs 111*3/(350*3)--->329/1050 vs 333/1050 --->|| > | and these fractions are closer than | and ||| ---> |<||<||| ----> option A
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Bunuel
Rank those three in order from smallest to biggest.

I. 47/150
II. 111/350
III.299/900


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II


Multiplying 47/150 by 6/6, we have 47/150 x 6/6 = 282/900. So 47/150 < 299/900.

Multiplying 47/150 by 7/7 and 111/350 by 3/3, we have 47/150 x 7/7 = 329/1050 and 111/350 x 3/3 = 333/1050. So 47/150 < 111/350.

From the above, we see that 47/150 is the smallest of the three fractions. To compare 111/350 and 299/900, we will not make both to equivalent fractions of the same denominator. We will argue that, although both are less than 1/3, 299/900 is much closer to 1/3 than 111/350 is (observe that 1/3 = 300/900 and 1/3 = 111/333). So 299/900 must be the largest of the three.

Answer: A
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Hi Bunuel,

Do we have any shortcut for such questions. It took me a little long to solve.

Regards,
Amulya
Bunuel
devarshi9283
Bunuel
Rank those three in order from smallest to biggest.

I. 47/150
II. 111/350
III.299/900


(A) I, II, III
(B) I, III, II
(C) II, I, III
(D) II, III, I
(E) III, I, II

Dear Bunuel Please correct the OA..it should be A
________________________
Edited. Thank you for noticing.
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