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20 Oct 2023, 01:30
Kudos
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
69% (01:50) correct
31%
(01:51)
wrong
based on 88
sessions
History
Date
Time
Result
Not Attempted Yet
If the following numbers are arranged in order from the smallest to the largest, what will be their correct order?
I. \(\frac{9}{13}\)
II. \(\frac{13}{9}\)
III. \(70\)%
IV. \(\frac{1}{0.70}\)
(A) II, I, III, IV
(B) III, II, I, IV
(C) III, IV, I, II
(D) II, IV, III, I
(E) I, III, IV, II
I. \(\frac{9}{13}\)
II. \(\frac{13}{9}\)
III. \(70\)%
IV. \(\frac{1}{0.70}\)
(A) II, I, III, IV
(B) III, II, I, IV
(C) III, IV, I, II
(D) II, IV, III, I
(E) I, III, IV, II
20 Oct 2023, 23:43
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Bunuel
Using options
We have two numbers with their reciprocals => I & II AND III & IV
So two will be lesser than 1 and two greater than 1.
I and III are <1, so should occupy first two positions.
Only E has them in first two places.
Comparison
Let us compare the two >1
\(\frac{13}{9},\frac{10}{7}\)
\(\frac{13*7}{9*7},\frac{10*9}{7*9}\)
Now, the denominator is the same, so the larger fraction will have greater numerator. 13*7>10*9 => Thus \(\frac{13}{9}>\frac{10}{7}\)
II > IV
Also a>b means \(\frac{1}{b}>\frac{1}{a}\) or
III > I
Increasing order = I, III, IV, II
E
rey1009
Joined: 20 Oct 2023
Last visit: 31 Jul 2024
Posts: 16
Location: India
Concentration: Technology, Entrepreneurship
Schools: Cornell '26 Tepper '26 NUS '26
GPA: 3.74
WE:Engineering (Computer Software)
21 Oct 2023, 03:42
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I. 9/13 = 0.68
II. 13/9 = 1.44
III. 70% =0.7
IV. 1/0.70 =10/7 =1.42
clearly, correct order is E.
II. 13/9 = 1.44
III. 70% =0.7
IV. 1/0.70 =10/7 =1.42
clearly, correct order is E.
