Wiley the official guide for GMAT Episode 1 Part 9

Tham khảo tài liệu 'wiley the official guide for gmat episode 1 part 9', ngoại ngữ, ngữ pháp tiếng anh phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | Data Sufficiency Answer Explanations Now assume both 1 and 2 . From 1 it follows that - 1 or 5c a c and so a 4c. a c 5 From 2 it follows that io or 10c 3b 3c and so 7c 3b and b 3 c. Since 4c -3 c from the statements it can be deduced that c 0 it follows that a b. Therefore 1 and 2 together are sufficient. The correct answer is C both statements together are sufficient. 90. If k m and t are positive integers and 6 mm I2 do t and 12 have a common factor greater than 1 1 k is a multiple of 3. 2 m is a multiple of 3. Arithmetic Properties of numbers Using a common denominator and expressing the 2k 3m t sum as a single fraction gives y Th . 12 12 12 Therefore it follows that 2k 3m t. Determine if t and 12 have a common factor greater than 1. 1 Given that k is a multiple of 3 then 2k is a multiple of 3. Since 3m is also a multiple of 3 and a sum of two multiples of 3 is a multiple of 3 it follows that t is a multiple of 3. Therefore t and 12 have 3 as a common factor SUFFICIENT. 2 If k 3 and m 3 then m is a multiple of 3 and t 15 since 2 3 3 3 . 6 n 12 12 so t and 12 have 3 as a common factor. However if k 2 and m 3 then m is a multiple of 3 and t 13 since 1 13 12 12 12 12 so t and 12 do not have a common factor greater than 1 NOT sufficient. The correct answer is A statement 1 alone is sufficient. ------------ -------------- -------------- A B C D 91. In the figure above is CD BC 1 AD 20 2 AB CD Geometry Lines 1 Information is given about the total length of the segment shown which has no bearing on the relative sizes of CD and BC NOT sufficient. 2 Here AB and CD are equal which also has no bearing on the relative sizes of BC and CD NOT sufficient. It cannot be assumed that the figure is drawn to scale. Considering 1 and 2 together if lengths AB and CD were each a little larger than pictured for example I----------------20---------------- A 8 B 4 C 8 D then BC CD. But if the reverse were true and lengths AB and CD were instead a little smaller than .

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