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Published: 29/05/2017In the figure to the right, O is the center of the circle. If the area of the circle is 9, then the perimeter of the sector PRQO is Attachment: mmnn.jpg (A) \(\frac{}{2}\)+ 18 (B) \(\frac{}{2}\) + 6 ( ... Read moreSource: gmatclub.com


Published: 29/05/2017shailabh wrote: Brief : Indian, Male, 31 years, Engineer, and an MBA from top 15 Indian B school Work experience : Corporate Banking (6 years); IT (1 year, pre Indian MBA) Current Designation : Associ ... Read moreSource: gmatclub.com

Published: 29/05/2017In the figure to the right, what is the greatest number of regions into which two straight lines will divide the shaded region? Attachment: MMMM.jpg (A) 1 (B) 2 (C) 3 (D) 4 (E) 5 Nova GMAT Read moreSource: gmatclub.com

Published: 29/05/2017The perimeter of the outside of truck tire R is 7/8 the perimeter of the outside of truck tire T. What is the area of the outside of truck tire T? (1) The diameter of the outside of truck tire R is 1. ... Read moreSource: gmatclub.com

Published: 29/05/2017ankitsaroha wrote: Clearly, k=2 alone is insufficient as we know nothing of x and y. Therefore, B and D choices are gone. Let us take z = 3^k  2^k If x = 3^(k  1) and y = 2^(k  1) then x  y = 3^k/ ... Read moreSource: gmatclub.com

Published: 29/05/2017Focus shifts from X to Y , where X and Y MUST be  . A) textbooks instructing the children to educating the children with teachers > Not  B) textbooks instructing the children to the children ... Read moreSource: gmatclub.com

Published: 29/05/2017Hi, Thank you for the time to evaluate my profile. Background: Male, 22 year old (too young? ) Nationality: Indian Undergrad: graduated in 2016, from a UK Russel group (tier 1) in aerospace engineerin ... Read moreSource: gmatclub.com

Published: 29/05/2017Statement 1. Square root of an integer might or might not be an integer. We don't know if even x is an integer or not. So Insufficient . Statement 2. Square of an odd integer is odd, and square of an ... Read moreSource: gmatclub.com

Published: 29/05/2017Since, the ship A leaves three hours early and travels 15 mph, it would have traveled 45 miles. Now the speed relative to Ship B, which travels at 25 mph is 10 miles/hour Since the distance it need to ... Read moreSource: gmatclub.com

Published: 29/05/2017utkarshthapak wrote: I know that difficulty level is decided automatically by system as per the no. of correct attempts but this question was difficult. I got the right answer but it took me 2.30 min. ... Read moreSource: gmatclub.com

Published: 29/05/2017mynamegoeson wrote: Please correct me if i am wrong M1, M2, M3, F1, F2, F3 can be arranged in 6! ways Sent from my iPhone using GMAT Club Forum Hi when you say 6!, you are listing ALL the possible arr ... Read moreSource: gmatclub.com

Published: 29/05/2017Using venn diagram, we know student taking only french = 42, only chemistry = 27 and both = 21. Add all three categories, we get 90 student who have taken either french, or chemistry or both. So, stud ... Read moreSource: gmatclub.com

Published: 29/05/2017Let them meet 'n' hours after 4 pm. From 1 to 4 pm, ship A has already travelled = 15*3 = 45 miles So in 'n' hours, A has total travelled = (45 + 15n) miles and B has travelled = 25n miles Since they ... Read moreSource: gmatclub.com

Published: 29/05/2017mynamegoeson wrote: There will be 4C3 ways in which any three out of A B C D can be picked . there are 2C1 ways in which A can be picked . 4C1 ways in which B can be picked. 1C1 ways in which C can be ... Read moreSource: gmatclub.com

Published: 29/05/2017Walkabout wrote: After driving to a riverfront parking lot, Bob plans to run south along the river, turn around, and return to the parking lot, running north along the same path. After running 3.25 mi ... Read moreSource: gmatclub.com

Published: 29/05/20175 pages in 2 sec. So 2.5 pages in 1 sec. 150 pages in 60 sec or 1 min. therefore, 1050 pages in 7 mins. Option E should be correct Read moreSource: gmatclub.com

Published: 29/05/2017For each friend: he/she can take either 1 or 2 or 3 or 4 or 5 coins to the beggar. Since the coins are all identical: 1 coin can be chosen in = 1 way 2 coins can be chosen in = 1 way 3 coins can be ch ... Read moreSource: gmatclub.com

Published: 29/05/20171 to 1000 we have one thousand positive integers. We don't require 1 here so remaining integers will be 999. Read moreSource: gmatclub.com
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