21. The number of flights leaving a certain airport doubles during every one-hour period between 9 A.M. and noon; after noon, the number of flightsleaving from the airport doubles during every two-hour period.If 4 flights leftfrom the airport between 9 and 10 A. M., how many flights left the airportbetween 2 and 4 P. M. ?

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21. C9 and 10 am..........4 flights left10 and 11 am.........8 flights left11 and noon..........16 flights leftnoon and 2pm……32 flights left2pm and 4pm……64 flights left

22. The ratioto is equal to(A)to 2(B)24 to 7(C)14 to 9(D)9 to 14(E)7 to 24

22. DFactoras= , which simplifies to=Second Expression=

22. D=, which yields= , and reduces to= , or 9 to 14 ratio

23.List I: {y,2,4,7,10,11}List II:{3,3,4,6,7,10}If the median of List I is equal to the sum of the median of list II and themode of List II, thenyequals(A)5(B)7(C)8(D)9(E)10

23. DList II, the mode is 3 and median is 5.Their sum is 8.Median of List 1 must also be 8. There are 6 elements inList I. Median will be the average of 3rdand 4thnumbers.Those two numbers must add up to 16.Since 7 is already listed, to get a median of 8 is if y is 9.

24. 5 years ago, Joe was twice as old as Jessica.If in 4 years Joewill be one and a half times as old as Jessica, then how old is Joenow?(A)23(B)25(C)27(D)29(E)31

24. A

25. At a local office, each trainee can stuffas many envelopes per day as a full time worker.If there areas many trainees as there are full time workers, then what fraction of all theenvelopes stuffed in a day did the trainees stuff?(A)(B)(C)

25. CLet the number of full-time workers be 5.There areas many as trainee asthere are full-time workers, so there are5 = 2 traineesLet the number of envelops each full-time worker can stuff per day be 3.So each trainee can stuff3 = 2 envelops per dayFull-time workers can stuff 53 = 15 envelops, trainees stuff 22 = 4envelopesTwo groups stuffs 15 4 = 19 envelopsTrainee stuff 4 envelops, so they stuffedof the envelops.

26. If Set X contains 10 consecutive integers and the sum of the 5least members of the set is 265, then what is the sum of the 5greatest members of the set?(A)290(B)285(C)280(D)275(E)270

26. AWe assign x to represent the smallest integer, then x + (x+1)+ (x+2) + (x+3) + (x+4) =265, so 5x + 10 = 265, 5x = 255,and x = 51. The five least consecutive integers are thus 51through 55.The five greatest integers are 56 through 60. The total is 290.

27. If only people who paid deposits attended the Rose Seminar, how many people attendedthis year?(1)70 people sent in deposit to attend the Rose Seminar this year.(2)60% of the people who sent deposits to attend the Rose Seminar this year actually went.(A)Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.(B)Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.(C)BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE issufficient.

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Term

Winter

Professor

Prof Dr Tun Aung

Tags

Prime number