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08-02-2005, 08:41 PM
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#1 | | Member
Join Date: Mar 2005
Threads: 27
Posts: 691
| SAT Math question
From blue book, pg. 473
7. If 18sqrt(18) = Rsqrt(t), where R and t are positive integers and R > t, which of the following could be the value of rt?
(A) 18
(B) 36
(C) 108
(D) 162
(E) 324
Is there a quick way to do this one? | 
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08-02-2005, 08:48 PM
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#2 | | Senior Member
Join Date: Aug 2004 Location: Student @ JHU (Baltimore, MD) / Houston, TX
Threads: 26
Posts: 2,131
| Break up the number in the square root:
18sqrt(18) = 18*sqrt(9*2) = 18*3*sqrt(2) = 54sqrt(2)
R = 54; t = 2
Check the stated conditions:
"R and t are positive integers and R > t"? Yup, it fits
So rt = 108 | |
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08-02-2005, 08:54 PM
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#3 | | Junior Member
Join Date: Jun 2005
Threads: 18
Posts: 159
| The answer is
C. 108
Given:
18sqrt(18) = Rsqrt(t)
sqrt 18 = 3sqrt2
18(3)sqrt(2) = Rsqrt(t)
18*3= 54
so,
54sqrt(2) = Rsqrt(t)
R=54 t=2
R>t? yes.
Tanman you beat me to it...ha!
R*t = 54*2
= 108 | |
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08-02-2005, 09:32 PM
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#4 | | Member
Join Date: Mar 2005
Threads: 27
Posts: 691
| Thanks, gentlemen. Would you like to solve another one. I know you would. Here goes
There are 5 cards: a b c d e
How many 5 card arrangements can be made if card c cannot be on either end | |
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08-02-2005, 09:49 PM
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#5 | | Junior Member
Join Date: Jun 2005
Threads: 18
Posts: 159
| 72.
Any more? | |
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08-02-2005, 09:50 PM
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#6 | | Junior Member
Join Date: Jul 2005
Threads: 21
Posts: 237
| There are 5 cards: a b c d e
How many 5 card arrangements can be made if card c cannot be on either end
c cannot be at either end, so for slot a there are 4 possibilites (all the cards except c)
now go to the other end (slot e) - there are 3 possibilites because one of them was used up on the first slot
now go back to slot b - there are 3 possibilities - all 5 cards minus the 2 that are already placed at either end
2 possibilites for slot c
1 possibility for slot d
so there are 4 x 3 x 2 x 1 x 3 = 72 different arrangements. | |
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08-02-2005, 09:52 PM
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#7 | | Member
Join Date: Mar 2005
Threads: 27
Posts: 691
| yesss. thanks guys. I finally figured it out. And your explanation confirms it. Much love. | |
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08-02-2005, 09:53 PM
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#8 | | Senior Member
Join Date: Oct 2004
Threads: 33
Posts: 1,580
| 4*3*3*2*1=72
There are 4 possibilities for the first end, then three for the other end, then 3 for one of the middles, then 2 for the next middle, then 1 for the final middle (idk if that makes sense....lol)
Edit: man you people are fast... | |
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08-02-2005, 10:00 PM
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#9 | | Junior Member
Join Date: Jun 2005
Threads: 18
Posts: 159
| Here is a straight foward explanation; This is a simple permutation problem...
5 cards arranged in random order (no repetition):
5! = 5*4*3*2*1 = 120 total combinations
Subtract the number of combinations possible with C at the end.
That value = 4! because C is fixed and there are 4 cards left to be arranged in random order.
4*3*2*1 = 24
Multiply this by 2 to account for the other end,
24*2 = 48
120 total - 48 with C at the end
120 - 48 = 72 | |
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08-02-2005, 10:04 PM
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#10 | | Junior Member
Join Date: Jul 2005
Threads: 21
Posts: 237
| That is so cool. | |
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08-02-2005, 10:08 PM
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#11 | | Junior Member
Join Date: Jun 2005
Threads: 18
Posts: 159
| Anything else? I love helping people understand math. | |
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08-02-2005, 10:11 PM
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#12 | | Member
Join Date: Mar 2005
Threads: 27
Posts: 691
| haha... not at this time. but stay tuned... | |
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08-03-2005, 01:03 PM
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#13 | | Member
Join Date: Jun 2005 Location: NY
Threads: 35
Posts: 488
| Ask something other than probability please. Next challenge question.......... | |
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08-03-2005, 02:45 PM
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#14 | | Junior Member
Join Date: Jun 2005 Location: Earth
Threads: 26
Posts: 96
| If it takes 4 men 8 hours to dig a hole, how many hours will it take 2 men to dig a half of a hole? | |
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08-03-2005, 02:56 PM
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#15 | | Member
Join Date: May 2005
Threads: 45
Posts: 883
| i may be wrong but is it half an hour? | |
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