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A circular rim 28 inches in diameter rotates the same number

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Intern
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A circular rim 28 inches in diameter rotates the same number [#permalink] New post 09 Dec 2007, 10:21
A circular rim 28 inches in diameter rotates the same number of inches
per second as a circular rim 35 inches in diameter. If the smaller
rim makes y revolutions per second, hom many revolution per minute
does the larger rim make in terms of y
------------------------------------------------------------------------

The ratio of two quantities is 3 to 4. If each of the quantities is
increased by 5, what is the ratio of these two new quantities

------------------------------------------------------------------------

If m is an integer, is m odd?
1) m/2 is not an even integer
2) m-3 is an even integer

-------------------------------------------------------------------------

1234
1243
....
....
+4321
------

The addition problem above shows four of the 24 different integers that can
be formed by using each of the digits 1, 2, 3 and 4 exactly once in each
integer. What is the sum of these 24 integers
Intern
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 [#permalink] New post 10 Dec 2007, 04:21
First problem:
First let's find the number of inches the small and large rim rotate in one revolution - it is circumference length or:
smaller rim: 2piR=28pi (which is around 84 but we won't need this number).
larger rim: 2piR=35pi
Let x be number of revolutions per minte the larger rim makes => x/60 - number of revolutions per second.
If a smaller rim makes y revolutions per second then it rotates 28pi*y number of inches per second which equals number of inches per second of larger rim or 35pi*x/60. Solve it for x:
28piy=36pix/60 => 36x=1680y => x=140/3*y
That's what I got... strange answer. I wonder if it is in the answer choices?
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 [#permalink] New post 10 Dec 2007, 04:27
Second problem:
To my mind it's impossible to tell what will be the ratio of new quantities. If we take different numbers with the given ratio and check them the new ratios will be different:
1) If we take 3 and 4. 3+5=8, 4+5=9, so the ratio is 8/9.
2) If we take 6 and 8. 6+5=11, 8+5=13, so the ratio is 11/13
We get two different ratios, so we can't say for sure what will be the answer.
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 [#permalink] New post 10 Dec 2007, 06:39
Third:
If m is an integer, is m odd?
1) m/2 is not an even integer
2) m-3 is an even integer

I would say B
(i) can't say m=6 and m=5 both don't have even m/2
(ii) m-3=even => m=even +3=> m=odd
================================
Fourth:
1234
1243
....
....
+4321

24 permutations means that on the ones we will have 6x1's, 6x2's,6x3's,6x4's
6+12+18+24=60
same on the tens, hundredths and thousands
so 60+600+6000+60000=66660
=================================
where did you get these questions from
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Re: I need help with these problems [#permalink] New post 10 Dec 2007, 09:27
gmat2006-8001 wrote:
A circular rim 28 inches in diameter rotates the same number of inches
per second as a circular rim 35 inches in diameter. If the smaller
rim makes y revolutions per second, hom many revolution per minute
does the larger rim make in terms of y
------------------------------------------------------------------------


Getting 48y.

For smaller rim, 1 rev = 28pi inches
If it makes y revs/second then it cover 28*pi*y inches/second
Since speed of big rim = speed of small rim, big rim also covers 28*pi*y inches/second
Since for the big rim, 1 rev = 35pi inches, 28*pi*y inches = 28*pi*y/35*pi = 28y/35 revs
Since the big rim takes 28y/35 revs in 1 sec, it will make 28y/35*60 = 48 y revs in 1 min.
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Re: I need help with these problems [#permalink] New post 10 Dec 2007, 09:38
gmat2006-8001 wrote:


1234
1243
....
....
+4321
------

The addition problem above shows four of the 24 different integers that can
be formed by using each of the digits 1, 2, 3 and 4 exactly once in each
integer. What is the sum of these 24 integers


Can someone explain how to solve this?
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 [#permalink] New post 10 Dec 2007, 10:24
> Can someone explain how to solve this?

GK_Gmat did you read my reply? The answer is there unless it is wrong ;-)
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 [#permalink] New post 10 Dec 2007, 23:12
Yes, sorry for mistake (in the end of my explanation I switched to 36*pi*x instead of 35*pi*x by mistake), I also get 48y.
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Re: I need help with these problems [#permalink] New post 10 Dec 2007, 23:48
GK_Gmat wrote:
gmat2006-8001 wrote:


1234
1243
....
....
+4321
------

The addition problem above shows four of the 24 different integers that can
be formed by using each of the digits 1, 2, 3 and 4 exactly once in each
integer. What is the sum of these 24 integers


Can someone explain how to solve this?





1, 2, 3, 4 and orders matter so all possibilities = 4! = 24

for each of the digitd, there are six 1s, six 2s, six 3s, and six 4s.
so sum = 66,660
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 [#permalink] New post 11 Dec 2007, 03:04
alexperi wrote:
Fourth:
1234
1243
....
....
+4321

24 permutations means that on the ones we will have 6x1's, 6x2's,6x3's,6x4's
6+12+18+24=60
same on the tens, hundredths and thousands
so 60+600+6000+60000=66660
=================================
where did you get these questions from


very nice. THanks,
  [#permalink] 11 Dec 2007, 03:04
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