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1) For every positive even integer n, function h(n) is

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1) For every positive even integer n, function h(n) is [#permalink] New post 24 Nov 2006, 13:25
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1) For every positive even integer n, function h(n) is defined to be the product of all of the even integer from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is:

a) Between 2 and 10
b) Between 10 and 20
c) Between 20 and 30
d) Between 30 and 40
e) Greater than 40


2) For which of the following function f is f(x) = f(1 – x) for all x?

a) f(x) = 1 – x
b) f(x) = 1 – x^2
c) f(x) = (x^2) – (1 – x^2)
d) f(x) = (x^2)(1 – x^2)
e) f(x) = (x) / (1- x)


3) Last year a certain bond with a face value of $5000, yielded 8% of its face value in interest. If that interest was approximately 6.5% of the bond’s selling price, approximately what was the bond’s selling price? Answer is $6154


4) A certain company employs 6 senior officers and 4 junior officers. If company wants to make up 3 senior officers and 1 junior officers as a group. How many different groups are possible?
Answer is 80
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 [#permalink] New post 24 Nov 2006, 14:17
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2)
I believe there is something wrong with answer choice D. I think answer D should be f(x) = (x^2)(1 – x)^2.
If x=2 then f(x)=f(1-x) --> f(2)=f(-1) and option D yields 4 with x=2 and x=-1

3)
5000(8%)=400
400=6.58%(x) --> x=6,154 = (4000/65)(100) =(800/13)(100) --> 6.154

4)
6C3=20
20*4=80
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 [#permalink] New post 24 Nov 2006, 16:03
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I still do not understand #4. could you please explain number 4?
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 [#permalink] New post 24 Nov 2006, 16:23
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4.

6C3 * 4C1 = 80

The way to select 3 senior officers from 6 is 6C3.
The way to select 1 junior officer from 4 is 4C1.

Order does not matter, hence combination.
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 [#permalink] New post 24 Nov 2006, 17:40
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how did you get 80 in Problem 4?
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 [#permalink] New post 24 Nov 2006, 18:04
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g1m2a3t406 wrote:
how did you get 80 in Problem 4?


Let’s say that the Seniors are: ABCDEF and the Juniors are: xywz

First focus on the Seniors. You can make 20 possible 3 groups of the total 6 --> (6!)/[(3!)(3!)] = 20
ABC
ABD
ABE
ABF
BCD
BCE
.
.
.
.
20 combinations

Now focus on the 3 seniors group + 1 junior.
The ABC group can have the x OR y OR w OR z --> so for ABC you have 4 possible juniors and for ABD also 4 possible juniors and so on…….

So,
From the 20 seniors groups you have 4 possible juniors:
=(4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4+4) = (20)(4) = 80
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 [#permalink] New post 24 Nov 2006, 21:59
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2nd question has no answer. u posted it wrongly i guess.

agree with the rest.
  [#permalink] 24 Nov 2006, 21:59
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