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Math problems

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Math problems [#permalink] New post 09 Oct 2007, 08:42
Could you pls help with these questions.

1 ) -

(x^2-5x-1)^2 - 25 = (x-a) * (x-b) * (x-c) * (x-d). What is the value of

abcd?

2. is xyz > 0

a. xy > 0 ; b. yz > 0

3. A committe of 4 will be selected out of 6 men and 2 women. How many such commities can be there so that atleast 1 woman will be present.

4. is the range for these numbers greater than 2?

3,4,5,5,x?

a. the median is greater than the avg.

b. the median is 4

Thanks a lot friends
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 [#permalink] New post 09 Oct 2007, 08:56
Posted: Tue Oct 09, 2007 8:42 am Post subject: Math problems

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

Could you pls help with these questions.

1 ) -

(x^2-5x-1)^2 - 25 = (x-a) * (x-b) * (x-c) * (x-d). What is the value of

abcd?

difference between 2 squares. ie:

((x^2-5x-1)-5)((x^2-5x-1)+5) = (x^2-5x-6)(x^2-5x+4) = (x-3)(x-2)(x-4)(x-1), thus abcd = 3*2*4*1 = 24

is xyz > 0

a. xy > 0 ; b. yz > 0

question asks if xyz is +ve , this can happen if x,y,z are all positive or one of them is +ve and the other two are -ve.

xy>0 ..not suff z could be -ve or +ve

yz>0...not suff x could be -ve or +ve

both togehter.... insuff
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Re: Math problems [#permalink] New post 09 Oct 2007, 09:16
just4u1530 wrote:
Could you pls help with these questions.

1 ) -

(x^2-5x-1)^2 - 25 = (x-a) * (x-b) * (x-c) * (x-d). What is the value of

abcd?



#1:

Recognize the form x^2 - y^2 = (x+y)(x-y), the rest is working until things "fall into place"... here we go...

[(x^2-5x-1)+5][(x^2-5x-1)-5] = (x-a) * (x-b) * (x-c) * (x-d)
[x^2-5x+4][x^2-5x-6] = (x-a) * (x-b) * (x-c) * (x-d)
[(x-4)(x-1)][(x-6)(x+1)] = (x-a) * (x-b) * (x-c) * (x-d)

looky looky... now both sides look suspiciously similar, and abcd must equal:

(4)(1)(6)(-1) = -24 [careful with the negatives on this one!]

Please post answer choices so we know we're in the right ball-park with the solution.

Ans #1: -24
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 [#permalink] New post 09 Oct 2007, 09:18
yezz wrote:
Posted: Tue Oct 09, 2007 8:42 am Post subject: Math problems

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

Could you pls help with these questions.

1 ) -

(x^2-5x-1)^2 - 25 = (x-a) * (x-b) * (x-c) * (x-d). What is the value of

abcd?

difference between 2 squares. ie:

((x^2-5x-1)-5)((x^2-5x-1)+5) = (x^2-5x-6)(x^2-5x+4) = (x-3)(x-2)(x-4)(x-1), thus abcd = 3*2*4*1 = 24

is xyz > 0

a. xy > 0 ; b. yz > 0

question asks if xyz is +ve , this can happen if x,y,z are all positive or one of them is +ve and the other two are -ve.

xy>0 ..not suff z could be -ve or +ve

yz>0...not suff x could be -ve or +ve

both togehter.... insuff


Yezz: in your factorization (-2)(-3) make +6, we need -6...
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Re: Math problems [#permalink] New post 09 Oct 2007, 09:34
just4u1530 wrote:
Could you pls help with these questions.

2. is xyz > 0

a. xy > 0 ; b. yz > 0

3. A committe of 4 will be selected out of 6 men and 2 women. How many such commities can be there so that atleast 1 woman will be present.



2. Agree with E. xy could be ++ or --, yz could be ++ or --, if xyz is ---, ans -ve, if xyz +++, ans +ve. Not suff.

3. All possibilities of at least one woman is same as [all combinations - combinations with zero women (only men)]. That said...

All combos = 8C4 = 70.

Only men = 6*5*4*3 / 4! = 15
(need to divie by 4! because order doesn't matter and we are filling 4 spots).

Solution = All - Only Men = 70 - 15 = 55.

Ans #3: 55
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Re: Math problems [#permalink] New post 09 Oct 2007, 11:24
[quote="just4u1530"]Could you pls help with these questions.

1 ) -

(x^2-5x-1)^2 - 25 = (x-a) * (x-b) * (x-c) * (x-d). What is the value of

abcd?

2. is xyz > 0

a. xy > 0 ; b. yz > 0

3. A committe of 4 will be selected out of 6 men and 2 women. How many such commities can be there so that atleast 1 woman will be present.

4. is the range for these numbers greater than 2?

3,4,5,5,x?

a. the median is greater than the avg.

b. the median is 4

Thanks a lot friends[/quote]


The first one, as everyone correctly pointed out, is -24.

However, you can skip 1 step. Since you know that abcd is the coefficient without any x term, you can simply multiply the constants when you've broken them down into the two quadratic multiples.

For #3, I did:

(2C2 x 6C2) + (2C1 x 6C3) = 55


#4:

if x is 10, then the median is smaller than the average, if x is 4, the median is still smaller than the average so st1 is INSUFF

st2 is also INSUFF because for median to be 4, the number has to be 4 and below.

combining statement 1 and 2 leaves us with the numbers 1,2,3 - however, you can't pick 3 because then the median will equal the average.

so BOTH together are also INSUFF.

I'd pick E for #4
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 [#permalink] New post 09 Oct 2007, 15:51
2. is xyz > 0

a. xy > 0 ; b. yz > 0

This is E. could be +'s or -'s
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 [#permalink] New post 09 Oct 2007, 16:02
4. is the range for these numbers greater than 2?

3,4,5,5,x?

a. the median is greater than the avg.

b. the median is 4


S1: X could be 2. Which means the avg is 19/5 thus less than 4 (which is the median. This means the range is greater than 2.

Or X could be 5 which means the avg is 22/5 thus less than 5 (which is the median)

Insuff.

S2: X could be 2 or 3.

Insuff.

1&2:

X cannot be 3 or the mean = median. Any number less than 3 gives a mean less than the mean. X cannot be greater than 3 b/c 4 would no longer be the median.

So x<3. this means that the range is greater than 2.


C.
  [#permalink] 09 Oct 2007, 16:02
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