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Five racers in a competition, no tie, how many possibilities

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Five racers in a competition, no tie, how many possibilities [#permalink] New post 18 Aug 2004, 13:47
Five racers in a competition, no tie, how many possibilities A is ahead of
B.

A) 24 B) 30 C) 60 D) 90 E) 120


2.Is quadrangle ABCD a square. ?
1) AB=BC=CD=DA
2) angle abc+angle bcd=180

3.ds |x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero

4.A circle is inscribed a rectangle, Find the area of the rectangular ?
1) The circle the radius is 8;
2) The length of AD (rectangular length of side) is 5

5. S-SCALE and R-SCALE MEASUREMENT are different,but is in the linear
expression.In S-SCALE 6 and 24 corresponds to 60 and 30 of R-SCALE . What is the value in S-SCALE if R-SCALE is 100?
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 [#permalink] New post 18 Aug 2004, 19:21
question 1:

Since A and B must occupy 2 positions where A is in front, that leaves 3 places that can be rotated among 3 runners
so 3! = 6
But this is only for A and B in a pacticular position, for example, A is first and B is second.
We also need to account for A and B being 2nd and 3rd, 3rd and 4th and 4th and 5th respectively.
That's 4 combinations, so 4*6 = 24

question 2:

1) not sufficient. a square must have 4 equal sides and 4 equal angles. we are only told it has 4 equal sides.
2) not suficient, it only says one pair of sides are parallel, but not equal in length
(1) and (2) together will be sufficient, since both will let us know they are equal in lenght, and have equal angles
so (C) is my answer

question 3:

i. sufficient. for |x+1| = 2|x-1|, x is 4. |4| = 4 > 1. so we can answer the question.
ii not sufficient. for |x-3| to be not equal to zero, x cannot be 3, but it can still take values of -1, -2, 0,1,2,etc. so we can't tell for sure if x>2.
(A) is my answer.

question 4:

1. not sufficient. raidus only gives us one side of the rectangle which is 16
2. not sufficient. same as above, only one side is given
1 and 2 are both sufficient, as 2 sides are given so we can find the area
(C) is my answer

question 5:

80? If 80 is correct i'll show my working.
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 [#permalink] New post 18 Aug 2004, 20:01
Please post 1 question per thread to ease the discussion process. Thank you.
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 [#permalink] New post 18 Aug 2004, 21:49
ywilfred wrote:
question 1:

Since A and B must occupy 2 positions where A is in front, that leaves 3 places that can be rotated among 3 runners
so 3! = 6
But this is only for A and B in a pacticular position, for example, A is first and B is second.
We also need to account for A and B being 2nd and 3rd, 3rd and 4th and 4th and 5th respectively.
That's 4 combinations, so 4*6 = 24

.


But you don't consider cases like when A is in a position 2 and B is in position 4? As the question does not say that they need to be next to each other..Am I right?
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 [#permalink] New post 18 Aug 2004, 21:59
ah.. yes you are right.. let me recalculate and post again..
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 [#permalink] New post 18 Aug 2004, 23:55
If A is first, then B can be 2nd, 3rd, 4th or 5th --- 4 combinations
If A is 2nd, then B can be 3rd, 4th or 5th --- 3 combinations
If A is 3rd, then B can be 4th or 5th --- 2 combinations
If A is 4th, then B must be 5th -- 1 combinations

So total of 10 combinations

Three places for 3 runners to rotate aobut - 3! = 6
So total combinations = 10*6 = 60
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 [#permalink] New post 19 Aug 2004, 00:22
Quote:
question 3:

i. sufficient. for |x+1| = 2|x-1|, x is 4. |4| = 4 > 1. so we can answer the question.
ii not sufficient. for |x-3| to be not equal to zero, x cannot be 3, but it can still take values of -1, -2, 0,1,2,etc. so we can't tell for sure if x>2.
(A) is my answer.

I think answer is C
I gives us, x+1= 2(x-1) or x+1= 2(1-x)
Therefore x= 3 or 1/3
II gives us x not equal to 3
Hence combining both we know x=1/3.. and we can answer the question.
Correct me if iam wrong on some concept. But I checked substituing values also and it proves my method
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Circle inscribed in a rectangle [#permalink] New post 19 Aug 2004, 02:00
4.A circle is inscribed a rectangle, Find the area of the rectangular ?
1) The circle the radius is 8;
2) The length of AD (rectangular length of side) is 5



In this question,I just wanted to clear one concept.

When it is mentioned that a circle is inscribed in a quadrilateral, does it mean that the circle touches all the 4 sides of the quadrilateral or can it just touch two of the sides .

In case, it has to touch all the 4 sides, then I guess the only rectangle in which this is possible is one with equal length and breadth -- that in case of a square

So, if the circle touched all the 4 sides, then in the above questions, both 1) and 2) are sufficient and answer would be D. (although we would get different values for the areas from 1) and 2) )

Correct me guys if I am wrong.......
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Re: combination [#permalink] New post 19 Aug 2004, 04:15
2.Is quadrangle ABCD a square. ?
1) AB=BC=CD=DA
2) angle abc+angle bcd=180

1) not sufficient, a regular romboid can also fit this
2) not sufficient, any trapezoid with opposite sides parallel to each other also fulfills this condition

1+2) Not sufficient, a romboid can have all sides equal

Answer E
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Re: combination [#permalink] New post 19 Aug 2004, 04:36
awab75 wrote:
3.ds |x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero


|x|<1 means -1<x<1, therefore x is a fraction with denominator> numerator or 0.

i. Not sufficient, there are two possible solutions: x=3 and x=1/3. One fulfils the condition, but the other doesn't.

ii. Not sufficient. It means that x is any number except 3 and -3, including fractions

i+ ii. Statement ii excludes 3 from the solutions of statement i, leaving only one solution: x=1/3 which is <1

Answer C
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Re: Circle inscribed in a rectangle [#permalink] New post 19 Aug 2004, 04:54
crackgmat04 wrote:
4.A circle is inscribed a rectangle, Find the area of the rectangular ?
1) The circle the radius is 8;
2) The length of AD (rectangular length of side) is 5



In this question,I just wanted to clear one concept.

When it is mentioned that a circle is inscribed in a quadrilateral, does it mean that the circle touches all the 4 sides of the quadrilateral or can it just touch two of the sides .

In case, it has to touch all the 4 sides, then I guess the only rectangle in which this is possible is one with equal length and breadth -- that in case of a square

So, if the circle touched all the 4 sides, then in the above questions, both 1) and 2) are sufficient and answer would be D. (although we would get different values for the areas from 1) and 2) )

Correct me guys if I am wrong.......


I agree with this. The problem is what "inscribed" means: is it touching 1, 2, 3 or 4 sides of the rectangle? I think this is not clear, so a figure would have helped.
In any case:

1) Not sufficient (unless inscribed means touchin all possible sides, so it would be a square with side =16)
2) Not sufficient in any case

1 + 2) The problem now is that these 2 statements conflict with each other: it is impossible to inscribe a circle (touching 1, 2, 3 or 4 sides) with radius 8 inside a rectangle where one of the sides is 5. A circle with radius 8 requires that the smaller of the sides of the rectangle be =16 (is circle touching 3 or 4 sides) or >16 (if it is touching 1 or 2 sides)

Answers A, B, C and D are discarded. I would go for answer E, but strictly this is not correct since although more information is needed, when combined with one of the 2 statements there will be two different results (whatever this information is).

Last edited by artabro on 19 Aug 2004, 05:29, edited 1 time in total.
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Re: combination [#permalink] New post 19 Aug 2004, 05:25
awab75 wrote:
5. S-SCALE and R-SCALE MEASUREMENT are different,but is in the linear
expression.In S-SCALE 6 and 24 corresponds to 60 and 30 of R-SCALE . What is the value in S-SCALE if R-SCALE is 100?


Great question!!

My answer is -18.

18 units of S-scale = 30 units of R-scale, and when R-scale increases, S-scale decreases.

A value of 100 in the R-scale means 100 units of R-scale above 0 in the R-scale
100-60=40 units of R, this is the value in the R-scale is 40 units beyond the 60 unit. 40 units of R-scale are 24 units in S-scale ((40*18)/30=24). So the value we want in S-scale is 24 units beyond one of the known points of S: the point that is equivalent to 60 in R-scale, which is 6. And, since S-scale decreases when R-scale increases: 6-24= -18.

-18 in Scale is equivalent to 100 in the R-Scale.

It is easy if using two parallel lines to represent the scales: choose two points in both lines that are at the same distance from each other in both lines (eg 1 inch). Draw 6 (in S) above 60 (in R) and 1 inch to the right draw 24 (in S) above 30 (in R). It becomes clear that S increases to the right and R increases to the left.
This 1 inch is equivalent to 18 units of S-scale (24-6=18) and to 30 units of R-scale (60-30=30). Value 100 in R is 40 units to the left of 60. So we need to move also the same distance to the left in S. Since 40 units of R are the same as 24 units of S ((40/30)*18=24), we need to count 24 units of S to the left starting at 6, counting to the left: 6-24=-18.
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Re: combination [#permalink] New post 19 Aug 2004, 05:36
awab75 wrote:
Five racers in a competition, no tie, how many possibilities A is ahead of
B.

A) 24 B) 30 C) 60 D) 90 E) 120


Answer :C

If A is first, all combinations are possible for the rest of competitors: 4!=24
If A is second: all combinations of the other 4 – combinations where B is 1st = 4! – 1x3!=18
If A is third: all combinations of the other 4– combinations where B is 1st of 2nd= 4! – 2x3!=12
If A is fourth: all combinations of the other 4 – combinations where B is 1st , 2nd or 3rd= 4! – 3x3!=6

24+18+12+8=60
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 [#permalink] New post 19 Aug 2004, 06:39
HI

For races i got 60
For R scale -18
FOr Ds A
For others i do nto know

thanks
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 [#permalink] New post 19 Aug 2004, 06:41
hi
why do not you consider that lx+1l can be -ve or +ve ,you seem to change the signs of rhight hand side only why not left .thanks





i. sufficient. for |x+1| = 2|x-1|, x is 4. |4| = 4 > 1. so we can answer the question.
ii not sufficient. for |x-3| to be not equal to zero, x cannot be 3, but it can still take values of -1, -2, 0,1,2,etc. so we can't tell for sure if x>2.
(A) is my answer. [/quote]
I think answer is C
I gives us, x+1= 2(x-1) or x+1= 2(1-x)
Therefore x= 3 or 1/3
II gives us x not equal to 3
Hence combining both we know x=1/3.. and we can answer the question.
Correct me if iam wrong on some concept. But I checked substituing values also and it proves my method[/quote]
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 [#permalink] New post 19 Aug 2004, 07:00
Quote:
Five racers in a competition, no tie, how many possibilities A is ahead of
B.

A) 24 B) 30 C) 60 D) 90 E) 120


Just for the record, there are 120 ways all together, right? And if you think about it, A will always be either in front of B, or the other way around. And since there's no reason why one would have prominance over the other, it must be exactly half one way and half the other way. So the answer is 60.

Quote:
4.A circle is inscribed a rectangle, Find the area of the rectangular ?
1) The circle the radius is 8;
2) The length of AD (rectangular length of side) is 5


How could a circle with a radius of 8 be in a rectangle that has a side of 5? This is impossible.
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 [#permalink] New post 19 Aug 2004, 11:29
Quote:
I think answer is C
I gives us, x+1= 2(x-1) or x+1= 2(1-x)
Therefore x= 3 or 1/3
II gives us x not equal to 3
Hence combining both we know x=1/3.. and we can answer the question.
Correct me if iam wrong on some concept. But I checked substituing values also and it proves my method

Ian:
Is my approach for question 3 correct? Generally speaking, can you explain how to approach these kind of questions..
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 [#permalink] New post 19 Aug 2004, 12:24
Hi

I think that what you did is fine.

I actually answered this in another post just today (I didn't realize they were the same question)

http://www.gmatclub.com/phpbb/viewtopic.php?t=8965

hope that helps!
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Re: combination [#permalink] New post 19 Aug 2004, 21:10
As Paul suggested, let us post one Q per topic. It is too difficult to follow, since every member talks about different question.
Ans: C,E,C,E and 18

awab75 wrote:
Five racers in a competition, no tie, how many possibilities A is ahead of B.
A) 24 B) 30 C) 60 D) 90 E) 120

Total number of ways = 5! = 120
Number of palces he is ahead = 50% = 60
In another 50% cases he is behind.
C.
awab75 wrote:
2.Is quadrangle ABCD a square. ?
1) AB=BC=CD=DA
2) angle abc+angle bcd=180

I is insuff, since this can be a rhombus.
II is insufficient, since this can be any 4 sided figure.
Together is also insiffucient, since it may 140+40
E
awab75 wrote:
3.ds |x|<1?
i |x+1|=2|x-1|
ii |x-3| is not equal to zero

Solving I, (x+1)=2(x-1) OR (x+1)=-2(x-1), x=3 or 1/3. Not suff
II, Not suff
Together, we can say Yes, so sufficient.
C.
awab75 wrote:
4.A circle is inscribed a rectangle, Find the area of the rectangular ?
1) The circle the radius is 8;
2) The length of AD (rectangular length of side) is 5

E. We cannot fit a 8-rad-circle in a 5 sided rectangle.
awab75 wrote:
5. S-SCALE and R-SCALE MEASUREMENT are different,but is in the linear expression.In S-SCALE 6 and 24 corresponds to 60 and 30 of R-SCALE . What is the value in S-SCALE if R-SCALE is 100?

I found the equation of the line joining S scale as x axis and R scale as y axis, using the formula:
(y-y1)/(y2-y1) = (x-x1)/x2-x1)
(y-60)/30-60 = (x-6)/(24-6)
3y+5x=210
Substituting, y = 100, we get x = -18.
18.
Re: combination   [#permalink] 19 Aug 2004, 21:10
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