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The possibility that the value of stock A will increase is [#permalink]
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03 Dec 2005, 23:31
4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32
5.Integer n is a factor of integer p. Both n and p cannot be divided by 8.Is p/n odd?
1) p can divided by 4
2) n can divided by 4
Answer: B
If n is a positive integer, what is the hundreds digit of 30^n?
(1) 30^n > 1000.
(2) n is a multiple of 3.
Reference answer: D
. Two positive integers A and B have a least common multiple of 120. What is the greatest possible value of their common factor?
(a) 20
(b) 18
(c) 15
(d) 12
(e) 10
Reference answer: A,
33.N is a positive integer and smaller than 10,what is N?
(1) N is the tenths digit of 1/N.
(2) N is the hundredths digit of 1/N.
Reference answer: A
35. The sequence X1, X2, X3,..., Xn , is such that Xn=1/n  1/(n+1),what is the sum of the first 100 terms of the sequence?
Reference answer: 100/101
38.On a number line, the points A, B, C, D, E, F, G and H distribute equidistantly from left to right. If A =2^6, D=2^8, H=?
. If x=10^100, x^x=?
Reference answer: [10^102]
64.If h (x) is the product of the integers from 1 to x, the least prime factor of h (100)+1 is in which of the following ranges?
2 to 10 10 to20 20 to 30 30 to 40 above 40
Reference answer: above 40
75.X=k^4, k is a positive integer and x is divisible by 32. When k is divided by 32, what is the remainder?
2 4 6
Reference answer: 4
85.What is the value of the integer n?
(1) n*(n+2)=15
(2) (n+2)^n=125
230.DS: If X2=Y2, X+Y=?
(1). X is not equal to Y
(2). X>2 and Y<2.
Answer: D
236.Is z1>1?
1) z1>1
2) z=1/3
286.In a sequence, the Nth item is described as Xn=(1/n)  1/(n+1), what is the sum of the first 100 items of the sequence?
OA>[100/101]
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Quote: 4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32
0,34+0,68both+neither=1
neither is biggest when both is biggest !
0,34+0,680,34+neither=1
neither=0,32
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Re: PLz explain these.... [#permalink]
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04 Dec 2005, 03:12
almighty1 wrote: 5.Integer n is a factor of integer p. Both n and p cannot be divided by 8.Is p/n odd?
1) p can divided by 4
2) n can divided by 4
Answer: B
1. p=12, n=3 >p/n=4, even >answer no
p=12, n= 4 > p/n=3 , odd >answer yes
> insuff
2. n can be divided by 4>n= 4k ( k is integer) > p= n*l = 4*k*l ( l is also integer) , since p is not divided by 8 > k*l can't be even, in other words, k*l must be odd> both k AND l are odds.
p/n= 4kl/ 4k= l , since l is odd > p/n is odd>answer yes.
>suff



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Re: PLz explain these.... [#permalink]
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04 Dec 2005, 03:18
almighty1 wrote: If n is a positive integer, what is the hundreds digit of 30^n?
(1) 30^n > 1000.
(2) n is a multiple of 3.
Reference answer: D
1. 30^n > 1000 >900 > n> 2 or n>=3 > n=3+r (r is positive integer or 0) >
30^n = 30^ (3+r) = 27000* 30^r = N000 ( N stands for other digits rather than 100th and 10th and unit digit) > the hundreds digit of 30^n is 0>suff
2. n is a positive integer and n is a multiple of 3 > n>=3 >similar to 1 > suff



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Quote: Two positive integers A and B have a least common multiple of 120. What is the greatest possible value of their common factor?
(a) 20
(b) 18
(c) 15
(d) 12
(e) 10
to calc the LCM u have to take ALL primes of the 2 numbers in their highest degrees. to calc the HCF u have to take ONLY the COMMON factors of the 2 numbers in their lowest degrees. 120=2^3*3*5. all of these primes must be in both numbers A and B. the HCF is 120 ! thats not given ! the HCF of the given answers is 20 !
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Re: PLz explain these.... [#permalink]
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04 Dec 2005, 03:31
almighty1 wrote: 33.N is a positive integer and smaller than 10,what is N?
(1) N is the tenths digit of 1/N.
(2) N is the hundredths digit of 1/N.
Reference answer: A
1. > 1/N= 0.Nabcd.... = N.abcd/10 > N*N.abcd... =10
> N<= 3 coz if N>=4 > N^2>10 . Since the statement says "tenth digit of 1/N" > we eliminate N=1 ...check N=1,2 , we get N=3 is correct >suff
2. N=3 > 1/N= 0.33333... , this N=3 is ok
N=6 > 1/N= (1/3)/2= (0.3333....)/2= 0.1666666...... > N=6 is also ok
>insuff



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Re: PLz explain these.... [#permalink]
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04 Dec 2005, 03:38
almighty1 wrote: 230.DS: If X2=Y2, X+Y=? (1). X is not equal to Y (2). X>2 and Y<2. Answer: D
x2= y2 > x2= y2 OR x2=2y
1. since x<>y > the first case doesn't happen > x2=2y
>x+y= 4 >suff
2. x>2> x2= x2
y<2> y2 = 2y
> similar to 1 >suff



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Quote: 230.DS: If X2=Y2, X+Y=? (1). X is not equal to Y (2). X>2 and Y<2.
this question seems to be wrong b/c u dont even need the stms to solve it=> x2=y2=> (x2)^2=(y2)^2 => x^2y^2=4x4y => x+y=4 ! sufficient to answer the question !
alternative approach:
1st case: x2=y2=>xy=0
2nd case: x+2=y2=>x+y=4
3rd case: x2=y+2=> x+y=4
4th case: x+2=y+2=>yx=0
...sufficient to answer the question !
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Last edited by christoph on 05 Dec 2005, 00:48, edited 1 time in total.



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christoph wrote: Quote: Two positive integers A and B have a least common multiple of 120. What is the greatest possible value of their common factor?
(a) 20
(b) 18
(c) 15
(d) 12
(e) 10 to calc the LCM u have to take ALL primes of the 2 numbers in their highest degrees. to calc the HCF u have to take ONLY the COMMON factors of the 2 numbers in their lowest degrees. 120=2^3*3*5. all of these primes must be in both numbers A and B. the HCF is 120 ! thats not given ! the HCF of the given answers is 20 !
i didn't understand, can you explain again, thanks



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christoph wrote: Quote: Two positive integers A and B have a least common multiple of 120. What is the greatest possible value of their common factor?
(a) 20
(b) 18
(c) 15
(d) 12
(e) 10 to calc the LCM u have to take ALL primes of the 2 numbers in their highest degrees. to calc the HCF u have to take ONLY the COMMON factors of the 2 numbers in their lowest degrees. 120=2^3*3*5. all of these primes must be in both numbers A and B. the HCF is 120 ! thats not given ! the HCF of the given answers is 20 !
What about if the number are 30 and 40 there LCM is 120 ....and there hight common factor is 10 ....but this is not the OA ......can you pls explain your answer little bit more



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Quote: 38.On a number line, the points A, B, C, D, E, F, G and H distribute equidistantly from left to right. If A =2^6, D=2^8, H=?
H= 2^9



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cool_jonny009 wrote: Quote: 38.On a number line, the points A, B, C, D, E, F, G and H distribute equidistantly from left to right. If A =2^6, D=2^8, H=? H= 2^9
2^9 is correct.
A is 64 keep adding 64 till you reach H which will be 512 = 2^9



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christoph wrote: Quote: 4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32 0,34+0,68both+neither=1 neither is biggest when both is biggest ! 0,34+0,680,34+neither=1 neither=0,32
How did you get both = 0.34?



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rianah100 wrote: christoph wrote: Quote: 4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32 0,34+0,68both+neither=1 neither is biggest when both is biggest ! 0,34+0,680,34+neither=1 neither=0,32 How did you get both = 0.34? Yeap. how did you come out with 0.34? can you pls elaborate your calculation pls? 0,34+0,68both+neither=1
neither is biggest when both is biggest !
0,34+0,680,34+neither=1



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miweekend wrote: Quote: 4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32
Yeap. how did you come out with 0.34?
can you pls elaborate your calculation pls?
We don't know whether the stock prices are dependent or independent and how they are related if they are dependent. That is the reason the question asks for 'the biggest (really??) possibility'. For probability questions, make sure you remember this formula: P(A or B) = P(A) + P(B)  P(A and B) (You can establish this on your own from the similar Sets formula that you know.) P(A or B) = Probability that only A or only B or both will happen (At least one of A and B will happen) P(A) = Probability that A will happen P(B) = Probability that B will happen P(A and B) = Probability that A and B both will happen Probability that neither will happen = 1  Probability that only A or only B or both will happen = 1  P(A or B) To maximize 'Probability that neither will happen', we need to minimize P(A or B). We can infer this from the equation above. So here P(A or B) = P(A) + P(B)  P(A and B), we need to minimize the left hand side. P(A) and P(B) are given as .34 and .68. We cannot change them. P(A and B) is not given. To minimize P(A or B), we need to maximize P(A and B). What is the maximum value of P(A and B) i.e. the probability that both will happen? Since the probability of P(A) is .34, probability that both happen cannot exceed .34. The maximum value that P(A and B) can take is .34. This is the situation where we can say that whenever the price of stock A increases, the price of stock B increases too. As we said before, we do not know the relation between the two stock prices but this relation gives us the maximum value of P(A and B). So we use this to minimize 'neither increases'. Minimum value of P(A or B) = .34 + .68  .34 = .68 Maximum value of 'Probability that neither will happen' = 1  P(A or B) = 1  .68 = .32
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Re: PLz explain these.... [#permalink]
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26 Jul 2011, 00:09
Is there a real need to calculate? cant we just say that when both is maximized  probability of (A) is all "contained" into probability of (b) and than non will be 1(b)? Isn't that the easiest way?
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VeritasPrepKarishma wrote: miweekend wrote: Quote: 4.The possibility that the value of stock A will increase is 0.34 and the possibility that stock B will increase is 0.68.What is the biggest possibility that neither will happen?
Answer: 0.32
Yeap. how did you come out with 0.34?
can you pls elaborate your calculation pls?
We don't know whether the stock prices are dependent or independent and how they are related if they are dependent. That is the reason the question asks for 'the biggest (really??) possibility'. For probability questions, make sure you remember this formula: P(A or B) = P(A) + P(B)  P(A and B) (You can establish this on your own from the similar Sets formula that you know.) P(A or B) = Probability that only A or only B or both will happen (At least one of A and B will happen) P(A) = Probability that A will happen P(B) = Probability that B will happen P(A and B) = Probability that A and B both will happen Probability that neither will happen = 1  Probability that only A or only B or both will happen = 1  P(A or B) To maximize 'Probability that neither will happen', we need to minimize P(A or B). We can infer this from the equation above. So here P(A or B) = P(A) + P(B)  P(A and B), we need to minimize the left hand side. P(A) and P(B) are given as .34 and .68. We cannot change them. P(A and B) is not given. To minimize P(A or B), we need to maximize P(A and B). What is the maximum value of P(A and B) i.e. the probability that both will happen? Since the probability of P(A) is .34, probability that both happen cannot exceed .34. The maximum value that P(A and B) can take is .34. This is the situation where we can say that whenever the price of stock A increases, the price of stock B increases too. As we said before, we do not know the relation between the two stock prices but this relation gives us the maximum value of P(A and B). So we use this to minimize 'neither increases'. Minimum value of P(A or B) = .34 + .68  .34 = .68 Maximum value of 'Probability that neither will happen' = 1  P(A or B) = 1  .68 = .32 can you elaborate on that equation? perhaps its name? I'm trying to understand conceptually how that equation was created? it just looks strange because P(AorB) would include the probability of P(AandB) since AorB includes the case AandB. But adding both P(A) and P(B) would intuitively be too large.



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pinchharmonic wrote: can you elaborate on that equation? perhaps its name? I'm trying to understand conceptually how that equation was created? it just looks strange because P(AorB) would include the probability of P(AandB) since AorB includes the case AandB. But adding both P(A) and P(B) would intuitively be too large.
Have you done Set Theory? Do you know n(A or B) = n(A) + n(B)  n(A and B)? Attachment:
Ques3.jpg [ 7.09 KiB  Viewed 1340 times ]
Since 'A or B' means all the elements that are in A or in B or both, when we add number of elements in A to number of elements in B, we count the 'both' elements twice. Hence, we subtract the n(A and B) once so that we don't end up double counting the common area. The probability equation is exactly the same concept. When we add P(A) to P(B), we have included P(A and B) twice because both P(A) and P(B) include the probability that 'A and B' happen. Hence we subtract it out once. This gives us the probability that A happens or B happens or both happen. Now think, if the question is: The probability that it will rain on any given day is 30% and the probability that the sun will shine on any given day is 60%. If the probability that it will rain and the sun will shine on the same day is 10%, what is the probability that on any given day, either it will rain or the sun will shine but not both?
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