Form 8767: #25, #30 and #50

Forum for the GRE subject test in mathematics.
YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Form 8767: #25, #30 and #50

Postby YKM » Mon Aug 29, 2011 5:23 am

#25
Let x and y be positive integers such that 3x + 7y is divisible by 11. Which of the following is true?

a. 4x + 6y
b. x + y + 5
c. 9x + 4y
d. 4x - 9y
e. x + y -1

The answer is d. 4x - 9y

My logic:
Since x and y are positive integers and 3 and 7 are relative primes, then 11 | 3x + 7y would imply that 11 | x and 11 | y. Thus, any linear combination of x and y should be divisible by 11.

I know my logic is wrong, can anyone tell me what's wrong?

#30
The improper integral, from b to a, f(x)*f'(x) dx is

The answer is a. necessarily zero.

My logic:
If I let f(x) = x^2, then f'(x) = 2x, meaning f(x)f'(x) = 2x^3. The integral, from b to a, 2x^3 dx is 1/2 * (b^4 - a^4), which is not zero.

I know my logic is wrong, can anyone tell me what's wrong?

#50
In a game two players take turns tossing a fair coin, the winner is the first one to toss a head. The probability that the player who makes the first toss wins the game is?

The answer is D: 2/3.

My logic:
Let the two player be called A and B. Suppose A go first, and toss a head, the game is done and the probability of this event is 1/2. Suppose A go first and does not toss a head, then B's turn, and toss a head, the game is done, and the probability of this event is 1/2 * 1/2. Since the name A and B are generic, the sum of the probability of these two disjoint events should be the answer, which is 3/4.

I know my logic is wrong, can anyone tell me what's wrong?

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Form 8767: #25, #30 and #50

Postby YKM » Mon Aug 29, 2011 10:14 am

for #25, my logic is clearly wrong as 11 | (7 + 4).

By some trials and error, I found that 11 |(3*4 + 7*3) = 33, thus the answer.

Does anyone have a more general way? as trials and error take too much time in exam.

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #25, #30 and #50

Postby blitzer6266 » Mon Aug 29, 2011 2:33 pm

For 25, the statement is equivalent to

3x+7y = 0 (mod 11)

You want to see what this looks like when you have a 4x instead of a 3x, so you can multiply both sides by 5, giving

15x + 35y = 0 (mod 11)

This is equivalent to

4x+2y = 0 (mod 11) or
4x-9y = 0 (mod 11)

For 30, I believe you're missing some information in the problem.. It probably uses the fact that the antiderivative is (f(x))^2

For 50, say the probability of playing A winning is p. Then there is a 1/2 chance player A wins in the first round, and 1/2 that he doesn't. Consider the 2nd case. Conditional on player A not winning the first round, player b has a probability p of winning. Thus, player A's probability of winning is:

p= 1/2 + (1/2)(1-p)

(3/2)p=1

p=2/3

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #25, #30 and #50

Postby blitzer6266 » Mon Aug 29, 2011 2:34 pm

For 30, I meant ((f(x))^2)/2 but that's probably inconsequential

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #25, #30 and #50

Postby blitzer6266 » Mon Aug 29, 2011 2:39 pm

Annnd I found the problem you're looking at which says f is a graph of a semicircle with endpoints (a,0) and (b,0). So the integral becomes (f(b))^2/2 - (f(a))^2/2 = 0 - 0 = 0

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Form 8767: #25, #30 and #50

Postby YKM » Mon Aug 29, 2011 9:35 pm

blitzer6266 wrote:Annnd I found the problem you're looking at which says f is a graph of a semicircle with endpoints (a,0) and (b,0). So the integral becomes (f(b))^2/2 - (f(a))^2/2 = 0 - 0 = 0


Thanks, but the function f in the question should be in general, not any specific function. The statement should work for any general function.

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #25, #30 and #50

Postby blitzer6266 » Mon Aug 29, 2011 10:13 pm

Absolutely not. Reread that page that has question 30. The top says 28-30 refer to f(x) which is the graph of a semicircle with endpoints (a,0) (b,0)

YKM
Posts: 45
Joined: Mon Aug 08, 2011 10:25 am

Re: Form 8767: #25, #30 and #50

Postby YKM » Mon Aug 29, 2011 11:34 pm

Thank you very much, I miss that.

Topoltergeist
Posts: 44
Joined: Tue Aug 09, 2011 6:18 pm

Re: Form 8767: #25, #30 and #50

Postby Topoltergeist » Sat Sep 10, 2011 10:32 pm

For #50, your logic supposes that the game will end fairly quickly. The game can go on arbitrarily long.

To solve this problem, write the probability that player A wins as a sum of the probabilities that he/she wins on the Nth round, with N ranging from 1 to infinity. The sum will be a geometric series, so computing it is tractable.

Bitterfish
Posts: 1
Joined: Wed Sep 21, 2011 3:49 pm

Re: Form 8767: #25, #30 and #50

Postby Bitterfish » Wed Sep 21, 2011 3:53 pm

Alternative solution to number 30:

f(x) is even about the line x = (b+a)/2. f'(x) is odd about the same line. In general the integral of an odd function times an even function is always zero.

Vince
Posts: 4
Joined: Sat Mar 17, 2012 4:36 pm

Re: Form 8767: #25, #30 and #50

Postby Vince » Thu Apr 05, 2012 8:24 am

blitzer6266 wrote:Annnd I found the problem you're looking at which says f is a graph of a semicircle with endpoints (a,0) and (b,0). So the integral becomes (f(b))^2/2 - (f(a))^2/2 = 0 - 0 = 0


Do you mean integration by parts?

Integral (f(x)*f'(x)) = f^2 - integral (f(x)*f'(x)), so integral (f(x)*f'(x)) = (f^2)/2 ?

blitzer6266
Posts: 61
Joined: Sun Apr 04, 2010 1:08 pm

Re: Form 8767: #25, #30 and #50

Postby blitzer6266 » Thu Apr 05, 2012 2:42 pm

That works but another way to do it is just a u-sub with u=f(x) then du=f'(x)

Vince
Posts: 4
Joined: Sat Mar 17, 2012 4:36 pm

Re: Form 8767: #25, #30 and #50

Postby Vince » Thu Apr 05, 2012 4:35 pm

Thanks!

goatman2743
Posts: 31
Joined: Mon Sep 02, 2013 7:29 pm

Re: Form 8767: #25, #30 and #50

Postby goatman2743 » Thu Sep 12, 2013 3:23 pm

I'm confused about all of these answers to 30. We're dealing with an improper integral as f' approaches infinity/-infinity at the endpoints, so we can't just say the integral of f*f' is f^2/2 and evaluate at endpoints, correct? I see that if the integrals exist they must be 0 because of symmetry, but how do we know they exist?

DDswife
Posts: 58
Joined: Thu Aug 14, 2014 5:29 pm

Re: Form 8767: #25, #30 and #50

Postby DDswife » Thu Aug 14, 2014 8:01 pm

50. Think of the possible winning events. A begins, B is the second person

Case 1 A gets H. A wins, B doesn't get to play.
Even (H), Probability = P = 1/2

Case 2 Winning event (T,T,H) is the only winning event

Possible cases = 2^3
Winning cases= 1

P = 1/(2^3)

Case 3 (T,T,T,T,H), P = 1/(2^5)

Etc

All these will be of the form P = 1/[2^(2n+1)]

Now add these numbers and use the sum of the geometric series. Calculate the limit as n goes to infinity.

Hope this helps.




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