|
Author |
Message |
|
TAGS:
|
|
|
Manager
Joined: 16 Jan 2011
Posts: 87
Followers: 5
Kudos [?]:
8
[0], given: 12
|
If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]
 06 May 2012, 09:39
Question Stats:
40% (02:12) correct
60% (01:03) wrong based on 0 sessions
If two of the four expressions a+b, 3a+b, a-b, and a-3b are chosen at random, what is the probability that their product will be of the form of a^2 + kb^2, where k is a positive integer? A. 1/2 B. 1/6 C. 1/4 D. 1/3 E. 1/2 Guys,
could you please tell me whether my logic was correct or not?
i solved this task in this way:
4*3=12 --> total number of outcomes
but only two of the expressions will result in desired form -->so the probability of choosing the right pair is 2/12=1/6
but the official explanation is a bit different:
First, determine how many possible combinations there are of the four expressions. Since there are four expressions and we'll choose two of them, we can use the combinations formula:
c=4!/(2!2!)=6
Now determine how many of those combinations fit the requirements. The product must result in an a^2, which rules out 3a+b, since 3a times a will result in 3a^2. The product must also result in a positive k, since the form is +k and k must be positive. That rules out any combination where one of the b terms is positive and the other is negative. The only remaining pair is: (a-b)(a-3b)
That's just one, so the probability of choosing that one is 1/6 , choice (B)
|
|
|
|
|
|
|
Manager
Joined: 29 Oct 2009
Posts: 212
Followers: 49
Kudos [?]:
418
[0], given: 18
|
Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]
07 May 2012, 09:18
If the question states that k is a POSITIVE integer, then no combination of the equations will give us the form a^2 + kb^2This is because 5 out of the 6 possible combinations of two terms give you a result with an ab term which does not fit the required form. The only one that fits the form is (a+b)*(a-b) = a^2 - b^2 in which case k = -1. Therefore for the question to be valid, it should not specifically state that k is a POSITIVE integer. Now assuming k can be negative, your logic to select the probability is correct as only 1 term out of the possible 6 satisfies the requirement.
_________________
Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942
Word Problems Made Easy!
1) Translating the English to Math : word-problems-made-easy-87346.html
2) 'Work' Problems Made Easy : work-word-problems-made-easy-87357.html
3) 'Distance/Speed/Time' Word Problems Made Easy : distance-speed-time-word-problems-made-easy-87481.html
|
|
|
|
|
|
Senior Manager
Joined: 23 Oct 2010
Posts: 335
Location: Azerbaijan
Followers: 6
Kudos [?]:
68
[0], given: 67
|
Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]
08 May 2012, 10:19
the only possible choice is (a+b)(a-b) so we get 1 out of 6 (2c4=6)
_________________
Happy are those who dream dreams and are ready to pay the price to make them come true
|
|
|
|
|
|
|
Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are
[#permalink]
08 May 2012, 10:19
|
|
|
|
|
|
|
|
|
Similar topics |
Author |
Replies |
Last post |
|
Similar
Topics:
|
 |
|
|
In the xy coordinate system, if (a,b) and (a+3, b+k) are two
|
desiguy |
4 |
09 Nov 2005, 17:51 |
|
|
|
|
In the xy-coordinate system, if (a,b) and (a+3, b+k) are two
|
ggarr |
3 |
07 Oct 2006, 21:35 |
|
|
|
|
In the XY coordinate system, if (a,b) and (a+3,b+k) are two
|
fawreel902 |
6 |
10 Mar 2008, 20:06 |
|
|
|
|
In the xy-coordinate system, if (a,b) and (a+3,b+k) are two
|
alimad |
4 |
21 Apr 2008, 16:49 |
|
|
1
|
 |
In the xy-coordinate system,if (a,b) and (a+3, b+k) are two
|
thevenus |
4 |
27 Jun 2012, 15:25 |
|
|
|
|
|
|
close
