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If x is a randomly chosen integer between 1 and 20
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If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4? (A) 1/4 (B) 1/3 (C) 3/8 (D) 7/16 (E) ½ Please use the timer and let us know the time. The quest for the fastest way !
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Originally posted by nusmavrik on 09 Aug 2010, 11:36.
Last edited by Bunuel on 07 Jul 2013, 06:15, edited 1 time in total.
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Re: Hard one !
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09 Aug 2010, 12:20
nusmavrik wrote: Please use the timer and let us know the time. The quest for the fastest way !
If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4? (A) 1/4 (B) 1/3 (C) 3/8 (D) 7/16 (E) ½ Let's find the probability of an opposite event and subtract this value from 1. There are three cases xy NOT to be a multiple of 4: A. both x and y are odd > 1/2*1/2=1/4; B. x is odd and y is even but not multiple of 4 > 1/2*1/4=1/8; C. y is odd and x is even but not multiple of 4 > 1/2*1/4=1/8. \(P(xy=multiple \ of \ 4)=1(\frac{1}{4}+\frac{1}{8}+\frac{1}{8})=\frac{1}{2}\) Answer: E.
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Re: Hard one !
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09 Aug 2010, 12:07
1.7 minutes There are 400 combinations, 4 choices have denominator that satisfy this 1/4 3/8 1/2 7/16 Even x Even is divisible by 4 so 10 x 10 = 100 , clearly we have more than 100, as we haven't counted odd x even. So 1/4 is out. I see 3/8 7/16 1/2. 1/2 is the highest, if I get 200 then I am done. For even x odd; I have 4 6 8 10... 20 so 5 x 10 oddin 21 to 40 = 50 and another 50 the other way round. Will go for 100+ 50+ 50 = 200/400 = 1/2 Posted from my mobile device
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Re: Hard one !
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09 Aug 2010, 12:35
There are two ways that I see of getting a multiple of 4
1) x and/or y is a multiple of 4 2) x is a multiple of 2 and y is a multiple of 2
for 1 we have
P(option 1) = (5/20)(20/20) + (5/20)(20/20)  (5/20)(5/20) = 1/2  1/16
since 5 of the numbers in each range are divisible by 4, and if either x or y is selected to be divisible by 4 the selection of the other variable does not matter. Notice the last subtraction term which corrects the fact that we have counted the possibility of both being divisible by 4 twice (e.g. P(A union B) = P(A) + P(B)  P(A and B)).
Notice at this point that the probability of option one is already 7/16, the second largest available option. If we can come up with any other possibilities for getting xy divisble by 4 we must go with option E.
Of course this is the case because if we select a multiple of 2 that is not a multiple of 4 from both ranges of numbers, the product is divisible by 4.
We can formally compute this as:
P(option 2) = (5/20) * (5/20) = 1/16
since we do not wish to double count possibilities that fall into category 1, we only count the probability of selecting even numbers from each set that are not divisible by 4 (e.g. the other 5 even integers in each set).
Total probability is then P(option 1) + P(option 2) = 1/2



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Re: Hard one !
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10 Aug 2010, 02:05
I believe Bunuel's explanation is workable within 60 secs. Thanks guys for the awesome explanations ! All of you got the same answer but I posted this one to get the "fastest" solution LOL. I did this question by approximately calculation. I will like to share  We have { 4,8,12,16,20 } and { 24,28,32,36,40 } So we have 5 choices in each group that are divisible by 4. Approx choices not divisible = 15 * 15 = 225 Approx choices divisible = 400  225 = 175 Probability of being div by 4 = 175/400 = 7/16 (approx). But NOT 7/16. So pick E. My method is certainly NOT reliable. 50/50 chances. If I go deeper, I know 225 is problematic since I have to subtract 25 more. 5 evens in group 1 and 5 in group 2 (alone they are not divisible by 4) So exact calculation is (225  25) / 400 = 1/2. Bunuel wrote: nusmavrik wrote: Please use the timer and let us know the time. The quest for the fastest way !
If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4? (A) 1/4 (B) 1/3 (C) 3/8 (D) 7/16 (E) ½ Let's find the probability of an opposite event and subtract this value from 1. There are three cases xy NOT to be a multiple of 4: A. both x and y are odd > 1/2*1/2=1/4; B. x is odd and y is even but not multiple of 4 > 1/2*1/4=1/8; C. y is odd and x is even but not multiple of 4 > 1/2*1/4=1/8. \(P(xy=multiple \ of \ 4)=1(\frac{1}{4}+\frac{1}{8}+\frac{1}{8})=\frac{1}{2}\) Answer: E.



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Re: Hard one !
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10 Aug 2010, 04:49
nusmavrik wrote: I believe Bunuel's explanation is workable within 60 secs.
there is no doubt that he is geek ...math machine BUT in this question i think lots of things has been hidden or mind work (we call encapsulation in programming) for eg B. x is odd and y is even but not multiple of 4 > 1/2*1/4=1/8; you need some mind work to get that 1/4 i dont say it is very difficult to do that but will require some mind work which will consume time.... so may be in normal situation may not be below 60 seconds....



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Re: Hard one !
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10 Aug 2010, 16:59
the way i approached it Case 1 x is a multiple of 4 5/20 ( 5 multiples of 4 in that range ) = 1/4 it does not matter what y is XY will be a multiple of 4 Case 2 same as case 1 execpt it does not matter what x is so 1/4 total P = 1/4 + 1/4 = 1/2
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Re: Hard one !
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10 Aug 2010, 19:19
first set will have 10 elements all multiple of 2 second will have all 20 elements as
4* 21 , 4*23 .... are going to be divisible by 4
total ways to choose x = 20 total ways to choose y = 40 21 +1 = 20
total ways to chose element (xy) divisible by four = 10 * 20
10 * 20/ 20 * 20 = 1/2



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Re: Hard one !
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11 Aug 2010, 03:45
from the first row we have 5 digits divisible by 4 (4,8,12,16,20) probability to choose: 5/20 the same situation is for the second row 5/20 + 5/20 = 1/2 (E)



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Re: Hard one !
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29 Sep 2010, 08:04
rxs0005 wrote: the way i approached it
Case 1 x is a multiple of 4 5/20 ( 5 multiples of 4 in that range ) = 1/4
it does not matter what y is XY will be a multiple of 4
Case 2 same as case 1 execpt it does not matter what x is so 1/4
total P = 1/4 + 1/4 = 1/2 May be I am wrong but I think this is not a full proof method. Lets say x=6 (not multiple of 4) and y=22 (not multiple of 4) but xy = 132 (which is multiple of 4). Although you got the answer but cannot be justified by the theory.



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Re: Hard one !
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29 Sep 2010, 08:13
ulm wrote: from the first row we have 5 digits divisible by 4 (4,8,12,16,20) probability to choose: 5/20 the same situation is for the second row 5/20 + 5/20 = 1/2 (E) I think what you are trying to do here is find the probability of picking a number between 140 which is divisible by 4. But the question is to find the probability of picking 2 number from different set and the multiple will be divisible by 4. for ex. 2 from set 1 (not divisible by 4) and 26 from other set (which is not divisible by 4) but the product 52 is divisible by 4.



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Re: Hard one !
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29 Sep 2010, 08:18
No1EzPerfect wrote: rxs0005 wrote: the way i approached it
Case 1 x is a multiple of 4 5/20 ( 5 multiples of 4 in that range ) = 1/4
it does not matter what y is XY will be a multiple of 4
Case 2 same as case 1 execpt it does not matter what x is so 1/4
total P = 1/4 + 1/4 = 1/2 May be I am wrong but I think this is not a full proof method. Lets say x=6 (not multiple of 4) and y=22 (not multiple of 4) but xy = 132 (which is multiple of 4). Although you got the answer but cannot be justified by the theory. It is incorrect, the answer is only correct because of luck x is multiple of 4 and y is odd : 1/4* (1/2) = 1/8 y is multiple of 4 and x is odd : 1/4* (1/2) = 1/8 x and y are both even : 1/2*1/2 = 1/4 So net probability = 1/2
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Re: Hard one !
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11 Nov 2010, 00:29
RXS005
Your answer is correct. But my doubt is , since the question asked is what is probability of (xy) be a multiple of 4 and if one chooses 'x' as 2 and 'y' as 22 then we do get 44 which is a multiple of 4. But the choice made by you for x and y are both not multiples of 4. So in such case it would be wrong. Please correct me.



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Need to understand this solution
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28 Jun 2011, 09:06
If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4?
(A) 1/4 (B) 1/3 (C) 3/8 (D) 7/16 (E) 1/2
The solution is given below. However i want to know to solve it by not talking the "neither" approach.
Answer: There are two ways xy can be a multiple of 4. First, if either x or y (or both) is a mulitple of 4, it doesn't matter what the other number is: a multiple of 4 times anything is a mulitple of 4. Second, if neither of the numbers are multiples of 4, but both are even (for instance, 2 and 22), the product will be a multiple of 4. We need to find the probability of each of those possibilities.
To find the probability that either x, y, or both is a multiple of 4, it's easiest to find the probability that NEITHER are multiples of 4. The probability that x is NOT a multiple of 4 is 3/4 (1/4 of numbers are multiples of 4), and the probability that y is NOT a multiple of 4 is also 3/4. Thus, the probability that NEITHER is a multiple of 4 is (3/4)(3/4) = 9/16. Thus, the probability that one or both of the numbers is a multiple of 4 is 1  9/16 = 7/16.
That leaves us to solve for the other possibility: that both numbers are even but not multiples of 4. In any sequence of 4 consecutive integers, one of the 4 will be an even number that is not a multiple of 4. Thus, 1/4 of the numbers between 1 and 20 (or 21 and 40) is an even nonmultiple of 4. The probability that BOTH numbers have these characteristics is (1/4)(1/4) = 1/16.
The probability that the product is a multiple of 4, then, is the sum of our two probabilities: 7/16 + 1/16 = 8/16 = 1/2



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Re: Hard one !
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28 Jun 2011, 21:53
I got the "straight" approach. 10 even nos each between 1 and 20 and 2140 Out of the 10 in each sequence 5 are multiples of 4. 1.) Find the probability of either or both the nos from the sequences being multiples of 4. [4's multiples from group1] *[nos tht are not 4's multiples from group2] + [nos tht are not 4's multiples from group1]*[4's multiples from group2] + [4's multiples from group1] * [4's multiples from group2]
5/20*15/50 + 15/20*5/20 + 5/20*5/20 = 7/16
2.) The probability of nos selected from both sequences to be even and none of those nos being multiples of 4
[even nos excluding 4's multiples from group1] * [even nos excluding 4's multiples from group2]
5/20*5/20 = 1/16
Adding 7/16 + 1/16 = 1/2



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Re: Hard one !
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29 Jun 2011, 09:38
x is a multiple of 4, y is anything : 5*20 = 100 x is even not multiple of 4, y is even : 5*10 = 50 x is odd, y is a multiple of 4 : 10*5 = 50 200/400 = 1/2
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Re: Hard one !
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29 Jun 2011, 11:03
There are 5 multiples of 4 in the range from 1 to 20 and 21 to 40. For xy to be divisible by 4, we can pick up one of the 5 multiples from one series and any number from the other series.
P = (5c1. 20c1 + 5c1.20c1)/ 20c1.20c1 = 200/400 = 1/2



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Re: Hard one !
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08 Aug 2011, 13:11
Bunuel wrote: nusmavrik wrote: Please use the timer and let us know the time. The quest for the fastest way !
If x is a randomly chosen integer between 1 and 20, inclusive, and y is a randomly chosen integer between 21 and 40, inclusive, what is the probability that xy is a multiple of 4? (A) 1/4 (B) 1/3 (C) 3/8 (D) 7/16 (E) ½ Let's find the probability of an opposite event and subtract this value from 1. There are three cases xy NOT to be a multiple of 4: A. both x and y are odd > 1/2*1/2=1/4; B. x is odd and y is even but not multiple of 4 > 1/2*1/4=1/8; C. y is odd and x is even but not multiple of 4 > 1/2*1/4=1/8. \(P(xy=multiple \ of \ 4)=1(\frac{1}{4}+\frac{1}{8}+\frac{1}{8})=\frac{1}{2}\) Answer: E. +1. Brilliant solution!



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Re: Hard one !
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08 Aug 2011, 19:06
There are three cases xy NOT to be a multiple of 4: A. both x and y are odd > 1/2*1/2=1/4; B. x is odd and y is even but not multiple of 4 > 1/2*1/4=1/8; C. y is odd and x is even but not multiple of 4 > 1/2*1/4=1/8.
Well i seem to be doing something wrong for sure for not arriving at 1/2*1/4 equation..
Can someone pls help with the working..



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Re: Hard one !
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19 Aug 2011, 15:42
pkmme wrote: There are 5 multiples of 4 in the range from 1 to 20 and 21 to 40. For xy to be divisible by 4, we can pick up one of the 5 multiples from one series and any number from the other series.
P = (5c1. 20c1 + 5c1.20c1)/ 20c1.20c1 = 200/400 = 1/2 your 5c1 * 20c1 for X,Y respectively would include, for example, 4*24 your 5c1 * 20c1 for Y,X respectively would again include, 24*4 so you ended up double counting some items, and you were fortunate enough to double count those that you didn't include in the first place where X is a factor of 2, Y is a factor of 2 ie, 2*22 doesn't show up in any of your combinations, but is divisible by 4




Re: Hard one ! &nbs
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