If x is a randomly chosen integer between 1 and 20

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

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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.

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....

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

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

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)

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.

I think what you are trying to do here is find the probability of picking a number between 1-40 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.

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

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.

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 non-multiple 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

I got the "straight" approach.
10 even nos each between 1 and 20 and 21-40
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

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

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

+1. Brilliant solution!

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..

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