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by pm » Thu Jun 30, 2011 3:47 am
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
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by testprepDublin » Thu Jun 30, 2011 3:57 am
x can be a multiple of 4, y can be anything : 5*20=100

x can be an even non-multiple of 4, y can be even : 5*10 = 50

x can be odd, y can be a multiple of 4 : 10*5 = 50

(100+50+50)/20*20 = 1/2
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by Frankenstein » Thu Jun 30, 2011 3:58 am
Hi,
xy is multiple of 4 if
1)x is multiple of 4
2)y is multiple of 4
3)x and y are both even
In order not to count repetitions we make it:
1) x is multiple of 4 and y is odd
Probability that x is multiple of 4 y is odd is (5/20)*(10/20) = 1/8
2) y is multiple of 4 and x is odd
Probability that y is multiple of 4 is (5/20)*(10/20) = 1/8
3) x and y are both even
Probability that both are even is (1/2)*(1/2) = 1/4

So, total probability is 1/8 + 1/8 + 1/4 = 1/2

Hence, E
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by GMATGuruNY » Fri Jul 01, 2011 8:44 am
pm wrote: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
I received a PM asking me to comment.

Case 1: even*even = multiple of 4.
1/2 the numbers in each range are even.
P(x is even AND y is even) = 1/2 * 1/2 = 1/4.

Case 2: multiple of 4 * odd = multiple of 4.
Since every other even number is a multiple of 4, and 1/2 the numbers in each range are even, 1/4 of the numbers in each range are multiples of 4.
1/2 the numbers in each range are odd.
Thus:
P(x is a multiple of 4 AND y is odd) = 1/4 * 1/2 = 1/8.
P(x is odd AND y is a multiple of 4) = 1/2 * 1/4 = 1/8.

Since any of the outcomes above would yield a mulitple of 4, we add the fractions:
1/4 + 1/8 + 1/8 = 1/2.

The correct answer is E.
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by Scott@TargetTestPrep » Sat Jan 20, 2018 6:12 am
pm wrote: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
We see that xy is a multiple of 4 if both x and y are multiples of 2 (but not multiples of 4), OR x is a multiple of 4, OR y is a multiple of 4, or . Let's look at the three cases.

Case 1: Both x and y are multiples of 2 but neither is a multiple of 4

Thus x could be 2, 6, 10, 14, or 18 and y could be 22, 26, 30, 34, or 38. We see that there are 5 choices for x and 5 choices for y; thus, the number of integers xy that are a multiple of 4 is 5 x 5 = 25.

Case 2: x is a multiple of 4

If x is a multiple of 4 (i.e., x could be 4, 8, 12, 16, or 20), then y could be any integer. Thus, the number of integers xy that are a multiple of 4 is 5 x 20 = 100.

Case 3: y is a multiple of 4

If y is a multiple of 4 (i.e., y could be 24, 28, 32, 36, or 40), then x could be any integer. Thus, the number of integers xy that are a multiple of 4 is 20 x 5 = 100.

However, there are overlaps in cases 2 and 3. For example, x = 4 and y = 24 is an overlap in both cases. We need to subtract the overlaps. The overlaps occur when both x and y are multiples of 4. Since there are 5 choices for x being a multiple of 4 and another 5 choices for y being a multiple of 4, the total number of overlaps is 5 x 5 = 25.

Thus, the total number of integers xy that are a multiple of 4 is 25 + 100 + 100 - 25 = 200. Since the total number of integers xy that can be formed is 20 x 20 = 400, the probability that the product xy is a multiple of 4 is 200/400 = 1/2.

Answer: E

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