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QotD #24 Set X consists of the positive multiples of 5, and
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07 Oct 2016, 13:01
07 Oct 2016, 13:01
Expert Reply
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Question Stats:
34% (02:02) correct
65% (02:36) wrong based on 86 sessions
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Set X consists of the positive multiples of 5, and set Y consists of the odd prime numbers less than 20. If set Z consists of every distinct integer less than 100 that is the product of one element from set X and one element from set Y, then set Z consists of how many elements?
A. 12
B. 14
C. 15
D. 16
E. 18
A. 12
B. 14
C. 15
D. 16
E. 18
Drill 1
Question: 9
Page: 284
Question: 9
Page: 284
Re: QotD #24 Set X consists of the positive multiples of 5, and
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14 May 2022, 08:29
14 May 2022, 08:29
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First, let us define the set -
Set A = {5n} where n = 1,2,3, .....
Set B = {3 , 5 , 7 , 11 , 13 , 17}
Now Set Z is just a product of all elements of A and B.
So,
Set Z = {15n , 25n , 35n , 55n , 65n , 85n)
To satisfy the condition of Z < 100
15n < 100 .... can give us max 6 numbers
25n < 100 .... can give us max 3 numbers
35n < 100 .... can give us max 2 numbers
55n < 100 .... can give us max 1 number
65n < 100 .... can give us max 1 number
85n < 100 .... can give us max 1 number
so Ans = 6+3+2+1+1+1 = 14
Ans B
Set A = {5n} where n = 1,2,3, .....
Set B = {3 , 5 , 7 , 11 , 13 , 17}
Now Set Z is just a product of all elements of A and B.
So,
Set Z = {15n , 25n , 35n , 55n , 65n , 85n)
To satisfy the condition of Z < 100
15n < 100 .... can give us max 6 numbers
25n < 100 .... can give us max 3 numbers
35n < 100 .... can give us max 2 numbers
55n < 100 .... can give us max 1 number
65n < 100 .... can give us max 1 number
85n < 100 .... can give us max 1 number
so Ans = 6+3+2+1+1+1 = 14
Ans B
General Discussion
Re: QotD #24 Set X consists of the positive multiples of 5, and
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07 Oct 2016, 13:05
07 Oct 2016, 13:05
2
Expert Reply
Explanation
Solve this problem by brute force, but be systematic about it. Set Y has a finite number of elements, so list them out and start finding the products when those elements are multiplied by positive multiples of 5. Set Y = {3, 5, 7, 11, 13, 17, 19}, so multiplying by 5—the first positive multiple of 5—yields 15, 25, 35, 55, 65, 85, and 95; that’s 7 elements for set Z thus far.
Multiplying by 10—the next positive integer multiple of 5—yields 3 more products, 30, 50, and 70. Multiplying by 15 yields two new products, 45 and 75; multiplying by 20 yields only one new product, 60. That’s a total of 13 elements for set Z so far. You already have 75 as a member of set Z, so multiplying by 25 yields no new products; multiplying by 30 yields the final new product, 90. Set Z thus consists of 14 elements: set Z = {15, 25, 30, 35, 45, 50, 55, 60, 65, 70, 75, 85, 90, 95}. If you got choice (E), you may have mistakenly included 2 as an element of set Y.
Hence Option B is the correct option.
Solve this problem by brute force, but be systematic about it. Set Y has a finite number of elements, so list them out and start finding the products when those elements are multiplied by positive multiples of 5. Set Y = {3, 5, 7, 11, 13, 17, 19}, so multiplying by 5—the first positive multiple of 5—yields 15, 25, 35, 55, 65, 85, and 95; that’s 7 elements for set Z thus far.
Multiplying by 10—the next positive integer multiple of 5—yields 3 more products, 30, 50, and 70. Multiplying by 15 yields two new products, 45 and 75; multiplying by 20 yields only one new product, 60. That’s a total of 13 elements for set Z so far. You already have 75 as a member of set Z, so multiplying by 25 yields no new products; multiplying by 30 yields the final new product, 90. Set Z thus consists of 14 elements: set Z = {15, 25, 30, 35, 45, 50, 55, 60, 65, 70, 75, 85, 90, 95}. If you got choice (E), you may have mistakenly included 2 as an element of set Y.
Hence Option B is the correct option.
