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How many integers between 50 and 100, inclusive, are divisible by 2 o
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07 Jun 2018, 01:43
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[GMAT math practice question] How many integers between 50 and 100, inclusive, are divisible by 2 or 3? A. 35 B. 37 C. 42 D. 47 E. 52
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Re: How many integers between 50 and 100, inclusive, are divisible by 2 o
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07 Jun 2018, 01:59
MathRevolution wrote: [GMAT math practice question]
How many integers between 50 and 100, inclusive, are divisible by 2 or 3?
A. 35 B. 37 C. 42 D. 47 E. 52 First integer divisible by \(2\) on the list is \(50\) and last is \(100\). Divide them by \(2\): \(25\) and \(50\) respectively. So, there are \(5025+1=26\) integers that are divisible by \(2\). First integer divisible by 3 on the list is 51 and last is 99. Divide them by \(3\): \(17\) and \(33\) respectively. So, there are \(3317+1=17\) integers that are divisible by \(3\). Pay attention that some integers are divisible by both \(2\) and \(3\). So, they are divisible by \(6\). We have to exclude this overlap. Thus, first integer divisible by 6 on the list is 54 and last is 96. Divide them by \(6\): \(9\) and \(16\) respectively. So, there are \(169+1=8\) integers that are divisible by \(6\). \(268=18\) integers are divisible by \(2\), but not by \(3\). \(178=9\) integers are divisible by \(3\), but not by \(2\). \(8\) integers are divisible by both \(2\) and \(3\). Thus, overall \(18+9+8=35\) integers are divisible by \(2\) or \(3\). It took about two minutes to solve the question. Answer: A
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Re: How many integers between 50 and 100, inclusive, are divisible by 2 o
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07 Jun 2018, 02:10
MathRevolution wrote: [GMAT math practice question]
How many integers between 50 and 100, inclusive, are divisible by 2 or 3?
A. 35 B. 37 C. 42 D. 47 E. 52 Question stem: Required no of integers(divisible by 2 or 3)=No of integers divisible by 2+ No of integers divisible by 3 No of integers divisible by both(2 and 3) [( A or B= A+BA and B)] Now, No of integers divisible by 2= \(\frac{(10050)}{2}\)+1=26 No of integers divisible by 3=\(\frac{(9951)}{3}\)+1=17 No of integers divisible by both(2 and 3) or No of integers divisible 6= \(\frac{(9654)}{6}\)+1=8 Therefore, Required no of integers(divisible by 2 or 3)=26+178=35 So, Answer option(A).
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Re: How many integers between 50 and 100, inclusive, are divisible by 2 o
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08 Jun 2018, 10:22
MathRevolution wrote: [GMAT math practice question]
How many integers between 50 and 100, inclusive, are divisible by 2 or 3?
A. 35 B. 37 C. 42 D. 47 E. 52 For this problem, we must recall that to calculate the number of integers in a range of values that are divisible by a certain number n, we use the formula: (largest multiple  smallest multiple)/n + 1. For example, to calculate the number of integers between 10 and 29 that are divisible by 3, we see that 27 is the largest multiple of 3 in that range of numbers, and 12 is the smallest multiple of 3. Thus, we have (27  12)/3 + 1 = 15/3 + 1 = 6. The number of integers from 50 to 100, inclusive, that are divisible by 2 is: (100  50)/2 + 1 = 26 The number of integers from 50 to 100, inclusive, that are divisible by 3 is: (99  51)/3 + 1 = 17 Now we need to subtract the overlap, that is, we have to subtract the number of multiples of 6 because multiples of 6 were counted as both multiples of 2 and multiples of 3. The number of integers from 50 to 100, inclusive, that are divisible by 6 is: (96  54)/6 + 1 = 8 Thus, the number integers between 50 and 100, inclusive, that are divisible by 2 or 3 is 26 + 17  8 = 35. Answer: A
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Re: How many integers between 50 and 100, inclusive, are divisible by 2 o
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10 Jun 2018, 17:34
=> Let A be the set of integers between 50 and 100 that are divisible by 2. Let B be the set of integers between 50 and 100 that are divisible by 3. Let C be the set of integers between 50 and 100 that are divisible by both 2 and 3. This is the same as the set of integers between 50 and 100 that are divisible by 6. Then \(A = { 50, 52, …, 100 }\) \(B = { 51, 54, …, 96, 99 }\) \(C = { 54, 60, …, 96 }\) The number of elements of the set A is \(A = \frac{( 100 – 50 )}{2 + 1} = 26.\) The number of elements of the set B is \(B = \frac{( 99 – 51 )}{3 + 1} = 17.\) The number of elements of the set C is \(C = \frac{( 96 – 54 )}{6 + 1} = 8.\) Using a Venn diagram, we can see that we need to find \(A + B  C\) as the integers in the intersection of sets A and B are counted twice. Attachment:
6.11.png [ 5.62 KiB  Viewed 952 times ]
\(A + B  C = 26 + 17 – 8 = 35.\) Therefore, the answer is A. Answer : A
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