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If W is the set of all the integers between 49 and 99, inclusive, that

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If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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27 Jun 2015, 03:44
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Question Stats:

62% (02:22) correct 38% (02:28) wrong based on 208 sessions

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If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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18 Jul 2015, 05:47
1
# multiples of 2 = (98-50)/2 + 1 = 25
# multiples of 3 = (99-51)/3 + 1 = 17
Total numbers = 25 + 17 - common elements
Way to find common elements: multiples of 3 starts at 51 and ends at 99. They alternate between an even an odd number, eg. 51 is odd and 54 is even.
Commeon elemets = even elements = (17-1)/2 = 8

Total number of elements = 25 + 17 - 8 = 34.
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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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18 Jul 2015, 06:00
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reto wrote:
If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

Multiples of 2 from 49 to 99 = Multiples of 2 from 1 to 99 - Multiples of 2 from 1 to 48 = [99/2] - [48/2] = 49 - 24 = 25

Multiples of 3 from 49 to 99 = Multiples of 3 from 1 to 99 - Multiples of 3 from 1 to 48 = [99/3] - [48/3] = 33 - 16 = 17

Multiples of 2 and 3 both i.e.6 from 49 to 99 = Multiples of 6 from 1 to 99 - Multiples of 6 from 1 to 48 = [99/6] - [48/6] = 16 - 8 = 8

These 8 Numbers have been counted twice in both the above calculation while calculating multiples of 2 and 3

i.e. Total Numbers in W = 25 + 17 - 8 = 34

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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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18 Jul 2015, 06:11
What you will notice is that 4 out of 6 numbers will satisfy the criteria.

49 - NO
50 - YES
51 - YES
52 - YES
53 - NO
54 - YES

55 - NO
56 - YES
57 - YES
58 - YES
59 - NO
60 - YES

So, 2/3 of the numbers up to 96 will follow the pattern, then add 2 for 98 and 99. How many numbers are there? 96 - 49 + 1 = 48, 2/3(48) = 32

32 + 2 = 34
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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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05 Sep 2015, 00:15
1
Official Solution:

Number of multiples of 3

Step 1. Subtract the extreme multiples of 3 within the range (The greatest is 99, the smallest is 51): 99 - 51 = 48
Step 2. Divide by 3: 48 / 3 = 16
Step 3. Add 1: 16 + 1 = 17.

So there are 17 multiples of 3 within the range: examples are 51, 54, 57, 60, etc.

Number of multiples of 2

Step 1. Subtract the extreme multiples of 2 within the range (The greatest is 98, the smallest is 50): 98 - 50 = 48
Step 2. Divide by 2: 48 / 2 = 24
Step 3. Add 1: 24 + 1 = 25.

So there are 25 multiples of 2 within the range: examples are 50, 52, 54, 56, 58, 60 etc.

Add the 17 multiples of 3 and the 25 multiples of 2: 25+17=42. However, by adding the multiples of 2 and the multiples of 3, we are effectively counting several numbers twice: for example, 54 and 60 are parts of both the lists above. So we can't just take 25+17=42. Find the Number of multiples of 6 (which are the double counted, as 6 is divisible by both 2 and 3), and subtract it from 42:

Step 1. Subtract the extreme multiples of 6 within the range (The greatest is 96, the smallest is 54): 96 - 54 = 42
Step 2. Divide by 6: 42 / 6 = 7
Step 3. Add 1: 7 + 1 = 8.

So there are 8 multiples of 6 within the range: we counted 8 numbers twice.
Subtract the 8 multiples of 6 from the sum of the multiples of 2 and 3:

= 17 + 25 - 8
= 42 - 8
= 34

Therefore, the final number of multiples of 2, 3 or 6 is 34.

Hence, this is the correct answer. (D)
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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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05 Sep 2015, 10:15
reto wrote:
If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

Multiples of 2:
first multiple of 2 is 50 and the last multiple is 98
so the number of multiples of 2 between 50 and 98 inclusive will be
$$t_n$$=a+(n-1)d
or 98 = 50 + (n-1)2
or 48 = (n-1)2
or n-1=24
or n= 25

Multiples of 3:
first multiple of 3 is 51 and the last multiple is 99
so the number of multiples of 3 between 51 and 99 inclusive will be
99=51+(n-1)3
or 48=(n-1)3
or 16=n-1
or n=17

Total number of multiples of 2 and 3 is 25+17 = 42
But this number also contains common multiples of 2 and 3.
We will have to subtract these multiples from total to get the desired number of integers in set W

Common multiples of 2 and 3 or multiples of 6:
first multiple of 6 will be 54 and last multiple will be 96
so the number of multiples of 6 between 54 and 99 inclusive will be
96=54+(n-1)6
or 42=(n-1)6
or 7=n-1
or n=8

So,
number of integers in set W will be
42-8=34

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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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14 Dec 2016, 15:09
reto wrote:
If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

my approach
between 49 and 99, there are 50 numbers, 25 of which are even. so definitely >25 or 26. A is out.
now, multiples of 3:
51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 -> 54, 60, 66, 72, 78, 84, 90, 96 are already counted in the 25 from above.
remains 51, 57, 63, 69, 75, 81, 87, 93, 99 -> 9
25+9 = 34.
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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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14 Dec 2016, 17:30
mvictor wrote:
reto wrote:
If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

my approach
between 49 and 99, there are 50 numbers, 25 of which are even. so definitely >25 or 26. A is out.
now, multiples of 3:
51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 -> 54, 60, 66, 72, 78, 84, 90, 96 are already counted in the 25 from above.
remains 51, 57, 63, 69, 75, 81, 87, 93, 99 -> 9
25+9 = 34.

When it is said, between numbers 49 n 99, should we include them as well?

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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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14 Dec 2016, 21:27
sudhirc wrote:
mvictor wrote:
reto wrote:
If W is the set of all the integers between 49 and 99, inclusive, that are either multiples of 3 or multiples of 2 or multiples of both, then W contains how many numbers?

A. 26
B. 32
C. 33
D. 34
E. 35

my approach
between 49 and 99, there are 50 numbers, 25 of which are even. so definitely >25 or 26. A is out.
now, multiples of 3:
51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99 -> 54, 60, 66, 72, 78, 84, 90, 96 are already counted in the 25 from above.
remains 51, 57, 63, 69, 75, 81, 87, 93, 99 -> 9
25+9 = 34.

When it is said, between numbers 49 n 99, should we include them as well?

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In this question ,if you notice , it is explicitly mentioned " inclusively ",,if only between is mentioned , we no need to consider boundary value...most of the time inclusive or exclusive will be mentioned in question stem..
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Re: If W is the set of all the integers between 49 and 99, inclusive, that  [#permalink]

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23 Mar 2018, 20:02
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Re: If W is the set of all the integers between 49 and 99, inclusive, that &nbs [#permalink] 23 Mar 2018, 20:02
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