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# Set I consists of the integers from 11 through 100, inclusiv

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Set I consists of the integers from 11 through 100, inclusiv [#permalink]  05 Dec 2017, 12:42
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Question Stats:

85% (00:45) correct 14% (01:01) wrong based on 14 sessions
Set I consists of the integers from 11 through 100, inclusive.

 Quantity A Quantity B 4 times the number of integers in set T that are multiples of 4 5times the number of integers in set T that are multiples of 5

A. Quantity A is greater.
B. Quantity B is greater.
C. The two quantities are equal
D. The relationship cannot be determined from the information given.
[Reveal] Spoiler: OA

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Manager
Joined: 03 Dec 2017
Posts: 65
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Kudos [?]: 15 [1] , given: 20

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  06 Dec 2017, 23:52
1
KUDOS
how do I check how many muｌtiple of 4 between 11-100?
Can anyone give me the formula, thank you
Director
Joined: 20 Apr 2016
Posts: 762
Followers: 6

Kudos [?]: 522 [1] , given: 94

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  07 Dec 2017, 02:23
1
KUDOS
wongpcla wrote:
how do I check how many muｌtiple of 4 between 11-100?
Can anyone give me the formula, thank you

Definitely it has a formula-

Multiple of x in the range = $$\frac{(last multiple of "x" in the range - first multiple of "x" in the range)}{x}$$ +1

Now in the first case we need the multple of 4 in the range from 11 to 100, so applying the formula

Multiple of 4 = $$\frac{(96-12)}{4}+ 1$$ (Last multiple of 4 in the range from 11 to 100 is 96 and the first multiple in the range is 12)

i.e equal to 22 nos.

In second case

Multiple of 5 = $$\frac{(100-15)}{5}+ 1$$

i.e equals to 17 nos.

Hence A > B
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Intern
Joined: 13 Oct 2017
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Kudos [?]: 1 [0], given: 0

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  22 Jan 2018, 06:34
why Last multiple of 4 in the range from 11 to 100 is not 100 but 96

pranab01 wrote:
wongpcla wrote:
how do I check how many muｌtiple of 4 between 11-100?
Can anyone give me the formula, thank you

Definitely it has a formula-

Multiple of x in the range = $$\frac{(last multiple of "x" in the range - first multiple of "x" in the range)}{x}$$ +1

Now in the first case we need the multple of 4 in the range from 11 to 100, so applying the formula

Multiple of 4 = $$\frac{(96-12)}{4}+ 1$$ (Last multiple of 4 in the range from 11 to 100 is 96 and the first multiple in the range is 12)

i.e equal to 22 nos.

In second case

Multiple of 5 = $$\frac{(100-15)}{5}+ 1$$

i.e equals to 17 nos.

Hence A > B
Director
Joined: 20 Apr 2016
Posts: 762
Followers: 6

Kudos [?]: 522 [1] , given: 94

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  22 Jan 2018, 07:27
1
KUDOS
HsienYi wrote:
why Last multiple of 4 in the range from 11 to 100 is not 100 but 96

Yes you are right it is 100 and not 96.

So after calculation
A= 23 and B=17

Hence A>B
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Director
Joined: 07 Jan 2018
Posts: 560
Followers: 4

Kudos [?]: 485 [0], given: 85

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  22 Jan 2018, 21:28
HsienYi wrote:
why Last multiple of 4 in the range from 11 to 100 is not 100 but 96

pranab01 wrote:
wongpcla wrote:
how do I check how many muｌtiple of 4 between 11-100?
Can anyone give me the formula, thank you

Definitely it has a formula-

Multiple of x in the range = $$\frac{(last multiple of "x" in the range - first multiple of "x" in the range)}{x}$$ +1

Now in the first case we need the multple of 4 in the range from 11 to 100, so applying the formula

Multiple of 4 = $$\frac{(96-12)}{4}+ 1$$ (Last multiple of 4 in the range from 11 to 100 is 96 and the first multiple in the range is 12)

i.e equal to 22 nos.

In second case

Multiple of 5 = $$\frac{(100-15)}{5}+ 1$$

i.e equals to 17 nos.

Hence A > B

I guess multiples of 5 should 18 instead of 17 (100-15)/5 = 85/5 = 17 + 1 = 18
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Manager
Joined: 02 Jan 2018
Posts: 66
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Kudos [?]: 30 [0], given: 0

Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink]  23 Jan 2018, 09:21
T{ 12,16,......100}
100=12+(n-1)4
n=23

T{15,20,25....100}
100=15+(n-1)5
n=18

Therefore A is greater
Re: Set I consists of the integers from 11 through 100, inclusiv   [#permalink] 23 Jan 2018, 09:21
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# Set I consists of the integers from 11 through 100, inclusiv

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