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TAGS: Intern Joined: 11 Jan 2019
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Set I consist of the integers from 11 through 100, inclusive [#permalink]
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Question Stats: 50% (00:51) correct 50% (01:18) wrong based on 12 sessions
Set I consist 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 5 times 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

Last edited by Carcass on 12 Jan 2019, 01:20, edited 1 time in total.
Edited by Carcass
Founder  Joined: 18 Apr 2015
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Kudos [?]: 1338 , given: 6306

Re: Set I consist of the integers from 11 through 100, inclusive [#permalink]
Expert's post
Hi,

when posting a new question, please, post it under the right forum and not in general quant section.

Also, format it , is easy https://greprepclub.com/forum/qq-how-to ... -2357.html

Thank you so much
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Intern Joined: 07 Jan 2019
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Kudos [?]: 20 , given: 11

Re: Set I consist of the integers from 11 through 100, inclusive [#permalink]
To count how many integers there are between two integers, we have certain rules.

The given question comes under the rule that:

If both end points are included: subtract and then add 1.

The number of integers from 11 through 100 inclusive = (100-11)+1 = 90

Now,
The number of integers in set T that are the multiples of 4 = 23 ( from 12 to 100)
The number of integers in set T that are the multiples of 5 = 20 (from 15 to 100)

Therefore,
the expression A = 4 * 23 = 92
the expression B = 5 * 18 = 90
Thus, A>B.

Please rectify if the process is not appropriate.
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Joined: 01 Nov 2017
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Kudos [?]: 111 , given: 4

Re: Set I consist of the integers from 11 through 100, inclusive [#permalink]
Expert's post
lw657 wrote:
Set I consist 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 5 times 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.

Main, we have to find the number of integers in each

(I) the number of integers in set T that are multiples of 4..
Quick way... There are $$\frac{100}{4}=25$$multiples. But take out 4 and 8, so 2 of them
Total 25-2=23..
A=4*23=92

(II) the number of integers in set T that are multiples of 5..
Quick way... There are $$\frac{100}{5}=20$$multiples. But take out 5 and 10, so 2 of them
Total 20-2=18..
B=5*18=90

A>B

A
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Some useful Theory.
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5. Absolute Modulus and Inequalities : https://greprepclub.com/forum/absolute-modulus-a-better-understanding-11281.html Director Joined: 20 Apr 2016
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Kudos [?]: 668  , given: 147

Re: Set I consist of the integers from 11 through 100, inclusive [#permalink]
1
KUDOS
lw657 wrote:
Set I consist 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 5 times 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.

Explanation::

To find out the multiples of a given number from a set = (Last number of the set i.e a multiple of the given number - First number of the set i.e a multiple of the same given number)/(Multiple of the given number)+ 1

Option A : multiple of 4 from 11 to 100, therefore : $${\frac{(100 - 12)}{4}}+ 1 = 23$$ i.e. no. of integer that are multiples of 4 from 11 to 100 is 23)

Therefore 4 times the number of integers in set = $$4 * 23 = 92$$

Option B: multiple of 5 from 11 to 100, therefore :$${\frac{(100 - 15)}{5}}+ 1 = 18$$ i.e. no. of integer that are multiples of 4 from 11 to 100 is 18)

Therefore 5 times the number of integers in set = $$5 * 18 = 90$$

Hence Option A is the answer
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Kudos [?]: 45  , given: 22

Re: Set I consist of the integers from 11 through 100, inclusive [#permalink]
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KUDOS
Expert's post
Use the arithmetic series formula

the first multiple of 4 from 11 to 100 is 12, and 100 is the last multiple

100=12+4(n-1)
88=4(n-1)
22= n-1
23=n
23*4=92

15 is the first multiple of 5 and 100 is the last multiple
100= 15+5(n-1)
85=5(n-1)
17=n-1
18=n
18*5= 90

92>90, A is greater Re: Set I consist of the integers from 11 through 100, inclusive   [#permalink] 13 Jan 2019, 21:20
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