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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] New post   05 Dec 2017, 12:42
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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.
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A
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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   06 Dec 2017, 23:52
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how do I check how many muｌtiple of 4 between 11-100?
Can anyone give me the formula, thank you
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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   07 Dec 2017, 02:23
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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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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   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
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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   22 Jan 2018, 07:27
1
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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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   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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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   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
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Re: Set I consists of the integers from 11 through 100, inclusiv [#permalink] New post   04 Mar 2023, 07:28
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