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How many multiples of 4 are there between 12 and 96, inclusive?

21 22 23 24 25

My answer was 21 and that's incorrect.

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

In the original case: \(\frac{96-12}{4}+1=22\).

If the question were: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;

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Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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08 Jan 2012, 17:49

1

This post was BOOKMARKED

Bunuel wrote:

marcusaurelius wrote:

How many multiples of 4 are there between 12 and 96, inclusive?

21 22 23 24 25

My answer was 21 and that's incorrect.

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

In the original case: \(\frac{96-12}{4}+1=22\).

If the question were: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).

Hope it helps.

I have one concern regarding this How many multiples and how many integers?

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1

but the answer is only 128,How is that possible?????

As per our knowledge we know that

->If the question says both inclusive we add +1 at the last, For eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1

->If the question says both not inclusive we add -1 at the last, For eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128-1

->If the question says one inclusive we only subtract two extremes, for Eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128, so 128 is the answer.....

Anything wrong with my concept???????Please explain i am totally confused with this...............

If anything wrong,please explain how to with these kind of problems, Like How many integers between two numbers? and How many integers between two multiples?
_________________

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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08 Jan 2012, 18:07

You forgot to add 1. It goes like this - (96-12)/4 + 1 = 22
_________________

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DS - If negative answer only, still sufficient. No need to find exact solution. PS - Always look at the answers first CR - Read the question stem first, hunt for conclusion SC - Meaning first, Grammar second RC - Mentally connect paragraphs as you proceed. Short = 2min, Long = 3-4 min

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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03 Sep 2012, 15:33

1

This post received KUDOS

1

This post was BOOKMARKED

marcusaurelius wrote:

I have one concern regarding this How many multiples and how many integers?

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1

but the answer is only 128,How is that possible?????

As per our knowledge we know that

->If the question says both inclusive we add +1 at the last, For eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128+1

->If the question says both not inclusive we add -1 at the last, For eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128 so we need to include both digits so we add 1 to total so it is 128-1

->If the question says one inclusive we only subtract two extremes, for Eg

How many of the three-digit numbers are divisible by 7?

Generally we do--->999-100= 899

so 899/7=128, so 128 is the answer.....

Anything wrong with my concept???????Please explain i am totally confused with this...............

If anything wrong,please explain how to with these kind of problems, Like How many integers between two numbers? and How many integers between two multiples?

three digit numbers divisible by 7 or multiples of 7 are from 105 to 994. Now if u do (994-105)/7 + 1 { as both are inclusive } you get 127 + 1 = 128 as the answer...

the problem in your approach is that neither 100 nor 999 are multiples of 7 and hence you need to solve (999-100)/7 with both NOT INCLUSIVE and you get the correct answer 128
_________________

If something helps, you must appreciate! I'll regard Kudosas appreciation.. Thanks

Why add one to the final result? I can count from 12-96 by four and come up with 22 that way, but I want to know the logic behind it.

Set of consecutive multiples of 4 is an evenly spaced set (arithmetic progression).

If the first term of arithmetic progression is \(a_1\) and the common difference of successive members is \(d\), then the \(n_{th}\) term of the sequence is given by:

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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06 Oct 2014, 22:47

Bunuel wrote:

marcusaurelius wrote:

How many multiples of 4 are there between 12 and 96, inclusive?

21 22 23 24 25

My answer was 21 and that's incorrect.

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

In the original case: \(\frac{96-12}{4}+1=22\).

If the question were: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).

Hope it helps.

Hi Bunuel,

can i use this formula for not inclusive? But when i use this formula I got different result. Last multiple of 7 - First Multiple of 7/ 7 - 1

How many multiples of 4 are there between 12 and 96, inclusive?

21 22 23 24 25

My answer was 21 and that's incorrect.

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

In the original case: \(\frac{96-12}{4}+1=22\).

If the question were: how many multiples of 5 are there between -7 and 35, not inclusive?

Last multiple of 5 IN the range is 30; First multiple of 5 IN the range is -5;

\(\frac{30-(-5)}{5}+1=8\).

OR: How many multiples of 7 are there between -28 and -1, not inclusive? Last multiple of 7 IN the range is -7; First multiple of 7 IN the range is -21;

\(\frac{-7-(-21)}{7}+1=3\).

Hope it helps.

Hi Bunuel,

can i use this formula for not inclusive? But when i use this formula I got different result. Last multiple of 7 - First Multiple of 7/ 7 - 1

There are examples with "not inclusive" cases above. Please be more specific.
_________________

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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07 Oct 2014, 00:24

# of multiples of x in the range = (Last multiple of x in the range - First multiple of x in the range)/x - 1. Is this formula correct to calculate no of non inclusive multiple of an integer?

# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x} - 1. Is this formula correct to calculate non inclusive multiple of an integer?

Yes. Again, examples in my post above have not inclusive ranges! _________________

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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21 Apr 2016, 16:41

Bunuel wrote:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple.

For example multiples of 11 in 1000, [(990-11)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps?
_________________

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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22 Apr 2016, 10:03

redfield wrote:

Bunuel wrote:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple.

For example multiples of 11 in 1000, [(990-11)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps?

Bunuel's approach is perfect...

Here you are trying to calculate the numbers divisible by 11 from 0 - 1000

But say in a situation like -

Quote:

Find the total no of numbers between 900 - 990 , (both inclusive) which are divisible by 11

Here your formula will not work .

Actually we have 2 sets -

1. 0 - 899

2. 900 - 990

We are interested in the 2nd set

So, U can go this way ( Bunnels approach)

\(Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to \ 990 - Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to 899\) + \(The \ number \ 990\ itself \ (\ which \ is \ divisible \ by\ 11)\)

Hope I am clear, please feel free to revert in case of any doubt. _________________

Thanks and Regards

Abhishek....

PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS

Re: How many multiples of 4 are there between 12 and 96, inclusive? [#permalink]

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22 Apr 2016, 10:11

Abhishek009 wrote:

redfield wrote:

Bunuel wrote:

\(# \ of \ multiples \ of \ x \ in \ the \ range = \frac{Last \ multiple \ of \ x \ in \ the \ range \ - \ First \ multiple \ of \ x \ in \ the \ range}{x}+1\).

I know this is a really old thread but isn't this an unnecessary step for this formula? I've seen you use it on similar problems including a couple on the GMAT Club tests and I don't understand why you subtract by the first multiple.

For example multiples of 11 in 1000, [(990-11)/11] + 1 is the same as (990/11) so why add the additional subtraction and addition steps?

Bunuel's approach is perfect...

Here you are trying to calculate the numbers divisible by 11 from 0 - 1000

But say in a situation like -

Quote:

Find the total no of numbers between 900 - 990 , (both inclusive) which are divisible by 11

Here your formula will not work .

Actually we have 2 sets -

1. 0 - 899

2. 900 - 990

We are interested in the 2nd set

So, U can go this way ( Bunnels approach)

\(Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to \ 990 - Total \ no \ of \ Numbers \ divisible \ by \ 11 \ up \ to 899\) + \(The \ number \ 990\ itself \ (\ which \ is \ divisible \ by\ 11)\)

Hope I am clear, please feel free to revert in case of any doubt.

I feel compelled to say this But this ain't no formula actually This is just an extension of the basic AP series formula => An = A+(n-1)D Regards Stone Cold
_________________