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The sum of the multiples of 4 less than 100 or The sum of

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The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 22 Oct 2017, 06:14
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Question Stats:

71% (01:03) correct 28% (01:08) wrong based on 96 sessions
Quantity A
Quantity B
The sum of the multiples of 4 less than 100
The sum of the multiples of 5 less than 100


(A) The quantity in Column A is greater
(B) The quantity in Column B is greater
(C) The two quantities are equal
(D) The relationship cannot be determined from the information given


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[Reveal] Spoiler: OA
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 22 Oct 2017, 08:48
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We can proceed by steps:

1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).

2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).

We conclude that A is greater!
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 09 Jul 2018, 18:48
Bunuel wrote:
Quantity A
Quantity B
The sum of the multiples of 4 less than 100
The sum of the multiples of 5 less than 100


(A) The quantity in Column A is greater
(B) The quantity in Column B is greater
(C) The two quantities are equal
(D) The relationship cannot be determined from the information given


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My Answer is A
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 10 Jul 2018, 08:19
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Sum of multiples of 4 = 4(1+2+..+25)=4*(25+1)*25/2=50*26

Sum of multiples of 5 = 5(1+2+..+20)=5*(1+20)*20/2=50*21

-> QA > QB -> A.
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 05 Aug 2018, 15:33
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IlCreatore wrote:
We can proceed by steps:

1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).

2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).

We conclude that A is greater!


Hi, for step 1, is there a simplified formula that is easier to remember for exam?
Also, it says less that 100, so shouldnt we consider multiples less than 100? Like is 100 still included?

And thanks for the explanation btw
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 09 Aug 2018, 00:53
it should be given that multiples are of what kind...positive or negative,and inclusive or exclusive....
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 09 Aug 2018, 05:19
Expert's post
ragini123 wrote:
it should be given that multiples are of what kind...positive or negative,and inclusive or exclusive....


Multiples are always considered as positive unless explicitly mentioned.

Less than 100 means that 100 should not be considered for calculation in this case.
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 17 Apr 2019, 19:21
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IlCreatore wrote:
We can proceed by steps:

1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).

2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).

We conclude that A is greater!



I doubt the above explanation, kindly provide some feed back.

As we need to find the multiples of 4 & 5 less than 100, i.e. 100 exclusive

QTY A :: since the number has to be less than 100

The number of terms = \(\frac{96-4}{4}+1 = 24\) ( last multiple of 4, less than 100 - first multiple of 4 )

The average = \(\frac{{99 + 1}}{2} = 50\)

Hence the multiples of 4 less than 100 = 50 * 24

QTY B::

The number of terms = \(\frac{95-5}{5}+1 = 19\)


The average = \(\frac{{99 + 1}}{2} = 50\)

Hence the multiples of 5 less than 100 = 50 * 19


Therefore QTY A > QTY B
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 27 Aug 2019, 00:18
pranab01 wrote:
IlCreatore wrote:
We can proceed by steps:

1) number of elements in the two sets: we have to take the first and the last multiple, then compute last - first, divide by the number whose multiples are of interest and sum 1. In our case, in column A, we have \(\frac{100-4}{4}+1 = 25\), while for column B we get \(\frac{100-5}{5}+1 = 20\).

2) Compute the sum. The formula for the sum of an arithmetic progression is \(sum = \frac{n}{2}(first+last)\), where n is the number of elements in the progression and first and last are the first and the last elements. Thus, for column A, \(\frac{25}{2}(4+100) = 1300\), while column B equates \(\frac{20}{2}(5+100) = 1050\).

We conclude that A is greater!



I doubt the above explanation, kindly provide some feed back.

As we need to find the multiples of 4 & 5 less than 100, i.e. 100 exclusive

QTY A :: since the number has to be less than 100

The number of terms = \(\frac{96-4}{4}+1 = 24\) ( last multiple of 4, less than 100 - first multiple of 4 )

The average = \(\frac{{99 + 1}}{2} = 50\)

Hence the multiples of 4 less than 100 = 50 * 24

QTY B::

The number of terms = \(\frac{95-5}{5}+1 = 19\)


The average = \(\frac{{99 + 1}}{2} = 50\)

Hence the multiples of 5 less than 100 = 50 * 19


Therefore QTY A > QTY B


I got the same!
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 11 Aug 2020, 12:08
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We know that: Sum of integers from 1 to n is (n)(n+1)/2

Quantity A = Sum of multiples of 4 till 100
= 4 + 8 + 12 ...... 96 + 100
= 4*(1 + 2 + 3 ...... 24 + 25)
= 4*(25)(25+1)/2
= 4*25*13 (divided 26 by 2)
= 1300

Quantity B = Sum of multiples of 5 till 100
= 5 + 10 + 15 ...... 95 + 100
= 5*(1 + 2 + 3 ...... 19 + 20)
= 5*(20)(20+1)/2
= 5*10*21 (divided 20 by 2)
= 1050

So, A > B
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Re: The sum of the multiples of 4 less than 100 or The sum of [#permalink] New post 07 Sep 2020, 07:40
but n/2[2a+(n-1)d] is also valid? but why i am not getting right answer?
Re: The sum of the multiples of 4 less than 100 or The sum of   [#permalink] 07 Sep 2020, 07:40
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