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The sum of all the multiples of 3 between 250 and 350

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The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 30 Jul 2018, 10:30
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Question Stats:

68% (01:07) correct 31% (00:46) wrong based on 22 sessions
Quantity A
Quantity B
The sum of all the multiples of 3 between 250 and 350
9990


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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Re: The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 07 Aug 2018, 20:28
The first number divisible by \(3\) after \(250\) is \(252\)
The first number divisible by \(3\) before \(350\) is \(348\)
Hence there are altogether \(\frac{348 - 252}{3} + 1 = 33\) numbers
Since the 33 numbers are equally spread apart with a difference of 3 the avg of those numbers = \(\frac{252 + 348}{2} = 300\)
Hence the sum is \(300 * 33 = 9900\)
option B > option A
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Re: The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 12 Aug 2018, 06:00
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Explanation

First, find the smallest multiple of 3 in this range: 250 is not a multiple of 3 (2 + 5 + 0 = 7, which is not a multiple of 3). The smallest multiple of 3 in this range is 252 (2 + 5 + 2 = 9, which is a multiple of 3).

Next, find the largest multiple of 3 in this range. Since 350 is not a multiple of 3 (3 + 5 + 0 = 8), the largest multiple of 3 in this range is 348.

The sum of an evenly spaced set of numbers equals the average value multiplied by the number of terms. The average value is the midpoint between 252 and 348: (252 + 348) ÷ 2 = 300. To find the number of terms, first subtract 348 – 252 = 96. This figure represents all numbers between 348 and 252, inclusive.

To count only the multiples of 3, divide 96 by the 3: 96 ÷ 3 = 32. Finally, “add 1 before you’re done” to count both end points of the range: 32 + 1 = 33.

The sum is 300 × 33 = 9,900. Since 9,900 is smaller than 9,990, Quantity B is greater.
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Re: The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 25 Oct 2018, 01:37
it that answer true?
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Re: The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 25 Oct 2018, 18:13
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Re: The sum of all the multiples of 3 between 250 and 350 [#permalink] New post 28 Oct 2018, 23:36
Sum of Set = n[(last-first)/2) where n = [(last-first)/k]+1 and the value of K= the distance between numbers in a set.

Step 1: Find K.

The question is a sum of all multiples of 3, which would mean 3,6,9,12... the distance between each is (6-3)=3 or (9-6)=3 and so on. As such, the value of k for this problem is 3.

Step 2: determine the first and last values that fit the parameters of the question.

While we are given a range of 250 to 350, the first and last values are not 250 and 350 becuase they are NOT multiples of 3. The shorthand way to determine if something is a multiple of 3 is to add up all the digits. If the number you get from doing that is a number divisible by 3, then the larger original number is ALSO divisible by 3.

EX: 3312 is divisible by 3 because 3+3+1+2 =9 and 9 is clearly divisible by 3. Similarly, 23451321 is ALSO divisible by 3 because 2+3+4+5+1+3+2+1 = 21, a number you know is divisible by 3.

Anyway, 250 = 7, which is not divisible by 3 and 350 = 8 which is not divisible by 3. By testing numbers one by one above 250 and below 350 (within the range) we see that the first multiple of 3 is actually 251 and the last multiple of 3 in our range is 348.

Step 3: Solve for n

n = [(348-252)/3]+1 .... this is equal to (96/3)+1, which becomes 32+1 = 33

Step 4: With our n value in tow, we can now solve the original equation.

sum of set = n[(first+last)/2] = 33*[(252+348)/2] = 33*[300] = 9900.

Step 5: Compare Quantity A and B

A = 9900. B = 9990. Therefore, B is greater than A and the answer is B
Re: The sum of all the multiples of 3 between 250 and 350   [#permalink] 28 Oct 2018, 23:36
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