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If N = 255 is the lowest of a set of 23 consecutive multiple

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by BTGmoderatorDC » Mon Jul 29, 2019 5:46 pm

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If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

(A) 315
(B) 330
(C) 345
(D) 360
(E) 375

OA B

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by GMATGuruNY » Mon Jul 29, 2019 6:59 pm
BTGmoderatorDC wrote:If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

(A) 315
(B) 330
(C) 345
(D) 360
(E) 375
Range = biggest - smallest = 23rd value - 1st value.
Since the set is composed of consecutive multiples of 15, the difference between one value and the next = 15.

To get from the 1st value to the 2nd value, we must add 15 once.
Thus:
(2nd value) - (1st value) = 15*1 = 15

To get from the 1st value to the 3rd value, we must add 15 twice.
Thus:
(3rd value) - (1st value) = 15*2 = 30

By extension:
To get from the 1st value to the 23rd value, we must add 15 twenty-two times
Thus:
Range = (23rd value) - (1st value) = 15*22 = 330

The correct answer is B.
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Re: If N = 255 is the lowest of a set of 23 consecutive multiple

by Scott@TargetTestPrep » Sun Dec 13, 2020 4:49 am
BTGmoderatorDC wrote: ↑
Mon Jul 29, 2019 5:46 pm
 If N = 255 is the lowest of a set of 23 consecutive multiples of 15, what is the range of this set?

(A) 315
(B) 330
(C) 345
(D) 360
(E) 375

OA B

Solution:

We are given that 255 is the lowest of a set of 23 consecutive multiples of 15. That is, the subsequent multiples of 15 after 255 are 255 + 15(1), 255 + 15(2), etc. This pattern indicates that we have an arithmetic sequence, with first term a_1 = 255 and common difference d = 15. We need to determine the n = 23rd term in the sequence.

Using the formula for the nth term of an arithmetic sequence a_n = a_1 + (n -1)d, we calculate the value of the 23rd term in the sequence as:

255 + 15(23 - 1) = 255 + 15(22)

The range of a set is the difference between the highest and lowest numbers in the set. Therefore, the range is [255 + 15(22)] - 255 = 15(22) = 330.

Answer: B

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