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The sum of the positive integers from 1 to 27 is equivalent

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The sum of the positive integers from 1 to 27 is equivalent

by AAPL » Wed May 23, 2018 5:46 am

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The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from?

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

The OA is D.

Sum of the integers from 1 to 27 is the average of first and last terms multiplied by # of terms. Applying that to the question stem gives you
(28/2) * (27)
-(14) * (27)
-(3 * 3 * 3 * 2 * 7).

Choice A is basically the sum of 27 - 54 which will be too big
Choice B is tempting but after applying the formula you will get (14) * (28)
Choice C will be too big as well, there are 30 terms and each term is bigger than it's the corresponding term in the original sequence

Choice D looks like it could work, so apply the formula.

=(84/2) * (46 - 38 + 1)
=(42) * (9)
=(3 * 3 * 3 * 2 * 7)

Choice D is the answer.

Has anyone another strategic approach to solve this PS question? Regards!
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by Jay@ManhattanReview » Thu May 24, 2018 2:53 am
AAPL wrote:The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from?

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54

The OA is D.

Sum of the integers from 1 to 27 is the average of first and last terms multiplied by # of terms. Applying that to the question stem gives you
(28/2) * (27)
-(14) * (27)
-(3 * 3 * 3 * 2 * 7).

Choice A is basically the sum of 27 - 54 which will be too big
Choice B is tempting but after applying the formula you will get (14) * (28)
Choice C will be too big as well, there are 30 terms and each term is bigger than it's the corresponding term in the original sequence

Choice D looks like it could work, so apply the formula.

=(84/2) * (46 - 38 + 1)
=(42) * (9)
=(3 * 3 * 3 * 2 * 7)

Choice D is the answer.

Has anyone another strategic approach to solve this PS question? Regards!
We see that from 1 to 27, there are 27 terms. Before calculating the sum of the positive integers from 1 to 27, let's start analyzing the options.

Let's see each term one by one.

A. -27 to 54: The sum of negative integers from -27 to -1 would get canceled with the sum of the positive integers from 1 to 27. Thus, the option A basically is the sum of 28 to 54.

From 28 to 54, there are (54 -28) + 1 = 27 terms

We see that the number of terms (27) is equal to the number of terms from 1 to 27. Since the smallest value of the terms from 28 to 54 is 28 >> 1, its sum would also be far greater than the sum from 1 to 27. This cannot be the answer.

B. 0 to 28: This option basically is from 1 to 28. Compared to the terms given in the question (1 to 27), there is one extra term (28); thus, its sum would also be greater than the sum from 1 to 27. This cannot be the answer.

C. 15 to 45: From 15 to 45, there are (45 -15) + 1 = 31 terms

From the same reasoning as used in option A, this cannot be the answer.

D. 38 to 46: There are only 9 terms. The sum = Average of the two terms * the number of terms = [(38 + 46)/2]*9 = 42*9

Similarly, the sum from 1 to 27 = Average of the two terms * the number of terms = [(1 + 27)/2]*27 = 14*27 = 14*3*9 = 42*9

This is the correct answer.

E. 48 to 54: No need to discuss this now.

The correct answer: D

Hope this helps!

-Jay
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by Scott@TargetTestPrep » Thu May 24, 2018 5:08 pm
AAPL wrote:The sum of the positive integers from 1 to 27 is equivalent to the sum of the integers from?

A. -27 to 54
B. 0 to 28
C. 15 to 45
D. 38 to 46
E. 48 to 54
Let's analyze each choice.

A. -27 to 54

The sum of the integers from -27 to 27 is 0. Thus we only need to add the integers from 28 to 54. However, we can see that, though there are 27 integers from 28 to 54 (as many as from 1 to 27), each integer from 28 to 54 is greater than any of the integers from 1 to 27, so the sums can't be the same.

B. 0 to 28

The integers from 0 to 28 include those from 1 to 27, plus two more integers: 0 and 28. Though adding an extra 0 will not change the sum, adding an extra 28 will. So the sums can't be the same.

C. 15 to 45

There are 31 integers from 15 to 45 and if we exclude the last 4 integers, each of these is greater than the corresponding ones from 1 to 27. So the sum of the integers from 15 to 45 will be greater than the sum of the integers from 1 to 27. So the sums can't be same.

D. 38 to 46

Though each integer from 38 to 46 seems to be much larger than those from 1 to 27, there are fewer integers from 38 to 46 than from 1 to 27 (9 integers vs. 27 integers). So the sums could be equal. The only way we can know that for sure is to calculate each sum. To do that, we can use the formula: sum = average x quantity.

Sum of integers 1 to 27 = (1 + 27)/2 x 27 = 14 x 27 = 378

Sum of integers 38 to 46 = (38 + 46)/2 x 9 = 42 x 9 = 378.

We see that they do have an equal sum. Thus, the answer is D.

Answer: D

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by deloitte247 » Fri May 25, 2018 12:38 pm
Sum of the consecutive positive integers
$$n\left(\frac{n+1}{2}\right)$$
$$=\ 27\ \cdot\ \frac{28}{2}$$
27 * 14 = 378
eliminating the options we have
Option A - 27 to 54
-27 + 27 = 0
28 to 54 is greater than 378.
Option B 0 to 28
the sum of 1 to 28 is greater than 1-27
Option C 15 to 45
total number is 31, median number is 30
30 * 31 = 910 greater than 378
Option D 38 to 46; there are 9 numbers here and the median = 42 * 9 = 378.

Option e 48 to 54; there are 7 numbers here and the median = 51 * 7 = 357 it is less than 378.
Option D is the right answer.
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