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Bunuel
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56

We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E
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Bunuel
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56

We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E

Hi Scott,

Why did we group this ?? Is there any other method I can attempt this ?? I couldn't understand the explanation.

Please help.
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0- 5=0
6-11=1*6
12-17=2*6
18-23=3*6
24-28=4*5

total is 56.option E
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ScottTargetTestPrep
Bunuel
For all real numbers, [x] = the greatest integer less than or equal to x.

What is the sum of the 28 terms in the series [1/6] + [2/6] + [3/6] + … + [28/6]?

A. 48
B. 50
C. 52
D. 54
E. 56

We can group the terms by groups of 6, except the first and last groups will have 5 terms each.

First group (5 terms): [1/6], [2/6], …, [5/6]
Second group (6 terms): [6/6], [7/6], …, [11/6]
Third group (6 terms): [12/6], [13/6], …, [17/6]
Fourth group (6 terms): [18/6], [19/6], …, [23/6]
Fifth group (5 terms): [24/6], [25/6], …, [28/6]

We see that each term in the first group is 0, in the second group 1, in the third group 2, in the fourth group 3, and in the fifth group 4. Therefore, the sum is 5 x 0 + 6 x 1 + 6 x 2 + 6 x 3 + 5 x 4 = 6 + 12 + 18 + 20 = 56.

Answer: E

Hi Scott,

Why did we group this ?? Is there any other method I can attempt this ?? I couldn't understand the explanation.

Please help.

We group them because each term in any one particular group has the same value. The only other alternative I see is to add each term, one by one, meaning carrying out the addition 0 + 0 + … + 0 + 1 + 1 + … + 1 + 2 + … etc. ; but there are 28 terms and doing that would be a huge waste of time. However, if you started this laborious task, you would see the pattern emerging that would result in the groupings shown in my original solution.
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My logic.
1. Multiple of 7 since its 28
2. Second figure had to be divisible by 4 and given that we're rounding off, half of it would be a (-1) and another half (+1), so 4 x 2 =8
3. 7 x 8 = 56
Not sure if my pattern recognition helped.
I didnt allow the 1/6 to affect me so much cause its a round off situation so I tried to understand the logic behind it.
I could be veryyyy wrong here.
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when you add up the whole amounts by summing up the greatest and smallest number (1/6+28/6, 2/6+27/6 ..) you will have a summation of 14*(29/6) which will be close to 58 but since we know [x] is greatest number less than or equal to x it will naturally be the closest number to 58, that is, 56.
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