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Bob is training for a fitness competition. In order to

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by BTGmoderatorDC » Mon Oct 08, 2018 5:48 pm

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Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?

(A) 55

(B) 150

(C) 270

(D) 275

(E) 325

OA C

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by Jay@ManhattanReview » Tue Oct 09, 2018 5:24 am
BTGmoderatorDC wrote:Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?

(A) 55

(B) 150

(C) 270

(D) 275

(E) 325

OA C

Source: Magoosh
We have a set of certain consecutive numbers in decreasing order.

Set of # of pull-ups: {25, 24, 23, ..., 12, 11}

The sum of the elements of the set can be found in many ways. I provide one of those ways.

From 25 to 11, there are (25 - 11) + 1 = 15 numbers

Average of the consecutive numbers from 11 to 25 = (11 + 25)/2 = 18

Thus, the sum of numbers from 11 to 25 = Total number of elements * Average of numbers = 15*18 = 270 pull-ups.

The correct answer: C

Hope this helps!

-Jay
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by swerve » Tue Oct 09, 2018 11:09 am
In order to find the total number of pull-ups that Bob does for the fitness competition, we need to find the sum of the numbers from 11 to 25.

This can be done by subtracting the sum of the first 10 natural numbers from the sum of the first 25 natural numbers.

$$\text{Sum: } \frac{25(26)}{2}-\frac{10(11)}{2}=325-55=270.$$
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by [email protected] » Tue Oct 09, 2018 7:23 pm
Hi All,

We're told that Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. We're asked for the total number of pull-ups that Bob completes. This question can be solved in several different ways, including by 'bunching.'

In simple terms, we are adding up all of the integers from 11 to 25, inclusive. Since there are 25 integers from 1 to 25 - and we are NOT including the first 10 integers, that means we are adding 25 - 10 = 15 individual numbers together. We can 'bunch' those numbers into seven groups of 2 and one left over number:

11 + 25 = 36
12 + 24 = 36
13 + 23 = 36
...
17 + 19 = 36
and there's one number left over: an 18

Thus, we have 7(36) + 18 = 252 + 18 = 270

Final Answer: C

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Rich
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by Scott@TargetTestPrep » Fri Oct 12, 2018 6:54 am
BTGmoderatorDC wrote:Bob is training for a fitness competition. In order to increase his maximum number of pull-ups, he follows the following routine: he begins with 25 pull-ups, rests for thirty seconds, and then does 24 pull-ups and rests, dropping one pull-up each time (25, 24, 23, etc.) until his final set of 11 pull-ups. How many total pull-ups does Bob do?

(A) 55

(B) 150

(C) 270

(D) 275

(E) 325
The number of pull-ups Bob does is the sum of integers from 11 to 25, inclusive. Thus, the total number of pull-ups he does is

15 x (11 + 25)/2 = 15 x 18 = 270

Alternate Solution:

The sequence of the integers from 11 to 25, inclusive, is an evenly-spaced set. Our goal is to find the sum of the set. We first determine the average value of this set:

average = (smallest + largest)/2 = (11 + 25) /2 = 38/2 = 18

We also see that there are 25 - 11 + 1 = 15 integers in the set. We can now use the formula:

sum = average x number

sum = 18 x 15 = 270.

Answer: C

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