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# Bob is training for a fitness competition. In order to increase his ma

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Math Expert
Joined: 02 Sep 2009
Posts: 53066
Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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28 Jul 2017, 07:58
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Difficulty:

5% (low)

Question Stats:

90% (01:55) correct 10% (01:53) wrong based on 83 sessions

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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

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Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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28 Jul 2017, 09:58
The sum of n natural numbers is $$\frac{n(n+1)}{2}$$

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 the sum of the first 10 natural numbers from the sum of the first 25 natural numbers.

Sum : $$\frac{25(26)}{2} - \frac{10(11)}{2} = 325 - 55 = 270$$(Option C)
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Re: Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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29 Jul 2017, 10:14
in case u r not aware of the formula,,,
adding will help,,,25 + 24 +....11

hope this helps
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Re: Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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29 Jul 2017, 19:59
1
1
sum of consecutive integers = mean * number of terms
mean = (first term + last term) / 2 = (25 + 11)/2=18
number of terms = 25 - 11 + 1 (11 inclusively) = 15
18 * 15 = 270
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Re: Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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24 Jun 2018, 17:04
Bunuel 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(11 + 25)/2 = 15 x 18 = 270

Alternate Solution:

We need to calculate the sum of the consecutive integers from 11 to 25, inclusive. Rather than calculate 11 + 12 + 13 + … + 24 + 25, we can calculate the average of the integers and then multiply the average by the number of integers to get the sum.

We know that there are (25 – 11) + 1 = 14 + 1 = 15 integers.

To calculate the average of the 15 integers, we use (smallest + largest)/2 = (11 + 25)/2 = 18.

The average is 18. Thus, the sum is 18 x 15 = 270.

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Re: Bob is training for a fitness competition. In order to increase his ma  [#permalink]

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09 Oct 2018, 22:01
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

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Re: Bob is training for a fitness competition. In order to increase his ma   [#permalink] 09 Oct 2018, 22:01
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