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

tagged by: M7MBA

This topic has 1 expert reply and 0 member replies

### Top Member

M7MBA Master | Next Rank: 500 Posts
Joined
29 Oct 2017
Posted:
162 messages
1

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

Mon Nov 06, 2017 6:21 am
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 OA is C.

I can do it making the whole sum: 25+24+23+22+...+11, but, is there another way (faster or easier) to make it?

Experts, can you give me a hand here?

### GMAT/MBA Expert

ErikaPrepScholar Master | Next Rank: 500 Posts
Joined
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Posted:
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GMAT Score:
770
Mon Nov 06, 2017 9:11 am
Great question - there is a faster way! When we have any group of terms, we can find the sum of all of those terms with:

$$mean\ \cdot\ number\ of\ terms$$

When we have a group of consecutive terms, we can find the mean of those terms with:

$$\frac{first\ term+last\ term}{2}$$

Combining those two equations, we can find the sum of a group of consecutive terms with:

$$\frac{first\ term+last\ term}{2}\cdot\ number\ of\ terms$$

Here, this gives:

$$\frac{25\ +\ 11}{2}\ \cdot\ 15\ =\ 270$$

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