A gym offers 11 levels of fitness classes, and in an attempt to reward

An alternate way is to consider to cost of level 11 =x
since this is the value that needs to be determined, this would save on the last step "x-500=175" and also save some people of the mistake of reporting 675 as the answer. The rest of the solution majorly remains the same. Given below for reference.

Let the cost of a course in level 11 = x
=> Cost of level 10 = x+(50*1)... and so on
=> cost of level 1 = x+(50*10) = x+500

=> Total cost of 1 course in each of the 11 levels = x+(x+50)+(x+100)+.....+(x+500)
= 11x + 50 (1+2+....+10)
= 11x + (50*55)
=> 11x+2750 = 4675
11x = 1925
x= 175

Answer A

Hi Bunuel,
Good question! But don't you think the question should have mentioned 'equally priced fitness levels'? Here we have to assume that each level is at same price to solve the problem.

Ambarish

It's basically a question on arithematic progression in which we are given
common difference,d=-50 (negative because the value is decreasing)
the sum of all terms= $4675

we know that sum of all terms,
Sn=n/2(2a+(n-1)d)
4675=11/2(2a+(11-1)(-50))
on soving giving a=675

we know nth term in AP is, an= a+(n-1)d
so 11th, ie the least amout to be paid is $150

It's possible to do it very quickly if you are comfortable with Arithmetic Progressions (AP). This is an AP with common difference as -50 and 11 total terms. We know the following things about an AP:
\(Sum = n/2 [2a + (n-1)*d]\)
\(Last term = a + (n-1)*d\)

\(Sum*2/n = 2a + (n-1)*d\)

Divide the equation by 2 to get: \(Sum/n = a + (n-1)*d/2\)

Add (n-1)*d/2 on both sides to get: \(Sum/n + (n-1)*d/2 = a + (n-1)*d = Last term\)

We know that Sum = 4675, n = 11 and d = -50. Plug in the values to get

\(4675/11 + (11 - 1)*(-50)/2 = 175 = Last term\)

Answer (A)

For more on AP and the related formulas, check: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2012/03 ... gressions/

[url][/url]

VERITAS PREP OFFICIAL SOLUTION:

One important consideration makes this problem much more solvable in under two minutes:

This is an evenly-spaced set!

Which, of course, means that the median (the middle number) equals the average. So if you divide 4675 by 11 to find the average you'll also have the middle number. 4675/11 isn't the nicest math, but if you break it up:

4400/11 = 400

and

275/11 = 25

So the mean/median is $425. Then think about the increments of $50; if $425 is the sixth of eleven values, then you need to move five increments of $50, a total of $250, to get to the target value. And here's where it pays to double-check the question. They want the highest LEVEL of course which then would have the LOWEST price, so subtract $250 from $425 to get the correct answer, $175.

my approach:
there are 10 courses, each less 50$
so
11th x-10*50
10th x-9*50
9th x-8*50
8th x-7*50
7th x-6*50
etc.

so there is (1+2+3+4+5+6+7+8+9+10)*50
sum of all from 1 to 10 is 10*11/2 = 55
55*50 = 2750
11x = 4675 + 2750
11x = 7450
x=675
11th course is 675 - 10*50 = 175.

We can let x = the cost for a course at the highest level. Since that is the cheapest course, the cost for a course at the second highest level is x + 50, that at the third highest level is x + 100, and so on. Finally the most expensive course, which is at the lowest level, will cost x + 500. Therefore, we can create the equation:

x + (x + 50) + (x + 100) + … + (x + 500) = 4675

Notice there are 11 terms on the left hand side of the equation, and the terms are even spaced; therefore, we can use the formula for the sum of an evenly spaced set, which is average * quantity = sum. We note that the average of an evenly spaced set is found by the formula: (smallest value + largest value) / 2. Combining these two formulas, we can create the equation:

[x + (x + 500)]/2 * 11 = 4675

[2x + 500]/2 = 425

x + 250 = 425

x = 175

Answer: A

From the lowest-level class to the highest, the 11 prices decrease in increments of 50 and thus constitute an EVENLY SPACED SET.

For any EVENLY SPACED SET:
\(median = \frac{sum}{count}\)

Thus
median of the 11 prices \(= \frac{sum}{count} =\frac{ 4675}{11} = 425\)

As the class level INCREASES, the price of the class DECREASES.
Since the five prices below the median decrease in increments of 50, the price of the highest-level class is five $50 increments below the median price of $425:
\(425 - (5*50) = 175\)