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A gym offers 11 levels of fitness classes, and in an attempt to reward
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05 May 2015, 01:37
Question Stats:
61% (02:45) correct 39% (02:57) wrong based on 157 sessions
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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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05 May 2015, 18:33
Let the cost of a course in level 1 = x => Cost of level 2 = x(50*1)... and so on => cost of level 11 = x(50*10) = x500
=> Total cost of 1 course in each of the 11 levels = x+(x50)+(x100)+.....+(x500) = 11x 50 (1+2+....+10) = 11x  (50*55) => 11x2750 = 4675 11x = 7425 x= 675 x500=175
Answer A



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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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05 May 2015, 23:57
akhilbajaj wrote: Let the cost of a course in level 1 = x => Cost of level 2 = x(50*1)... and so on => cost of level 11 = x(50*10) = x500
=> Total cost of 1 course in each of the 11 levels = x+(x50)+(x100)+.....+(x500) = 11x 50 (1+2+....+10) = 11x  (50*55) => 11x2750 = 4675 11x = 7425 x= 675 x500=175
Answer A Is there a faster way? Hardly possible to do this in < 2minutes



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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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06 May 2015, 17:24
noTh1ng wrote: Is there a faster way? Hardly possible to do this in < 2minutes 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 "x500=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



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A gym offers 11 levels of fitness classes, and in an attempt to reward
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06 May 2015, 17:42
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
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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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06 May 2015, 18:00
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+(n1)d) 4675=11/2(2a+(111)(50)) on soving giving a=675 we know nth term in AP is, an= a+(n1)d so 11th, ie the least amout to be paid is $150
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A gym offers 11 levels of fitness classes, and in an attempt to reward
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07 May 2015, 22:31
noTh1ng wrote: akhilbajaj wrote: Let the cost of a course in level 1 = x => Cost of level 2 = x(50*1)... and so on => cost of level 11 = x(50*10) = x500
=> Total cost of 1 course in each of the 11 levels = x+(x50)+(x100)+.....+(x500) = 11x 50 (1+2+....+10) = 11x  (50*55) => 11x2750 = 4675 11x = 7425 x= 675 x500=175
Answer A Is there a faster way? Hardly possible to do this in < 2minutes 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: \(Sum = n/2 [2a + (n1)*d]\) \(Last term = a + (n1)*d\) \(Sum*2/n = 2a + (n1)*d\) Divide the equation by 2 to get: \(Sum/n = a + (n1)*d/2\) Add (n1)*d/2 on both sides to get: \(Sum/n + (n1)*d/2 = a + (n1)*d = Last term\) 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: http://www.veritasprep.com/blog/2012/03 ... gressions/
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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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11 May 2015, 04:16
[url][/url] Bunuel wrote: A gym offers 11 levels of fitness classes, and in an attempt to reward those who progress toward higher levels of fitness it charges $50 less per course for each level of fitness. Jessica completed all 11 levels by taking one course at each level, and her total cost was $4675. What is the cost for a course at the gym's highest level?
A. $175 B. $245 C. $425 D. $675 E. $725
Kudos for a correct solution. VERITAS PREP OFFICIAL SOLUTION:One important consideration makes this problem much more solvable in under two minutes: This is an evenlyspaced 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 doublecheck 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.
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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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27 Mar 2016, 19:31
Bunuel wrote: A gym offers 11 levels of fitness classes, and in an attempt to reward those who progress toward higher levels of fitness it charges $50 less per course for each level of fitness. Jessica completed all 11 levels by taking one course at each level, and her total cost was $4675. What is the cost for a course at the gym's highest level?
A. $175 B. $245 C. $425 D. $675 E. $725
Kudos for a correct solution. my approach: there are 10 courses, each less 50$ so 11th x10*50 10th x9*50 9th x8*50 8th x7*50 7th x6*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.



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Re: A gym offers 11 levels of fitness classes, and in an attempt to reward
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03 Apr 2018, 02:23
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