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) = 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 = 7425
x= 675
x-500=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:
\(Sum = n/2 [2a + (n-1)*d]\)
\(Last term = a + (n-1)*d\)
\(Sum*2/n = 2a + (n-1)*d\)
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:
http://www.veritasprep.com/blog/2012/03 ... gressions/
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Karishma
Veritas Prep GMAT Instructor
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