Multiple choice

At IIM Bangalore class of 1995, Sonali, a first year student has taken 10 courses, earning grades A (worth 4 points each), B (worth 3 points each) and C (worth 2 points each). Her grade point average is 3.2, and if the course in which she get C's were deleted, her GPA in the remaining courses would be 3.333. How many A's, B's and C's did she get?

  1. 3, 1 and 6

  2. 1, 3 and 6

  3. 3, 6 and 1

  4. 1, 6 and 3
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let A, B, C be the number of courses. A+B+C = 10. (4A+3B+2C)/10 = 3.2 -> 4A+3B+2C = 32. If C's are removed, (4A+3B)/(A+B) = 3.333 = 10/3. 12A+9B = 10A+10B -> 2A = B. Substitute B=2A into A+B+C=10 -> 3A+C=10. Substitute into 4A+3(2A)+2C=32 -> 10A+2C=32 -> 5A+C=16. Subtracting: 2A=6 -> A=3. Then B=6, C=1.

AI explanation

Her total grade points for the 10 courses is 3.2 multiplied by 10, which equals 32. If the courses with grade C are removed, the average of the remaining courses is 3.333, or 10 divided by 3, meaning the number of remaining courses must be a multiple of 3 to yield an integer total. Assuming the total remaining courses is 9, their total points would be 30, meaning the single C course is worth 2 points. This leaves 9 courses with A and B grades summing to 30 points, so solving the equations a + b = 9 and 4a + 3b = 30 results in 3 A's, 6 B's, and 1 C.

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