Question: There are 3 sections A, B and C for eighth standard in XYZ school. In a test the average marks of sections A and B together is 56. The average marks of sections B and C together is 60. The average marks of sections C and A together is 64. Find the range of average marks of all the three sections combined?
Explanation:
Let the average marks of the three sections A, B and C be a, b and c respectively.
Let the number of students in the three sections A, B and C be na , nb and nc respectively.
Average marks of sections A and B combined is 56.
⇒ 56 = a × n a + b × n b n a + n b
⇒ ana + bnb = 56na + 56nb ...(1)
Similarly, average marks of sections B and C combined is 60.
⇒ bnb + cnc = 60nb + 60nc ...(2)
Similarly, average marks of sections C and A combined is 64.
⇒ cnc + ana = 64nc + 64na ...(3)
Adding (1), (2) and (3), we get
⇒ 2ana + 2bnb + 2cnc = 120na + 116nb + 124nc
⇒ ana + bnb + cnc = 60na + 58nb + 62nc
Now, average of all three sections combined = a × n a + b × n b + c × n c n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 60 n a + 58 n b + 62 n c n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 58 ( n a + n b + n c ) + 2 n a + 4 n c n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 58 ( n a + n b + n c ) n a + n b + n c + 2 n a + 4 n c n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 58 + (positive number)
⇒ a × n a + b × n b + c × n c n a + n b + n c > 58
Also,
⇒ a × n a + b × n b + c × n c n a + n b + n c = 60 n a + 58 n b + 62 n c n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 62 ( n a + n b + n c ) - 2 n a - 4 n b n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 62 ( n a + n b + n c ) n a + n b + n c - 2 n a + 4 n b n a + n b + n c
⇒ a × n a + b × n b + c × n c n a + n b + n c = 62 - (positive number)
⇒ a × n a + b × n b + c × n c n a + n b + n c < 62
∴ 58 < a × n a + b × n b + c × n c n a + n b + n c < 62
Hence, option (c)