Discussion

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×na+b×nbna+nb
⇒ 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×na+b×nb+c×ncna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 60na+58nb+62ncna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 58(na+nb+nc)+2na+4ncna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 58(na+nb+nc)na+nb+nc + 2na+4ncna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 58 + (positive number)

a×na+b×nb+c×ncna+nb+nc > 58

Also, 

a×na+b×nb+c×ncna+nb+nc = 60na+58nb+62ncna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 62(na+nb+nc)-2na-4nbna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 62(na+nb+nc)na+nb+nc - 2na+4nbna+nb+nc

a×na+b×nb+c×ncna+nb+nc = 62 - (positive number)

a×na+b×nb+c×ncna+nb+nc < 62

∴ 58 < a×na+b×nb+c×ncna+nb+nc < 62

Hence, option (c)

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