A) 80
B) 81
C) 81.5
D) 80.5
Correct Answer: C
Solution :
| [c] Let the number of students in classes A, B and C be x, y and z, respectively. Then, Total score of class A \[=83\times x=83x\] Total score of class B \[=76\times y=76y\] Total score of class C \[=85\times z=85z\] According to the question, \[\frac{83x\times 76y}{(x+y)}=79\]\[\Rightarrow \]\[83x+76y=79x+79y\] \[\therefore \] 4x = 3y ...(i) and \[\frac{76y+85z}{y+1}=81\] \[\Rightarrow \] \[76y+85z=81y+81z\] \[\therefore \] \[=4z=5y\] ...(ii) From Eqs. (i) and (ii), \[x:y:z=3:4:5\] Now again, let \[x=3k;\]\[y=4k\]and \[z=5k\] \[\therefore \] Required average score of A, B and C \[=\frac{83x+76y+85z}{x+y+z}\] \[=\frac{83\times 3k+76+4k+85\times 5k}{3k+4k+5k}\] \[=\frac{249+304+425}{12}\] \[=\frac{978}{12}=81.5\] |
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