A) 42.33%
B) 43.34%
C) 41.32%
D) 39.30%
Correct Answer: A
Solution :
Let the marks obtained by first student be\[x.\] \[\therefore \]Marks obtained by second student \[=x+9\] Sum of their marks \[=2x+9\] As given, \[x+9=56%\]of \[(2x+9)\] \[\Rightarrow \] \[x+9=\frac{56}{100}\times (2x+9)\] \[\Rightarrow \] \[x+9=\frac{14}{25}\times (2x+9)\] \[\Rightarrow \] \[25x+225=28x+126\] \[\Rightarrow \] \[3x=225-126\]\[\Rightarrow \]\[x=\frac{99}{3}=33\] \[\therefore \] Marks obtained are 42 and 33.You need to login to perform this action.
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