A) 40%
B) 45%
C) 42%
D) 48%
Correct Answer: C
Solution :
Let minimum pass marks = x Then, \[x+\frac{3x}{100}=515\] \[\frac{103x}{100}=515\] \[x=500\] Now, required % marks above the pass marks \[=\frac{710-500}{500}\times 100\] \[=\frac{210}{500}\times 100\] \[=\frac{210}{5}=42%\]You need to login to perform this action.
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