
In the half-yearly exam, only \[70\%\] of the students were passed. Out of those(passed in the half-yearly exam), only \[60\%\] passed in the annual exam. Out of those who did not pass the half-yearly exam, \[80\%\] passed the annual exam. What percent of the students passed the annual exam?
[a] \[42\%\]
[b] \[56\%\]
[c] \[66\%\]
[d] \[38\%\]
Answer
532.6k+ views
Hint: Assume that the total number of students be x. Hence find the total number of students passing the half-yearly exam and the total number of students failing the half-yearly exam. Using this information calculate the total number of students who passed the half-yearly as well as the annual exam and the total number of students who passed the annual exam but failed the half-yearly exam. Hence find the percentage of the students passing the final exam.
Complete step-by-step answer:
Let the total number of students be x.
Since \[70\%\] of the total students passed the half-yearly exam, we have
The total number of students who passed the half-yearly exam = 0.7x
and the total number of students who failed the half-yearly exam = 0.3x
Now since only $60\%$ of the students who passed the half-yearly exam also passed the annual exam, we have
The total number of students who passed both half-yearly and the annual exam $=0.6\times 0.7x=0.42x$
Also, since $80\%$ of the students who failed the half-yearly exam passed the annual exam, we have
The total number of students who failed the half-yearly exam but passed the annual exam $=0.8\times 0.3x=0.24x$
Hence the total number of students who passed the annual exam $=0.42x+0.24x=0.66x$
Hence the percentage of total students who passed the annual exam $=\dfrac{0.66x}{x}\times 100=66\%$
Hence $66\%$ of the total students passed the annual exam.
Hence option [c] is correct.
Note: We can compute directly the percentage of total students passing the final exam.
We have the percentage of total students passing the half-yearly exam $=70\%$ and the percentage of students failing the half-yearly exam $=100-70=30\%$
Now the percentage of the total students passing the annual exam who passed the half-yearly exam $=60\%\text{ of }70\%=0.6\times 70\%=42\%$
Similarly the percentage of total students passing the annual exam who failed the half-yearly exam $=80\%\text{ of }30\%=0.8\times 30\%=24\%$
Hence the percentage of the total students passing the annual exam $=24\%+42\%=66\%$.
Hence 66% of the total students pass the annual exam.
Complete step-by-step answer:
Let the total number of students be x.
Since \[70\%\] of the total students passed the half-yearly exam, we have
The total number of students who passed the half-yearly exam = 0.7x
and the total number of students who failed the half-yearly exam = 0.3x
Now since only $60\%$ of the students who passed the half-yearly exam also passed the annual exam, we have
The total number of students who passed both half-yearly and the annual exam $=0.6\times 0.7x=0.42x$
Also, since $80\%$ of the students who failed the half-yearly exam passed the annual exam, we have
The total number of students who failed the half-yearly exam but passed the annual exam $=0.8\times 0.3x=0.24x$
Hence the total number of students who passed the annual exam $=0.42x+0.24x=0.66x$
Hence the percentage of total students who passed the annual exam $=\dfrac{0.66x}{x}\times 100=66\%$
Hence $66\%$ of the total students passed the annual exam.
Hence option [c] is correct.
Note: We can compute directly the percentage of total students passing the final exam.
We have the percentage of total students passing the half-yearly exam $=70\%$ and the percentage of students failing the half-yearly exam $=100-70=30\%$
Now the percentage of the total students passing the annual exam who passed the half-yearly exam $=60\%\text{ of }70\%=0.6\times 70\%=42\%$
Similarly the percentage of total students passing the annual exam who failed the half-yearly exam $=80\%\text{ of }30\%=0.8\times 30\%=24\%$
Hence the percentage of the total students passing the annual exam $=24\%+42\%=66\%$.
Hence 66% of the total students pass the annual exam.
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