question_answer

Direction: Study the information carefully and answer the question given below

From a school 500 students appeared in all in an examination consisting of three papers\[{{P}_{1}},{{P}_{2}}\]and \[{{P}_{3}}.52.6%\]students passed in \[{{P}_{1}},57%\]in \[{{P}_{2}}\]and 50.4% in \[{{P}_{3}}.18%\]of students passed only in\[{{P}_{1}}\]and\[{{P}_{2}},10.2%\]passed only in \[{{P}_{1}},\]and \[{{P}_{3}}\]and 84% passed in all three papers. The number of girls passed in Paper\[{{P}_{1}}\]only is 60% of the number of boys who passed only in Paper \[{{P}_{1}}.\]The ratio of the number of boys to the number of girls who passed only in Paper \[{{P}_{2}}\]is 7:6. The number of girls who passed in Paper\[{{P}_{3}}\]only is 6% of the total number of students who appeared in the examination.

Answer the following questions based on the above information.

What percentage of students passed only in Paper \[{{P}_{2}}?\,\]

A)  14.2%                         

B)  15%        

C)  15.6%             

D)  16%        

E)  16.4%

Correct Answer: C

Solution :

\[z=263-(90+51+42)=263-183=80\] \[x+k=285-(90+42)=153\]         ?(i) \[y+k=252-(51+42)=159\]         ?(ii) \[x+y+k=500-(90+42+51+80)=237\]?(iii) Solving eqn (i), (ii) and (iii)             \[x=78,y=84,\]  \[k=75\] Let the number of boys who passed in \[{{P}_{1}}\]only be B. The number of girls who passed in \[{{P}_{1}}\]only will be\[\frac{3B}{5}.\] \[\therefore \]      \[B+\frac{3B}{5}=80\] \[\therefore \] B= 50, so number of girls \[=\frac{3B}{5}=30\] The number of students who passed only in \[{{P}_{2}}\]is 78 where B: G = 7:6, ie B + G = 78 ...,(i)       And 6B = 7G Solving these two eqns, B = 42, G = 36. Number of girls who passed only in \[{{P}_{3}}=6%\]of = 500 \[=\frac{500\times 6}{100}=30\] So the number of boys who passed only in \[{{P}_{3}}\]is \[84-30=54\]  Reqd% \[=\frac{78}{500}\times 100=15.6%\]