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 is the ratio of the number of students who passed only in Paper \[{{P}_{1}}\]and \[{{P}_{2}}\]to the number of boys who passed only in Paper \[{{P}_{3}}?\]

A)  5:4                              

B)  7:4        

C)  5:3                              

D)  9:5        

E)  8:5

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 \[84-30=54\]  \[=\text{Ratio}\,\text{=}\frac{90}{54}=\frac{5}{3}=5:3\]