4. In an examination, 3/5h of the students who appearedfailed by 10 ma

1.	4. In an examination, 3/5h of the students who appearedfailed by 10 marks and 1/5h of the students gotmarks above the pass mark. Each of the remainingstudents got 20 marks above the pass mark. Studentswho gave the exam scored 62 marks on an average.The pass mark is(A) 64 (B) 66 (C) 62 (D) 56
Answer» 3/5th of the students that appeared failed by 10 marks Hence 3x/5 students get marks = y−10 1/5th of the students got 10 marks above the pass marks Hence x/5 students get marks = y+10 Remaining students = x−(3x/5 +x/5) =x/5 Marks of each of the remaining students got 20 marks above the pass mark. Hence x/5 students get marks = y+20 Average marks of the students={(3x/5)×(y−10)+(x/5)×(y+10)+ x/5)×(y+20)}/x={(3/5)×(y−10)+(1/5)×(y+10)+ 1/5)×(y+20)} The average marks of the students = 62Hence {(3/5)×(y−10)+(1/5)×(y+10)+ (1/5)×(y+20)}=62 →(3y−30)+(y+10)+y(+20)}=310 or 5y = 310 or y = 62

4. In an examination, 3/5h of the students who appearedfailed by 10 marks and 1/5h of the students gotmarks above the pass mark. Each of the remainingstudents got 20 marks above the pass mark. Studentswho gave the exam scored 62 marks on an average.The pass mark is(A) 64 (B) 66 (C) 62 (D) 56

Answer»

3/5th of the students that appeared failed by 10 marks

Hence 3x/5 students get marks = y−10

1/5th of the students got 10 marks above the pass marks

Hence x/5 students get marks = y+10

Remaining students = x−(3x/5 +x/5) =x/5

Marks of each of the remaining students got 20 marks above the pass mark.

Hence x/5 students get marks = y+20

Average marks of the students={(3x/5)×(y−10)+(x/5)×(y+10)+ x/5)×(y+20)}/x={(3/5)×(y−10)+(1/5)×(y+10)+ 1/5)×(y+20)}

The average marks of the students = 62Hence {(3/5)×(y−10)+(1/5)×(y+10)+ (1/5)×(y+20)}=62

→(3y−30)+(y+10)+y(+20)}=310

or 5y = 310

or y = 62