In a school, an exam was held for 80 marks in which 32 was the passing marks. Who among the following scored the minimum marks?
[Note: Round off up to one decimal only.]
- AD, who got
19
12
of the total marks - BC, who got 160% of the passing marks
- CA, who got 72% marks in the exam
- DB, who got
11
7
of the total marks
Solution & Step-by-step Explanation
Let's calculate the exact marks obtained by each person:
A's marks: 72% of total marks (80)
Marks=
100
72
×80=57.6
B's marks:
11
7
of total marks (80)
Marks=
11
7
×80=
11
560
≈50.9
C's marks: 160% of passing marks (32)
Marks=
100
160
×32=1.6×32=51.2
D's marks:
19
12
of total marks (80)
Marks=
19
12
×80=
19
960
≈50.5
Comparing the calculated marks:
50.5 (D)<50.9 (B)<51.2 (C)<57.6 (A)
Thus, D scored the minimum marks.
A's marks: 72% of total marks (80)
Marks=
100
72
×80=57.6
B's marks:
11
7
of total marks (80)
Marks=
11
7
×80=
11
560
≈50.9
C's marks: 160% of passing marks (32)
Marks=
100
160
×32=1.6×32=51.2
D's marks:
19
12
of total marks (80)
Marks=
19
12
×80=
19
960
≈50.5
Comparing the calculated marks:
50.5 (D)<50.9 (B)<51.2 (C)<57.6 (A)
Thus, D scored the minimum marks.