Question
Download Solution PDFComprehension
The following table shows the details about (i) the number of students studying in five different classes A-E of a school, and (ii) participating neither in chess nor in carrom (iii) the ratio of students who participate in Chess to Carrom. The remaining students play either chess or carrom. Based on the data in the table answer questions:
Class-wise Participation in Games
|
Class |
Total Number |
Number of Students |
Ratio of Students |
|
A |
420 |
119 |
4 ∶ 3 |
|
B |
330 |
88 |
7 ∶ 4 |
|
C |
240 |
110 |
8 ∶ 5 |
|
D |
125 |
45 |
2 ∶ 3 |
|
E |
390 |
130 |
8 ∶ 5 |
What is the ratio of students participating in Carrom from B and C together to the students participating in Chess from A and D together?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFThe total number of students participating in Carrom from B and C together is:
For B: 330 - 88 = 242 students play either chess or carrom.
For C: 240 - 110 = 130 students play either chess or carrom.
The ratio of students participating in Chess to Carrom is 7:4 for B and 8:5 for C, respectively.
So, the number of students participating in Carrom from B and C together is:
For B: (4/11) x 242 = 88 students
For C: (5/13) x 130 = 50 students
Therefore, the total number of students participating in Carrom from B and C together is 88 + 50 = 138.
The total number of students participating in Chess from A and D together is:
For A: (4/7) x (420 - 119) = 168 students
For D: (3/5) x (125 - 45) = 48 students
Therefore, the ratio of students participating in Carrom from B and C together to the students participating in Chess from A and D together is 138:216, which simplifies to 23 : 36.
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