To start, it's not known how many students play all three games (denoting cricket=C, football=F, hockey=H), and we introduce this as a variable \(x\). Those who play exclusively C&F, F&H and C&H are \(15-x\), football \(12-x\), and hockey \(10-x\) respectively. Solve for \(x\) by setting an equation below
\(33=20+25+18-(15-x+12-x+10-x+2x)\), \(x=7\).
Hence, \(C&F=15-7=8\) and \(C&H=10-7=3\)
The number of students who play only cricket is \(F-C&F-C&H-C&F&H=20-8-3-7=2\).
Answer must be
C
KarunMendiratta, thanks for posting
KarunMendiratta wrote:
In a class of 33, every student plays at least one game.
20 students play cricket, 25 play football and 18 play hockey.
15 play both cricket and football, 12 play football and hockey and 10 play cricket and hockey.
Quantity A |
Quantity B |
Students who play only cricket |
2 |
A. Quantity A is greater
B. Quantity B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given