A) 1100 and 900
B) 1200 and 800
C) 1300 and 700
D) 1500 and 300
Correct Answer: B
Solution :
| \[x+y=2000\] ...(1) |
| \[x+2y=2800\] ...(2) |
| From eq.(1), \[x=2000-y\] ...(3) |
| Substituting this value of x in eq. (2), we get, |
| \[2000-y+2y=2800\] |
| or \[2000+y=2800\] |
| \[y=2800-2000=800\] |
| Therefore, \[x=2000-800=1200\] |
| We find that \[1,200\] children and 800 adults bought tickets to the park on that day. |
| So, option [b] is correct. |
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