In a club there are certain number of males and females. If 15 females are absent then number of males will be half of females. If 45 males are absent then female strength will be 5 times that of males. Number of males actually present is
Step-by-step Solution:
This problem can be solved by setting up a system of two algebraic equations based on the two scenarios provided in the question.
- Let the actual number of males be M and the actual number of females be F.
- We have two equations and two variables. The easiest way to solve is to substitute the second equation into the first one to eliminate F:
2M = [ 5(M - 45) ] - 15
- Now, we can solve this single equation for M:
2M = 5M - 225 - 15
2M = 5M - 240
240 = 5M - 2M
240 = 3M
M = 240 / 3 = 80.
- If there are 80 males, we can find the number of females using the second equation: F = 5(80 - 45) = 5(35) = 175 females.
- Let's check Scenario 1: If 15 females are absent, there are 175 - 15 = 160 females. Is the number of males (80) half of 160? Yes. The answer holds true.
Final Answer: The number of males actually present is 80.