1) In a club there are certain no. of males and females. If
15 females are absent then no. of males will be twice that
of females. If 45 males are absent then female strength
will be 5 times that of males. Find no. of males actually
present.
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Answer Posted / khalid usman
m = 2*(f-15) -> equ I
f - 15 = 5*(m-45) -> equ II
=> f = 5*(m-45) + 15
=> m = 2*((5*(m-45)+15)-15)
=> m = 2*(5m - 225 +15 - 15)
=> m = 2*(5m-225)
=> m = 10m - 450
=> 9m = 450
=> m = 50
so from equation II
f - 15 = 5*(m - 45)
=> f - 15 = 5*(50-45)
=> f - 15 = 5*(5)
=> f = 25 + 15
f = 40
so originally there are 40 females 50 males in club.
| Is This Answer Correct ? | 3 Yes | 6 No |
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