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.

2 Answers on this Question

1. Let the number of males be given the name M.
   Let the number of females be given the name F.
   If 15 females are absent, then M will be twice that of
   present females.
   This means that M = 2 * (F – 15)
   M = 2 * F – 30.
   or 2 * F – M = 30.
   Now if in addition to the 15 females being absent, we also
   have 45 males being absent,
   then this gives the equation,
   (F – 15) = 5 * (M – 45)
   which simplifies to
   F – 15 = 5 * M – 225
   5 * M – F = 210
   Pulling the equations together, we get
   5 * M – F = 210
   -M + 2 * F = 30
   Multiply the first equation by 2, and keep the second
   equation as is.
   10 * M – 2 * F = 420
   – M + 2 * F = 30
   Add the equations.
   9 * M = 450
   M = 50
   Verify answer.
   Calculate F
   from – M + 2 * F = 30
   -50 + 2 * F = 30
   2 * F = 30 + 50
   F = 40.
   If 15 females are absent, then number of males will be twice
   that of females.
   40 – 15 = 25.
   50 = 2 * 25. Confirmed.
   If also 45 males were absent, then female strength would be
   5 times that of males.
   Female strength is 25 due to the 15 females being absent.
   50 – 45 = 5.
   25 = 5 * 5. Confirmed.

   Reply

2. m f
   m=2(f-15)
   f= 5(m-45)= 5m-225
   m=2(5m-225)-30
   m=10m-480
   9m=480
   m=53.3
   approximately 53

   Reply