Answer / nila

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

Answer / kermit rose

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.

Answer / srv

F-15=2M. (1)
5(M-45)=F (2)
Putting value of F in equation (1) we get
5M-225-15=2M
-240= -3M
M=80

Putting vale of M in equation (2) we get
F=5(80-45)
F=5*35
F=175

Answer / kermit rose

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 have 45 males being absent,

then this gives the equation,

F = 5 * (M - 45)= 5 m - 225


which simplifies to

F = 5 * M - 225

5 * M - F = 225

Pulling the equations together, we get

5 * M - F = 225
-M + 2 * F = 30

Multiply the first equation by 2, and keep the second
equation as is.

10 * M - 2 * F = 450
- M + 2 * F = 30

Add the equations.

9 * M = 480

M = 480/9 is not an integer.


The question is not well stated.

In order to get an integral answer, the question should have been stated as:


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 15 females and 45 males are absent then female strength will be 5 times that of males. Find no. of males actually present.
In this form of the question the number of females will be 40 and the number of males will be 50.
(40 - 15) = 25 and 50 = 2 * 25

(50 - 45) * 5 = (40 - 15)

Answer / 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.

Answer / dhruv

let no of male=X
and no of female=Y
now according to question,
2(Y-15)=X;
and
5(X-45)=Y;
after solving the two equations we get,
X=53;
Y=40;

Answer / guest

30 males

Answer / mathi

m f


m=2(f-15)

f= 5(m-45)

m=2(5(m-45))
m= 2(5m-225)
m= 10m -450
9m=450
m=50

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