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.

Interview Candidate
Mar 21st, 2006
11
7015

Showing Answers 1 - 11 of 11 Answers

Deepak

Apr 5th, 2006

let m=male, f=female

m=2(f-15)
m=2f-30
m-2f=-30 -- 1

f=5(m-45)
5m-f=225 -- 2
m - 2f=-30
10m- 2f=450
------------
9m=480
Approximately m=53

Deepakbtech
Profile Answers by Deepakbtech

Apr 5th, 2006

let female be f
and m be male
m=2(f-15)
m-2f=30 --1

f=5(m-45)
5m-f=225 --2
solving 1 and 2
we get m=53(approximately)

sravani811
Profile Answers by sravani811

Jun 3rd, 2008

app 53!

anand3186
Profile Answers by anand3186

Jul 20th, 2009

480/9 -> 53.33.

so,

ans is 54 approx

suhana
Profile Answers by suhana Questions by suhana

Jul 23rd, 2009

the answer is 53.
but i've doubt bcoz actual ans is:53.33
as we know people are not in points.the must be in fixed no.
i think there is a wrong in question.