the population in atown is 10000 .in a certain year the men increase by 4% ans women by 6%. increased population is 10470. what number of women in the town have increased.. ?
vimalraj:
the above question is wrong....coz population increased by 4% men and 6% women will be 11000
No Vimal....it says men increased by 4% on the existing number of men and women increased by 6% on the existing number of women. ....check my answer.
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Let number of men be x and no of women be y
So, x + y = 10000 ......(1)
4% of men increased and 6% women increased
Now, men = 1.04 x and women = 1.06 y
Increase population = 10470
Therefore, 1.04x + 1.06y = 10470...........(2)
Multiply the equ(1) by 1.04
We get,
1.04x + 1.04y = 10400 ....(3)
Solve two equations (2) and (3)
1.04x + 1.06y = 10470
1.04x + 1.04y = 10400 (Sub)
--------------------------------
0.02y = 70
y = 3500
No. of women were 3500
Increased no. of women = 6% of 3500 = 210
So, x + y = 10000 ......(1)
4% of men increased and 6% women increased
Now, men = 1.04 x and women = 1.06 y
Increase population = 10470
Therefore, 1.04x + 1.06y = 10470...........(2)
Multiply the equ(1) by 1.04
We get,
1.04x + 1.04y = 10400 ....(3)
Solve two equations (2) and (3)
1.04x + 1.06y = 10470
1.04x + 1.04y = 10400 (Sub)
--------------------------------
0.02y = 70
y = 3500
No. of women were 3500
Increased no. of women = 6% of 3500 = 210
my options are a. 60 b.70 c.80 d.90
that means, either some information in the question missing or the options are incorrect. As far as my understanding about the question, my solution is correct.
thank u.. !
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