Puzzle:
A contractor had employed 100 labourers for a flyover construction task. He did not allow any woman to work without her husband. Also, atleast half the men working came with their wives. He paid five rupees per day to each man, four rupees to each woman and one rupee to each child. He gave out 200 rupees every evening. How many men, women and children were working with the constructor?
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Solution:
There were 16 men, 12 women and 72 children working with the constructor.
Let's assume that there were X men, Y women and Z children working with the constructor. Hence,
X + Y + Z = 100
5X + 4Y + Z = 200
Eliminating X and Y in turn from these equations, we get
X = 3Z - 200
Y = 300 - 4Z
As if woman works, her husband also works and atleast half the men working came with their wives; the value of Y lies between X and X/2. Substituting these limiting values in equations, we get
if Y = X,300 - 4Z = 3Z - 200
7Z = 500
Z = 500/7 i.e. 71.428
if Y = X/2,
300 - 4Z = (3Z - 200)/2
600 - 8Z = 3Z - 200
11Z = 800
Z = 800/11 i.e. 72.727
But Z must be an integer, hence Z=72. Also, X=16 and Y=12
12 comments:
16men
12women
72 children......................
...............rajesh
19,8,73
19 * 5 = 95
8 * 4 = 32
73 * 1 = 73
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100(W)---200 Rupees
2nd anonymous :)...
The number of women is atleast equal to half the number of men...So I think you aint correct...
19 8 73 isn't the answer, though it's what you come at after the first step...
(1) 5X+4Y+Z=200
(2) X+Y+Z=100
(3) X>Y
(4) Y>X/2
(1)-(2)=4X+3Y=100
X=25-3/4Y
since X should be a whole number (not fraction) Y should be divisible by 4. Possibilities:
Y=4,8,12,16,...
X=22,19,16,13,...
Only first three options are valid because of (3). Only Y=12, X=16 is valid because of (4). From (2):
Z=100-(X+Y)
X=16, Y=12, Z=72
Men=16
Women=8
Children=88
Men earn = 16*5 = 80
Women earn= 8*4 = 32
TOTAL = 112 rupees
Rest = 200-112 = 88 rupees
1 Rupee per child = 88 children
* Half of(16)men = so many women
16 men, 8 women
I liked Suren's way of doing it.
Thanks,
Hemant
child labor!!!!
no the above ans is wrong as no of women isnt half of men..
let no of men b X
chldrn=Y
now acc to given puzzle
x/2+ 2* x/2 + y= 100
(m multiplying 2 as half of men will their wife with equal num)
for d money
5x + 4x/2 +y= 200
solve both the equatn u get
x= 27 approx
y =60
and as women s half d men it 13 women
total wud b =100 then
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28,14 and 4 .
28*5=140
14*4= 56
4*1 = 4
200 Rs. in total .
Good math puzzle here.
With the rules laid out, you can make 2 equations and 2 inequalities.
M = number of men working
W = number of women working
C = number of children working
Since the contractor hires 100 people
M + W + C = 100.
And since he pays out 200 Rupees per day, with 5 rupees to each man, 4 rupees to each woman and 1 rupee to each child we know
5M + 4W + C = 200.
And since a woman can only work with her husband and at least half the men brought their wives,
M > W >= M/2.
Subtracting the first equation from the second equation gives
4M + 3W = 100.
From here I just guessed values of M and W that satisfied the above equation and the above inequalities.
The first one that worked for me was
M = 16
W = 12.
Now, plugging those values into the second equation gives
C = 72.
And checking those values for M, W, and C in the first equation, everything works out in this hilariously discriminatory puzzle!
M = 16
W = 12
C = 72.
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