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

## 11 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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