Men At A Construction Site

Let's assume that there were M men, W women and C children working with the constructor. Hence,

M + W + C = 100

5M + 4W + C = 200

Eliminating M and W in turn from these equations, we get

M = 3C - 200

W = 300 - 4C

As if woman works, her husband also works and atleast half the men working came with their wives; the value of W lies between M and M/2. Substituting these limiting values in equations, we get

if W = M,

300 - 4C = 3C - 200

7C = 500

C = 500/7 i.e. 71.428

if W = M/2,

300 - 4C = (3C - 200)/2

600 - 8C = 3C - 200

11C = 800

C = 800/11 i.e. 72.727

But C must be an integer, hence C=72. Also, M=16 and W=12

Let us consider Number of men to be x, Number of women to be y and number of children to be z Firstly , they all add up to 100 so, x + y + z = 100 Next considering their daily wages in a respective order we get the equation , 5x + 4y + z = 200

Subtracting both the equations we get, 4x + 3y = 100 , now let us consider the integral solutions of this equation, we get 6 solutions ; (22,4) (19,8) (16,12) (13,16) (10,20) (7,24) Since, women are not allowed to work without their husbands (y) cannot be greater than (x) . Hence we can eliminate the last three solutions. Also, atleast half the men came with their wives. Therfore, we can eliminate the first two solutions, leaving behind (16,12) as the right answer. Hence the right answer is 16 men.

The correct answer should be 0, not 16!

The contractor goes to jail for paying sexist wages and using child labor!

PS: Posting a note does not take a problem away!

NOTE: I didn't mean to offend the guy who posted the question!

Let the numbers of man, women and child employed be $a$ , $b$ , and $c$ . Therefore,

$a+b+c=100\quad 5a+4b+c = 200$

Subtracting the two equations, we have: $\quad 4a + 3b = 100$

Dividing by $4$ throughout, $\quad \Rightarrow a + \frac {3}{4}b = 25$

We know that $\frac {3}{4} b$ must be an integer, which means that $b$ must be a multiple of $4$ , and that $\frac {1}{2} a \le b \le a$ .

From $a = 25 - \frac {3}{4} b$ , the possible values of $(a,b)$ :

$(a,b) = (22,4),(19,8),(16,12),(13,16),(10, 20), (7,24),(4,28), (1,32)$

The only pair that satisfies $\frac {1}{2} a \le b \le a$ is $(a,b) = (16,12)\quad \Rightarrow a = \boxed {16}$

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 C = 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 C=72. Also, M=16 and W=12

We can simplify the conditions given as follows: - From "all women are not allowed to work without their husbands", we can infer that all women here are married. This statement also means that there is always more men than women, because for every woman working a man must be working. This is important, because it means M>W. - "At least half the men bring their wives with them" means that there is for x number of men, there is at least x/2 number of women. I did not use this fact in my calculations though.

M + W + C = 100   (eq1)
5M + 4W + C = 200  (eq2)

(eq2) - (eq1) eliminates C and gives : 4M + 3W = 100.

From here on we know that : - M>W - M and W are integers.

Spam a few numbers and you'll get the right combination.

Looking back, the solving process makes sense; the number of children is not used in the equation since it does not correlate with any other quantities.

Let total man = x, total women = y, and total chid = z. Now, x+y+z=100 --------(1) and also, 5x+4y+z=200 --------(2) Subtracting (1) from (2), we have 4x+3y=100 => x=25-3y/4 We can see from the above equation that y have to be a multiple of 4, for x to be an integer.. We get the set of solutions (22,4), (19,8), (16,12)... It is given that atleast half of the men come with their wives and 12x2>16 So we take number of men as 16

let there be x women at work than they come with x man also suppose y men are working solo, than we can say that 100 - 2x - y children are at work, now the contractor has to pay 5(x + y ) + 4 x + 1(100 - 2x - y ) = 200, solving this equation we get x = ( 100 - 4y)/ 7, since y has to be a natural number, by trial and error method we conclude that x gets a whole number for y = 4, thus x=16 and hence the answer.

16 men, 12 women and 72 children were 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

There were 16 men, 12 women and 72 children working with the

well my solution is kind of different and a bit messed up but here it is.

let us assume the following:

x=number of men

y=number of women

z=number of children

we have three equations/inequalities derived from the problem:

(1) 5x+4y+z=200

(2) x+y+z=100

and since it is said that there are AT LEAST half of the men with their wives, and that NO women are working without their husbands (or at least they were not allowed to do so), we derive the inequality:

(3) x>=y>=x/2

using elimination method (cancelling z for this case) for equations 1 and 2, we derive:

(4) 4x+3y=100

using equation 4, we solve for x:

4x=100-3y

x=(100-3y)/4

first we set the limitations x>=y by substituting the value of x

(100-3y)/4>=y

100-3y>=4y

-4y-3y>=-100

7y<=100

y<=100/7

the same goes for solving y>=x/2, and in that limitation, we get y>=100/11

with those limitations, we now have:

100/7>=y>=100/11

100/7=~14.29 and 100/11=~9.09

since we cannot get a fraction-human, we only have to get the whole numbers in the range y=[10,14]

and we go back to equation 4 and substitute 10, 11, 12, 13, 14 for y, and the only whole number that we can get for the value of x from those substitution is by using y=12:

4x+3(12)=100

4x+36=100

4x=64

x=16

thus the number of men working at the construction site is 16.

So, let there be x men with their wives, x wives, y men without wives, 100-2x-y children. Therefore, 9x+5y+100-2x-y=200(total salary paid) After rearrange, 7x+4y=100 Then, keep in mind that x>y since more men with their wives than those without Also, x has to be a multiple of 4 When x=4 or 8, y>x thus don't work when x=12, y=4. This solution works. So, there are x+y=16 men.

Let M = men without wives.

Let N = men with wives.

Let W = women.

Let C = children

5M + 5N + 4W + C = 200 (eq - 1)

M + N + W + C = 100 (eq - 2)

Subtracting eq-2 from eq-1

4M + 4N + 3W = 100 (eq-3)

Since women work with husbands, we have

N = W

Substituting N = W in eq-3

4M + 7W = 100

Dividing both sides by 4 and rearranging,

M = 25 - 7/4 x W (eq-4)

Since almost half of the total number of men work with wives, we select W such that M is a minimal whole number. The fraction 7/4 in eq-4 suggests that the substituted number be a multiple of 4.

With W = 12, we get

M = 4

N = 12

C = 72

Total number of men = M + N = 4 + 12 = 16 (Final answer)

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 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 ,Z=72. Also, X=16 and Y=12....thus we can get the full answer.16

M+W+C=100(M=>Men, W=>Women,C=>Children) 5 M+4 W+1 C=200(Rs.5/Men, Rs.4/Women,Rs.1/children) solving=>4 M+3*W=100 Now use these conditions: i)M,W are integers ii)Half of n(Men) <= n(Women) =>U can get only one solution: M=16 and W=12