Men At A Construction Site

Algebra Level 3

A contractor had employed 100 laborers for a flyover construction task. He did not allow any woman to work without her husband. Also, at least 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 were working on this construction?

Note: This problem neither promotes child labor nor gender discrimination.

Image credit: Voanews


The answer is 16.

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14 solutions

Prakkash Manohar
Mar 5, 2014

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

the question didnt state out the condition for children so i think that there should have multiple answers tell me if anything wrong of my thought :)

Zhen Hui Chong - 7 years, 2 months ago

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Well there's actually only 1 answer and apparently the proportion of children is needed to answer this question;)

Seraph Yang - 7 years, 2 months ago

Let the number of men working be x(who are single) Let the number of men who are married be y and let z be the number of children. According to the pay scale, 5x+(5+4)y+z=200 (note that we are classifying the married and unmarried men to avoid any confusion.) Further,given that total laborers are 100.i.e. x+2y+z=100(2y implicates the number of couples,for simplicity) solving above 2 equations, 4x+7y=100 there are 3 possible answers to this equation (18,4)(11,8)(4,12) It is given that number of men with their wife > number of single men.(y>x) (4,12)satisfies the equation. Thus, no of single laborers=4 no. of married laborers=12 thus no of wives=12 no. of children=100-(4+12+12)=72 Hope this solution satisfies all the given conditions.

kushal ghate - 7 years, 2 months ago

Somebody requested to clarify "The problem doesn't tell about the child's proportion to the men and women working. ".

I didn't quite understand it. What is the need of the proportion of the number of children to the number of men and women working. It is not necessary that the children working there belong to the men and women working there.

Prakkash Manohar - 7 years, 2 months ago

three unknown but 2 equation but you have a limitation so you can do this buy this: first: M + W + C = 100 then second: 5M + 4W + C = 200: you cancel the Child(C) so you can get 4M + 3W = 100 now he said that at least half so use it as the limitation(M/2---->): 4M +3(M/2) = 100; M will get 18.18 so it means limitation M > 18.18; use integer ofcourse, lets check it one by one the y must be integer, start with 19 first: 4(19)+ 3y = 100; y = 8; y = integer: check! M = 18

Franz Gerrard Dimaapi - 7 years, 2 months ago

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If there are 19 men, then, there must be at least 10 women (half of 19 = 9.5 which is not possible) but here, women are 8.

Prakkash Manohar - 7 years, 2 months ago

Don't u think we do have 3 equations As q says atleast half the men came with their wives So if m is for men and w for women then equation shld.be m/2 =w Wat say According above two equations and this equation i got 18 men as an answer Please clarify

Anshuman Singh - 6 years, 11 months ago

Bhaiya , ye to sir ke upar se hi gaya ...thoda easy question dala karo

Abhishek Chauhan - 7 years, 1 month ago

@Prakkash Manohar i did it a little bit with hit n trial but your solution is quite good :-)

Navdeep Agarwal - 7 years, 1 month ago

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Thanx Navdeep

Prakkash Manohar - 7 years, 1 month ago

nice solution i solved it the same way.

Uttam Bhadauriya - 7 years, 2 months ago

why did you round it down?

Eka Kurniawan - 7 years, 2 months ago

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because the value of W lies between M and M/2.

if W = M, C = 500/7 i.e. 71.428

if W = M/2, C = 800/11 i.e. 72.727

So, C lies between 71.428 and 72.727. Also, C is an integer. So, C can only be 72.

Prakkash Manohar - 7 years, 2 months ago

Haha......because he cannot employ a fraction of workers.......

Satvik Golechha - 7 years, 2 months ago

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no, i mean why he didn't round it up.

Eka Kurniawan - 7 years, 2 months ago

16 men, 12 women and 72 children could be correct answer. In between, 4M+3W=100 is also satisified by these numbers. 5M+4W+nC=200; M+W+nC=100. Subtracting you get the proposed equation.

Narayan Garimella - 7 years, 2 months ago

what's wrong with 20M, 10W, 60C?

Danilo Mante - 7 years, 2 months ago

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20 + 10 + 60 = 90 laborers but there are 100 laborers as given in the question.

Prakkash Manohar - 7 years, 2 months ago

where is the equation W=M/2 Satisfied

NARENDRA SINGH - 7 years, 2 months ago

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'at least'

Ankit Agrawal - 7 years, 2 months ago

W doesn't have to equal to m/2 but is equal to OR more than m/2. It says 'at least'.

Seraph Yang - 7 years, 2 months ago

Yes, Zhen Hui Chong you are correct. The question didn't state out any condition for children. But, then, what can be a possible answer to this problem?

Prakkash Manohar - 7 years, 2 months ago

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13 men, 16 women, 71 children... Unless I'm missing something

Krystal Shimabukuro - 5 years, 9 months ago

Well, I think if anything, the question wasn't clear enough xD

Gabba-gabba Jin - 6 years, 8 months ago

The answer from what I got could also be 13 men, 16 women, 71 children. I don't remember seeing a rule women could only go with husbands. What if they're not married :-p

Krystal Shimabukuro - 5 years, 9 months ago

It doesn't satisfy the question. At least half of the men came with their wife but your final answer is 16 and 12 respectively. Where is the " at least half of the men came with their wives"? Is 12 at least half of 16?

Carlito Carrascal - 7 years, 2 months ago

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atleast half of the men=> 8 or more men can come with heir wives. Here it is 12.

nagendra babu - 7 years, 2 months ago

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definitely not..I think the answer is still incorrect..where does 8 or more men came from?

Carlito Carrascal - 7 years, 2 months ago

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@Carlito Carrascal There are 16 men, of which 12 of them had their wives come along, which is why there are 12 women.

Calvin Lin Staff - 7 years, 2 months ago

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@Calvin Lin How is 8W, 19M, 73C is unfit?

Rahul Deshpande - 7 years, 2 months ago

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@Rahul Deshpande Check it against the conditions in the question.

For example, "at least half the men working came with their wives."

Calvin Lin Staff - 7 years, 2 months ago

12 out of 16 men came with their wives.. 2*12>16.. simple as that..

Ayan Bhuyan - 7 years, 2 months ago

What about (8, 19, 73)? W+M+C = 8+19+73 = 100 4W+5M+1C = 32 + 95 +73 =200

Rahul Deshpande - 7 years, 2 months ago

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1f there are 19 men and at least half brought their wives then there must be at least 10 women. 8 is too few.

Liam Ashby - 7 years, 2 months ago

at least half woman

Ankit Agrawal - 7 years, 2 months ago

The question says #of women is at least half of the # of men. here, 8<19/2, so it can't be.

Billal Rony - 7 years, 2 months ago

check dis out... it satisfies... Men=22 Women=4 Children=74

Sushant Jain - 7 years, 2 months ago

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It doesn't follow "At least half the men working came with their wives."

Prakkash Manohar - 7 years, 2 months ago

If 72.727 is the no. Of childrew the no rounding off shld come as 73 whch will give 18 men and 9 women

Anshuman Singh - 6 years, 11 months ago
Kartikay Kaushik
Mar 20, 2014

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.

Interesting take on the question instead of the typical lengthy approach.

Venkata Subramanian R - 7 years, 2 months ago
Debanjan Basu
Mar 20, 2014

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!

That guy is paying men about 8 cents per day. I have a feeling fair labour is not one of their priorities.

A Former Brilliant Member - 7 years, 2 months ago

Posting a note does not take an offence away, but I won't offend you!

Prakkash Manohar - 7 years, 2 months ago
Chew-Seong Cheong
Nov 23, 2014

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

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

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

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

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

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

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

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

Yogesh Kumar
Apr 2, 2014

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.

Ayan Bhuyan
Mar 24, 2014

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

Lucky Leo Ambal
Mar 21, 2014

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.

Leo Jiang
Mar 20, 2014

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.

Maqsood Alam
Mar 20, 2014

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)

Anand Shah
Mar 19, 2014

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

Nagendra Babu
Mar 19, 2014

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

men 22 women 22 child 02 may be a solution

Arindam Marik - 7 years, 2 months ago

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Not possible because 22+22+02 != 100(total)

nagendra babu - 7 years, 2 months ago

this question has multiple solutions. Apart from the ones above M 15 W 10 C 85 is also a solution.

Mankaran Ahluwalia - 6 years, 10 months ago

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