Workers

At a construction site, Jorge is in charge of hiring skilled workers for the project. Out of 80 candidates that he interviewed, he found that

  • 45 were painters,
  • 50 were electricians,
  • 50 were plumbers,
  • 15 had skills in all three areas, and
  • all of them had skills in at least one of these areas.

If he hired everyone who was skilled in exactly 2 areas, how many candidates were hired?

35 45 55 65

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4 solutions

Naveen Rome
Jun 16, 2014

Since 15 are skilled in all 3. Subtract 15 from all three to get a total with Single skilled and double skilled workers including the duplicates

Painters = 45 - 15 = 30 Elect. = 50 - 15 = 35 Plumb. = 50 - 15 = 35

Total = 100. --> This is a total set of Single and double skilled workers including duplicates.

Now coming to real number, Out of 80, 15 were skilled in three areas.

So, 80 - 15 = 65 (Actual no of workers skilled with single and both skills)

Now the difference between the number with out duplicates (65) and with duplicates (100)

100 - 65 = 35

So, Bingo, 35 are Dual skilled.

Solved it in exactly the same way

Tasneem Khaled - 6 years, 7 months ago

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It that according to inclusive and exclusive principle?

loyal girl - 1 year ago

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It's just problem solving, but it relies on the same underlying concept as PIE.

wesley johnson - 1 year ago

45+50+50-2(15)=115 Then 115-80=35

Willa Waliyadin - 4 years, 1 month ago
Unstable Chickoy
Jun 14, 2014

let

x x - plumbers and electricians

y y - painters and electricians

z z - painters and and plumbers

50 ( x + y + 15 ) 50 - (x + y + 15) --->electricians only

50 ( x + z + 15 ) 50 - (x + z +15) --->plumbers only

44 ( y + z + 15 ) 44 - (y + z +15) --->painters only

80 = [ 50 ( x + y + 15 ) ] + [ 50 ( x + z + 15 ) ] + [ 44 ( y + z + 15 ) ] + x + y + z + 15 80 = [50 - (x + y + 15)] + [50 - (x + z +15)] + [44 - (y + z +15)] + x + y + z + 15

x + y + z = 35 x + y + z = \boxed{35}

It is AuBuC=A+B+C-(AintB+BintC+CintA) or x +AintBintC 80=45+50+50-x+15=>x=80 Hence ppl who were exactly skilled in two=80-3*15=35

Debtanu Bhuiya - 6 years, 12 months ago

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most logical and easiest of all!

Wrushabh Warshe - 6 years, 11 months ago
Aditya Pandey
Jun 14, 2014

great question.let no. of electricians be x no. of plumbers be y nd no. of painters be z. now since 15 of them were workers who had skills in all 3 works we want to remove it from 80 because the no. of workers were skilled in only 2areas.so x+y= 20 y+z=15 x+z= 15 we got x+y=20 because out of 65, 45 were painters and so happens for the rest 2 equations so after doing the math we get the values of x,y and z then we add these values and then subtract it from 65 so we get 35

easier solution with a venn diagram... three circles intersect with the common point where all three do=15 where any two meet= a b and c ....rest of un intersected areas left out can be x, y and z... then obtain 3 equations like this a+ b + x+15 = 45 a+ c + y +15= 50 b+c + z +15 = 50

now add all three

2(a+b+c) +x+y+z=100

since total people are 80 so last question is a+b+c+x+y+z+15 =80

substitute x+y+z u get a+b+c = 35

which is the area intersected by any two circles

aroop kundu - 6 years, 12 months ago
Prabir Mondal
Nov 30, 2014

Naveen Rome's solve is my choice.

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