Workers

Probability Level 2

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

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

It that according to inclusive and exclusive principle?

loyal girl - 1 year ago

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

$x$ - plumbers and electricians

$y$ - painters and electricians

$z$ - painters and and plumbers

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

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

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

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

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

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

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.

Workers

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