In an office, where working in at least one of the three departments i

Question

In an office, where working in at least one of the three departments is mandatory, 78% of the employees are in operations, 69% are in finance and 87% are in HR. Whats is the maximum percentage of employees that could have been working in all three departments.

A. 69
B. 45
C. 66
D. 67
E. 34

A. 69
B. 45
C. 66
D. 67
E. 34

Hal90 · Accepted Answer

Maximum % working in all 3 = assume no one works in 2 departments

Assume x = % working all 3

100% = (78% - x%) + (69% - x%) + (87% - x%) + x%

100% = 234% - 2x%

134% = 2x%

X = 67%

IMO Ans is D

Posted from my mobile device

nick1816 · Answer

My method is almost similar to yours 
Let x percent works for one department, y percent works for 2 departments and z percent works for 3 departments
x+y+z=100.....(1)
and x+2y+3z= 78+69+87=234
x+2y+3z=234........(2)
subtract equation (1) from equation (2), we get
y+2z=134
We have to maximize z, hence put y=0
2z=134
z=67

Chethan92 · Answer

Concept:

X_{min} = A+B+C-2*Total

If  X_{min}  is less than zero, then  X_{min}  of the set is zero.

For   X_{max} ≤ Min(A,B,C)

If  X_{max}  is more than the total,i.e, 100% in this problem, then  X_{max}  is the Min(A,B,C)

If  X_{max}  is less than the total, we have to make an adjustment.

In this example, Min( Op, HR, Fin) = 69.
And the other values in the set for only Operations is 78-69 = 9
Only in HR = 87-69 = 18
As all the other values in the set should be zero.

69+9+18 = 96, which is less than a total of 100%.
Hence we need to make an adjustment.
100-96 = 4, this has to be equally divided into the set of only operations and HR. 4/2 = 2% should be added to OP and HR or 2% should be subtracted from the max of X. 69-2 = 67%

D is the answer.

exc4libur · Answer

three-overlapping sets: 
T=A+B+C-both-2(mid)+neither
T=A+B+C-both-2(mid)+0 ("working in at least one of the three departments is mandatory")
1=.78+.69+.87-both-2mid
-1.34=-both-2mid (minimize both to 0 to maximize mid)
-1.34/-2=mid
mid=0.67=67%

Answer (D)

Fdambro294 · Answer

Pretend there is 100 people instead of 100 percent.

All 100 people are lined up in front of you. Every time a person is part of 1 of the 3 groups they raise their hand and they get a piece of candy.

After the 3 questions are asked (who works in OP? Who works in HR? Who works in FIN?):

The total number of times a hand gets raised is = (78 + 69 + 87) = 234 times

There are only 100 people. All 100 do at least one of the three jobs. Obviously some people did more than 1 job.

(1st)Satisfy the Minimum Condition

We have 100 people ———234 pieces of candy to hand out

Since each person must do at least one job, we hand our 1 piece for each of the 100 people.

Right now: 100 ppl —— have 1 candy

234 - 100 = 134 pieces left to hand out

(2nd)we want to maximize the amount of people who do all 3 jobs

The upper limit is given by the smallest set total (69). Since only 69 do finance, we can’t have more than 69 do all 3

Right now we still have 134 pieces to hand out to the 100 ppl. To MAX the number of ppl who get 3 in total, we will hand out +2 pieces to as many people as possible

(134 left) / 2 ————> 67

67 ppl can do all 3 jobs

Remaining 33 do 1 job only

Answer: 67%

Posted from my mobile device

TarunKumar1234 · Answer

100 = 78 + 69 + 87 - n(AUB)- n(BUC)- n(AUC) + A∩B∩C
or, 100 = 234 - (only n(AUB)+ A∩B∩C) + (only n(BUC)+ A∩B∩C) - (only n(AUC)+ A∩B∩C) + A∩B∩C

To maximize A∩B∩C, we need to consider only n(AUB) = only n(BUC) = only n(AUC) = 0

So, 2*A∩B∩C = 134 or, A∩B∩C = 67.

So, I think D.

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