In a group of civil, mechanical and electrical engineers, all but 20

Question

In a group of civil, mechanical and electrical engineers, all but 20 are civil engineers, all but 15 are mechanical engineers and all but 25 are electrical engineers. How many of them are electrical engineers?

A. 5
B. 10
C. 15
D. 20
E. None of these

Show Answer

Official Answer

A

A. 5
 B. 10
 C. 15
 D. 20
 E. None of these

Sabuj40040 · Accepted Answer

M+E=20 ->(1)
C+E=15 ->(2)
C+M=25 ->(3)
(3)-(2)-> M-E=10 ->(4)
(1)+(4)-> M=15 Therefore 15+E=20.
Therefore, E=5

DJ209 · Answer

All but 20 are civil engineers → So 20 are not civil → These 20 are mechanical or electrical

All but 15 are mechanical engineers → So 15 are not mechanical → These 15 are civil or electrical

All but 25 are electrical engineers → So 25 are not electrical → These 25 are civil or mechanical

So:

Not civil = 20

Not mechanical = 15

Not electrical = 25

We are being asked: How many are electrical engineers?
Let’s add all "not" values:

Not Civil+Not Mechanical+Not Electrical=20+15+25=60
Not Civil+Not Mechanical+Not Electrical=20+15+25=60
But each person misses only one category (they are in 2 categories), so:
If the total number of people is 
x
x, then:
Each person is counted twice in the 60⇒
2x=60 ⇒ x=30
Each person is counted twice in the 60⇒2x=60⇒x=30
So total engineers = 30
Now:
Electrical engineers = total - not electrical = 
30−25=5

​

subhamkmr390 · Answer

Let total of X engineers in the group.

Therefore, Civil = X-20, Mech = X-15, Electrical = X-25.

Also, 
Civil + Mech + Electrical = X 
or, X-20 + X-15 + X-25 = X 
or, X=30.

Therefore, Civil = 30 - 25 = 5.(ANS A)

SizeLog · Answer

Why there can't be an overlap. Like somebody all of three or Electrical or Mechnical?

Bunuel · Answer

The  intended  meaning is that civil, mechanical, and electrical engineers form three non-overlapping groups, but the question never states this.

Therefore, the answer cannot be determined unless the question specifies:

“Each engineer belongs to exactly one of the three categories.”

With that condition added, the answer is A.

egmat · Answer

Hi SizeLog,

Great instinct to question the setup, because the whole solution DJ209 posted only works  because  of one hidden assumption, and it's worth making it explicit.

The assumption is that  each engineer is exactly one type  - civil, mechanical,  or  electrical, never two or three at once. The problem names three separate professions and asks us to split one group of people among them. In these "all but" counting problems, the categories are treated as  mutually exclusive : every person sits in exactly one bucket.

Watch how the wording depends on that. " All but 20  are civil" means: take everyone, remove the civil ones, and exactly  20  are left. Those  20  leftovers have to be  something  - and the only other options are mechanical or electrical. So  M + E = 20 . That sentence only makes clean sense if a person can't be civil  and  mechanical at the same time. If overlaps were allowed, "the non-civil ones" wouldn't be a single clean number you could split between just two groups.

So the three equations -
-  M + E = 20 
-  C + E = 15 
-  C + M = 25

- all rely on that same one-person-one-type rule. Adding them gives  2(C+M+E) = 60 , so the total is  30 , and electrical  = 30 - 25 = 5 .

A quick way to feel it:  imagine a room of only civil and mechanical engineers, no overlap. If "all but  4  are civil," then exactly  4  people are mechanical - simple, because each person is one or the other. The moment you let someone be  both , that clean count falls apart and the puzzle has no single answer. That's exactly why the problem intends no overlap.

So the answer stays  A (5) .

Answer: A

In a group of civil, mechanical and electrical engineers, all but 20

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In a group of civil, mechanical and electrical engineers, all but 20

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