gig92 wrote:1) In a referendum about three proposals, 78% of the people were against at least one of the proposals. 50% of the people were against proposal n°1, 30% against proposal n°2, and 20% against proposal n°3. If 5% of the people were against all the three proposals, what percentage of people were against more than one of the three proposals?

What do you think?

78% were against at least one; and
against 1&2&3=5

against 1 + against 2 + against 3 + against 1&2 + against 2&3 + against 1&3 + against 1&2&3 = 78
against 1 + against 2 + against 3 + against 1&2 + against 2&3 + against 1&3 = 73 --- eqn (1)


against 1 + against 1&2 + against 1&3 + against 1&2&3 = 50
against 2 + against 1&2 + against 2&3 + against 1&2&3 = 30
against 3 + against 1&3 + against 2&3 + against 1&2&3 = 20

against 1 + against 1&2 + against 1&3 = 45
against 2 + against 1&2 + against 2&3 = 25
against 3 + against 1&3 + against 2&3 = 15

add the three equations

against 1 + against 2 + against 3 + 2 (against 1&2 + against 1&3 + against 2&3) = 85 ---eqn 2

from eqn 1 and 2

against 1&2 + against 1&3 + against 2&3 = 85 - 73 = 12

% against more than 1 of 3 = against 1&2 + against 1&3 + against 2&3 + against 1&2&3 = 12+5 = 17

gig92 wrote:1) In a referendum about three proposals, 78% of the people were against at least one of the proposals. 50% of the people were against proposal n°1, 30% against proposal n°2, and 20% against proposal n°3. If 5% of the people were against all the three proposals, what percentage of people were against more than one of the three proposals?

I suggest you use a very easy-justifiable equation: AuBuC = A+B+C - (Sum of Exactly 2 of them) - 2.(Exactly 3 of them)

Then you will get (Sum of Exactly 2 of them) = 12, therefore the answer will be 12+5 = 17. Justify all that!

Regards,
Fabio.

gig92 wrote:2) In a club, all the members are free to vote for one, two or three of the candidates. 20% of the members did not vote. 38% of the total members voted for at least two candidates. What percentage of the members voted for either one or three candidates if 10% of the total members who voted, voted for all the three candidates?

Hi there!

Let us consider 100 members; focus: (Sum of Exactly 1) + (Exactly 3)

a) Only 80 of them voted.

b) (Sum of Exactly 2) + (Exactly 3) = 38

c) (Exactly 3) = 10% of 80 = 8 members

Therefore:

>> (Sum of Exactly 2) = 38 - 8 = 30

80 = (Sum of Exactly 1) + (Sum of Exactly 2) + (Exactly 3) , then:

>> (Sum of Exactly 1) = 80 - 38 = 42

Answer: 42 + 8 = 50 (%).

Let's work with the formula:
G1 + G2 + G3 - G12 - G13 - G23 - 2(G123) +N = T

where G! = # in group 1
G12 = number in both 1 and 2
G123 = number in all three
N = number not in any group
T = total

Where we are dealing only with percents, T = 100

For the first problem, with 78% voting against at least one, we are left with N = 22
G1 = 50, G2 = 30, G3 = 20 and G123 = 5

Plugging in we get

50 + 30 + 20 - G12 - G13 - G23 -2(5) + 22 = 100
simplifying we see G12 + G13 + G23 = 12
Total against more than one = the 12% that voted against 2 proposals plus the 5% that voted against all 3 = 17%.


Problem 2 is similar:

G1 = voted for one; G2 = voted for 2; G3 = voted for 3; N = voted for none.
(there is no G12 because no member can vote for both one and two)

G1 + G2 + G3 + N = 100
N = 20
G2 + G3 = 38 (at LEAST 2 candidates means either 2 or 3)
G3 = 10 therefore G2 = 28
G1 + 28 + 10 + 20 = 100
G1 = 42
Question: how many voted for one or three: G1 + G3 = 42 + 10 = 52


Question 3:

The largest possible number that could like both is limited by the number who like cheese = 63%

To determine the least: G1 (apples) + G2 (cheese) - both + neither = 100%

Assume everyone likes something. (If some liked neither, that would only increase the number who had to like both and we want to find its minimum.)

76 + 63 - both + 0 = 100

Let's work with the formula:
G1 + G2 + G3 - G12 - G13 - G23 - 2(G123) +N = T

where G! = # in group 1
G12 = number in both 1 and 2
G123 = number in all three
N = number not in any group
T = total

Where we are dealing only with percents, T = 100

For the first problem, with 78% voting against at least one, we are left with N = 22
G1 = 50, G2 = 30, G3 = 20 and G123 = 5

Plugging in we get

50 + 30 + 20 - G12 - G13 - G23 -2(5) + 22 = 100
simplifying we see G12 + G13 + G23 = 12
Total against more than one = the 12% that voted against 2 proposals plus the 5% that voted against all 3 = 17%.


Problem 2 is similar:

G1 = voted for one; G2 = voted for 2; G3 = voted for 3; N = voted for none.
(there is no G12 because no member can vote for both one and two)

G1 + G2 + G3 + N = 100
N = 20
G2 + G3 = 38 (at LEAST 2 candidates means either 2 or 3)
G3 = 10 therefore G2 = 28
G1 + 28 + 10 + 20 = 100
G1 = 42
Question: how many voted for one or three: G1 + G3 = 42 + 10 = 52


Question 3:

The largest possible number that could like both is limited by the number who like cheese = 63%

To determine the least: G1 (apples) + G2 (cheese) - both + neither = 100%

Assume everyone likes something. (If some liked neither, that would only increase the number who had to like both and we want to find its minimum.)

76 + 63 - both + 0 = 100
Both = 39

Range = 39 to 63

Some prefer to use Venn diagrams for these, but they are impossible to do with typing

gig92 wrote:3) In a survey shows that 63% of people in a town like cheese whereas 76% like apples. If x% of people like both cheese and apples then find the range of x.

The maximum value of x occurs when all 63% of the people that like cheese also like apples, therefore xMax = 63%.

From the fact that: AuB = A+B - (A intersection B), then x = 139 - AuB.

We want to find xMin, therefore we need to maximize AuB... take AuB = 100 (in other words, every person likes cheese or apples)!

Therefore xMin = 39%, and the range asked is xMax - xMin = 63% - 39% = 24%

Hi, Tani.

In question 2 there was a trap...

Tani Wolff - Kaplan wrote:G3 = 10 therefore G2 = 28

The question stem says 10% of the voters, not of all people involved.

Therefore G3 = 10% of 80 = 8, not 10.

Regards,
Fabio.

AHA! A trap - and I fell into it!

Tani Wolff - Kaplan wrote:AHA! A trap - and I fell into it!

Yep, but that does not invalidade your good arguments.

See you in other posts!

Cheers,
Fabio.

fskilnik wrote:

gig92 wrote:3) In a survey shows that 63% of people in a town like cheese whereas 76% like apples. If x% of people like both cheese and apples then find the range of x.

The maximum value of x occurs when all 63% of the people that like cheese also like apples, therefore xMax = 63%.

From the fact that: AuB = A+B - (A intersection B), then x = 139 - AuB.

We want to find xMin, therefore we need to maximize AuB... take AuB = 100 (in other words, every person likes cheese or apples)!

Therefore xMin = 39%, and the range asked is xMax - xMin = 63% - 39% = 24%

The answer is:

[spoiler]0<= x <=39%[/spoiler]

How can we explain it?

gig92 wrote:The answer is: 0<= x <=39%
How can we explain it?

Range means maximum value - minimum value, therefore we are looking for a NUMBER, not for an interval. The question was ill-posed, just to start with the discussion.

As far as the INTERVAL OF POSSIBLE VALUES for x is concerned,

Take:

x = 43%
only cheese = 20%
only apples = 33%
not cheese not apples = 4%

This is a possible scenario (total = 100%), satisfying all conditions posted in the question stem, therefore x COULD BE over 39%. In other words, I´ve proven to you that your "official answer" is wrong, simple as that.

Another good reason to be sure that the "official answer" you have provided is wrong: if x could be zero percent (as you believe), then (only cheese) + (only apples) = 63 + 76 > 100 (%), what is (also) absurd.

Regards,
Fabio.

fskilnik wrote:

gig92 wrote:The answer is: 0<= x <=39%
How can we explain it?

Range means maximum value - minimum value, therefore we are looking for a NUMBER, not for an interval. The question was ill-posed, just to start with the discussion.

As far as the INTERVAL OF POSSIBLE VALUES for x is concerned,

Take:

x = 43%
only cheese = 20%
only apples = 33%
not cheese not apples = 4%

This is a possible scenario (total = 100%), satisfying all conditions posted in the question stem, therefore x COULD BE over 39%. In other words, I´ve proven to you that your "official answer" is wrong, simple as that.

Another good reason to be sure that the "official answer" you have provided is wrong: if x could be zero percent (as you believe), then (only cheese) + (only apples) = 63 + 76 > 100 (%), what is (also) absurd.

Regards,
Fabio.

Thanks, you're right. One cannot always believe a book

3) In a survey shows that 63% of people in a town like cheese whereas 76% like apples. If x% of people like both cheese and apples then find the range of x.


c= cheese
a=apples
ca= like cheese & apples

63%= c+ca
76%=a+ca
_________
139 = c+a+2ca (equation 1 -from adding the 2 first statements)

100% = c+a+ca (equation 2- from knowing that the whole of the population likes cheese, apples or both only)

________
39% = ca (equation 1 minus equation 2)

gig92 wrote:Thanks, you're right. One cannot always believe a book

My pleasure, gig92.

That´s it! As I tell my students, the "only" problem in "blindly trusting" official answers is to forget that a human (therefore imperfect) being was responsible for solving the problem and (perhaps even another one) was responsible for typing its solution!

Just one detail: the range could also be interpreted as the interval of possible values, therefore I should NOT have bothered on that... in other words, if your book answer were "39% <= x% <= 63%", then we could consider it perfect.

Regards,
Fabio.