A Question of Rover Numbers

SydneyRoverP6B

SydneyRoverP6B

Well-Known Member
Staff member
In a local Rover car club there are 63 members, 22 of which own a P6, 26 own a P6B and 25 own an SD1. 18 members own both a P6B and an SD1, 4 members own both a P6 and a P6B, 3 members own a P6 and an SD1, while one member owns all three,..ie a P6, a P6B and an SD1.

How many members own only a P6?

How many members own a P6B or an SD1?

Ron.
 
OK, I make it 14 that only own a P6 and you've already said in the question that 26 members have a P6B and 25 members have an SD1.

Otherwise, it's 3 that only own a P6B and another 3 that only own an SD1
 
testrider wrote,..
I make it 14 that only own a P6. and.....it's 3 that only own a P6B and another 3 that only own an SD1
Click to expand...

Hmmm...sorry Paul :( neither answer is correct. Think about it very carefully.

Ron.
 
norsewood wrote,..
One last try, 16 P6, 19 P6B and 5 SD1?
Click to expand...

Hello John,

How many members own only a P6? 16 is indeed the correct answer... :D

How many members own a P6B or an SD1? 19 and 5...sorry John :(

Ron.
 
SydneyRoverP6B said:
norsewood wrote,..
One last try, 16 P6, 19 P6B and 5 SD1?
Click to expand...

Hello John,

How many members own only a P6? 16 is indeed the correct answer... :D

How many members own a P6B or an SD1? 19 and 5...sorry John :(

Ron.
Click to expand...

How can there be 16 people only own a P6?

22 own a P6, 4 members own both a P6 and a P6B, 3 members own a P6 and an SD1, while one member owns all three,..ie a P6, a P6B and an SD1.

There are 4+3+1 = 8 members who own another car as well as a P6. So, 22-8 own only a P6, 22-8 = 14.

If 16 is the correct answer, then surely there should be 24 P6 owners?!
 
darth sidious wrote,..
How can there be 16 people only own a P6?

22 own a P6, 4 members own both a P6 and a P6B, 3 members own a P6 and an SD1, while one member owns all three,..ie a P6, a P6B and an SD1.

There are 4+3+1 = 8 members who own another car as well as a P6. So, 22-8 own only a P6, 22-8 = 14.

If 16 is the correct answer, then surely there should be 24 P6 owners?!
Click to expand...


Hello Darth,

Sorry but there is just 16 members who own only a P6. The 22 members who own a P6 don't own only that type of Rover. The numbers all listed in the question are all correct.

Ron.
 
darth sidious said:
How can there be 16 people only own a P6?


There are 4+3+1 = 8 members who own another car as well as a P6.
Click to expand...

Hi Darth,

it's 16 because the 1 is already in the four and the three as he owns all three types of Rover - lucky sod!

John
 
norsewood said:
darth sidious said:
How can there be 16 people only own a P6?


There are 4+3+1 = 8 members who own another car as well as a P6.
Click to expand...

Hi Darth,

it's 16 because the 1 is already in the four and the three as he owns all three types of Rover - lucky sod!

John
Click to expand...

I've just spotted that I made the same mistake in my first calculations and ended up with 14 too, 16 is indeed correct.

Now to fathom the second part of the conundrum........
 
I don't think that all the club members have a car! There are 63 members (73 cars if you add it up) and 16 of them have only a P6 then it's easy to say that the other 47 (63-16) have a P6B or and SD1 but Ron has already said this is wrong.

So, if there are 51 P6B or SD1 cars (26+25) in the club and 18 of them are owned by the same people then there are 33 members who own them (51-18).

Or am I barking up the wrong tree?
 
testrider wrote,..
I don't think that all the club members have a car! There are 63 members (73 cars if you add it up) and 16 of them have only a P6 then it's easy to say that the other 47 (63-16) have a P6B or and SD1 but Ron has already said this is wrong.

So, if there are 51 P6B or SD1 cars (26+25) in the club and 18 of them are owned by the same people then there are 33 members who own them (51-18).

Or am I barking up the wrong tree?
Click to expand...

Hello Paul,

Well done,...you have the correct answer... :D However, the process that you used to obtain the answer is actually incorrect.. :( All of the members in the club do own a car. As we know there are 16 members who own only a P6. Now there are also 5 members who just own an SD1 and another 5 members who just own a P6B. These members are all distinct. Now we have the group of members who own more than one car. There are 4 members who own a P6 and a P6B. There are 18 members who own a P6B and an SD1 and there are 3 members who own both a P6 and an SD1 and one member who owns all three.

Now there are 33 members who own a P6B or an SD1. An explanation on the meaning and the application of "or" is in order here. In the english language the word "or" has two interpretations. Inclusive or and exclusive or. In the case of the former this means one or the other or both, and this application of "or" is always used in mathematics unless otherwise advised. In the case of the other "or" the exclusive or, this means one or the other but not both. So the number of members who own a P6B or an SD1 are those members who own either a P6B or an SD1 or they own both a P6B and an SD1.

The best way to answer this question is by using a Venn Diagram.

Well done!

Ron.
 
Hello Darth,

Draw up a Venn diagram, rectangle enclosing three overlapping circles and fill in the information, and then solve for the missing values. :wink:

You'll kick yourself when you see it really is 16!

Ron.
 
In a local Rover car club there are 63 members, 22 of which own a P6, 26 own a P6B and 25 own an SD1. 18 members own both a P6B and an SD1, 4 members own both a P6 and a P6B, 3 members own a P6 and an SD1, while one member owns all three,..ie a P6, a P6B and an SD1.

How many members own only a P6?

How many members own a P6B or an SD1?

Ron.
Click to expand...

Ok, I think I see how to come to 16 owning only a P6:-

4 members own a P6 and P6B, 3 members own a P6 and SD1, and 1 lucky git owns a P6, P6B and an SD1.

The lucky git is included in the 4 who own a P6 and P6B, and also included in the 3 who own a P6 and SD1, so there are 3 who own a P6 and P6B only, 2 who own a P6 and SD1 only, and 1 lucky git who owns all three.

There are therefore 3 + 2 + 1 = 6 people who own a P6 with something else. If 22 own a P6 and 6 own a P6 with something else, there are 22 - 6 = 16 people who only own a P6. I get the 16 answer now.

There are 26 with a P6B, 25 with an SD1, and 18 with a P6B and an SD1, (# with P6B) + (# with SD1) - (# with both P6B and SD1) = 26 + 25 - 18 = 51 - 18 = 33

P6B OR SD1 = 33, P6 only = 16.
 
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