Sets? "56 students got part 1 of a driving test correct..."

[miamivince]

56 students got part 1 of a driving test correct, 41 part 2, and 40 part 3. 37 got 2 or more correct, 13 got all 3 parts correct. Nobody failed all 3 parts. How many people took the tests ?

I deduced that 24 got exactly 2 correct. I drew 3 circles which intersected each other . t, V U are the values that got only part 1, par t2 and part 3 correct. N, P and R are the values of the number of people that got 1 and 2, 2 and 3, 1 & 3 correct.

n - t = 43, p-v = 28, r-u =27. This is where I got stuck. The soln is 87 ? Thanks.

 

[Ishuda]

56 students got part 1 of a driving test correct, 41 part 2, and 40 part 3. 37 got 2 or more correct, 13 got all 3 parts correct. Nobody failed all 3 parts. How many people took the tests ?

I deduced that 24 got exactly 2 correct. I drew 3 circles which intersected each other . t, V U are the values that got only part 1, par t2 and part 3 correct. N, P and R are the values of the number of people that got 1 and 2, 2 and 3, 1 & 3 correct.

n - t = 43, p-v = 28, r-u =27. This is where I got stuck. The soln is 87 ? Thanks.

Draw three overlapping circles. The part where they all overlap [all passed] contains 13. The three parts where 1 & 2 overlap, 1&3 overlap, and 2&3 overlap [those that got only two correct] is 37-13^*. So now count up the individual parts that got part 1 correct [56], part two correct [41], and part 3 [40] correct and subtract the overlaps.




^*Are you sure that it 37 who got 2 or more correct, and not 37 who got only 2 correct.

 

[Dale10101]

Another point of view

56 students got part 1 of a driving test correct, 41 part 2, and 40 part 3. 37 got 2 or more correct, 13 got all 3 parts correct. Nobody failed all 3 parts. How many people took the tests ?

I deduced that 24 got exactly 2 correct. I drew 3 circles which intersected each other . t, V U are the values that got only part 1, par t2 and part 3 correct. N, P and R are the values of the number of people that got 1 and 2, 2 and 3, 1 & 3 correct.

n - t = 43, p-v = 28, r-u =27. This is where I got stuck. The soln is 87 ? Thanks.

From this point of view:

On list L1 are the 56 names of the individuals that passed part 1.
On list L2 are the 41 names of the individuals that passed part 2.
On list L3 are the 40 names of the individuals that passed part 3.

All students have their name on at least one of the lists since no student failed all three parts of the test.

Counting all names on the three lists gives 56 + 41 + 40 = 137 names, but, some names are on two lists and some are on all three lists.

If 37 names are on two lists, i.e we counted them twice, we subtract 37 from 137, 137 - 37 = 100.

Now the problem is that we still have 13 people who have a name that must be removed a second time since they are still duplicated on the overall list (that is, we originally counted them three times rather then once or twice). 100 - 13 = 87.

(I suppose that if more then 37 people were on the list twice the maximum number of duplicates would be 137/2 = 68 matches with the remaining person unmatched, 137 - 68 = 69. One would still need to subtract 13 to get 69 - 13 = 56. I think.)

 

[Ishuda]

From this point of view:

On list L1 are the 56 names of the individuals that passed part 1.
On list L2 are the 41 names of the individuals that passed part 2.
On list L3 are the 40 names of the individuals that passed part 3.

All students have their name on at least one of the lists since no student failed all three parts of the test.

Counting all names on the three lists gives 56 + 41 + 40 = 137 names, but, some names are on two lists and some are on all three lists.

If 37 names are on two lists, i.e we counted them twice, we subtract 37 from 137, 137 - 37 = 100.

Now the problem is that we still have 13 people who have a name that must be removed a second time since they are still duplicated on the overall list (that is, we originally counted them three times rather then once or twice). 100 - 13 = 87.

(I suppose that if more then 37 people were on the list twice the maximum number of duplicates would be 137/2 = 68 matches with the remaining person unmatched, 137 - 68 = 69. One would still need to subtract 13 to get 69 - 13 = 56. I think.)

That was the reason for my question ^*Are you sure that it 37 who got 2 or more correct, and not 37 who got only 2 correct." If it were 37 who got only two correct, then as you mention we did count those twice and need to subtract from 137 to get 100 people. But we also need to subtract the 13 who got three right since e also counted them twice and, as you mention, 100-13=87.

 

[Dale10101]

Yikes

That was the reason for my question ^*Are you sure that it 37 who got 2 or more correct, and not 37 who got only 2 correct." If it were 37 who got only two correct, then as you mention we did count those twice and need to subtract from 137 to get 100 people. But we also need to subtract the 13 who got three right since e also counted them twice and, as you mention, 100-13=87.

Now I think I see three ways the problem could be interpreted, well, I think the bush is well beaten anyway.