Is there an easier way to solve this problem?

[priyankadhawan223]

This is what I did:

>Total amount of students: 68
History: 25 students
Math: 25 students
English: 34 students
All three classes: 3 students

That means that there are a total of:
25-3=22 people taking history
25-3=22 people taking math
34-3=31 people taking English
and 3 people taking all three
This gives us a total of 78 enrollments (22+22+31+3 = 78)

So we have 78 enrollments, but 68 people. We know that (78-68 = 10) 10 people must be enrolled in at least 2 classes. Therefore, the answer is 10.

[priyankadhawan223]

Total amount of students: 68
>History: 25 students
>Math: 25 students
>English: 34 students
>All three classes: 3 students

That means that there are a total of:
>25-3=22 people taking history
>25-3=22 people taking math
>34-3=31 people taking English
>and 3 people taking all three
>This gives us a total of 78 enrollments (22+22+31+3 = 78)

So we have 78 enrollments, but 68 people. We know that (78-68 = 10) 10 people must be enrolled in at least 2 classes. Therefore, the answer is 10.

[jnelson0612]

Total amount of students: 68
>History: 25 students
>Math: 25 students
>English: 34 students
>All three classes: 3 students

That means that there are a total of:
>25-3=22 people taking history
>25-3=22 people taking math
>34-3=31 people taking English
>and 3 people taking all three
>This gives us a total of 78 enrollments (22+22+31+3 = 78)

So we have 78 enrollments, but 68 people. We know that (78-68 = 10) 10 people must be enrolled in at least 2 classes. Therefore, the answer is 10.

Very nice! Clear and efficient. :-)

[tomas.nmr]

Hi,
I followed the following approach and got the right ans. not sure whether i will end up with the correct ans every time using this approach.

1st of all, I assumed all the students took one course each.
so the number of students becomes= 84
but the given number of students=68
(84-68)=16, these are the students who took more than one courses.
3 students took 3 courses each (2 additional courses each)
So they took 6 additional courses in total
16-6=10
this is the number of students who took one additional course (that means who took exactly two courses)

I love this explanation, thank you so much. I just though i would clarify one small thing for other students. I added some bold in the citation above and I would like to complement you wording there. The number 16, imo, is the amount of extra courses taken, not the students who took more than one course. When you subtract 68 from 84, you count all students as taking one course. You then find that 16 courses are taken by people also taking other courses. You could make an equation like

16=1a +2b where a is the number or students taking one extra course and b is the number of students taking 2 extra courses. Inserting the known number b=3 --> 10=1a --> a=10

You could also solve this problem from the beginning with
"(25+25+34)=68+1a+2b" where a is the number of people taking one extra course and b is the number of people taking two extra courses.

I hope this helps someone. I struggled with this problem so writing this peace really helped me understand the thinking.

[nocheivyirene]

This is easier to answer with Venn Diagram.

let A be the students in History and Math
let B be the students in History and English
let C be the students in Math and English

22-A-B + 22-A-C + 31-B-C + 3 + A + B + C=68
10 = A + B + C

Answer: B

Done in a little less than 2 minutes... with helpful drawing on your work sheet...

[jlucero]

Correct. Just to clarify, you chose 22, 22, & 31 as they were the # of students in each topic minus the ones already counted in all three classes (i.e. history - all three topics = 25-3).

A different usage of the formula that others mentioned above.

[divineacclivity]

Hi, I was using the following approach and made a mistake somewhere along the way. Would be very grateful if any one were to help me figure this out.
Thanks,
Maria

68= A+B+C - A&B-B&C-A&C + A&B&C= 25+25+34 - (all students in 2 classes)+3

thus: 68-84-3=all students in 2 classes -> all students in 2 classes=19

?

hey Maria, the following shd help:

68 = A + B + C - A&B - B&C -C&A - 2 * A&B&C
68 = 84 - (A&B + B&C +C&A) - 2 * 3
(A&B + B&C +C&A) = 84 - 68 -6 = 10

[-2 * A&B&C because while adding A,B,C A&B&C was counted thrice, so to balance subtract it 2 times; for better understanding, please refer to the venn diagram at: http://www.google.co.in/imgres?imgurl=h ... A&dur=1150]

[tim]

i can't tell if there's a question here. if so, please make it more explicit..

[divineacclivity]

No, no, I just wanted to answer. No questions on this one from my side :)
I just edited to make my post look like an answer to an earlier post :)

[jnelson0612]

No, no, I just wanted to answer. No questions on this one from my side :)
I just edited to make my post look like an answer to an earlier post :)

Oh, good! We love answers and students helping other students! :-)

[divineacclivity]

:)

[tim]

:)