In a class of 100 students, 80 passed Physics, 70 passed Chemistry, an

In a class of 100 students, 80 passed Physics, 70 passed Chemistry, and 40 passed Math. If 10 students failed in all the three subjects, at least how many of the students passed all the three subjects?

(A) 0
(B) 5
(C) 10
(D) 20
(E) 25


Out of 100 students 10 failed in all 3 subjects. So, 90 students passed at least one of the subjects - Physics, Chemistry or Math.
We need to find the minimum value of the students passed in all 3 subs or the min. value of the overlapping of the 3 sets having - 80, 70 and 40 students, where the total sh be 90.
80 : 10
|-----------------------------------------------------|---------| 90

20 : 70
|---------------|-----------------------------------------------| 90

40 : 50
|----------------------------|----------------------------------| 90

The min. overlap of 80 and 70 is: 80+70-90 = 60.
The min. overlap of the common 60 and 40 is: 60+40-90 = 10.
This is the answer. C.

In a class of 100 students, 80 passed Physics, 70 passed Chemistry, and 40 passed Math. If 10 students failed in all the three subjects, at least how many of the students passed all the three subjects?

Total = A + B + C - (Sum of 2 overlaps) - 2(all 3 overlaps) + Neither
100 = 80 + 70 + 40 - (Sum of 2 overlaps) - 2(all 3 overlaps) + 10
100 = 200 - (Sum of 2 overlaps) - 2(all 3 overlaps)

(Sum of 2 overlaps) + 2(all 3 overlaps) = 100 --- (1)
Also, max value of (Sum of 2 overlaps) + (all 3 overlaps) = 90 --- (2)
(1) - (2)
All 3 overlaps = 10.

Number of students who passed all 3 subjects = 10

Answer: C

Please refer to the attached Picture for the solution







Saunak Dey

Total students passed in atleast one subject = Total number of students - students failed in all 3 subjects
= 100 - 10 = 90
To find the least students passed in all 3 subjects (I used process eliminating each answer)
see attached venn diagrams


Answer (C)

Total Students = 100

Total passed = Total - failed = 100 - 10 = 90

Now the question is what is the minimum number of students passed in all three subjects?

Here our aim is to plug in such nummbers so that sum of total students passed is 90 and students passed in a subject individually remains what is given in question.

We notice, Choices A to E are in ascending order and we need the least number. so we will start plugging in the values from A to E.


Option - A, Passed in All three subjects = 0

1. Passed in Physics + Chemistry = 50
2. Passed in Chemistry + Maths = 20
3. Passed in Maths + Physics = 20
4. Passed in All three subjects (P,C,M) = 0

In this we notice that students passed in Physics = 1 + 3 + 4 = 50 + 20 + 0 = 90. But It is given that In physics only 80 passed.
Hence, "A" is wrong.

Option - B, Passed in All three subjects = 5

1. Passed in Physics + Chemistry = 50
2. Passed in Chemistry + Maths = 15
3. Passed in Maths + Physics = 20
4. Passed in All three subjects (P,C,M) = 5

In this we notice that students passed in Physics = 1 + 3 + 4 = 50 + 20 + 5 = 75. But It is given that In physics only 80 passed.
Hence, "B" is wrong.

Option - C, Passed in All three subjects = 10

1. Passed in Physics + Chemistry = 50
2. Passed in Chemistry + Maths = 10
3. Passed in Maths + Physics = 20
4. Passed in All three subjects (P,C,M) = 10

In this we notice that students passed in Physics, chemistry and maths are equal to given values and sum of all passed students is 90.
Hence, "C" is Correct.

We do not need to test D and E because question states "at least".

My solution: ( C is the correct answer)

80 +70 + 40 - (numbers of those passed only 2 subjects (A) + 2* numbers of those passed 3 subjects (B)) = 100 - 10 =90
=> A+2*B = 100

Trials and errors: The question asks about the least possible value, so we start from the smallest.

(1) B = 0 => A = 100. But with 80, 70, and 40 students passed Physics, Chemistry, and Maths, the largest number of those passed only 2 subjects is 95 (max A =95) (80+15), as below

=> (1) INCORRECT

(2) B=5 => A=90. Since 5 students passed all 3 subjects, the remains are 75 Physics, 65 Chemistry, and 35 Maths. Same to (1), the max A = 87 (75+12), as below

=> (2) INCORRECT

(3) B=10 => A=80. Since 10 students passed all 3 subjects, the remains are 70 Physics, 60 Chemistry, and 30 Maths. Now it is possible to have 80 students passed only 2 subjects (70+10), as below


So (C) is CORRECT

QUESTION #13:

In a class of 100 students, 80 passed Physics, 70 passed Chemistry, and 40 passed Math. If 10 students failed in all the three subjects, at least how many of the students passed all the three subjects?

(A) 0
(B) 5
(C) 10
(D) 20
(E) 25

Answer C

100 = P + C + M - (Sum of 2 subjects passed) - 2(all three subject passed) + Neither
100 = 80 + 70 + 40 - (2 subject) - 2(3 subject) + 10
(2 subject) + 2(3 subject) = 100

Also, passed in at least one subject is 90, i.e., passed in 1 subject + passed in 2 subject + passed in 3 subject = 90
To maximize pass in at least 2 subject, passed in 1 subject = 0, i.e., passed in 2 subject + passed in 3 subject = 90

From the above two equation, we can get passed in all three subject = 10.

In a class of 100 students, 80 passed Physics, 70 passed Chemistry, and 40 passed Math. If 10 students failed in all the three subjects, at least how many of the students passed all the three subjects?

(A) 0
(B) 5
(C) 10
(D) 20
(E) 25

Ans:C ahh I marked the wrong one again ---I marked D in a hurry.
The answer can be arrived with the help of a formula--->
Total=A+B+C -(sum of exactly 2 groups overlap)- 2(All three) +Neither
100=80+70+40- (A)-2(B)+ 10
-100=-A-2B
100=A+2B
we need to find B. It is to found out at least how many students passed in all.
Sum of exactly two groups overlaps can be maximum 80 because it cannot be greater than the number of students passing in a single subject.
so, 100=80+2B
B=10
Answer is C.
Will I get points for correct explanation, even my answer was wrong.

Answer: option C

Detailed explanation is as follows

Can anyone please explain the area "Also, max value of (Sum of 2 overlaps) + (all 3 overlaps) = 90 --- (2)" How can we say this?

If you tend to get confused with formulas then refer the following explanation

Data:
Number of students passed in each subject:
Physics = 80
Chemistry = 70
Maths = 40

Number of students failed in all subjects = 10

Solution:

Total number of students passed = 100 - 10 = 90

Failed students in each subject:
Physics = 90 - 80 = 10
Chemistry = 90 - 70 = 20
Maths = 90 - 40 = 50

Total students who failed in at least one subject = 80

Number of students who passed in all subjects = 90 - 80 = 10

Answer is C.

Posted from my mobile device

Out of 100 people, 10 fail all 3 subjects

That means the remaining 90 must pass at least one subject.

Further, we have (80 + 70 + 40) = 190 recorded “passes” that we have to account for among the 90 people.


(1st) satisfy the minimum condition.

Here, since we know the 90 remaining students have to pass at least one course, we can assign 90 of the 190 recorded “passes” to every student.

Right now:
# of students who pass only 1 = 90

And

Remaining # of recorded “passes” we need to account for = (190 - 90) = 100 left


(2nd) To minimize the number of students who pass all 3, we want to maximize the number of students who pass exactly 2 subjects.

We can give all 90 students another (+1) recorded “pass” each.

Right now:

# of students who passed exactly 1 course = 0 (shifted to exactly 2)

# of students who passed exactly 2 courses = 90

Remaining # of recorded “passes” we need to account for = (100 - 90) = 10

The only place we can put these 10 more recorded “passes” is with 10 of the students who already have exactly 2.


Minimum number of students who must have passed all 3 = 10

(And 80 will pass exactly 2)

Posted from my mobile device

GMATinsight
Can u tell me plz how did u get the first line 2(a+b+c)+d+e+f=80?

Posted from my mobile device

Bunuel
KarishmaB
chetan2u
GMATinsight

What if the question asked "at most how many of the students passed all the three subjects?"

Thank You!

Posted from my mobile device

Hi,
I will use the official solution to answer your query.

We will represent the total number of 100 students on the number line as shown



we will start from the left to right and place the group with largest number .i.e physics in this case and the place the group with second largest number .i.e chemistry in this case from the right to left. Also it is mentioned that 10 students did not pass in any subject so we will keep them separate



Now, we need to place the 40 students who passed the math subject among the 90 students as shown in the number line so that the over lap of all the three subjects is the MAXIMUM. We can place these 40 from 20 to 80 so that all these 40 students are part of the group consisting of students passing in all three object. Thus, MAX is 40.

The question can be easily solved if we think in terms of number of people and no of instances.
We have 100 people of whom 10 belong to none of the three sets. So 90 people belong to at least one set.

No of instances of passes = 80+70+40 = 190 instances
Since 90 people will take up 90 instances of one pass, we are left with 100 extra instances for overlap. When we give one of these instances to one of the 90 people, we get an overlap of 2 sets (since each person already has one instance). When we give 2 of these instances to one of the 90 people, we get an overlap of 3 sets.

Minimize overlap of all 3:
To minimise overlap of 3, we must maximise the overlap of 2 by giving 90 of the 100 instances to 90 people. We are still left with 10 instances which need to be given to 10 people. This means at least 10 people must belong to all 3 sets.

Maximize overlap of all 3:
Of the 100 extra instances, if we were to give 2 each to 50 people, 50 people would belong to all three sets. But note that only 40 people belong to Math. This means we must give 80 extra instances, 2 each to 40 people so that maximum 40 people belong to all 3 sets and 20 leftover extra instances will overlap with 20 other people.