In a class of 120 students numbered 1 to 120, all even numbered studen

Total = 120

Total who study Physics = 120/2 = 60
Total who study Chemistry = 120/5 = 24
Total who study Math = 120/7 = 17

Students who study all 3 subjects = LCM (2, 5, 7) = 70x = 1
Students who study only Physics and Chemistry = LCM (2,5) = 10x = 12 - 1(students who study all 3) = 11
Students who study only Physics and Math = LCM(2, 7) = 14x = 8 - 1 = 7
Students who study only Chemistry and Math = LCM(5, 7) = 35x = 3 - 1 = 2

Students who study only Physics = 60 - 11 - 7 - 1 = 41
Students who study only Chemistry = 24 - 11 - 2 - 1 = 10
Students who study only Math = 17 - 7 - 2 - 1 = 7

Total who study atleast one subject = 1 + 11 + 7 + 2 + 41 + 10 + 7 = 79

None = Total - Total who study atleast one subject = 120 - 79 = 41

Answer: D

Total number of students: 120
even: 60 -> Odd= 60
Divisible by 5: 120/5=24. Half of them are odd=12.
Divisible by 7: 120/7=17 ( we only care about the integer value). The number of odd numbers is either 8 or 9. To check, multiply 17*7. Since the product is odd, we know that the number of odd numbers is 9.
Divisible by both 7 and 5: 35,70,105. We only want the number of odd numbers=2.

Odd numbers=60
Odd numbers that are divisible by 5=12
Odd numbers that are divisible by 7=9
Odd numbers that are divisible by 7 and 5=2
Number of people who did not take any of the 3 classes=60-12-9+2=41

Our goal is to find the number of students who do not opt for any of the three subjects. We first can find the number of students who do not opt for physics (i.e., eliminate the number of students who opt for it). Then, from those students, we eliminate those who opt for chemistry. Finally, from those who are left (after eliminating physics and chemistry), we eliminate those who opt for math. Thus, the students who are left are those who do not opt for any of the three subjects.

Since all even-numbered students (60 students) opt for physics, we know the odd-numbered students (the other 60 students) do not opt for physics. That is, the students numbered 1, 3, 5, …, 119 do not opt for physics. From these students, we see that the odd multiples of 5 (5, 15, 25, …, 115) opt for chemistry, and thus we have to eliminate them. The number of these students is:

(115 - 5)/10 + 1 = 12

Thus, we have 60 - 12 = 48 students left who do not opt for either physics or chemistry (or both). From these students, we need to eliminate those who are multiples of 7, since they opt for math. The numbers must be odd multiples of 7, namely, 7, 21, 35, 49, 63, 77, 91, 105, and 119. There are 9 such numbers. However, we see that we’ve already counted 35 and 105 since they are odd multiples of 5. Thus, there are 7 odd multiples of 7 that are not odd multiples of 5, and we have to exclude them. Thus, we have 48 - 7 = 41 students left and these students do not opt for any of the three subjects.

Answer: D
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My reasoning:

Total even integers from 1-120 = 60

Total numbers divisible by 5 = 120/5 = 24

Total numbers divisible by 7 = 17 (17*7 = 119 which is the largest number smaller than 120)

Number of even integers that are divisible by 5 is 12. 5 is either odd or even. So in the set of 24, half will be even.

Number of even integers divided by 7 - we must find multiples of 14. So 14 * 8 = 112 (the largest number smaller than 120)

Number of integers divisible by 7 and 5. We need to find multiples of 35. So there's 35, 70, and 105. 3 values total.

Number of even integers divisible by 7 and 5. This is asking for the multiple of 2, 7, and 5. There is only 1 value (70).

From here we can find the students who attend exactly one class. which is 12-1, 8-1, and 3-1, or 11 + 7 + 2 = 20

We can now solve for the Neither value:

60 + 24 + 17 - 20 - 2*1 + Neither = 120

Neither = 41 (Answer choice D)

Total = 120

Physics = 120/2 = 60
Chemistry = 120/5 = 24
Math = 120/7 = 17

all 3 subjects = LCM (2, 5, 7) = 70 = 1
only Physics and Chemistry = LCM (2,5) = 10 multiples till 120 = 12 - 1(3 sub) = 11
only Physics and Math = LCM(2, 7) = 14 multiples till 120 = 8 - 1(3 sub) = 7
only Chemistry and Math = LCM(5, 7) = 35 multiples = 3 - 1(3 subs) = 2

only Physics = 60 - 11 - 7 - 1 = 41
only Chemistry = 24 - 11 - 2 - 1 = 10
only Math = 17 - 7 - 2 - 1 = 7



None =120 - 79 = 41

Answer: D

P = 120/2 = 60 students
C = 120/5 = 24 students
M = 119/7 = 17 students

P&C = LCM(2,5) = 10x ---> 12 students
C&M = LCM(5,7) = 35x ---> 3 students
M&P = LCM(7,2) = 14x ---> 8 students

P&C&M = LCM(2,5,7) = 70x ---> 1 student

Now going by 3 set formula

P + C + M - P&C - C&M - M&P + P&C&M + None = 120
60+24+17-12-3-8+1+None = 120
None = 120 - (60+24+17-12-3-8+1) = 41
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In a class of 120 students numbered 1 to 120, all even numbered students opt for Physics, those whose numbers are divisible by 5 opt for Chemistry and those whose numbers are divisible by 7 opt for Math. How many opt for none of the three subjects?

The total no of students = 120
N(Phy)= 120/2 = 60
N(Chem) = 120/5 = 24
N (Math) = 120/7 = 17

N(Phy and Chem)= 120/lcm(2,5)= 120/10 = 12
N( Phy and Math)= 120/lcm(2,7)= 120/14 =8
N(Math and Chem)=120/lcm(5,7)= 120/35 = 3
N( Phy , Chem and Math)=120/lcm(2,5,7)= 120/70 = 1

By using 3 set formula
N( Phy U Chem U Math)= N(Phy) + N(Chem) + N (Math) - N(Phy and Chem) - N( Phy and Math) - N(Math and Chem) + N( Phy , Chem and Math)
= 60 + 24 + 17 - 12 -8 -3 +1
= 79

No of people who opt for none of the three subjects = Total - N( Phy U Chem U Math) = 120 - 79 = 41

Option D is the answer

Thanks,
Clifin J Francis,
GMAT SME
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We can take Euler's number of 120 directly to find the answer
Calculation is as follows

Euler number of 120 = 120 * (1-1/2) * (1-1/5) * (1-1/7) = 120 * (1/2) * (4/5) * (6/5) = 41.14
Therefore number of students opted for none of the subject will be 41