A class has 150 students numbered from 1 to 150. All even numbered stu

Question

Competition Mode Question

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

A. 37
B. 45
C. 51
D. 62
E. 99

nick1816 · Accepted Answer

Closest multiple of LCM(2,5,7) to 150 is 140

Number of co-primes of 2, 5 or 7 from 1 to 140 =  140(1-\frac{1}{2})*(1-\frac{1}{5}) * (1-\frac{1}{7}) = 48

Number of co-primes of 2, 5 or 7 from 141 to 150= 3 {141,143 and 149}

Number of the students are in none of these clubs = 51

OR

Number of students in the Math club =   = 75

Number of students in the Physics club =   = 30

Number of students in the Astronomy club =   = 21

Number of students in the Math club and Physics club =   = 15

Number of students in the Physics club and Astronomy club =   = 4

Number of students in the Math club club and Astronomy club =   = 10

Number of students in all 3 clubs =   = 2

Number of the students are in none of these clubs = 150- (75+30+21-15-4-10+2) = 51

eakabuah · Answer

Number of Students in Math Club = 150-2/2+1=75
Number of Students in Physics Club = 150-5/5+1=30
Number of Students in Astrology Club = 147-7/7+1=21
Number of Students in both Math and Physics Clubs = 150-10/10+1=15
Number of Students in both Math and Astrology Clubs = 140-14/14+1=10
Number of Students in both Physics and Astrology Clubs =140-35/35+1=4
Number of Students in all three clubs = 140-70/2+1=2

150=None+75+30+21+2-15-10-4
None=150-99 = 51

The answer is C.

TomGRB · Answer

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

Maths club = Even numbers (from 1 to 150) = 75
+
Physics club (excluding members also in maths club) = Odd multiples of 7 (from 1 to 150) = 11
+
Astronomy club (excluding members in maths or physics club) = Odd multiples of 5 not divisible by 7 (from 1 to 150)= 13

Total number of students in club = 75+11+13 = 99

Students in no clubs = 150-99= 51

Answer = C

freedom128 · Answer

Total class = 150
Math ppl (#2,4,6,...,150) = 150/2 = 75
Phys ppl (#5,10,...,150) = 150/5 = 30
Astro ppl (#7,14,21,...,147) = 150/7 = ~21

Ppl who do not take Math (#1,3,5,7,...,149) = 150 - 75 = 75

Ppl who do not take Math & Phys (eliminate #5,15,...,145) = 75 - 30/2 = 60

Ppl who do not take Math, Phys & Astro (eliminate # 7,21,...,147, but  remember that we have excluded no. 35 & 105 above ) = 60 - 21/2 + 2 = 50.5 = ~51.

Thus, ppl who do not take Math, Phys & Astro = 51

FINAL ANSWER IS (C)

Posted from my mobile device

ArunSharma12 · Answer

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

A. 37
B. 45
C. 51
D. 62
E. 99

MC = 150/2 = 75
PC = 150/5 = 30
AC = 150/7 = 21

total students in all three clubs =  75 + 30 + 21 - (\frac{150}{10}+\frac{150}{35}+\frac{150}{14}) + \frac{150}{2*5*7} 
total students in all three clubs = 128 - 29 = 99
number of students in none of clubs = 150 - 99 = 51
Ans: C

Krishh9119 · Answer

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

For three sets A, B and C, n(AᴜBᴜC) = n(A) + n(B) + n(C) – n(A∩B) – n(B∩C) – n(C∩A) + n(A∩B∩C)  
150 students
Set A: Math : All even numbered: 150/2=75
Set B: Physics: all divisible by 5: 150/5=30
Set C: Astronomy: all divisible by 7: 147/7= 21
n(A∩B): all number divisible by 10 i.e. (5*2)= 150/10=15
n(B∩C): all number divisible by 35 i.e. (5*7)= 150/35=4
n(C∩A): all number divisible by 14i.e. (2*7)= 150/14=10
n(A∩B∩C): all number divisible by 70.e. (2*5*7)= 150/70=2

n(AᴜBᴜC) = 75+30+21-15-4-10+2 =>126-29+2=>126-27=99

As we need none of the n(AᴜBᴜC), we subtract it from total:
150-99=51

C is our answer.

umasarath52 · Answer

Basically, question can be re-written as the number of digits that are not divisible by 2, 5 and 7 for the numbers between 1 to 150

Formula for counting numbers between to given numbers =  \frac{last - first}{difference} +1

Number of Odd numbers (1,3,5,7...149) =  \frac{149-1}{2} +1=75
Odd numbers that are divisible by 5 (5,15,25,....145) =  \frac{145-5}{10} +1 = 15
Odd numbers that are divisible by 7 (7,21, 35 ,49,63,77,91, 105 ,119,133,147) = 9 {striking off 35 and 105 as these are already removed from above list divisible by 5}

Number of students that are not part of any group is 75-15-9 = 51

Answer C

Archit3110 · Answer

out of 150 
total even in maths ; 150/2 ; 75
total Physics club ; 150/5 ; 30 i.e 15 odd multiples of 5
and total Astronomy ; 150/7 ; 21 ; remove 10 even and 2 multiple of 5 ; left with 9
so total 75+15+9 = 99
150-99 ; 51
OPTION C

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

A. 37
B. 45
C. 51
D. 62
E. 99

exc4libur · Answer

T=A+B+C-both-2mid+neither

math (A): div by 2, 148-2/2+1=75
psyh (B): div by 5, 145-5/5+1=30
astro (C): div by 7, 147-7/7+1=21

mid:
math/psyh/astro: div by 2*5*7, {70 140}=2

both (=intersection-g):
math/psyh: div by 2*5, 150-10/10+1=15-(2)=13
math/astro: div by 2*7, 140-14/14+1=10-(2)=8
psyh/astro: div by 5*7, {35 70 105 140}=4-(2)=2

150=75+30+21-(13+8+2)-2(2)+neither
150-126=-23-4+neither
neither=24+27=51

Ans (C)

unraveled · Answer

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

A. 37
B. 45
C. 51
D. 62
E. 99

Students with even numbers - Math Club =  \frac{150}{2}  = 75
Students left(with odd numbers) = 150 - 75 = 75
Students with odd numbers divisible by 5 - Physics Club =  \frac{150}{2*5}  = 15
Students now left = 75 - 15 = 60
Students with odd numbers divisible of 7 - Astronomy Club(excluding divisible by 5) =  \frac{147-7}{7}  + 1 - 10 - 2 = 9
Students left who are not member of any club = 60 - 9 = 51

Answer C.

monikakumar · Answer

A class has 150 students numbered from 1 to 150. All even numbered students are in a Math club; all with a number divisible by 5 are in a Physics club; and all with a number divisible by 7 are a Astronomy club. How many of the students are in none of these clubs?

A. 37
B. 45
C. 51
D. 62
E. 99

A. 37
B. 45
C. 51
D. 62
E. 99

75 numbers divisible by 2,
30 numbers divisible by 5, but there will be multiplier of 2 equally already considered, so 15
21 numbers divisible by 7, multiplier of 2 will be 10, multiplier of 5 will be 2, so 9
75+15+9=99
150-99=51
Ans C

lacktutor · Answer

Math  - 2,4,6...150 ( 75  students)
 Physics - 5,10,15...150 (  30  students)
 Astronomy  - 7,14,21...147 (  21  students)

---> LCM (Math, Physics)= LCM(2,5)= 10 
150/10 =  15  students (taking both  Math and Physics )

---> LCM (Math, Astronomy)= LCM(2,7)= 14 
150/14 ≈  10  students (taking both  Math and Astronomy )

---> LCM (Physics, Astronomy)= LCM(5,7)= 35 
150/35 ≈  4  students (taking both  Physics and Astronomy )

---> LCM (Math,Physics and Astronomy)= LCM(2,5,7)= 70 
150/70 ≈  2  students (taking all  Math, Physics and Astronomy  classes).

75+30 +21 -(15-2)- (10-2)- (4-2) -2*2+ Neither = 150
126- 13-8-2-4+ Neither = 150
--> Neither = 150 -99= 51

Answer (C).

unraveled · Answer

nick1816 
Can you shed some light on the highlighted text.

nick1816 · Answer

bhai it's a quicker way to know number of co-primes to a number that are less than that number.

N= a^x*b^y*c^z

φ(N) = N* (1-\frac{1}{a})(1-\frac{1}{b})(1-\frac{1}{c})

φ(N) = number of coprimes of N from 1 to N

a, b and c are prime factors of N.

unraveled · Answer

nick1816 Thanks for explaining that. I could a get sense out it. So, this is how i'm relating φ(N) to my understanding. (1-\frac{1}{a}) gets you half the odd numbers (1-\frac{1}{b}) gets you non-multiples of 5 (1-\frac{1}{c}) gets you non-multiples of 7 We are doing the same thing but this is a nicer and better way that'll come handy in the moment needed. Awesome

Kudos once again.

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A class has 150 students numbered from 1 to 150. All even numbered stu

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