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Well, This is the hardest maths question I ever came across anyone who gets it right is a damn genious

A class has less than 30 students
Exactly 3/4 of them own footballs
Exactly 7/8 of them own football boots.

i)How many students are there in the class?

ii)How many students own both football boots and footballs?

this really confusing ooo,buh Considering the fact the class have less than 30,..
Should be having 8,16,24..in the class, 6 should have ball, while 7 should have boot, which means 6+7-8 having both ball and boot..

Mathematical:
this really confusing ooo,buh Considering the fact the class have less than 30,..
Should be having 8,16,24..in the class, 6 should have ball, while 7 should have boot, which means 6+7-8 having both ball and boot..

Y d name "mathematical"

I)Total number of students in class is 24
Ii)Total number of students with both football boots and footballs is 15

mtprofessor5:
Y d name "mathematical"

Ion know sef

No of students < 30

Exactly 3/4 of them own footballs, therefore no of students is a multiple of 4 i. e 4,8,12,16,20,24,28
Exactly 7/8 of them own football boots therefore no of students is a multiple of 8 i. e 8,16,24.

If no of students = 8, 6 own balls and 7 own boots. This means that one person doesn't have boots. The 6 that own balls can be made up of all the people that have boots or 5 that own boots and 1 that doesn't. Meaning at least 5 students own both.

Using the same logic

If no of students = 16, 12 own balls and 14 own boots meaning at least 10 own both.

If no of students = 24, 18 own balls and 21 own boots meaning at least 15 own both.

I think there might be a part missing from the question.

olusolaj:
I)Total number of students in class is 24
Ii)Total number of students with both football boots and footballs is 15

how did u arrive at d conclusion dat 15students have both football and boots. thanks

CNN80:
No of students < 30

Exactly 3/4 of them own footballs, therefore no of students is a multiple of 4 i. e 4,8,12,16,20,24,28
Exactly 7/8 of them own football boots therefore no of students is a multiple of 8 i. e 8,16,24.

If no of students = 8, 6 own balls and 7 own boots. This means that one person doesn't have boots. The 6 that own balls can be made up of all the people that have boots or 5 that own boots and 1 that doesn't. Meaning at least 5 students own both.

Using the same logic

If no of students = 16, 12 own balls and 14 own boots meaning at least 10 own both.

If no of students = 24, 18 own balls and 21 own boots meaning at least 15 own both.

I think there might be a part missing from the question.

u na senpai

olusolaj:
I)Total number of students in class is 24
Ii)Total number of students with both football boots and footballs is 15

Friend, the question has a flaw,bn tryna solve it buh i keep stumbling off if 8 students: 7 own boots, 6 own balls. (at least 5 haveboth, maximum of 6)
16 students: 14 have boots, 12 havr balls. (at least 10 have both, maximum of 12)
24 students: 21 have boots, 18 own balls. (at least 15 own both, maximum of 18),which is wrong.. pls op solve it oo

Mathematical:
Friend, the question has a flaw,bn tryna solve it buh i keep stumbling off if 8 students: 7 own boots, 6 own balls. (at least 5 haveboth, maximum of 6)
16 students: 14 have boots, 12 havr balls. (at least 10 have both, maximum of 12)
24 students: 21 have boots, 18 own balls. (at least 15 own both, maximum of 18),which is wrong.. pls op solve it oo

I agree wit him. cnn80 has a detailed explanation too.

Ukeachu1:
I agree wit him. cnn80 has a detailed explanation too.

Ok.. still pondering

Mathematical:
Ok.. still pondering

easy lady. my f. maths teacher back then said maths gives wrinkles. u surely don't need it

Ukeachu1:
easy lady. my f. maths teacher back then said maths gives wrinkles. u surely don't need it

Loolz..

Ukeachu1:
how did u arrive at d conclusion dat 15students have both football and boots. thanks

Let the probability of student with both football boots and footballs be x
Then: (3/4 - x) + (7/8 - x) + x = 1
Then x=5/8
Therefore the of students with both football boots and footballs = 5/8 * 24
=15

olusolaj:
Let the probability of student with both football boots and footballs be x
Then: (3/4 - x) + (7/8 - x) + x = 1
Then x=5/8
Therefore the of students with both football boots and footballs = 5/8 * 24
=15

ok. dats tru. but I doubt d kid has been tot sets already. ur correct sha. u bad.

.Let the number of students be n
Let the number for which 30 exceeds n be x
therefore n+x =30,
n=30-x
since the class has two sets of people with footballs and football boots then 3/4n +7/8n=n,
substituting 30-x for n
3/4(30-x) +7/8(30-x)=30-x.
L.C.M
8x3/4(30-x)+8x7/8(30-x)=8(30-x)
6(30-x) +7(30-x)=240-8x
180-6x+210-7x=240-8x
390-13x=240-8x
C.L.T
390-240= - 8x+13x
150=5x
x=150/5
x=30
the number,x for which 30 is greater than the number of students in class,n is 30
but n=30 - x
so n=30 - 30
n=0
therefore,the solution for the number of students is UNDEFINED upon which question b has no solution

olusolaj:
Let the probability of student with both football boots and footballs be x
Then: (3/4 - x) + (7/8 - x) + x = 1
Then x=5/8
Therefore the of students with both football boots and footballs = 5/8 * 24
=15

I'm feeling you man! Nice try.

olusolaj:
Let the probability of student with both football boots and footballs be x
Then: (3/4 - x) + (7/8 - x) + x = 1
Then x=5/8
Therefore the of students with both football boots and footballs = 5/8 * 24
=15

This solution works if the question specified that all the students have at least a ball or boot.

28 it is.