[tex] \large\underbrace \red{\frak{QUESTION} }[/tex]
In a certain class, one-third of the students were absent. Half of the total strength attended the Maths test and one-fourth of the total strength attended the Physics test. If 6 students attended both the tests and every student who was present attended at least one of the two tests, then how many students were absent on that day?
(a) 16
(b) 18
(c) 24
(d) 32
[tex] \blue\star \: \large \blue{\rm{
No \: Spam}}[/tex]
[tex] \pink\star \large \pink{\rm{ With \: Explanation}}\ \textless \ br /\ \textgreater \ [/tex]
[tex] \orange\star \: \large \orange{\rm {No \: copied \: answer}}[/tex]
Share
Step-by-step explanation:
QUESTION:-
In a certain class, one-third of the students were absent. Half of the total strength attended the Maths test and one-fourth of the total strength attended the Physics test. If 6 students attended both the tests and every student who was present attended at least one of the two tests, then how many students were absent on that day?
SOLUTION:-
Let be the total number of students as x
Students who were absent=x/3
Students who were present=x-x/3=2x/3
Set A(Student's who attended maths test)
Set B( Students who attended physics test)
n(A)=x/2
n(B)=x/4
n(A∩B)=6
As we know that every student who was present attended atleast 1 paper
So, we can say n(A∪B)=2x/3
n(A∪B)=n(A)+n(B)-n(A∩B)
2x/3=x/2+x/4-6
2x/3=3x/4-6
2x/3-3x/4=-6
-x/12=-6
-x=-72
x=72
Total number of students=72
As we know number of students absent=x/3
72/3
24
Number of students absent=24
HOPE THIS HELPS YOU DEAR
HAVE A WONDERFUL DAY AHEAD
Verified answer
[tex]\Large\boxed{\bf(c)\ 24}[/tex]
[tex]\;[/tex]
[tex]\Large\textbf{Solution}[/tex]
Let's assume:-
The set of the total students as [tex]\bf U[/tex]
The set of the students who took the Maths test as [tex]\bf A[/tex]
The set of the students who took the Physics test as [tex]\bf B[/tex]
[tex]\;[/tex]
Let the number of the total students be [tex]\bf x[/tex].
We know:-
[tex]\;[/tex]
[tex]\boxed{\bf n(A\cup B)=n(A)+n(B)-n(A\cap B)}[/tex]
[tex]\bf\Longrightarrow \dfrac{2x}{3}=\dfrac{x}{2}+\dfrac{x}{4}-6[/tex]
[tex]\bf\Longrightarrow 12\times\dfrac{2x}{3}=12\times\left(\dfrac{x}{2}+\dfrac{x}{4}-6\right)[/tex]
[tex]\bf\Longrightarrow 8x=6x+3x-72[/tex]
[tex]\bf\Longrightarrow x=72[/tex]
[tex]\;[/tex]
[tex]\boxed{\bf n(U)-n(A\cup B)=n(A^{c}\cap B^{c})}[/tex]
Small note:-
The Venn diagram proves the formula.
[tex]\bf\Longrightarrow n(U)-n(A\cup B)=24[/tex]
[tex]\textbf{$\Longrightarrow$ 24 students were absent.}[/tex]
[tex]\;[/tex]
[tex]\Large\textbf{Learn More}[/tex]
The attached Venn diagram will help you understand :)