[tex]\huge\color{green}\boxed{\colorbox{black}{❥Questions ❤}} \ \textless \ br /\ \textgreater \ [/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]\gray\bigstar \large \red{\rm{ No \:Spam}}[/tex]
[tex]\gray\bigstar \large \red{\rm{ With \: Explanation}}\ \textless \ br /\ \textgreater \ [/tex]
[tex]\gray\bigstar \large \red{\rm {No \:copied \: answer}}[/tex]
Answers & Comments
[tex]\large\underline{\sf{Solution-}}[/tex]
Given that,
Let assume that
Given that,
One - third of the students were absent.
[tex]\rm\implies \:Number \: of \: students \: absent \: = \: \dfrac{x}{3} \\ [/tex]
So,
[tex]\rm\implies \:Number \: of \: students \: present \: = \:x - \dfrac{x}{3} = \dfrac{2x}{3} \\ [/tex]
It is further given that, half of the total strength attend the math test.
So,
[tex]\rm\implies \:Number \: of \: students \: appeared \: in \: maths \: = \dfrac{x}{2} \\ [/tex]
Further, given that one - fourth of total strength attend the Physics test.
[tex]\rm\implies \:Number \: of \: students \: appeared \: in \: physics \: = \dfrac{x}{4} \\ [/tex]
Number of students attend both the tests = 6
Let assume that
So, we have
[tex]\rm \: n(A\cup B) = \dfrac{2x}{3} \\ [/tex]
[tex]\rm \: n(A) = \dfrac{x}{2} \\ [/tex]
[tex]\rm \: n(B) = \dfrac{x}{4} \\ [/tex]
[tex]\rm \: n(A\cap B) = 6 \\ [/tex]
We know,
[tex]\rm \: n(A\cup B) = n(A) + n(B) - n(A\cap B) \\ [/tex]
[tex]\rm \: \dfrac{2x}{3} = \dfrac{x}{2} + \dfrac{x}{4} - 6 \\ [/tex]
[tex]\rm \: \dfrac{2x}{3} = \dfrac{2x + x - 24}{4} \\ [/tex]
[tex]\rm \: \dfrac{2x}{3} = \dfrac{3x - 24}{4} \\ [/tex]
[tex]\rm \: 8x = 3(3x - 24) \\ [/tex]
[tex]\rm \: 8x = 9x -72 \\ [/tex]
[tex]\rm \: 8x - 9x = -72 \\ [/tex]
[tex]\rm \: - x = -72 \\ [/tex]
[tex]\bf\implies \:x = 72 \\ [/tex]
So,
[tex]\bf\implies \:Number \: of \: students \: absent \: = \: \dfrac{72}{3} = 24\\ [/tex]
So, option (c) is correct.
Verified answer
Answer:
Let the total strength be 'x'
[tex] \bf \pink{ absent \: student=x/3}[/tex]
Present student = 2x/3
A=students who attended the maths test
B= students who attended the physics test
[tex] \rm \: n(A)=x/2[/tex]
[tex] \tt \: n(B)=x/4 \\ \\ [/tex]
[tex] \rm \purple{ n(A ∩ B)=6} \\ \\ [/tex]
n(A∪B) no. of students present 2x/3
[tex] \tt \red {(A∪B)=n(A)+n(B)−n(A∩B)} \\ \\ \\ [/tex]
[tex] \bf\frac{2x}{3} = \frac{x}{2} + \frac{x}{4} - 6 \\ \\ \\ [/tex]
[tex] \bf \: x = 72 \\ \\ \\ \\ \\ [/tex]
[tex] \bf \: absent \: students = \frac{72}{3} = 24[/tex]
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