
In a seminar, the number of participants in Hindi, English and Mathematics are \[60,84\] and $108$ respectively. Find the minimum number of rooms required if in each room the same number of participants is to be seated and all of them being in the same subject.
Answer
590.1k+ views
Hint: The number of rooms can be minimised if each room accommodates maximum number of participants. Since there must be the same number of participants in each room and all of them must be of same subject, number of participants in each room must be the highest common factor of \[60,84\] and $108$.So all we need is to calculate the HCF of the given numbers.
Complete step-by-step answer:
It is given that the number of participants for Hindi, English and Mathematics are \[60,84\] and $108$ respectively.
We have to allocate rooms for them in such a way that each room contains the same number of participants and all of them belong to the same subject.
So we have to find the highest common factor of these numbers.
We can use prime factorisation for this.
Prime factorisation is writing a number as a product of prime numbers.
So we have,
$60 = 2 \times 2 \times 3 \times 5$
$84 = 2 \times 2 \times 3 \times 7$
$108 = 2 \times 2 \times 3 \times 3 \times 3$
Thus we can see \[2 \times 2 \times 3 = 12\] is the highest common factor of these three numbers.
So we can allocate $12$ persons to each room.
Then the number of rooms needed is obtained by dividing total number of persons by number of persons in each room.
Total number of persons $ = 60 + 84 + 108 = 252$
$ \Rightarrow {\text{Number of rooms needed = }}\dfrac{{252}}{{12}} = 21$
$\therefore $ The answer is $21$.
Note: Since $12$ is the common factor of all the three numbers, we can divide all of them by it. So we get $5$ rooms for Hindi, $7$ rooms for English and $9$ rooms for Mathematics. So there are $21$ rooms in total and each room consists of $12$ persons with the same subject.
Complete step-by-step answer:
It is given that the number of participants for Hindi, English and Mathematics are \[60,84\] and $108$ respectively.
We have to allocate rooms for them in such a way that each room contains the same number of participants and all of them belong to the same subject.
So we have to find the highest common factor of these numbers.
We can use prime factorisation for this.
Prime factorisation is writing a number as a product of prime numbers.
So we have,
$60 = 2 \times 2 \times 3 \times 5$
$84 = 2 \times 2 \times 3 \times 7$
$108 = 2 \times 2 \times 3 \times 3 \times 3$
Thus we can see \[2 \times 2 \times 3 = 12\] is the highest common factor of these three numbers.
So we can allocate $12$ persons to each room.
Then the number of rooms needed is obtained by dividing total number of persons by number of persons in each room.
Total number of persons $ = 60 + 84 + 108 = 252$
$ \Rightarrow {\text{Number of rooms needed = }}\dfrac{{252}}{{12}} = 21$
$\therefore $ The answer is $21$.
Note: Since $12$ is the common factor of all the three numbers, we can divide all of them by it. So we get $5$ rooms for Hindi, $7$ rooms for English and $9$ rooms for Mathematics. So there are $21$ rooms in total and each room consists of $12$ persons with the same subject.
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