# In a seminar, the number of participants in Hindi, English and Mathematics are \[60,84\] and \[108\]. Find the number of rooms required if in each room the same number of participants are to be seated and all of them being in the same subject.

Answer

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Hint: To find the number of rooms required to seat all the participants, firstly find the number of participants to be seated in each room. As an equal number of participants are to be seated in each room, find the HCF of the number of participants, i.e., HCF of \[60,84,108\]. Divide the HCF of three numbers by each of the numbers to get the number of rooms needed for participants of each subject. Add the number of rooms required for each subject to find the total number of rooms required to seat all the participants.

Complete step-by-step answer:

We have \[60,84,108\] participants of Hindi, English and Mathematics. We have to find the number of rooms required to seat all the participants such that the number of participants is the same in all the rooms and all the participants of the same subject are seated in one room. Firstly, we will calculate the number of rooms needed to seat all the participants of a particular subject. As the same number of participants are to be seated in each room, we will evaluate the HCF of \[60,84,108\].

To find the HCF of \[60,84,108\], we will write the prime factorization of each of the numbers. We will choose the common prime factors of all the three numbers. The number obtained by choosing the common prime factors of all the numbers will be the HCF of three numbers.

We will now write prime factorization of all the three numbers. The prime factorization of \[60\] is \[60={{2}^{2}}\times 3\times 5\]. The prime factorization of \[84\] is \[84={{2}^{2}}\times 3\times 7\]. The prime factorization of \[108\] is \[108={{2}^{2}}\times {{3}^{3}}\].

The terms common to prime factorization of all the three numbers is \[{{2}^{2}}\times 3=12\]. Thus, the HCF of \[60,84,108\] is \[12\].

We will now calculate the number of rooms required to seat all the participants of each subject. To do so, we will divide each of the numbers by their HCF.

Number of rooms required to seat Hindi participants \[=\dfrac{60}{12}=5\].

Number of rooms required to seat English participants \[=\dfrac{84}{12}=7\].

Number of rooms required to seat Mathematics participants \[=\dfrac{108}{12}=9\].

So, the total number of rooms required to seat all the participants \[=5+7+9=21\].

Hence, we need \[21\] rooms to seat all the participants such that \[12\] participants will be seated in all the rooms.

Note: To write the prime factorization of any number, start by dividing the number by the first prime number, which is \[2\] and then continue to divide by \[2\] until you get a number which is not divisible by \[2\] (which means that you get a decimal or remainder on dividing the number by \[2\]). Then start dividing the number by the next prime number which is \[3\]. Continue dividing the number by \[3\] until you get a number which is not divisible by \[3\]. Thus, continuing the process, keep dividing the numbers by series of prime numbers \[5,7,...\] until the only numbers left are prime numbers. Write the given number as a product of all the prime numbers (considering the fact to count each prime number as many times as it divides the given number) to get the prime factorization of the given number.

Complete step-by-step answer:

We have \[60,84,108\] participants of Hindi, English and Mathematics. We have to find the number of rooms required to seat all the participants such that the number of participants is the same in all the rooms and all the participants of the same subject are seated in one room. Firstly, we will calculate the number of rooms needed to seat all the participants of a particular subject. As the same number of participants are to be seated in each room, we will evaluate the HCF of \[60,84,108\].

To find the HCF of \[60,84,108\], we will write the prime factorization of each of the numbers. We will choose the common prime factors of all the three numbers. The number obtained by choosing the common prime factors of all the numbers will be the HCF of three numbers.

We will now write prime factorization of all the three numbers. The prime factorization of \[60\] is \[60={{2}^{2}}\times 3\times 5\]. The prime factorization of \[84\] is \[84={{2}^{2}}\times 3\times 7\]. The prime factorization of \[108\] is \[108={{2}^{2}}\times {{3}^{3}}\].

The terms common to prime factorization of all the three numbers is \[{{2}^{2}}\times 3=12\]. Thus, the HCF of \[60,84,108\] is \[12\].

We will now calculate the number of rooms required to seat all the participants of each subject. To do so, we will divide each of the numbers by their HCF.

Number of rooms required to seat Hindi participants \[=\dfrac{60}{12}=5\].

Number of rooms required to seat English participants \[=\dfrac{84}{12}=7\].

Number of rooms required to seat Mathematics participants \[=\dfrac{108}{12}=9\].

So, the total number of rooms required to seat all the participants \[=5+7+9=21\].

Hence, we need \[21\] rooms to seat all the participants such that \[12\] participants will be seated in all the rooms.

Note: To write the prime factorization of any number, start by dividing the number by the first prime number, which is \[2\] and then continue to divide by \[2\] until you get a number which is not divisible by \[2\] (which means that you get a decimal or remainder on dividing the number by \[2\]). Then start dividing the number by the next prime number which is \[3\]. Continue dividing the number by \[3\] until you get a number which is not divisible by \[3\]. Thus, continuing the process, keep dividing the numbers by series of prime numbers \[5,7,...\] until the only numbers left are prime numbers. Write the given number as a product of all the prime numbers (considering the fact to count each prime number as many times as it divides the given number) to get the prime factorization of the given number.

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