
During the commonwealth games contingent of three countries A, B and C included \[128,224,320\] athletes respectively. Find the number of rooms required to house them in the Games Village if in each room, the same number of participants are to be accommodated and all the athletes in a room belong to the same country.
Answer
591.6k+ views
Hint: Here we will be using the prime factorization method to calculate the number of rooms required to house them in the Games Village. In this, we will prime factorize the number of athletes from each country and the addition of uncommon factors will be the required number of rooms.
Complete step-by-step answer:
Given that:
Number of athletes from country A \[ = 128\]
Number of athletes from country B \[ = 224\]
Number of athletes from country C \[ = 320\]
As per the question, the number of athletes in each room should be the same. So let the number of athletes in each room be \[y\] .
Also, we know that the number of athletes in each room should be from the same country. So let us assume the number of rooms required for the athletes from country A, B and C be 1, m and n respectively.
Now,
Number of athletes from country A \[ = 1 \times y\]
Number of athletes from country B \[ = m \times y\]
Number of athletes from country C \[ = n \times y\]
Let’s put the values of the number of athletes here.
\[128 = 1 \times y\] ……………\[\left( 1 \right)\]
\[224 = m \times y\] …………\[\left( 2 \right)\]
\[320 = n \times y\] …………\[\left( 3 \right)\]
We will now factorize \[{\rm{128, 224\, and 320}}\] .
\[ \Rightarrow 128 = 4 \times 32\] ……………\[\left( A \right)\]
\[ \Rightarrow 224 = 7 \times 32\] ……………\[\left( B \right)\]
\[\Rightarrow 320 = 10 \times 32\]……………\[\left( C \right)\]
Now, by comparing equation (1) with (A), (2) with (B) and (3) with (C), we get
\[\begin{array}{l}l = 4\\m = 7\\n = 10\\y = 32\end{array}\]
Total number of rooms required for all the athletes \[ = l + m + n = 4 + 7 + 10 = 21\]
\[\therefore\] Number of rooms required is 21.
Note: We have used prime factorization to solve the question. Prime Factorization is the multiplication of prime numbers. Here, we can also use the HCF method to calculate the number of rooms. With the help of this, we can find the value of y which is HCF of the three numbers given in this case and once we get the value of y , values of l, m and n can be easily found out.
Complete step-by-step answer:
Given that:
Number of athletes from country A \[ = 128\]
Number of athletes from country B \[ = 224\]
Number of athletes from country C \[ = 320\]
As per the question, the number of athletes in each room should be the same. So let the number of athletes in each room be \[y\] .
Also, we know that the number of athletes in each room should be from the same country. So let us assume the number of rooms required for the athletes from country A, B and C be 1, m and n respectively.
Now,
Number of athletes from country A \[ = 1 \times y\]
Number of athletes from country B \[ = m \times y\]
Number of athletes from country C \[ = n \times y\]
Let’s put the values of the number of athletes here.
\[128 = 1 \times y\] ……………\[\left( 1 \right)\]
\[224 = m \times y\] …………\[\left( 2 \right)\]
\[320 = n \times y\] …………\[\left( 3 \right)\]
We will now factorize \[{\rm{128, 224\, and 320}}\] .
\[ \Rightarrow 128 = 4 \times 32\] ……………\[\left( A \right)\]
\[ \Rightarrow 224 = 7 \times 32\] ……………\[\left( B \right)\]
\[\Rightarrow 320 = 10 \times 32\]……………\[\left( C \right)\]
Now, by comparing equation (1) with (A), (2) with (B) and (3) with (C), we get
\[\begin{array}{l}l = 4\\m = 7\\n = 10\\y = 32\end{array}\]
Total number of rooms required for all the athletes \[ = l + m + n = 4 + 7 + 10 = 21\]
\[\therefore\] Number of rooms required is 21.
Note: We have used prime factorization to solve the question. Prime Factorization is the multiplication of prime numbers. Here, we can also use the HCF method to calculate the number of rooms. With the help of this, we can find the value of y which is HCF of the three numbers given in this case and once we get the value of y , values of l, m and n can be easily found out.
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