Quiz Level-1

Hint

Here, we need to find out the LCM of 48 and 300.
Prime factors of \(48=2 \times 2 \times 2 \times 2 \times 3=2^4 \times 3^1\)
Prime factors of \(300=2 \times 2 \times 3 \times 5 \times 5=2^2 \times 3^1 \times 5^2\)
List all numbers found, as many times as they occur, most often any one given number, and multiply them together to find the LCM.
Thus, \(2^4 \times 3^1 \times 5^2=1200\)
Hence, the LCM of 48 and 300 is 1200.

Hint

Let us find the prime factors of all the given numbers,
\(
\begin{aligned}
& 6=2 \times 3 \\
& 72=2^3 \times 3^2 \\
& 120=2^3 \times 3 \times 5
\end{aligned}
\)
Here, \(2^1\) and \(3^1\) are the smallest powers of the common factors 2 and 3 , respectively.
So, \(\operatorname{HCF}(6,72,120)=2^1 \times 3^1=2 \times 3=6\)
\(2^3, 3^2\), and \(5^1\) are the greatest powers of the prime factors 2,3 and 5 respectively involved in the three numbers.
So, \(\operatorname{LCM}(6,72,120)=2^3 \times 3^2 \times 5^1=360\).

Hint

Here, we need to find the LCM of \(12,18,30\).
Now, let us find the prime factors of the above three numbers.
The prime factors of \(12=2 \times 2 \times 3=2^2 \times 3^1\)
The prime factors of \(18=2 \times 3 \times 3=2^1 \times 3^2\)
The prime factors of \(30=2 \times 3 \times 5=2^1 \times 3^1 \times 5^1\)
Now, list all the prime numbers found as many times as they occur most often for anyone given number and multiply them together to find the LCM.
So, \(2 \times 2 \times 3 \times 3 \times 5=180\)
Using exponents instead, multiply together each of the prime numbers with the highest power
Thus, \(2^2 \times 3^2 \times 5^1=180\)
Hence, the LCM of \(12,18,30\) is 180 .

Hint

The ratio of two numbers \(5: 11\).
Let the numbers be \(5 x\) and \(11 x\).
Since \(5: 11\) is already the reduced ratio, ‘ \(x\) ‘ must be the HCF.
So, the numbers are \(5 \times 7=35\) and \(11 \times 7=77\)
Hence, the numbers are 35 and 77.

Hint

LCM of 510 and 92

Let us find the LCM of 510 and 92 using the prime factorization method.
The prime factors of \(510=2 \times 3 \times 5 \times 17=2^1 \times 3^1 \times 5^1 \times 17^1\)
The prime factors of \(92=2 \times 2 \times 23=2^2 \times 23\)
Now, we will find the product of only those factors with the
highest powers. This will be \(2^2 \times 3^1 \times 5^1 \times 17^1 \times 23^1=4 \times 3 \times 5 \times 17\) \(\times 23=23460\)
Therefore, the LCM of 510 and \(92=23460\)

HCF of 510 and 92

Since we know the 2 numbers and we also know the LCM of the two numbers we can find the HCF of 510 and 92 using the formula: \(\operatorname{HCF}(\mathbf{a}, \mathbf{b}) \times \operatorname{LCM}(\mathbf{a}, \mathbf{b})=\mathbf{a} \times \mathbf{b}\)
\(\mathrm{HCF} \times 23460=510 \times 92\)
Now, we can find the value of HCF by solving this equation.
\(\mathrm{HCF}=46920 / 23460\)
\(\mathrm{HCF}=2\)
Therefore, the HCF of 510 and \(92=2\)

Hint

Given fractions: \(2 / 5,4 / 7\) and \(6 / 11\)
As we know, the formula to find the LCM of fractions is:
LCM of fractions = LCM of Numerators/HCF of Denominators .. (1)
Thus, the LCM of Numerators \(=\operatorname{LCM}(2,4,6)=12\).
HCF of denominators \(=\operatorname{HCF}(5,7,11)=1\)
Now, substitute the values in (1), we get
LCM of fractions \(=12 / 1=12\)
Hence, the LCM of the fractions \(2 / 5,4 / 7\) and \(6 / 11\) is 12.

Hint

The prime factorisation of 54 is \(2 \times 3 \times 3 \times 3\).
The prime factorization of 60 is \(2 \times 2 \times 3 \times 5\).
Thus, the product of prime factors \(=2 \times 2 \times 3 \times 3 \times 3 \times 5=540\)
Hence, the LCM of 54 and 60 is 540.

Hint

To find: HCF \((408,1032)\).
The prime factorisation of 408 is \(2 \times 2 \times 2 \times 3 \times 17\)
The prime factorisation of 1032 is \(2 \times 2 \times 2 \times 3 \times 43\)
Thus, the product of the prime factors of \(2 \times 2 \times 2 \times 3\) is 24
Hence, the HCF of 408 and 1032 is 24.

Hint

To find: \(\operatorname{GCF}(18,48)\)
The factors of 18 are \(1,2,3, \mathbf{6}, 9\) and 18
The factors of 48 are \(1,2,3,4,6,8,12,16,24\) and 48 .
Thus, 6 is the greatest number that divides 18 and 48 completely.
Hence, the GCF/HCF of 18 and 48 is 6.

Hint

To find the HCF of 135 and 225 using the prime factorisation method, follow the below steps:
Thus, the prime factorisation of 135 is \(3 \times 3 \times 3 \times 5\)
The prime factorisation of 225 is \(3 \times 3 \times 5 \times 5\).
Hence, the prime factors that is common in both the numbers is \(3 \times 3 \times 5\)
Therefore, the product of prime factors of \(3 \times 3 \times 5\) is 45.
Hence, the HCF of 135 and 225 is 45.