Question Detail

Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}

Answer: Option B

Explanation:

HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
2 * 3*3 * 7

1. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

Answer: Option B

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number = \begin{aligned} \frac{693 \times 11}{99} \end{aligned} = 77

2. Find the HCF of \begin{aligned} \frac{2}{3}, \frac{4}{6}, \frac{8}{27} \end{aligned}

\begin{aligned} \frac{2}{27} \end{aligned}
\begin{aligned} \frac{8}{3} \end{aligned}
\begin{aligned} \frac{2}{3} \end{aligned}
\begin{aligned} \frac{8}{27} \end{aligned}

Answer: Option A

Explanation:

Whenever we have to solve this sort of question, remember the formula.
HCF = \begin{aligned} \frac{HCF of Numerators}{LCM of Denominators} \end{aligned}
So answers will be option 1

3. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

Answer: Option A

Explanation:

Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.

4. Find the HCF of 54, 288, 360

Answer: Option A

Explanation:

Lets solve this question by factorization method.

\begin{aligned}
18 = 2 \times 3^2, 288 = 2^5 \times 3^2, 360 = 2^3 \times 3^2 \times 5
\end{aligned}

So HCF will be minimum term present in all three, i.e.
\begin{aligned}
2 \times 3^2 = 18
\end{aligned}

5. There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

Answer: Option C

Explanation:

As given the questions these numbers are co primes, so there is only 1 as their common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;

So first number is : 551/29 = 19
Third number = 1073/29 = 37

So sum of these numbers is = (19 + 29 + 37) = 85

Question Detail

Find the HCF of \begin{aligned} 2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7 \end{aligned}

1. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

2. Find the HCF of \begin{aligned} \frac{2}{3}, \frac{4}{6}, \frac{8}{27} \end{aligned}

3. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is

4. Find the HCF of 54, 288, 360

5. There are three numbers, these are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. What will be the sum of three numbers :

Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}