Question Detail

HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is

\begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
\begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
\begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
\begin{aligned} 2 \times 3 \times 5 \end{aligned}

Answer: Option C

Explanation:

As in HCF we will choose the minimum common factors among the given.. So answer will be third option

1. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is

\begin{aligned} 2^4 \times 3^4 \times 5^3 \end{aligned}
\begin{aligned} 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned}
\begin{aligned} 2^2 \times 3^2 \times 5^2 \end{aligned}
\begin{aligned} 2 \times 3 \times 5 \end{aligned}

Answer: Option C

Explanation:

As in HCF we will choose the minimum common factors among the given.. So answer will be third option

2. The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number

Answer: Option D

Explanation:

Solve this question by using below formula.
Product of 2 numbers = product of their HCF and LCM

144 * x = 12 * 5040

x = (12*5040)/144 = 420

3. 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

4. LCM of two numbers is 7700 and HCF is 11. If one number is 308 then other number is

Answer: Option D

5. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

Answer: Option D

Explanation:

LCM of 2-4-6-8-10-12 is 120 seconds, that is 2 minutes.

Now 60/2 = 30
Adding one bell at the starting it will 30+1 = 31

Question Detail

HCF of \begin{aligned} 2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned} is

1. HCF of \begin{aligned} 2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11 \end{aligned} is

2. The H.C.F. and L.C.M. of two numbers are 12 and 5040 respectively If one of the numbers is 144, find the other number

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

4. LCM of two numbers is 7700 and HCF is 11. If one number is 308 then other number is

5. Six bells commence tolling together and toll at the intervals of 2,4,6,8,10,12 seconds resp. In 60 minutes how many times they will toll together.

HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is

1. HCF of
\begin{aligned}
2^2 \times 3^2 \times 5^2, 2^4 \times 3^4 \times 5^3 \times 11
\end{aligned} is