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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

Similar Questions :

1. 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 :

  • 80
  • 82
  • 85
  • 87

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

2. 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.

  • 15
  • 16
  • 30
  • 31

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

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. Reduce \begin{aligned}
\frac{803}{876}
\end{aligned} to the lowest terms.

  • \begin{aligned} \frac{11}{12} \end{aligned}
  • \begin{aligned} \frac{23}{24} \end{aligned}
  • \begin{aligned} \frac{26}{27} \end{aligned}
  • \begin{aligned} \frac{4}{7} \end{aligned}

Answer: Option A

Explanation:

HCF of 803 and 876 is 73, Divide both by 73, We get the answer 11/12

5. Find the greatest number that will divide 400, 435 and 541 leaving 9, 10 and 14 as remainders respectively

  • 19
  • 17
  • 13
  • 9

Answer: Option B

Explanation:

Answer will be HCF of (400-9, 435-10, 541-14)

HCF of (391, 425, 527) = 17

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