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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. Reduce \begin{aligned}
\frac{368}{575}
\end{aligned} to the lowest terms.
- \begin{aligned} \frac{30}{25} \end{aligned}
- \begin{aligned} \frac{28}{29} \end{aligned}
- \begin{aligned} \frac{28}{29} \end{aligned}
- \begin{aligned} \frac{16}{25} \end{aligned}
Answer: Option D
Explanation:
We can do it easily by in two steps
Step1: We get the HCF of 368 and 575 which is 23
Step2: Divide both by 23, we will get the answer 16/25
2. Find the largest number of four digits which is exactly divisible by 27,18,12,15
- 9700
- 9710
- 9720
- 9730
Answer: Option C
Explanation:
LCM of 27-18-12-15 is 540.
After dividing 9999 by 540 we get 279 remainder.
So answer will be 9999-279 = 9720
3. 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
4. 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
5. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
- 34
- 77
- 12
- 45
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
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