Question Detail

Find the LCM 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 B

Explanation:

Whenever we have to solve this sort of question, remember the formula.
LCM = \\begin{aligned} \\frac{HCF of Denominators}{LCM of Numerators} \\end{aligned}
So answers will be option 2,
Please also give attention to the difference in formula of HCF and LCM

1. 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

2. Find the HCF of 2.1, 1.05 and 0.63

0.44
0.64
0.21
None of above

Answer: Option C

Explanation:

To solve this question quickly, first remove decimal by multiplying each term with 100,
Then terms become 210, 105, 63
Then HCF of above terms is 21,
So Answer is 0.21

3. 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

4. Find the LCM 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 B

Explanation:

5. 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}