Question Detail

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}

1. 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.

2. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is

16cm
25cm
15cm
35cm

Answer: Option D

Explanation:

So by now, you must be knowing this is a question of HCF, right.
H.C.F. of (700 cm, 385 cm, 1295 cm) = 35 cm.

3. 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

4. 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

5. 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

Question Detail

Find the HCF of 54, 288, 360

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

2. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is

3. 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

4. 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.

5. Find the HCF of 2.1, 1.05 and 0.63

3. 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