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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

Similar Questions :

1. 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.

2. What will be the LCM of 8, 24, 36 and 54

  • 54
  • 108
  • 216
  • 432

Answer: Option C

Explanation:

LCM of 8-24-36-54 will be
2*2*2*3*3*3 = 216

3. An electronic device makes a beep after every 60 sec. Another device makes a beep after every 62 sec. They beeped together at 10 a.m. The time when they will next make a beep together at the earliest, is

  • 10:28 am
  • 10:30 am
  • 10:31 am
  • None of above

Answer: Option C

Explanation:

L.C.M. of 60 and 62 seconds is 1860 seconds
1860/60 = 31 minutes

They will beep together at 10:31 a.m.

Sometimes questions on red lights blinking comes in exam, which can be solved in the same way

4. 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

5. 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

Read more from - HCF and LCM Questions Answers
Post a comment

  • Roshan J 9 years ago

    hi mastguru could you please explain this answer:
    1. why LCM is followed?
    2. And why +1 is done at the end?

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