Question Detail

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

1. LCM of two numbers is 7700 and HCF is 11. If one number is 308 then other number is

Answer: Option D

2. 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

3. 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.

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 :

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. 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

Question Detail

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

1. LCM of two numbers is 7700 and HCF is 11. If one number is 308 then other number is

2. 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

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

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 :

5. 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.

2. 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