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. 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

2. 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

3. Find the greatest number which on dividing 1661 and 2045, leaves a reminder of 10 and 13 respectively

Answer: Option B

Explanation:

In this type of question, its obvious we need to calculate the HCF, trick is
HCF of (1661 - 10) and (2045 -13)
= HCF (1651, 2032) = 127

4. Which greatest possible length can be used to measure exactly 15 meter 75 cm, 11 meter 25 cm and 7 meter 65 cm

45cm
255cm
244cm
55cm

Answer: Option A

Explanation:

Convert first all terms into cm.
i.e. 1575 cm, 1125cm, 765cm.

Now whenever we need to calculate this type of question, we need to find the HCF. HCF of above terms is 255.

5. 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