Question Detail

If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other

Answer: Option B

Explanation:

For any this type of question, remember
Product of two numbers = Product of their HCF and LCM

So Other number = \begin{aligned} \frac{693 \times 11}{99} \end{aligned} = 77

1. 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

2. If the product of two numbers is 84942 and their H.C.F. is 33, find their L.C.M.

2574
2500
1365
1574

Answer: Option A

Explanation:

HCF * LCM = 84942, because we know

Product of two numbers = Product of HCF and LCM

LCM = 84942/33 = 2574

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. 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

5. Find the HCF 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 A

Explanation:

Whenever we have to solve this sort of question, remember the formula.
HCF = \begin{aligned} \frac{HCF of Numerators}{LCM of Denominators} \end{aligned}
So answers will be option 1