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Question Detail
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
1. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is
- 48
- 22
- 56
- 27
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. 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
3. Find the HCF of
\begin{aligned}
2^2 \times 3^2 \times 7^2, 2 \times 3^4 \times 7
\end{aligned}
- 128
- 126
- 146
- 434
Answer: Option B
Explanation:
HCF is Highest common factor, so we need to get the common highest factors among given values. So we got
2 * 3*3 * 7
4. Find the HCF of 54, 288, 360
- 18
- 36
- 54
- 108
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}
5. If LCM of two number is 693, HCF of two numbers is 11 and one number is 99, then find other
- 34
- 77
- 12
- 45
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
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