Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 29)
29.
| The H.C.F. of | 9 | , | 12 | , | 18 | and | 21 | is: |
| 10 | 25 | 35 | 40 |
Answer: Option
Explanation:
| Required H.C.F. = | H.C.F. of 9, 12, 18, 21 | = | 3 |
| L.C.M. of 10, 25, 35, 40 | 1400 |
Discussion:
31 comments Page 4 of 4.
Swarn Aabha Sao said:
4 weeks ago
HCF = HCF (9,12,18,21)/LCM(10,25,35,40) = 3/1400.
HCF --> common primes with smallest power.
HCF(9,12,18,21 )= 3.
9 = 3 x 3.
12 = 2 x 2 x 3.
18 = 2 x 3 x 3.
21 = 3 x 7.
Highest common factor in all four numbers = 3
LCM --> all primes with largest power
LCM(10,25,35,40)= 1400
10 = 2 x 5
25 = 5 x 5
35 = 5 x 7
40 = 2 x 2 x 2 x 5
So,the highest power of each prime = 2 x 2 x 2 x 5 x 5 x 7.
= 8 x 25 x 7.
= 200 x 7.
= 1400.
HCF --> common primes with smallest power.
HCF(9,12,18,21 )= 3.
9 = 3 x 3.
12 = 2 x 2 x 3.
18 = 2 x 3 x 3.
21 = 3 x 7.
Highest common factor in all four numbers = 3
LCM --> all primes with largest power
LCM(10,25,35,40)= 1400
10 = 2 x 5
25 = 5 x 5
35 = 5 x 7
40 = 2 x 2 x 2 x 5
So,the highest power of each prime = 2 x 2 x 2 x 5 x 5 x 7.
= 8 x 25 x 7.
= 200 x 7.
= 1400.
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