Quantitative Aptitude
HCF AND LCM MCQs
Problems On Hcf And Lcm, H.C.F And L.C.M. Of Numbers, Lcm & Hcf, Hcf And Lcm
Let the numbers be 3x, 4x and 5x.
Then, their L.C.M. = 60x.
So, 60x = 2400 or x = 40.
Therefore The numbers are (3 x 40), (4 x 40) and (5 x 40).
Hence, required H.C.F. = 40.
Given numbers are 1.08, 0.36 and 0.90. H.C.F. of 108, 36 and 90 is 18,
Threrefore H.C.F. of given numbers = 0.18.
Let the numbers 13a and 13b.
Then, 13a x 13b = 2028
ab = 12.
Now, the co-primes with product 12 are (1, 12) and (3, 4).
[Note: Two integers a and b are said to be coprime or relatively prime if they have no common positive factor other than 1 or, equivalently, if their greatest common divisor is 1 ]
So, the required numbers are (13 x 1, 13 x 12) and (13 x 3, 13 x 4).
Clearly, there are 2 such pairs.
L.C.M. of 6, 9, 15 and 18 is 90.
Let required number be 90k + 4, which is multiple of 7.
Least value of k for which (90k + 4) is divisible by 7 is k = 4.
Therefore Required number = (90 x 4) + 4 = 364.
2 | 24 - 36 - 40
--------------------
2 | 12 - 18 - 20
--------------------
2 | 6 - 9 - 10
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3 | 3 - 9 - 5
-------------------
| 1 - 3 - 5
L.C.M. = 2 x 2 x 2 x 3 x 3 x 5 = 360.
L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Therefore Number to be added = (60 - 37) = 23.
128352) 238368 ( 1
128352
---------------
110016 ) 128352 ( 1
110016
------------------
18336 ) 110016 ( 6
110016
-------
x
-------
So, H.C.F. of 128352 and 238368 = 18336.
128352 128352 ÷ 18336 7
Therefore, ------ = -------------- = --
238368 238368 ÷ 18336 13
L.C.M. of 5, 6, 7, 8 = 840.
Threrfore Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Threrfore Required number = (840 x 2 + 3) = 1683.
L.C.M. of 252, 308 and 198 = 2772.
So, A, B and C will again meet at the starting point in 2772 sec. i.e., 46 min. 12 sec.
Other number = \(\left(\frac{11\times7700}{275}\right) = 308\)