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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

Total Questions : 1401 | Page 8 of 141 pages
Question 71.

Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

  1.    40
  2.    80
  3.    120
  4.    200
 Discuss Question
Answer: Option A. -> 40

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.

Question 72.

The G.C.D. of 1.08, 0.36 and 0.9 is:

  1.    0.03
  2.    0.9
  3.    0.18
  4.    0.108
 Discuss Question
Answer: Option C. -> 0.18

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.

Question 73.

The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:

  1.    1
  2.    2
  3.    3
  4.    4
 Discuss Question
Answer: Option B. -> 2

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.

Question 74.

The least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18 is:

  1.    74
  2.    94
  3.    184
  4.    364
 Discuss Question
Answer: Option D. -> 364

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.

Question 75.

Find the lowest common multiple of 24, 36 and 40.

  1.    120
  2.    240
  3.    360
  4.    480
 Discuss Question
Answer: Option C. -> 360

 2 | 24  -  36  - 40


 --------------------


 2 | 12  -  18  - 20


 --------------------


 2 |  6  -   9  - 10


 -------------------


 3 |  3  -   9  -  5


 -------------------


   |  1  -   3  -  5


  


L.C.M.  = 2 x 2 x 2 x 3 x 3 x 5 = 360.  

Question 76.

The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:

  1.    3
  2.    13
  3.    23
  4.    33
 Discuss Question
Answer: Option C. -> 23

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.

Question 77.

Reduce \(\frac{128352}{238368}\) to its lowest terms.

  1.    \(\frac{3}{4}\)
  2.    \(\frac{5}{13}\)
  3.    \(\frac{7}{13}\)
  4.    \(\frac{9}{13}\)
 Discuss Question
Answer: Option C. -> \(\frac{7}{13}\)

 


 


 


          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
Question 78.

The least number which when divided by 5, 6 , 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is:

  1.    1677
  2.    1683
  3.    2523
  4.    3363
 Discuss Question
Answer: Option B. -> 1683

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.

Question 79.

A, B and C start at the same time in the same direction to run around a circular stadium. A completes a round in 252 seconds, B in 308 seconds and c in 198 seconds, all starting at the same point. After what time will they again at the starting point ?

  1.    26 minutes and 18 seconds
  2.    42 minutes and 36 seconds
  3.    45 minutes
  4.    46 minutes and 12 seconds
 Discuss Question
Answer: Option D. -> 46 minutes and 12 seconds

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.

Question 80.

The H.C.F. of two numbers is 11 and their L.C.M. is 7700. If one of the numbers is 275, then the other is:

  1.    279
  2.    283
  3.    308
  4.    318
 Discuss Question
Answer: Option C. -> 308

Other number = \(\left(\frac{11\times7700}{275}\right) = 308\)

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