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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 7 of 141 pages
Question 61.

  1. A heap of coconuts is divided into groups of 2, 3 and 5, and each time one coconut is left over. The least number of coconuts in the heap is

  1.    31
  2.    51
  3.    61
  4.    71
 Discuss Question
Answer: Option A. -> 31
Question 62.

  1. If A is the set of all integral multiples of 3 and B is the set of all integral multiples of 5, then the numbers common to both A and B form a set of all integral multiples of

  1.    5 – 3 
  2.    5 + 3
  3.    H.C.F.( 3, 5)
  4.    L.C.M.(3, 5)
 Discuss Question
Answer: Option D. -> L.C.M.(3, 5)
Question 63.

  1. Telegraph poles occur at equal distances of 220 m along a road and heaps of stones are put at equal distances of 300 m along the same road. The first heap is at the foot of the first pole. How far from it along the road is the next heap which lies at the foot of a pole?

  1.    300
  2.    3000
  3.    3300
  4.    3330
 Discuss Question
Answer: Option C. -> 3300
Question 64.

  1. The smallest number from which when 3000 is subtracted is exactly divisible by 7, 11 and 13 is

  1.    2001
  2.    3001
  3.    4001
  4.    5001
 Discuss Question
Answer: Option C. -> 4001
Question 65.

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.

  1.    4
  2.    7
  3.    9
  4.    13
 Discuss Question
Answer: Option A. -> 4

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)


     = H.C.F. of 48, 92 and 140 = 4.

Question 66.

The H.C.F. of two numbers is 23 and the other two factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

  1.    276
  2.    299
  3.    322
  4.    345
 Discuss Question
Answer: Option C. -> 322

Clearly, the numbers are (23 x 13) and (23 x 14).


Therefore  Larger number = (23 x 14) = 322.

Question 67.

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

  1.    4
  2.    10
  3.    15
  4.    16
 Discuss Question
Answer: Option D. -> 16

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.


So, the bells will toll together after every 120 seconds(2 minutes).


In 30 minutes, they will toll together \(\frac{30}{2}\) + 1 = 16 times

Question 68.

Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:

  1.    4
  2.    5
  3.    6
  4.    8
 Discuss Question
Answer: Option A. -> 4

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)


  = H.C.F. of 3360, 2240 and 5600 = 1120.


Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question 69.

The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:

  1.    9000
  2.    9400
  3.    9600
  4.    9800
 Discuss Question
Answer: Option C. -> 9600

Greatest number of 4-digits is 9999.


L.C.M. of 15, 25, 40 and 75 is 600.


On dividing 9999 by 600, the remainder is 399.


Threrefore Required number (9999 - 399) = 9600.

Question 70.

The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:

  1.    101
  2.    107
  3.    111
  4.    185
 Discuss Question
Answer: Option C. -> 111

Let the numbers be 37a and 37b.


Then, 37a x 37b = 4107


 ab = 3.


Now, co-primes with product 3 are (1, 3).


So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111).


Therefore Greater number = 111.

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