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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 9 of 141 pages
Question 81.

What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ?

  1.    196
  2.    630
  3.    1260
  4.    2520
 Discuss Question
Answer: Option B. -> 630

L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
----------------------------
= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
----------------------------
Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5
= 630.
Question 82.

The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is:

  1.    12
  2.    16
  3.    24
  4.    48
 Discuss Question
Answer: Option D. -> 48

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.

Question 83.

The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is:

  1.    1008
  2.    1015
  3.    1022
  4.    1032
 Discuss Question
Answer: Option B. -> 1015

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7


   = 1008 + 7


   = 1015

Question 84.

252 can be expressed as a product of primes as:

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

Clearly, 252 = 2 x 2 x 3 x 3 x 7.

Question 85.

Three numbers which are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:

  1.    75
  2.    81
  3.    85
  4.    89
 Discuss Question
Answer: Option C. -> 85

Since the numbers are co-prime, they contain only 1 as the common factor.


Also, the given two products have the middle number in common.


So, middle number = H.C.F. of 551 and 1073 = 29;


First number = \(\left(\frac{551}{29}\right) = 19\)


Third number = \(\left(\frac{1073}{29}\right) = 37.\)


Therefoer Required sum = (19 + 29 + 37) = 85.

Question 86.

Find the highest common factor of 36 and 84.

  1.    4
  2.    6
  3.    12
  4.    18
 Discuss Question
Answer: Option D. -> 18

36 = 22 x 32


84 = 22 x 3 x 7


Therefore H.C.F. = 22 x 3 = 12.

Question 87.

Which of the following fraction is the largest ?

  1.    \(\frac{7}{8}\)
  2.    \(\frac{13}{16}\)
  3.    \(\frac{31}{40}\)
  4.    \(\frac{43}{80}\)
 Discuss Question
Answer: Option A. -> \(\frac{7}{8}\)

L.C.M. of 8, 16, 40 and 80 = 80.


\(\frac{7}{8}=\frac{70}{80}; \frac{13}{16}=\frac{65}{80}; \frac{31}{40}=\frac{62}{80}\)  


   since,\(\frac{70}{80}>\frac{65}{80}>\frac{63}{80} so, \frac{7}{8}>\frac{13}{16}>\frac{69}{80}>\frac{31}{40}\)


 


 \(so,\frac{7}{8}is the largest\)


 


 

Question 88.

The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:

  1.    504
  2.    536
  3.    544
  4.    548
 Discuss Question
Answer: Option D. -> 548

Required number = (L.C.M. of 12, 15, 20, 54) + 8


   = 540 + 8


   = 548.

Question 89.

The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:

  1.    123
  2.    127
  3.    235
  4.    305
 Discuss Question
Answer: Option B. -> 127

Required number = H.C.F. of (1657 - 6) and (2037 - 5)


  = H.C.F. of 1651 and 2032 = 127.

Question 90.

Which of the following has the most number of divisors?

  1.    99
  2.    101
  3.    176
  4.    182
 Discuss Question
Answer: Option C. -> 176

99 = 1 x 3 x 3 x 11


101 = 1 x 101


176 = 1 x 2 x 2 x 2 x 2 x 11


182 = 1 x 2 x 7 x 13


So, divisors of 99 are 1, 3, 9, 11, 33, .99


Divisors of 101 are 1 and 101


Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176


Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.


Hence, 176 has the most number of divisors.

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