L.C.M and H.C.F : Aptitude Test

  • L.C.M and H.C.F
  • 1. The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is: 
   A.) 544
   B.) 548
   C.) 504
   D.) 536

Answer: Option 'B'

Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.

  • 2. Find the highest common factor of 36 and 84.
   A.) 4
   B.) 6
   C.) 12
   D.) 18

Answer: Option 'C'

36 = 22 x 32
84 = 22 x 3 x 7
H.C.F. = 22 x 3 = 12.

  • 3. 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: 
   A.) 85
   B.) 75
   C.) 65
   D.) 60

Answer: Option 'A'

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 = 551/29 = 19; Third number = 1073/29 = 37.
Required sum = (19 + 29 + 37) = 85.

  • 4. The greatest possible length which can be used to measure exactly the lengths 7 m, 3 m 85 cm, 12 m 95 cm is:
   A.) 15 cm
   B.) 35 cm
   C.) 25 cm
   D.) 52 cm

Answer: Option 'B'

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

  • 5. The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:
   A.) 127
   B.) 305
   C.) 235
   D.) 123

Answer: Option 'A'

Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.

  • 6. Which of the following has the most number of divisors? 
   A.) 176
   B.) 182
   C.) 99
   D.) 101

Answer: Option 'A'

Option A, 176 = 1 x 2 x 2 x 2 x 2 x 11.
Option B, 182 = 1 x 2 x 7 x 13. 
Option C, 99 = 1 x 3 x 3 x 11. 
Option D, 101 = 1 x 101.
Divisors of 99 are 1, 3, 9, 11, 33, 99.
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 have the most number of divisors..

  • 7. The L.C.M. of two numbers is 48. The numbers are in the ratio 2 : 3. Then sum of the number is:
   A.) 30
   B.) 40
   C.) 38
   D.) 48

Answer: Option 'B'

Let the numbers be 2x and 3x.
Then, their L.C.M. = 6x.
So, 6x = 48 or x = 8
The numbers are 16 and 24.
Hence, required sum = (16 + 24) = 40.

  • 8. 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:
   A.) 267
   B.) 318
   C.) 308
   D.) 279

Answer: Option 'C'

Other number = (11 x 7700/275) = 308.

  • 9. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30 ? 
   A.) 1260
   B.) 630
   C.) 2524
   D.) None of these

Answer: Option 'B'

L.C.M. of 12, 18, 21 30
= 2 x 3 x 2 x 3 x 7 x 5 = 1260.
Required number = (1260 ÷ 2)
= 630.

  • 10. The ratio of two numbers is 3 : 4 and their H.C.F. is 4. Their L.C.M. is: 
   A.) 12
   B.) 68
   C.) 48
   D.) 78

Answer: Option 'C'

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.

  • 11. The smallest number which when diminished by 7, is divisible 12, 16, 18, 21 and 28 is: 
   A.) 1008
   B.) 1032
   C.) 1015
   D.) 1022

Answer: Option 'C'

Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015

  • 12. 252 can be expressed as a product of primes as:
   A.) 2 × 3 × 3 × 3 × 7
   B.) 3 × 3 × 3 × 3 × 7
   C.) 2 × 2 × 2 × 3 × 7
   D.) 2 × 2 × 3 × 3 × 7

Answer: Option 'D'

Clearly, 252 = 2 × 2 × 3 × 3 × 7.

  • 13. A, B and C start at the same time in the same direction to run around a circular stadium. 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 ?
   A.) 26 minutes and 18 seconds
   B.) 45 minutes
   C.) 46 minutes and 12 seconds
   D.) 42 minutes and 36 seconds

Answer: Option 'C'

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

  • 14. The ratio of the two numbers is 9:11 and their L.C.M is 999999. The numbers are 
   A.) 99099,101101
   B.) 111111, 90990
   C.) 90909,111111
   D.) 90909,111110

Answer: Option 'C'

90909,111111

  • 15. 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: 
   A.) 1677
   B.) 2523
   C.) 1683
   D.) 3363

Answer: Option 'C'

L.C.M. of 5, 6, 7, 8 = 840.
Required number is of the form 840k + 3
Least value of k for which (840k + 3) is divisible by 9 is k = 2.
Required number = (840 x 2 + 3) = 1683.

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