RRB NTPC - L.C.M and H.C.F : Aptitude Test

1.

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.

2.

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.

3.

The L.C.M of two prime numbers x and y (x>y) is 203, the value of 2y-x 

   A.) -14
   B.) -15
   C.) 14
   D.) 15

Answer: Option 'B'

-15

4.

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.

5.

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:

   A.) 5
   B.) 6
   C.) 7
   D.) 4

Answer: Option 'D'

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

6.

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

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.) 28
   B.) 32
   C.) 40
   D.) 64

Answer: Option 'C'

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.

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?

   A.) 26 minutes and 18 seconds
   B.) 46 minutes and 12 seconds
   C.) 42 minutes and 36 seconds
   D.) 45 minutes

Answer: Option 'B'

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.

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.) 16
   C.) 48
   D.) 42

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.

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?

   A.) 4
   B.) 8
   C.) 12
   D.) 16

Answer: Option 'D'

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 30/2 + 1 = 16 times

12.

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.

13.

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

   A.) 4
   B.) 7
   C.) 9
   D.) 13

Answer: Option 'A'

Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43) = H.C.F. of 48, 92 and 140 = 4.

14.

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..

15.

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.

16.

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.) 1683
   C.) 1987
   D.) 2523

Answer: Option 'B'

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.

17.

Third of three fifth of one fourth of a number is 30. What is 80% of that number? 

   A.) 230
   B.) 250
   C.) 240
   D.) 235

Answer: Option 'C'

240

18.

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

   A.) 13
   B.) 3
   C.) 23
   D.) 33

Answer: Option 'C'

L.C.M. of 5, 6, 4 and 3 = 60.
On dividing 2497 by 60, the remainder is 37.
Number to be added = (60 - 37) = 23.

19.

Find the L.C.M of the fractions 7/3, 8/10, and 9/15: 

   A.) 4 90/5
   B.) 506
   C.) 504
   D.) 505

Answer: Option 'C'

504

20.

Which of the following has the most number of divisors ?

   A.) 99
   B.) 176
   C.) 101
   D.) 182

Answer: Option 'B'

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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