Instructions: This Monk test contains JK Bank HCF and LCM Problems with answer. This will help you in JK bank exam preparation.

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Results

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

Answer: Option A Explanation:
Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43) = H.C.F. of 48, 92 and 140 = 4.

#2. 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:

Answer: Option C

Explanation:

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

 Larger number = (23 x 14) = 322.

#3. 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 ?

Answer: Option D

Explanation:

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 1 = 16 times.
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#4. 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:

Answer: Option A

Explanation:

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

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

Answer: Option C

Explanation:

The 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.

 Required number (9999 – 399) = 9600.

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

Answer: Option C

Explanation:

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).

 Greater number = 111.

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

Answer: Option A

Explanation:

Let the numbers be 3x, 4x and 5x.

Then, their L.C.M. = 60x.

So, 60x = 2400 or x = 40.

 The numbers are (3 x 40), (4 x 40) and (5 x 40).

Hence, required H.C.F. = 40.

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

Answer: Option C

Explanation:

Given numbers are 1.08, 0.36 and 0.90.   H.C.F. of 108, 36 and 90 is 18,

 H.C.F. of given numbers = 0.18.

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

Answer: Option B

Explanation:

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.

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

Answer: Option D

Explanation:

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.

 Required number = (90 x 4) 4   = 364.

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

Answer: Option C

Explanation:

 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.

 

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

Answer: Option C

Explanation:

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.

#13. Reduce 128352/238368 to its lowest terms.

Answer: Option C

Explanation:

 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

#14. 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:

Answer: Option B

Explanation:

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.

#15. 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 ?

Answer: Option D

Explanation:

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.

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