Practice - HCF & LCM

Q1. If an integer n is divisible by 3, 5 and 12, what is the next larger integer divisible by all these numbers?

1) n + 3     2) n + 5     3) n + 12     4) n + 60     5) n + 15

Solution :

If n is divisible by 3, 5 and 12 it must a multiple of the LCM of 3, 5 and 12 which is 60.
n = 60 k
n + 60 is also divisible by 60 since
n + 60 = 60 k + 60 = 60(k + 1)
The answer is option 4.

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Q2. What is the smallest integer that is multiple of 5, 7 and 20?

1) 70     2) 35     3) 200     4) 280     5) 140

Solution :-

It is the LCM of 5, 7 and 20 which is 140.
The answer is option 5.

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Q3. 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

Solution:-

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

Larger number = (23 x 14) = 322. The answer is option 3.

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Q4. 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

Solution:-

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. One is added as for first time at t = 0 they tolled together.The answer is option 4.
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Q5. The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:(1)    9000    (2)     9400   (3)     9600        (4)    9800  (5) 9995 

Solution:- 

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.The answer is option 3.

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Q6.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 (5) None of these. 

Solution:- 

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

The answer is option 1. 

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Q7. Three number are in the ratio of 3 : 4 : 5 and their L.C.M. is 2400. Their H.C.F. is:

(1) 40      (2)  200 (3)   80 (4)  120     (5) None of these.

 Solution:- 

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.

The answer is option 1.

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Q8.  The G.C.D. of 1.08, 0.36 and 0.9 is:
(1)     0.03   (2)     0.9    (3)   0.18      (4)     0.108
Solution:-iven 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.
The answer is option 3.
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Q9. The product of two numbers is 2028 and their H.C.F. is 13. The number of such pairs is:  (1)     1   (2)     2    (3)   3      (4)     4
 Solution:-
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 co-prime 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.
The answer is option 2.

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 Q10. The least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3 is:
(1)   3  (2)    13   (3)   23      (4)    33
 Solution:-
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.
The answer is option 3.
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