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HCF And LCM Problems With Solutions, Formulas & Problem Solving Tricks

HCF And LCM Problems with Solutions

Dear students, here on this page we are going to provide you some HCF And LCM Problems With Solutions for those candidates who are always confused about this concept and looking for its problem solving tricks. By going through the below given details, aspirants may make their aptitude stronger. 

Applicants may scroll down this page and get detailed information about HCF And LCM Problems With Solutions like formulas to solve problems and problem solving tricks etc which is furnished by team of recruitmentresult.com.

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

Ans: 4

Explanation:

Required number = H.C.F. of (91 – 43), (183 – 91) and (183 – 43)

= H.C.F. of 48, 92 and 140 = 4.

2. 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?

Ans: 16

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 30/2+1 = 16 times.

3. 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:

Ans: 322

Explanation:

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

Larger number = (23 x 14) = 322.

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

Ans: 9600

Explanation:

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.

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:

Ans: 4

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

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

Ans: 40

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.

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

Ans: 111

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.

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

Ans: 2

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.

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

Ans: 0.18

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.

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

Ans: 364

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

Ans: 23

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.

12. 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:

Ans: 1683

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.

13. 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 ?

Ans: 46 minutes and 12 seconds

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.

14. 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:

Ans: 308

Explanation:

Other number = (11*7700 / 275) = 308

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

Ans: 48

Explanation:

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

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