business mathematics

Problems on hcf and lcm

Problems on hcf and lcm

1.The HCF of 2 numbers is 11 and LCM is 693.If one of numbers
is 77.find other.

Sol: Other number = 11 * 693/77=99.

2.Find largest number of 4 digits divisible by 12,15,18,27

Sol: The largest number is 9999.
LCM of 12,15,18,27 is 540.
on dividing 9999 by 540 we get 279 as remainder.
Therefore
number =9999 – 279 =9720.

3.Find least number which when divided by 20,25,35,40 leaves
remainders 14,19,29,34.

Sol: 20–14=6
25-19=6
35-29=6
40-34=6
Therefore number =LCM of (20,25,35,40) - 6=1394

4.252 can be expressed as prime as :
2 252
2 126
3 63
3 21
7
prime factor is 2 *2 * 3 * 3 *7

5.1095/1168 when expressed in simple form is
1095)1168(1
1095
------
73)1095(15
73
---------
365
365
---------
0
----------
So, HCF is 73
Therefore
1095/1168 = 1095/73/1168/73= 15/16

6.GCD of 1.08,0.36,0.9 is

Sol:
HCF of 108,36,90 36)90(2
72
----
18)36(2
36
----
0
----
HCF is 18.
HCF of 18 and 108 is 18
18)108(6
108
-------
0
--------
Therefore HCF =0.18

7.Three numbers are in ratio 1:2:3 and HCF is 12.Find numbers.

Sol:
Let the numbers be x.
Three numbers are x,2x,3x
Therefore
HCF is 2x)3x(1
2x
-----
x)2x(2
2x
--------
0
-------------

HCF is x so, x is 12
Therefore numbers are 12,24,36.

8.The sum of two numbers is 216 and HCF is 27.

Sol: Let numbers are 27a + 27 b =216
a + b =216/27=8
Co-primes of 8 are (1,7) and (3,5)
numbers=(27 * 1 ), (27 * 7)
=27,89

9.LCM of two numbers is 48..The numbers are in ratio 2:3. The sum
of numbers is

Sol:
Let the number be x.
Numbers are 2x,3x
LCM of 2x,3x is 6x
Therefore
6x=48
x=8.
Numbers are 16 and 24
Sum=16 +24=40.

10.HCF and LCM of two numbers are 84 and 21.If ratio of two numbers
is 1:4.Then largest of two numbers is

Sol:
Let the numbers be x,4x
Then x * 4x = 84 * 21
x2 =84 * 21 /4
x = 21
Largest number is 4 * 21.

11.HCF of two numbers is 23,and other factors of LCM are 13,14.
Largest number is

Sol:
23 * 14 is Largest number.

12.The maximum number of students among them 1001 pens and 910
pencils can be distributed in such a way that each student gets
same number of pens and pencils is ?

Sol:
HCF of 1001 and 910
910)1001(1
910
------------
91)910(10
910
--------
0
---------
Therefore HCF=91

13.The least number which should be added to 2497 so that sum is
divisible by 5,6,4,3 ?

Sol: LCM of 5,6,4,3 is 60.
On dividing 2497 by 60 we get 37 as remainder.
Therefore number to added is 60 – 37 =23.
Answer is 23.

14.The least number which is a perfect square and is divisible by
each of numbers 16,20,24 is ?

Sol: LCM of 16,20,24 is 240.
2 * 2*2*2*3*5=240
To make it a perfect square multiply by 3 * 5
Therefore 240 * 3 * 5=3600
Answer is 3600.