Find the least number which when divided by 12, 16 and 18 leaves 5 as remainder in each case.
Options:
148
149
150
151
LCM of 12, 16 and 18 = 144
The required least number = 144+5 = 149
Find the greatest 3 digit number which when divided by 6, 9 and 12 leaves 3 as remainder in each case.
965
975
985
995
The greatest 3 digit number is 999.
LCM of 6, 9 and 12 is 36.
When we divide 999 by 36 we get 27 as remainder.
So, the required 3 digit number is (999 - 27 + 3) i.e., 975.
Rocky has two numbers in the ratio 3:4. Find the LCM if their HCF is 4.
12
24
36
48
Let the two numbers be 3q and 4q where q is the HCF.
Given, HCF = 4.
So, the two numbers are 12 and 16.
Therefore, the required LCM = 48
If the HCF of two numbers is 8 then which of the following can never be their LCM.
60
56
LCM of two numbers is always a multiple of their HCF.
Given, HCF = 8
Now, from the list of options we see that 60 is not a multiple of 8.
Hence, 60 can never be the LCM of the two numbers.
The product of two coprime numbers is 117. Find their LCM.
117
111
171
None of these
HCF of two coprime numbers = 1.
We know, that Product of two numbers = HCF of the numbers x LCM of the numbers.
Therefore, 117 = 1 x LCM
or, LCM = 117
Or, we can say that LCM of two coprime numbers is equal to the product of the numbers.
Find the HCF of 1/2 and 3/4.
1/3
1/4
4/3
2/3
Required HCF = HCF of numerators / LCM of denominators
= HCF(1,3) / LCM(2,4)
= 1/4
Find the LCM of 1/3, 2/9, 5/6 and 4/27.
10/3
20/3
3/10
3/20
Required LCM = LCM of numerators / HCF of denominators
= LCM(1,2,5,4) / HCF(3,9,6,27)
= 20/3
Jai, Veeru and Dhanno are friends and they have 3 numbers 2a, 4a and 8a respectively. Find the LCM of their numbers.
2a
4a
8a
16a
LCM = 2x2x2xa = 8a