LCM and HCF Practice Test 5

The least number which when divided by 48, 64, 90, 120, will leave remainders 38, 54, 80, 110.
It is always observed that
(x−a)=(y−b)= (z−c)=(p−r)=k‌say
where x = 48, y = 64, z = 90, p = 120 and
a = 38, b = 54, c = 80, r = 110
(48−38)=(64−54)=(90−80)=(120−110)=k
So, k = 10
Now find the LCM of x, y, z and p.
LCM of 48, 54, 80, 110 is 2,880.
Required number = (LCM of x, y, z and p) - k
   = 2,880 - 10
   = 2,870