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
$\left(\mathit{x}-\mathit{a}\right)=\left(\mathit{y}-\mathit{b}\right)=$ $\left(\mathit{z}-\mathit{c}\right)=\left(\mathit{p}-\mathit{r}\right)=\mathit{k}\mathrm{say}$
where x = 48, y = 64, z = 90, p = 120 and
a = 38, b = 54, c = 80, r = 110
$\left(48-38\right)=\left(64-54\right)=\left(90-80\right)$$=\left(120-110\right)=\mathit{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