Question

If P = {factors of 36} and Q = {factors of 48}, find \[P\bigcup Q=K\], and if sum of each element in K is s, then \[\dfrac{s+3}{10}-10=\]

Answer 1

Hint: We will use the concept of factors and union of two sets to solve this question. Factors are numbers which we multiply together to get the desired number. Also the union of a collection of sets is the set of all elements in the collection.
Complete step-by-step answer:
We know that factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18, 36.
And factors of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48.
So, P is equal to { 1, 2, 3, 4, 6, 9, 12, 18, 36 } and Q is equal to { 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 }.
We also know that the union of two sets has all the elements of both the sets P and Q. So using this information we get,
\[P\bigcup Q=\{1,2,3,4,6,8,9,12,16,18,24,36,48\}........(1)\]
Now it is mentioned in the question that the sum of each element in K is s. So from equation (1) finding s we get,
\[\begin{align}
  & \Rightarrow s=1+2+3+4+6+8+9+12+16+18+24+36+48 \\
 & \Rightarrow s=187.....(2) \\
\end{align}\]
Now we have to find the value of \[\dfrac{s+3}{10}-10\]. So substituting the value of s from equation (2) in this expression we get,
\[\Rightarrow \dfrac{s+3}{10}-10=\dfrac{187+3}{10}-10=\dfrac{190}{10}-10=19-10=9\]
Hence the value of \[\dfrac{s+3}{10}-10\] is 9.

Note: Remembering the concept of factors and union of sets is the key here. We in a hurry can make a mistake in writing the union of P and Q in equation (1) by missing some elements and hence we need to be careful while solving this step.