Question

Let X and Y be the sets of all positive divisors of 400 and 1000 respectively (including 1 and the number) . Then $n\left( X\cap Y \right)=$ :
1) 4
2) 6
3) 8
4) 12

Answer 1

Hint: Here in this question we have been asked to find the value of $n\left( X\cap Y \right)$ given that $X$ and $Y$ be the sets of all positive divisors of 400 and 1000 respectively (including 1 and the number). For answering this question we will list out the elements in each set and find their intersection.

Complete step-by-step solution:
Now considering from the question we have been asked to find the value of $n\left( X\cap Y \right)$ given that $X$ and $Y$ be the sets of all positive divisors of 400 and 1000 respectively (including 1 and the number).
For answering this question we will list out the elements in each set and find their intersection.
The divisors of 400 are 1,2,4,5,8,10,16,20,25,40,50,80,100,200,400.
The divisors of 1000 are 1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000.
Hence we can say that the set $X$ is $\left\{ 1,2,4,5,8,10,16,20,25,40,50,80,100,200,400 \right\}$ .
Hence we can say that the set $Y$ is $\left\{ ~1,2,4,5,8,10,20,25,40,50,100,125,200,250,500,1000 \right\}$ .
Hence we can say that the intersection of both the sets that is $X\cap Y$ is $\left\{ 1,2,4,5,8,10,20,25,40,50,100,200 \right\}$ .
Therefore we can conclude that when it is given that $X$ and $Y$ be the sets of all positive divisors of 400 and 1000 respectively (including 1 and the number) then the value of $n\left( X\cap Y \right)$ will be given as 12.
Hence we will mark the option “D” as correct.

Note: In the process of answering questions of this type, we should carefully list out all the elements in the particular sets and clearly observe the number of common elements. If someone has missed out on any one element in a hurry then they will end up having a wrong conclusion.