
If we are given the sets as $A=\left\{ 2x:x\in N \right\}$, $B=\left\{ 3x:x\in N \right\}$ and $C=\left\{ 5x:x\in N \right\}$ then $\left( A\cap B\cap C \right)$ is equal to?
(a) {15, 30, 46, ……}
(b) {10, 20, 30, ……}
(c) {30, 60, 90, ……}
(d) {7, 14, 21, ……}
Answer
531.6k+ views
Hint: Find some of the elements of set A, B and C by substituting the values of x. Now, to find the elements present in the intersection of the three sets check for the common elements present in all of them. To determine these common elements take the L.C.M of 2, 3 and 5. The set containing the multiples of the obtained L.C.M will be our answer.
Complete step-by-step solution:
Here we have been provided with three sets $A=\left\{ 2x:x\in N \right\}$, $B=\left\{ 3x:x\in N \right\}$ and $C=\left\{ 5x:x\in N \right\}$ and we are asked to find the elements that will be obtained by considering the intersection of the three sets, i.e. $\left( A\cap B\cap C \right)$.
Now, let us see some of the elements of the three sets by substituting some values of x that belong to natural numbers according to the given question $\left( x\in N \right)$.
$A=\left\{ 2,4,6,8,10,.... \right\}$
$B=\left\{ 3,6,9,12,15,.... \right\}$
$C=\left\{ 5,10,15,20,25,.... \right\}$
Clearly we can see that set A contains multiples of 2, set B contains multiples of 3 and set C contains multiples of 5. So the intersection of these sets means that we have to consider all the elements that will be common in each of them. So, the first common element will be the L.C.M of 2, 3 and 5 and the rest elements will be the multiples of this L.C.M. Therefore, we need to find the L.C.M of 2, 3 and 5.
As we can see that 2, 3 and 5 are all prime numbers, so their L.C.M will be given as $2\times 3\times 5=30$. Therefore, the elements in the set $\left( A\cap B\cap C \right)$ will be given as: -
$\begin{align}
& \Rightarrow \left( A\cap B\cap C \right)=\left\{ 30x:x\in N \right\} \\
& \Rightarrow \left( A\cap B\cap C \right)=\left\{ 30,60,90,120,..... \right\} \\
\end{align}$
Hence, option (c) is the correct answer.
Note: Always remember that intersection of the sets means the set of elements common in all the given sets while union of sets means the set of elements present in any of the given sets. The symbol for the union of sets is $\cup $. Always remember the process to find the L.C.M of two or more numbers. Do not consider the H.C.F and its multiples as the answer here.
Complete step-by-step solution:
Here we have been provided with three sets $A=\left\{ 2x:x\in N \right\}$, $B=\left\{ 3x:x\in N \right\}$ and $C=\left\{ 5x:x\in N \right\}$ and we are asked to find the elements that will be obtained by considering the intersection of the three sets, i.e. $\left( A\cap B\cap C \right)$.
Now, let us see some of the elements of the three sets by substituting some values of x that belong to natural numbers according to the given question $\left( x\in N \right)$.
$A=\left\{ 2,4,6,8,10,.... \right\}$
$B=\left\{ 3,6,9,12,15,.... \right\}$
$C=\left\{ 5,10,15,20,25,.... \right\}$
Clearly we can see that set A contains multiples of 2, set B contains multiples of 3 and set C contains multiples of 5. So the intersection of these sets means that we have to consider all the elements that will be common in each of them. So, the first common element will be the L.C.M of 2, 3 and 5 and the rest elements will be the multiples of this L.C.M. Therefore, we need to find the L.C.M of 2, 3 and 5.
As we can see that 2, 3 and 5 are all prime numbers, so their L.C.M will be given as $2\times 3\times 5=30$. Therefore, the elements in the set $\left( A\cap B\cap C \right)$ will be given as: -
$\begin{align}
& \Rightarrow \left( A\cap B\cap C \right)=\left\{ 30x:x\in N \right\} \\
& \Rightarrow \left( A\cap B\cap C \right)=\left\{ 30,60,90,120,..... \right\} \\
\end{align}$
Hence, option (c) is the correct answer.
Note: Always remember that intersection of the sets means the set of elements common in all the given sets while union of sets means the set of elements present in any of the given sets. The symbol for the union of sets is $\cup $. Always remember the process to find the L.C.M of two or more numbers. Do not consider the H.C.F and its multiples as the answer here.
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