$A = {\frac{4}{7}, 1, \frac{5}{2}, 4, \frac{11}{2}, 7}$
$B = {\frac{4}{7}, \frac{7}{4}, 4, 7}$
If $n$ is a member of both set $A$ and set $B$ above, which of the following must be true?
I. $n$ is an integer
II. $4 n$ is an integer
III. $n = 4$

None

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

II only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I and II only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I and III only

No worries! We‘ve got your back. Try BYJU‘S free classes today!

I, II, and III

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B None
Given that $n$ is a member of both set $A$ and set $B$
The common members of set $A$ and set $B$ are $\frac{4}{7}, 4, 7$
So $n$ can be $\frac{4}{7}$ or $4$ or $7$
$n$ need not be integer $(\frac{4}{7}$ is not an integer $)$ .
$4 n$ need not be integer $(\frac{16}{7}$ is not an integer $)$
$n$ need not to be $4$ , it can be $4$ or $7$ or $\frac{4}{7}$
Therefore all the given statements are wrong
Therefore correct option is $A$

Suggest Corrections

Similar questions

= {\frac{4}{7}, 1, \frac{5}{2}, 4, \frac{1}{2}, 7}

= {\frac{4}{7}, \frac{7}{4}, 4, 7}

4 n

n = 4

Let A and B be two sets such that n(A)=20, n(AUB)=42, n(AintersectionB)=4.

Find i) n(B)

ii) n(A-B)

iii) n(B-A)

Let A and B be two sets such that : n (A) = 20 , $n (A \cup B) = 42 a n d n (A \cap B) = 4$ . Find

(i) n (B) (ii) n (A - B)

(iii) n (B - A)

n^{3}

n^{2}

n^{3}

Join BYJU'S Learning Program