Suppose $G$ is a group with exactly $8$ elements of order $10$. How many cyclic subgroups of order $10$ does $G$ have? In a cyclic subgroup of order $10$, there are $\phi(10)=4$ elements of order $10$. 
Since, there are exactly $8$ elements of order $10$, we can choose $4$ elements out of the $8$ elements of order $10$ in $^8C_4=70$ ways and for each way, we have a cyclic subgroup of order $10 (4$ elements will of order 4 and the rest 4 will come from group G). Thus, $G$ has $70$ nos. of cyclic subgroups of order $10$. 
But the answer to this question is $2$ since a cyclic subgroup of order $10$ can only have $4$ elements of order $10$ and hence there are $8/4$ cyclic subgroups only. But I don't understand why? There are $^8C_4=70$ ways to choose $4$ elements out of $8$ elements. So I think it should be $70$.  
Please help me understand. Thanks in advance.
PS: I know that in a finite group, no. of subgroups should divide the order of the group.  However, in my case since order of $G$ is not given, I am confused.
I claim that there is no group with exactly $8$ elements of order $10$, which makes the whole question logically meaningless (except possibly as a lemma in the proof that there is no such group).

Such a group $G$ would have two exactly two distinct subgroups $A$ and $B$ of order $10$. If they were not normal in $G$, then they would be conjugate and their normalizer would have index $2$ and contain both $A$ and $B$, so we could replace $G$ by this subgroup to get a smaller example.

So we can assume that $G = \langle A,B \rangle = AB$ with $A,B \unlhd G$.

Since $|{\rm Aut}(A)| = 4$, the element $g$ of order $5$ in $B$ must centralize $A$. Then, if $g \not\in A$, then $\langle g, A \rangle \cong C_5 \times C_{10}$ has $6$ subgroups of order $15$, contrary to assumption.

So $g \in A$, $|A \cap B| = 5$, and $|G| = |AB| = 20$.

Now, of the five (isomorphism types of) groups of order $20$, $C_{20}$, $D_{20}$ and the dicyclic group have a unique subgroup of order $10$, $C_2 \times C_{10}$ has three such subgroups, and the final group, which is a semidirect product $C_5 \rtimes C_4$ with faithful action has none.
As you have correctly pointed out that each cyclic subgroup will have exactly 4 elements of order 10.

claim:- if ${H_k}\cap{H_l}$ contains an element (say) $a_i$ of order 10 then $H_k=H_l$

Since ${a_i}\in{H_k}\cap{H_l}$ and order of $a_i$ is 10 then,
$&lt;a_i&gt;=H_k$ and $&lt;a_i&gt;=H_l$
Hence $H_k=H_l$

Did you got why there will be only two cyclic subgroup of order 10?