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A natural number n is such that $120≤ n ≤ 240.$ If HCF of $n$ and $240$ is $1,$ how many values of $n$ are possible?

$A)24$                   $B)32$                     $C)36$                     $D)40$
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HCF of $n$ and $240$ is $1$ 

$240 = 2^4 \times3\times 5$

so we need to find the numbers, such that, $120 \leq n \leq 240$ , and not divisible of $2$ or $3$  or $5$

Numbers divisible by $2$ such that $120 \leq n \leq 240 = 120 - 59 = 61$

Numbers divisible by $3$ such that $120 \leq n \leq 240 = 80  -39 = 41$

Numbers divisible by $5$ such that $120 \leq n \leq 240 = 48 - 23 = 25$

Numbers divisible by $2$ and $3$ such that $120 \leq n \leq 240 = 40 -19 = 21$

Numbers divisible by $2$ and $5$ such that $120 \leq n \leq 240 = 24 -11 = 13$

Numbers divisible by $3$ and $5$ such that $120 \leq n \leq 240 = 16 -7 = 9$

Numbers divisible by $2$ ,$3$,and $5$ such that $120 \leq n \leq 240 = 8 -3 = 5$

Numbers divisible by $2$ or $3$ or $5$ such that $120 \leq n \leq 240 = 61+41+25-21-13-9+5= 89$

Total numbers such that $120 \leq n \leq 240 =121$

Total numbers such that $120 \leq n \leq 240$ and not divisible by $2$ or $3$ or $5 = 121-89 = 32$

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