**The correct answer is d) 1020**

In the given numbers 1, 3,5,7,…, 147, 148, 151the numbers which are multiple of 5 are 5, 15, 25, 35,…, 145 which are an arithmetic sequence .

\(T_n=a+(n-1)d\)

\(=145=5+(n-1)10\)

\(\therefore n=15\)

and if total number of terms in the given sequence is m ,then

\(151=1+(m-1)2\)

\(\therefore m=76\)

so, the number of ways in which product is a multiple of 5=(both two numbers from 5, 15 , 25 ,.....,145) or (one number from 5, 15, 25 ,.....145 and one from remaining)

\(=^{15}C_2+^{15}C_1\times ^{76-15}C_1\)

\(=^{15}C_2+^{15}C_1\times ^{61}C_1\)

\(={ 15\times14\over 2}+15\times{61}\)

\(=105+915
=1020\)