after replacing an old member by a new member , it was found that the average age of 5 members of a club is the same as it was 3 years ago . what is the difference between the ages of the replaced and the new member ?
explanation : i ) let the ages of the five members at present be a , b , c , d & e years . and the age of the new member be f years . ii ) so the new average of five members ' age = ( a + b + c + d + f ) / 5 - - - - - - - ( 1 ) iii ) their corresponding ages 3 years ago = ( a - 3 ) , ( b - 3 ) , ( c - 3 ) , ( d - 3 ) & ( e - 3 ) years so their average age 3 years ago = ( a + b + c + d + e - 15 ) / 5 = x - - - - - ( 2 ) = = > a + b + c + d + e = 5 x + 15 = = > a + b + c + d = 5 x + 15 - e - - - - - - ( 3 ) iv ) substituting this value of a + b + c + d = 5 x + 15 - e in ( 1 ) above , the new average is : ( 5 x + 15 - e + f ) / 5 equating this to the average age of x years , 3 yrs , ago as in ( 2 ) above , ( 5 x + 15 - e + f ) / 5 = x = = > ( 5 x + 15 - e + f ) = 5 x solving e - f = 15 years . thus the difference of ages between replaced and new member = 15 years . answer : d
multiply(5, 3)