Jennifer, Bridget, Sandra, Robertina, and Joycelyn have their birthdays on consecutive days, but not necessarily in that order. Jennifer’s birthday is as many days before Sandra’s as Bridget’s is afte

Question

Jennifer, Bridget, Sandra, Robertina, and Joycelyn have their birthdays on consecutive days, but not necessarily in that order. Jennifer’s birthday is as many days before Sandra’s as Bridget’s is after Joycelyn’s. Robertina is two days older than Joycelyn. Sandra’s birthday this year is on a Wednesday. On what days of the week are the birthdays of the other four this year?

Answer 1

R=Jo-2 Robertyna is 2 days before Joycelyn, Jo = 3, 5, or 7, R=1, 3 or 5

S-Je = B-Jo = 1,2 or 3 Jen is as may days before Sandra's as Bridget is days after Joycelyn

S=4 (Wed is 4th day of the week)

4-Je = B- Jo

4-Je = 1,2 or 3, Je=3,2 or 1, Tuesday, Monday or Sunday

if Je =3 then B-Jo =1. Jo = 5 so B=6, R=3 but then Je = R

try Je=2 then b-Jo =2 Jo=5, so B= 7, R=5 but then Jo=R

try Je=3, then B-Jo=1, Jo=5, so B=6, R=7

Jen = 3 = Tuesday

Sandar 4 = = Wednesday

Joycelyn = 5 = Thursday

Bridget =6 = Friday

Robertina = 7 = Saturday

S-Je = B-Jo 4-3 =6-5

R-Jo =2 7-5=2