In A Survey Of Defaulted Payments Of Electrical Bills Of A Residentia

1.	In A Survey Of Defaulted Payments Of Electrical Bills Of A Residential Complex Of 125 Houses, It Is Found That 50 Houses Defaulted On Their Payment Of Electrical Bills In January, 60 In February And 40 In March. Houses Can Default In Consecutive Months Only. 20 Defaulted In January And February. 10 Defaulted In February And March. How Many Houses Defaulted In All The 3 Months?
Answer» We use formula for intersection of three sets, keeping in mind that Jan ∩ Mar does not exist, SINCE they are not consecutive months. Let X be the number of people defaulting in all 3 months. We GET the equation as : 125 = 50 + 60 + 40 – 20 – 10 + x. Solving we get x = 5. We use formula for intersection of three sets, keeping in mind that Jan ∩ Mar does not exist, since they are not consecutive months. Let x be the number of people defaulting in all 3 months. We get the equation as : 125 = 50 + 60 + 40 – 20 – 10 + x. Solving we get x = 5.

In A Survey Of Defaulted Payments Of Electrical Bills Of A Residential Complex Of 125 Houses, It Is Found That 50 Houses Defaulted On Their Payment Of Electrical Bills In January, 60 In February And 40 In March. Houses Can Default In Consecutive Months Only. 20 Defaulted In January And February. 10 Defaulted In February And March. How Many Houses Defaulted In All The 3 Months?

Answer»

We use formula for intersection of three sets, keeping in mind that Jan ∩ Mar does not exist, SINCE they are not consecutive months.

Let X be the number of people defaulting in all 3 months.

We GET the equation as : 125 = 50 + 60 + 40 – 20 – 10 + x. Solving we get x = 5.

We use formula for intersection of three sets, keeping in mind that Jan ∩ Mar does not exist, since they are not consecutive months.

Let x be the number of people defaulting in all 3 months.

We get the equation as : 125 = 50 + 60 + 40 – 20 – 10 + x. Solving we get x = 5.