A survey of 500 college students found that the percentage of students who went without using digital technology for up to 1 hr was 67. The survey also determined that the percentage of students who went without using digital technology for up to 30 min exceeded the percentage of students who went without using digital technology for over 1 hr by 17. Let x, y, and z represent the percentage of the students in the survey who went without using digital technology a for up to 30 min, b for more than 30 min but not more than 60 min, and c for more than 60 min, respectively. Find the values of x, y, and z

x = 50%

y = 17%

z = 33%

Explanation:

From the question, it is given that

x% represents those that don't use digital technology for up to 30 mins

y% represents those that don't use digital technology for more than 30 mins but less than 60 mins

z% represents those that don't use digital technology for more than 60 mins

Total percentage = 100% ofcourse

Then, percentage that don't use their phones for up to an hour, (x + y), is 67%

The percentage of students who went without using digital technology for up to 30 min (x%) exceeded the percentage of students who went without using digital technology for over 1 hr (z%) by 17%; x = z + 17

x + y + z = 100% (eqn 1)

x + y = 67% (eqn 2)

x = z + 17% (eqn 3)

Writing y and z as functions of x from eqn 2 and eqn 3

y = 67 - x

z = x - 17

Then substituting these into eqn 1

x + y + z = 100

x + (67 - x) + (x - 17) = 100

x + 67 - x + x - 17 = 100

x + 50 = 100

x = 50%

y = 67 - x = 67 - 50 = 17%

z = x - 17 = 50 - 17 = 33%