Question

A student's calculation was found to have a 35.5% error, and his experimental measurement was 15.6 cm. What are the two possible values for the actual measurement?

Answer 1

Let the two possible actual measurements be .x' and 'y', where x is a measurement less than 15.6 and y is a measurement more than 15.6 

x + 35.5% of x = 15.6
y - 35.5% of y

35.5% as decimal is 0.355

x + 0.355x = 15.6, and
y - 0.355y = 15.6

Solving x + 0.355x = 15.6
1.355x = 15.6
x = 15.6÷1.355 = 11.53

Solving y - 0.355x = 15.6
0.645y = 15.6
y = 15.6 - 0.645
y = 24.19 

Two possible measurements are 11.53 and 24.19