# Percentage Error

It is hard for one to take a measurement and get an exact result. Percentage error is a little degree in percent at which a known length is measured above/below its actual value.

The error that might be made during the measurement could be either positive or negative, in any ways just know that error still remains error provided that it has a value order than the actual result.

For instance if the actual length of a room is 3.0m. It is said that three students on their industrial Training measured it differently and they got:
2.85m, 3.01m and 2.91m respectively. What are their errors?

Error = (Wrong value – Actual value) or (Actual value – Wrong value)

The first student’s error
= 3.0 – 2.85 = 0.15m

Second student’s error
= 3.01 – 3.0 = 0.01m

Third student’s error
= 3.0 – 2.91 = 0.09m

Note that the ones we just did above are the errors. For one to calculate the percentage error, this formula will be used:

Percentage error (PE)
= Absolute error (AE) divided by Actual value (AV) and you multiply by 100.

(PE) = (AE/AV) x 100

Now, the percentage error of the example above are:

For the first student:

AE = 0.15m.
AV = 3.0m
PE = ?

PE = AE/AV x 100
= 0.15/3 x 100
= 15/3 = 5%

For the second student:

AE = 0.01m.
AV = 3.0m
PE = ?

PE = AE/AV x 100
= 0.01/3 x 100
= 1/3% = 0.33%

For the third student:

AE = 0.09m.
AV = 3.0m
PE = ?

PE = AE/AV x 100
= 0.09/3 x 100
= 9/3 = 3%

## QUESTIONS AND SOLUTIONS:

### Q1: A candidate was to subtract 15 from a certain number, but mistakenly added 25 and his answer was 145. Find the percentage error.

Solution:
Let the certain number be y.

He was meant to subtract 15 from it and mistakenly added 25.

y + 25 = 145
y = 120 = the certain number

Actual value = 120 – 15 = 105

Absolute Error = 145 – 105 = 40

PE = 40/105 x 100
= 38.1%

### Q2: The length of a wire is 6.35, a student measured it as 6.65. What is the percentage error to 1 decimal place?

AE = 6.65 – 6.35 = 0.3
AV = 6.35
PE = ?

PE = AE/AV x 100
= 0.3/6.35 x 100
= 4.7% (1 dp)

### Q3: If the age of a man 64 years is written as 71 years, calculate the percentage error to 3 significant figures.

Solution:
AE = 71 – 64 = 7 years
AV = 64
PE = ?

PE = AE/AV x 100
= 7/64 x 100
= 10.9% (3 SF)

### Q4: Find the percentage error in a piece of wood that was measured to be 1.26m whose actual length was 1.24m.

Solution:
AE = 1.26 – 1.24 = 0.02m
AV = 1.24
PE = ?

PE = AE/AV x 100
= 0.02/1.24 x 100
= 1.6%

### Q5: A man underestimated his expenses by 6.5% but actually spent #400.00. What was his estimate?

Solution:
Let his estimate be D
AV = #400
AE = (400 – D)
PE = 6.5%

PE = [(400 – D)/400] x 100 = 6.5

Solve for D.

PE = (400 – D) = 26
D = 400 – 26 = #374

### Q6: An error of 4% was made in finding the length of rope that was actually 25m. By how many metres was the measurement wrong?

Solution:

Let the wrong value be T
AV = 25m
AE = (T – 25)m
PE = 4%

PE = [(T – 25)/25] x 100 = 4

Solve for T.

100(T – 25) = 4 x 25
100(T – 25) = 100
T – 25 = 1
T = 1 + 25 = 26m

Therefore the measurement is ±1m wrong.

Students by now will be wondering why my absolute error (AE) is = (T – 25)m and not (25 – T)m. Note that in which ever way you place it, it is still the same provided that the question comes in that nature.

Now, let’s use the reverse (25 – T)m as our AE.

PE = [(25 – T)/25] x 100 = 4

100(25 – T) = 25 x 4
100(25 – T) = 100

Solve for T
25 – T = 1
-T = 1 – 25
-T = -24 = 24m

Actual length = 25m
So the measurement is ±1m wrong.

### Q7: What is the percentage error in an area of a lawn that actually measures 750m² but found to be 690m²?

Solution:
AV = 750m²
AE = 750 – 690 = 60m²
PE = ?

PE = 60/750 x 100
= 8%

### Q8: The percentage error in the measurement of the length of a rope was 6%. If the measurement was 35m, find the actual length of the rope and by how many metres was the measurement wrongto 1 decimal place.

Solution:
Let the actual length be x.

PE = 6%
AE = (x – 35)m

[(x – 35)/x] X 100 = 6

Solve for x.
100(x – 35) = 6x
100x – 3500 = 6x
94x = 3500
x = 37.2m

So, the measurement is 37.2 – 35 = ±2.2% wrong

Solution:
AE = 15cm
AV = 165cm
PE = ?

PE = AE/AV x 100
= (15/165) x 100
= 9.1%

### Q10: The length and breath of a rectangle was mistakenly measured as 40m and 35m instead of 42.5m and 34.2m respectively. Find the percentage error in:(a) the area(b) the perimeter

Solution:
(a) the area

Actual area = 40 x 35 = 1400m²

Wrong area = 42.5 x 34.2 = 1453.5m²

AV = 1400
AE = 1453.5 – 1400 = 53.5m

PE = (53.5/1400) x 100 = 3.8%

(b) the perimeter

Perimeter of the actual rectangular = 40 + 35 = 75m

Wrong Perimeter = 42.5 + 34.2 = 76.7m

AE = 76.7 – 75 = 1.7m

Therefore:
PE = (1.7/75) x 100
= 2.3%

### Q11: The length of a file 25cm was measured as 27.5cm. Calculate the percentage error.

Solution:

Actual value = 25cm
Wrong value = 27.5cm
Absolute Error =2.5cm
Percentage error = ?

PE = (2.5/25) x 100
= 250/25 = 10%