#### Topic Content:

- Percentage Error

### What is Percentage Error?

The percentage error of a measurement is defined as the percentage ratio of the absolute error to the actual value.

i.e. Percentage error = \( \frac {absolute\: error } {actual\: error} \scriptsize \: \times \: 100% \)

or Relative error = \( \frac {absolute\: error } {actual\: value} \scriptsize \: \times \: 100% \)

Note that absolute error is the difference between the measured value and the actual value.

### Example 6.5.1:

A salesperson gave a change of #1.15 to a customer instead of #1.25. Calculate the percentage error. (SSCE).

**Solution:**

error = 1.25 – 1.15 = 0.1

% error = \( \frac {error } {actual\: value} \scriptsize \: \times \: 100% \)

= \( \frac {0.1}{1.25} \scriptsize \: \times \: 100% \)

= 8%

### Example 6.5.2:

A gardener measures the length and breadth of a rectangular lawn as 59.6 m and 40.3 m respectively instead of 60 m and 40 m. What is the percentage error in his calculation of the perimeter of the lawn? (WAEC).

**Solution:**

Measured perimeter = 2 (59.6 + 40.3)

= 2 (99.9)

= 199.8 m

Actual Perimeter = 2(60 + 40)

= 2(100)

= 200 m

Therefore, error = 200 – 199.8

% error = \( \frac {error } {actual\: value} \scriptsize \: \times \: 100% \)

%error = \( \frac {0.2}{200} \scriptsize \: \times \: 100\% \)

= 0.1%

### Example 6.5.3:

A boy measured the length of a ladder as 7.96 m. The percentage error made is 5%. If this measurement is smaller than the exact length. Find the exact length of the ladder.

**Solution:**

Let the exact length of the ladder be y cm

% error = \( \frac {error } {actual\: value} \scriptsize \: \times \: 100% \)

% error = \( \frac {y \:- \:7.96}{y} \scriptsize \: \times \: 100% \)

⇒ 5 = \( \frac {y \: – \: 7.96}{y} \scriptsize \: \times \: 100% \) (multiply both sides by y)

⇒ 5y = (y – 7.96) × 100 (divide both sides by 5)

y = \( \frac{100}{5} \scriptsize (y \: – \: 7.96) \)

y = 20 (y – 7.96) (expand the brackets)

y = 20y – 159.2

19y = 159.2

y = \( \frac {159.2}{19} \)

y = 8.3789

The exact length is y = 8.38 m (2 d.p)

### Example 6.5.4:

A man bought 5 reams of duplicating paper, each of which was supposed to contain 480 sheets. The actual number of sheets in the packets were 435, 420, 405, 415, and 440.**1. **Calculate, correct to the nearest whole number, the average percentage error for the packets of paper.**2.** If the agreed price for a full ream was #35. Find, correct to the nearest naira, the amount by which the buyer was cheated.

**Solution: **

**1.** Expected number of sheets = 5 × 480 = 2400 sheets

Number of sheets found in the packet

= 435 + 420 + 405 +415 + 440

= 2115 sheets

⇒ error = 2400 – 2115

= 285 sheets

% error = \( \frac {error } {actual\: value} \scriptsize \: \times \: 100% \)

⇒ % error = \( \frac {285}{2400} \scriptsize \: \times \: 100% \)

= 11.875%

⇒ % error = 12% (to the nearest whole number)

**2. **Cost of 5 reams = 5 × 35 = #175

2400 sheets = #175

285 sheets = #x

⇒ #x = \( \frac {285}{2400} \scriptsize \: \times \: 175 \)

⇒ #x = 20.78125

The amount the buyer was cheated

= #21.00 (to the nearest Naira)

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