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## SS2: MATHEMATICS - 1ST TERM

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Lesson 6, Topic 5
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# Percentage Error

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#### 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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