A man bought 5 reams of duplicating paper, each of which are supposed to contain 480 sheets. The actual number of sheets in the packets were : 435, 420, 405, 415 and 440.
(a) Calculate, correct to the nearest whole number, the percentage error for the packets of paper;
(b) If the agreed price for a full ream was N35.00, find, correct to the nearest naira, the amount by which the buyer was cheated.
Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.
Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).
Total shortage \(= 2400 - 2115 = 285\) sheets.
(a) Percentage error.
\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]
To the nearest whole number, the percentage error \(= \mathbf{12\%}\).
(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.
The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:
\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]
To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).
Each ream is supposed to contain 480 sheets, so 5 reams should hold \(5 \times 480 = 2400\) sheets.
Actual sheets: \(435 + 420 + 405 + 415 + 440 = 2115\).
Total shortage \(= 2400 - 2115 = 285\) sheets.
(a) Percentage error.
\[\text{Percentage error} = \frac{\text{shortage}}{\text{supposed number}}\times 100 = \frac{285}{2400}\times 100 = 11.875\%.\]
To the nearest whole number, the percentage error \(= \mathbf{12\%}\).
(b) Amount cheated. A full ream of 480 sheets costs \(\text{N}35.00\), so one sheet is worth \(\dfrac{35}{480}\) naira.
The buyer paid for 2400 sheets but received only 2115, a shortage of 285 sheets:
\[\text{Amount cheated} = 285 \times \frac{35}{480} = \frac{9975}{480} = \text{N}20.78.\]
To the nearest naira, the buyer was cheated by \(\mathbf{\text{N}21}\).