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22 September, 08:29

Claudia buys 12 postcards, 12 stamps, and 1 pen. The postcards cost twice as much as the stamps. The pen costs $1.50. The total cost is $ 14.10. How much does each postcard cost? Show your work.

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Answers (2)
  1. 22 September, 10:17
    0
    the cost of a post card is $0.7

    Step-by-step explanation:

    Given

    Number of postcard = 12

    Number of stamp = 12

    Number of pen = 1

    Cost of a pen = $1.5

    Required

    Cost of each postcard ...

    Let P represent postcards, S represent stamps, Sp represented Pen.

    Since there's a total of 12 postcard and 12 staffs, we have:

    12P + 12S + 1.50 = $14.10

    From the question, we have that the postcards cost twice as much as the stamps.

    So, P = 2S.

    Substitute 2S for P

    12 * 2S + 12S + 1.5 = 14.10

    24S + 12S + 1.5. = 14.10

    36S + 1.5 = 14.10

    Collect like terms

    36S = 14.10 - 1.5

    36S = 12.6

    Divide through by 36

    S = 0.35

    Recall that P = 2S.

    So, p = 2 * 0.35

    P = 0.7

    Hence, the cost of a post card is $0.7
  2. 22 September, 10:24
    0
    Each postcard cost $0.7.

    Step-by-step explanation:

    In this problem we have two variables, the price of the postcard and the price of the stamp, since the price for the pain is known. We also have the total cost of the purchase that is $14.1. So we can create an equation as shown:

    12*postcards + 12*stamps + pen = 14.1

    This represents the total amount spent by Claudia, but we know that each pen costs $1.5, therefore:

    12*postcards + 12*stamps + 1.5 = 14.1

    12*postcards + 12*stamps = 14.1 - 1.5

    12*postcards + 12*stamps = 12.6

    We also know that the postcards cost twice as much as the stamps, so we have:

    postcards = 2*stamps

    Using this info we can solve for stams as shown below:

    12 * (2*stamps) + 12*stamps = 12.6

    24*stamps + 12*stamps = 12.6

    36*stamps = 12.6

    stamps = 12.6/36

    stamps = 0.35

    postcards = 2*stamps = 2*0.35 = 0.7

    Each postcard cost $0.7.
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