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5.5 integrated problems

# 5.5 integrated problems - 5.5 Integrated Problems Trade...

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5.5 Integrated Problems

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Discount D = d x L Net Price N = L – D N = L(1 – d) N = L (1 - d1) (1 - d2)…..(1 – dn) Discount Rate d = D / L d = 1 – [(1 - d1) (1 - d2)….(1 – dn)] Trade Discounts
Cash Discounts Amount Paid = Amount Credited x (1 – d) Amount Credited = Amount Paid (1 – d)

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Markup and Markdown M = E + P S = C + M S = C + E + P ROM C = M/C ROM S = M/S D = d x S Sale Price = S – D Sale Price = S(1 – d) d = D x 100
pg. 216, #2: A gas barbecue cost a retailer \$420 less 33 1/3 %, 20%, 5%. It carries a regular selling price on its price tag at a markup of 60% of the regular selling price. During the end-of-season sale, the barbecue is marked

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5.5 integrated problems - 5.5 Integrated Problems Trade...

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