A coffee vendor sells three types of coffee beverages: Latte, Cappuccino and Mocha. The cups of coffee come in three different sizes: Short, Tall and Grande. The number of cups for each type of coffee sold in a particular day is given in the following table.
ShortTallGrande
Latte3128
Cappuccino487
Mocha254
The price of a cup of coffee for each size is the same regardless of the type of coffee. The total amount collected on that day from the sale of Latte, Cappuccino and Mocha is $147.90, $120.10 and $70 respectively.

  1. (i)

    Find the price of a short, tall and grande cup of coffee respectively.

  2. (ii)

    The vendor offers a buy-one-get-one-free promotion if a customer buys a grande cup of Latte and gives a 10% discount if a customer buys a short or tall cup of Latte. Andrew wishes to have 1 short cup, 4 tall cups, and 4 grande cups of Latte. How much does Andrew need to pay in total?

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Worked solution

(a)

1. 设未知数并根据题意建立三元一次方程组

设小杯(Short)、中杯(Tall)和大杯(Grande)咖啡的单价分别为 $x\char36{}x、$y\char36{}y 和 $z\char36{}z。
根据题意,可列出如下三元一次方程组:
1)3x+12y+8z=147.901)\quad 3x + 12y + 8z = 147.90
2)4x+8y+7z=120.102)\quad 4x + 8y + 7z = 120.10
3)2x+5y+4z=70.003)\quad 2x + 5y + 4z = 70.00

2. 消元并求解方程组

将方程 (3)(3) 乘以 22,得:
4)4x+10y+8z=140.004)\quad 4x + 10y + 8z = 140.00
用方程 (4)(4) 减去方程 (1)(1):
(4x+10y+8z)−(3x+12y+8z)=140.00−147.90(4x + 10y + 8z) - (3x + 12y + 8z) = 140.00 - 147.90
x−2y=−7.90  ⟹  x=2y−7.90(5)x - 2y = -7.90 \implies x = 2y - 7.90\quad (5)
用方程 (4)(4) 减去方程 (2)(2):
(4x+10y+8z)−(4x+8y+7z)=140.00−120.10(4x + 10y + 8z) - (4x + 8y + 7z) = 140.00 - 120.10
2y+z=19.90  ⟹  z=19.90−2y(6)2y + z = 19.90 \implies z = 19.90 - 2y\quad (6)
将 (5)(5) 和 (6)(6) 代入方程 (3)(3):
2(2y−7.90)+5y+4(19.90−2y)=70.002(2y - 7.90) + 5y + 4(19.90 - 2y) = 70.00
4y−15.80+5y+79.60−8y=70.004y - 15.80 + 5y + 79.60 - 8y = 70.00
y+63.80=70.00  ⟹  y=6.20y + 63.80 = 70.00 \implies y = 6.20
将 y=6.20y = 6.20 代回 (5)(5) 和 (6)(6):
x=2(6.20)−7.90=4.50x = 2(6.20) - 7.90 = 4.50
z=19.90−2(6.20)=7.50z = 19.90 - 2(6.20) = 7.50

Answer

小杯(Short)、中杯(Tall)和大杯(Grande)咖啡的单价分别为 $4.50\char36{}4.50、$6.20\char36{}6.20 和 $7.50\char36{}7.50。

(b)

1. 计算 4 杯 Grande 杯拿铁在买一送一优惠下的费用

由第一问可知,Grande 杯单价 z=$7.50z = \char36{}7.50。
买一送一活动意味着买 2 杯只需付 1 杯的钱。
Andrew 需要 4 杯 Grande 杯拿铁,因此只需支付 2 杯的费用:
Grande 杯总费用=2×$7.50=$15.00\text{Grande 杯总费用} = 2 \times \$7.50 = \$15.00

2. 计算 1 杯 Short 杯和 4 杯 Tall 杯拿铁在 10% 折扣下的费用

由第一问可知,Short 杯单价 x=$4.50x = \char36{}4.50,Tall 杯单价 y=$6.20y = \char36{}6.20。
原价总计:
Short 与 Tall 原价总计=1×$4.50+4×$6.20=$4.50+$24.80=$29.30\text{Short 与 Tall 原价总计} = 1 \times \$4.50 + 4 \times \$6.20 = \$4.50 + \$24.80 = \$29.30
享受 10% 折扣(即按原价的 90% 结算):
折扣后费用=$29.30×(1−0.10)=$29.30×0.90=$26.37\text{折扣后费用} = \$29.30 \times (1 - 0.10) = \$29.30 \times 0.90 = \$26.37

3. 计算 Andrew 需要支付的总金额

将两部分的费用相加:
应付总额=$15.00+$26.37=$41.37\text{应付总额} = \$15.00 + \$26.37 = \$41.37

Answer

Andrew 总共需要支付 $41.37\char36{}41.37。

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