문제

Q1.svg

해설

2929 students and the teacher are going to buy the T-shirts, so the total number of people buying the T-shirts is 3030. Let a1a_{1} as the quantity of small-sized T-shirts, a2a_{2} as the quantity of medium-sized T-shirts, a3a_{3} as the quantity of large-sized T-shirts, a4a_{4} as the quantity of extra-large-sized T-shirts, and a5a_{5} as the quantity of extra-extra-large-sized T-shirts. Since a1, a2, a3, a4, a5a_{1},\:a_{2},\:a_{3},\:a_{4},\:a_{5} are in arithmetic progression and ∑k=15ak=30\displaystyle\sum_{k=1}^{5}{a_k}=30, the arithmetic mean would be a3=305=6a_{3}=\displaystyle\frac{30}{5}=6. Since a4=4a_{4}=4, the common difference would be d=−2d=-2. Also, according to the passage, the small-sized T-shirt costs £10 and a bigger size means an additional £2. Based on this information, the following table can be made:

Size S M L XL 2XL Total
Quantity 10 8 6 4 2 30
Price per unit 10 12 14 16 18 -
Total Price(£) 100 96 84 64 26 380

The number of students buying the red cap is 2727 (of the 2929 students buying the T-shirts) +2 + 2 (of the 44 students not buying the T-shirts) =29=29. Since the woman says she’s buying the red caps for her students, it can be inferred that the woman is a teacher and she is not going to buy the red cap. Therefore, the total number of people buying the red cap is 2929. According to the question, the price of the red cap is set as aa (an unknown value). Thus, the total price of the T-shirts and the red caps is £(380+29a)(380+29a).

Although applying the 10%-off coupon before the £20 discount coupon is cheaper, the question presumes that the man always applied the discounts in the wrong order. Thus, the £20 discount coupon was applied prior to the 10%-off coupon. The same mechanism applies to the £2 apology discount—it was applied before the two coupons. Therefore, the price after applying the two coupons and an additional discount would be £{(380+29a−2−20)×0.9}=\{(380+29a-2-20)\times 0.9\}=£(26.1a+322.2)(26.1a+322.2).

The next step would be to apply the 10% VAT, which makes the total price £(28.71a+354.42)(28.71a+354.42). This is equivalent to $(28.71a+354.42)×1.3=(37.323a+460.746)(28.71a+354.42)\times1.3=(37.323a+460.746). Only $600600 can be paid with the corporate card, which is $(600÷0.95)=1200019(600 \div 0.95)=\dfrac{12000}{19} before the 5%-off benefit of the corporate card. Therefore, the rest of the payment, which is $(37.323a+460.746−1200019)\left(37.323a+460.746-\dfrac{12000}{19}\right), was paid with US dollar cash.

The next step before deducting the quantity of each type of coin is to add the extra money the woman should pay. First of all, the $2 tip that the woman gave to the man should not be included in the coin quantity deduction process. This is because the man said the dimes, quarters, nickels, and the pennies are on the desk, while the $2 tip went to the man’s pocket. Next, the woman paid an extra £5 for the shipping fee, which is equivalent to $5×1.3=6.55\times1.3=6.5. If we add this to the money that the woman paid in cash, the total would be $(37.323a+467.246−1200019)\left(37.323a+467.246-\dfrac{12000}{19} \right). The woman then paid a $50 bill and 33 dimes, which is a total of $50.3. Therefore, using 46−3=4346-3=43 coins that include quarters, nickels, and pennies, the woman paid $(37.323a+416.946−1200019)\left(37.323a+416.946-\dfrac{12000}{19}\right). Let this value as nn. We will now find the maximum and minimum values of nn in order to specify aa.

If the woman paid 4141 quarters, 11 nickle, and 11 penny, n=41×0.25+1×0.05+1×0.01=10.31n=41\times0.25+1\times0.05+1\times0.01=10.31.

If the woman paid 11 quarter, 11 nickle, and 4141 pennies, n=1×0.25+1×0.05+41×0.01=0.71n=1\times0.25+1\times0.05+41\times0.01=0.71.

0.71≤37.323a+416.946−1200019≤10.310.71 \leq 37.323a+416.946-\dfrac{12000}{19} \leq 10.31
5.769⪅a⪅6.0265.769 \lessapprox a \lessapprox 6.026

Since aa is an integer, a=6a=6. If we substitute a=6a=6 in nn,

n=37.323×6+416.946−1200019≈9.305n=37.323\times6+416.946-\dfrac{12000}{19}\approx9.305. Since the minimum payment is 11 cent, the woman has to pay $9.319.31 to pay the rounded price. Let x1=x_{1}= the number of quarters the woman paid, x2=x_{2}= the number of nickles the woman paid (excluding the nickles she paid as a tip), and x3=x_{3}= the number of pennies the woman paid. We now have to solve the following indeterminate equation, given that x1, x2, x3∈Nx_{1},\: x_{2},\:x_{3} \in \mathbb{N}.

{x1+x2+x3=43 ⋯ (A)0.25x1+0.05x2+0.01x3=9.31⋯ (B)\begin{cases} x_{1}+x_{2}+x_{3}=43 \qquad\qquad\qquad\quad\,\cdots\: \textrm{(A)} \\ 0.25x_{1}+0.05x_{2}+0.01x_{3}=9.31\quad \cdots\: \textrm{(B)} \end{cases}

Multiplying 100100 at (B) gives

{x1+x2+x3=43⋯ (A)25x1+5x2+x3=931⋯ (C)\begin{cases} x_{1}+x_{2}+x_{3}=43\qquad\quad\cdots\:\textrm{(A)} \\ 25x_{1}+5x_{2}+x_{3}=931\quad\cdots\:\textrm{(C)} \end{cases}

Subtracting (A) and (C) leads to

24x1+4x2=88824x_{1}+4x_{2}=888
6x1+x2=2226x_{1}+x_{2}=222

(ⅰ) If x1≥37x_{1} \geq 37, then 6x1+x2≥2226x_{1}+x_{2}\geq 222; This does not satisfy x2≥1x_{2}\geq 1.

(ⅱ) If x1=36x_{1}=36, then x2=222−6×36=6x_{2}=222-6\times36=6, and x3=43−36−6=1x_{3}=43-36-6=1.

(ⅲ) If x1≤35x_{1}\leq 35, then x2=222−6x1x_{2}=222-6x_{1}, which leads to x1+x2=222−5x1≥47x_{1}+x_{2}=222-5x_{1}\geq 47. However, since x3≥1x_{3}\geq 1, this does not satisfy x1+x2+x3=43x_{1}+x_{2}+x_{3}=43.

Therefore, the woman paid 3636 quarters, 66 nickels, and 11 penny for the total price. Since she also tipped $2 with 2÷0.05=402\div 0.05=40 nickels,

bb (the total amount of nickels the woman paid) =6+40=46=6+40=46

∴ a+b=6+46=52\therefore \: a+b=6+46=\boxed{\bm{52}}

정답

52

Comment

영어듣기 문제를 풀다가 그나마 가장 어렵다고 하는 가격 계산 문제가 그래도 너무 쉬워서, 제대로 된 어려운 가격 계산 문제를 만들어보려다가 선을 넘어 버렸다.

History

  • 최초 출제일: 2022.04.14.
  • 해설 작성일: 2022.04.16.
  • 웹 업로드일: 2023.02.05.