IVE Foundation Mathematics I (ITE3705)

Unit: GCNU302A — Solve Problems of Variations (變數問題)

Hello class! Welcome to our tutorial session on Variations.

各位IVE同學好！歡迎來到我們「變數問題」的學習環節。

Today, we will go through the four core types of variations from your lecture notes: Direct Variation, Inverse Variation, Joint Variation, and Partial Variation.

今天，我們將按照筆記中的四大核心單元進行解構：正變、反變、聯變以及部分變。

Each section features two relatable examples combining Hong Kong local scenarios and gaming/anime tropes, followed by bilingual step-by-step solutions.

每個小節均包含兩個融入香港本土實況、遊戲機制與動漫元素的例子，並附有雙語對照的詳細解題步驟。

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Section 1: Direct Variation (正變)

Example 1.1 (Hong Kong Context: Octopus Card Rebate / 八達通回贈)

An MTR student commuter scheme offers a reward point system where points earned $P$P vary directly as the monthly Octopus spending $\mathrm{\$}S$$S.

港鐵學生乘車優惠計劃中，所獲得的積分 $P$P 與每月八達通乘車消費額 $\mathrm{\$}S$$S 成正變。

A student spends $\mathrm{\$}400$$400 in a month and receives $1,600$1,600 points.

一名同學某月乘車消費了 $\mathrm{\$}400$$400，共獲得 $1,600$1,600 積分。

(a) Express $P$P in terms of $S$S.

(a) 試以 $S$S 表示 $P$P。

(b) If a student spends $\mathrm{\$}650$$650 in the next month, how many points will they earn?

(b) 若該同學下月消費了 $\mathrm{\$}650$$650，他將獲得多少積分？

Solution / 解答:

(a)

Since $P$P varies directly as $S$S, we have $P=kS$P=kS, where $k$k is a non-zero variation constant.

由於 $P$P 與 $S$S 成正變，我們可設 $P=kS$P=kS，其中 $k$k 為非零常數。

Substitute $S=400$S=400 and $P=1600$P=1600:

代入 $S=400$S=400 及 $P=1600$P=1600：$$1600=k(400)$$1600=k(400)$$k=4$$k=4

Therefore, the equation is $P=4S$P=4S.

因此，所求的關係式為 $P=4S$P=4S。

(b)

When $S=650$S=650:

當 $S=650$S=650 時：$$P=4(650)=2600$$P=4(650)=2600

The student will earn $2,600$2,600 points.

該同學將會獲得 $2,600$2,600 積分。

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Example 1.2 (Gaming Context: Skill Damage Scaling / 技能傷害計算)

In a fantasy role-playing game, the physical damage $D$D dealt by an anime warrior’s ultimate skill varies directly as the square of the hero’s strength stat $S$S.

在一款奇幻角色扮演遊戲中，戰士終極技能所造成的物理傷害 $D$D，與該英雄的力量數值 $S$S 的平方成正變。

When the warrior’s strength is $30$30, the skill deals $4,500$4,500 damage.

當戰士的力量數值為 $30$30 時，該技能造成 $4,500$4,500 點傷害。

(a) Find the formula connecting $D$D and $S$S.

(a) 試求連接 $D$D 與 $S$S 的方程式。

(b) If the warrior levels up and their strength stat increases by $20\mathrm{\%}$20%, find the percentage increase in damage.

(b) 若戰士升級後力量數值增加了 $20\mathrm{\%}$20%，試求傷害輸出的百分增加。

Solution / 解答:

(a)

Let $D=k{S}^2$D=kS2, where $k$k is the variation constant.

設 $D=k{S}^2$D=kS2，其中 $k$k 為常數。

When $S=30$S=30, $D=4500$D=4500:

當 $S=30$S=30 時，$D=4500$D=4500：$$4500=k(30{)}^2$$4500=k(30)2$$4500=900k⟹k=5$$4500=900k⟹k=5

Hence, the formula is $D=5S^2$D=5S2.

因此，關係式為 $D=5S^2$D=5S2。

(b)

When $S$S increases by $20\mathrm{\%}$20%, the new strength stat is:

當 $S$S 增加 $20\mathrm{\%}$20% 時，新力量數值為：$$S^{\mathrm{\prime }}=(1+20\mathrm{\%})S=1.2S$$S′=(1+20%)S=1.2S

The new damage $D^{\mathrm{\prime }}$D′ is:

新的傷害輸出 $D^{\mathrm{\prime }}$D′ 為：$$D^{\mathrm{\prime }}=5(1.2S{)}^2=5(1.44S^2)=1.44(5S^2)=1.44D$$D′=5(1.2S)2=5(1.44S2)=1.44(5S2)=1.44D

Percentage increase in damage:

傷害輸出的百分增加為：$$\text{Percentage Increase}=\frac{D^{\mathrm{\prime }}-D}{D}\times 100\mathrm{\%}=\frac{1.44D-D}{D}\times 100\mathrm{\%}=44\mathrm{\%}$$Percentage Increase=DD′−D​×100%=D1.44D−D​×100%=44%

The damage increases by $44\mathrm{\%}$44%.

該技能的傷害輸出了增加了 $44\mathrm{\%}$44%。

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Section 2: Inverse Variation (反變)

Example 2.1 (Hong Kong Context: Minibus Waiting Time / 專線小巴班次與等候時間)

During peak hours outside Kwun Tong MTR station, the average passenger queue waiting time $T$T (in minutes) varies inversely as the number of green minibuses $n$n deployed on the route.

在觀塘港鐵站外的繁忙時間，乘客排隊等候的平均時間 $T$T（分鐘）與該路線行駛的綠色專線小巴數量 $n$n 成反變。

When $6$6 minibuses are operating, the average waiting time is $12$12 minutes.

當有 $6$6 輛小巴行駛時，平均等候時間為 $12$12 分鐘。

(a) Express $T$T in terms of $n$n.

(a) 試以 $n$n 表示 $T$T。

(b) If the transport operator increases the fleet to $9$9 minibuses, find the new average waiting time.

(b) 若營運商將小巴車隊增至 $9$9 輛，試求新的平均等候時間。

Solution / 解答:

(a)

Since $T$T varies inversely as $n$n, we can write $T=\frac{k}{n}$T=nk​, where $k$k is a constant.

由於 $T$T 與 $n$n 成反變，我們可設 $T=\frac{k}{n}$T=nk​，其中 $k$k 為常數。

When $n=6$n=6, $T=12$T=12:

當 $n=6$n=6 時，$T=12$T=12：$$12=\frac{k}{6}⟹k=72$$12=6k​⟹k=72

Therefore, $T=\frac{72}{n}$T=n72​.

因此，關係式為 $T=\frac{72}{n}$T=n72​。

(b)

When $n=9$n=9:

當 $n=9$n=9 時：$$T=\frac{72}{9}=8\text{ minutes}$$T=972​=8 minutes

The new average waiting time is $8$8 minutes.

新的平均等候時間為 $8$8 分鐘。

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Example 2.2 (Gaming Context: Skill Cooldown & Haste Stat / 技能冷卻與急速數值)

In a competitive MOBA game, the cooldown time $C$C (in seconds) of a mage’s spell varies inversely as the square root of the character’s haste attribute points $H$H.

在一款競技 MOBA 遊戲中，法師技能的冷卻時間 $C$C（秒）與角色的急速屬性點數 $H$H 的平方根成反變。

When the character has $16$16 haste points, the cooldown is $6$6 seconds.

當角色擁有 $16$16 點急速屬性時，冷卻時間為 $6$6 秒。

(a) Find the relation connecting $C$C and $H$H.

(a) 試求連接 $C$C 與 $H$H 的方程式。

(b) Find the haste points needed to reduce the cooldown time to $4$4 seconds.

(b) 若要將冷卻時間縮短至 $4$4 秒，需要多少點急速屬性？

Solution / 解答:

(a)

Let $C=\frac{k}{\sqrt{H}}$C=H​k​, where $k$k is a constant.

設 $C=\frac{k}{\sqrt{H}}$C=H​k​，其中 $k$k 為常數。

Substitute $H=16$H=16 and $C=6$C=6:

代入 $H=16$H=16 及 $C=6$C=6：$$6=\frac{k}{\sqrt{16}}=\frac{k}{4}⟹k=24$$6=16​k​=4k​⟹k=24

Hence, $C=\frac{24}{\sqrt{H}}$C=H​24​.

因此，關係式為 $C=\frac{24}{\sqrt{H}}$C=H​24​。

(b)

When $C=4$C=4:

當 $C=4$C=4 時：$$4=\frac{24}{\sqrt{H}}$$4=H​24​$$\sqrt{H}=\frac{24}{4}=6$$H​=424​=6$$H=6^2=36$$H=62=36

The character needs $36$36 haste points.

該角色需要 $36$36 點急速屬性。

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Section 3: Joint Variation (聯變)

Example 3.1 (Hong Kong Context: IVE Campus Data Transmission / 校園伺服器數據傳輸)

In the computer laboratory of an IVE campus, the total data transmission volume $V$V (in Gigabytes) varies jointly as the network bandwidth $B$B (in Gbps) and the connection duration $t$t (in hours).

在 IVE 校園的電腦實驗室中，伺服器數據傳輸總量 $V$V（吉字節 GB）與網絡頻寬 $B$B（Gbps）及連線時間 $t$t（小時）成聯變。

A system transfers $900\text{ GB}$900 GB when running at $2\text{ Gbps}$2 Gbps for $5$5 hours.

當頻寬為 $2\text{ Gbps}$2 Gbps 並運行 $5$5 小時，總傳輸量為 $900\text{ GB}$900 GB。

(a) Express $V$V in terms of $B$B and $t$t.

(a) 試以 $B$B 和 $t$t 表示 $V$V。

(b) If the network bandwidth is upgraded to $5\text{ Gbps}$5 Gbps, how many Gigabytes will be transmitted in $3$3 hours?

(b) 若網絡頻寬提升至 $5\text{ Gbps}$5 Gbps，在 $3$3 小時內可傳輸多少吉字節？

Solution / 解答:

(a)

Since $V$V varies jointly as $B$B and $t$t, let $V=kBt$V=kBt, where $k$k is a constant.

由於 $V$V 與 $B$B 及 $t$t 成聯變，設 $V=kBt$V=kBt，其中 $k$k 為常數。

When $V=900$V=900, $B=2$B=2, and $t=5$t=5:

當 $V=900$V=900、$B=2$B=2 及 $t=5$t=5 時：$$900=k(2)(5)$$900=k(2)(5)$$900=10k⟹k=90$$900=10k⟹k=90

Therefore, $V=90\text{形}Bt⟹V=90Bt$V=90形Bt⟹V=90Bt.

因此，關係式為 $V=90Bt$V=90Bt。

(b)

When $B=5$B=5 and $t=3$t=3:

當 $B=5$B=5 及 $t=3$t=3 時：$$V=90(5)(3)=1350\text{ GB}$$V=90(5)(3)=1350 GB

The total data transmitted is $1,350\text{ GB}$1,350 GB.

總傳輸量為 $1,350\text{ GB}$1,350 GB。

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Example 3.2 (Anime/Gaming Context: Guild Raid Boss EXP Reward / 公會討伐戰經驗值分配)

In a fantasy anime adventure guild, the bonus experience points $E$E rewarded to each player during a raid boss defeat varies directly as the monster’s level $L$L and inversely as the square root of the party size $n$n.

在一款奇幻動漫冒險公會中，玩家擊敗討伐首領所獲得的額外經驗值 $E$E，與首領等級 $L$L 成正變，並與冒險隊伍人數 $n$n 的平方根成反變。

When a team of $4$4 adventurers defeats a level $50$50 boss, each member receives $500\text{ EXP}$500 EXP.

當一支由 $4$4 人組成的隊伍擊敗一隻 $50$50 級的首領時，每位隊員獲得 $500\text{ EXP}$500 EXP。

(a) Express $E$E in terms of $L$L and $n$n.

(a) 試以 $L$L 和 $n$n 表示 $E$E。

(b) If a solo player ($n=1$n=1) defeats a level $80$80 boss, find the bonus experience points obtained.

(b) 若一名單人玩家（$n=1$n=1）單獨擊敗 $80$80 級的首領，試求其獲得的額外經驗值。

Solution / 解答:

(a)

According to the variation relation, let $E=\frac{kL}{\sqrt{n}}$E=n​kL​, where $k$k is a constant.

根據題意，設 $E=\frac{kL}{\sqrt{n}}$E=n​kL​，其中 $k$k 為常數。

Substitute $E=500$E=500, $L=50$L=50, and $n=4$n=4:

代入 $E=500$E=500、$L=50$L=50 及 $n=4$n=4：$$500=\frac{k(50)}{\sqrt{4}}=\frac{50k}{2}=25k$$500=4​k(50)​=250k​=25k$$k=\frac{500}{25}=20$$k=25500​=20

Hence, $E=\frac{20L}{\sqrt{n}}$E=n​20L​.

因此，關係式為 $E=\frac{20L}{\sqrt{n}}$E=n​20L​。

(b)

When $n=1$n=1 and $L=80$L=80:

當 $n=1$n=1 及 $L=80$L=80 時：$$E=\frac{20(80)}{\sqrt{1}}=1600\text{ EXP}$$E=1​20(80)​=1600 EXP

The solo player receives $1,600\text{ EXP}$1,600 EXP.

該單人玩家將獲得 $1,600\text{ EXP}$1,600 EXP。

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Section 4: Partial Variation (部分變)

Example 4.1 (Hong Kong Context: Local Food Delivery Fee / 香港外賣平台送遞收費)

The total delivery cost $\mathrm{\$}C$$C of ordering meal boxes in Sham Shui Po via a local courier app is partly constant and partly varies directly as the travel distance $d\text{ (km)}$d (km).

在深水埗使用外賣平台訂購餐盒時，運費總額 $\mathrm{\$}C$$C 一部分為固定基本收費，另一部分與運送距離 $d\text{（公里）}$d（公里） 成正變。

When the delivery distance is $2\text{ km}$2 km, the fee is $\mathrm{\$}26$$26.

當運送距離為 $2$2 公里時，運費為 $\mathrm{\$}26$$26。

When the distance is $5\text{ km}$5 km, the fee is $\mathrm{\$}41$$41.

當運送距離為 $5$5 公里時，運費為 $\mathrm{\$}41$$41。

(a) Find the formula connecting $C$C and $d$d.

(a) 試求連接 $C$C 與 $d$d 的方程式。

(b) If a customer orders from Mong Kok with a delivery distance of $6.5\text{ km}$6.5 km, what is the total delivery fee?

(b) 若顧客從旺角訂餐，外賣運送距離為 $6.5$6.5 公里，所需運費是多少？

Solution / 解答:

(a)

Let $C=k_1+k_{2}d$C=k1​+k2​d, where $k_1$k1​ and $k_2$k2​ are non-zero constants.

設 $C=k_1+k_{2}d$C=k1​+k2​d，其中 $k_1$k1​ 及 $k_2$k2​ 為非零常數。

When $d=2$d=2, $C=26$C=26:

當 $d=2$d=2 時，$C=26$C=26：$$26=k_1+2k_2\text{— (1)}$$26=k1​+2k2​— (1)

When $d=5$d=5, $C=41$C=41:

當 $d=5$d=5 時，$C=41$C=41：$$41=k_1+5k_2\text{— (2)}$$41=k1​+5k2​— (2)

Subtracting equation (1) from equation (2):

以方程式 (2) 減去方程式 (1)：$$(41-26)=(5k_2-2k_2)$$(41−26)=(5k2​−2k2​)$$15=3k_2⟹k_2=5$$15=3k2​⟹k2​=5

Substitute $k_2=5$k2​=5 back into equation (1):

把 $k_2=5$k2​=5 代入方程式 (1)：$$26=k_1+2(5)$$26=k1​+2(5)$$k_1=26-10=16$$k1​=26−10=16

Thus, the formula is $C=16+5d$C=16+5d.

因此，運費方程式為 $C=16+5d$C=16+5d。

(b)

When $d=6.5$d=6.5:

當 $d=6.5$d=6.5 時：$$C=16+5(6.5)=16+32.5=48.5$$C=16+5(6.5)=16+32.5=48.5

The total delivery fee is $\mathrm{\$}48.5$$48.5.

總外賣運費為 $\mathrm{\$}48.5$$48.5。

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Example 4.2 (Gaming Context: Anime Gacha Event Total Cost / 二次元抽卡活動總成本)

The total diamond consumption $D$D for hosting a limited gacha banner event in an online game consists of two parts: one part varies directly as the number of banner days $t$t, and the other part varies directly as the square of the number of featured rate-up characters $n$n.

某手遊營運在舉辦限定抽卡活動時，消耗的魔法水晶總成本 $D$D 分為兩部分：一部分隨活動開放天數 $t$t 成正變，另一部分隨活動限定角色數量 $n$n 的平方成正變。

When the banner runs for $7$7 days with $2$2 featured characters, the cost is $2,700$2,700 diamonds.

當活動開放 $7$7 天且有 $2$2 位限定角色時，總消耗為 $2,700$2,700 水晶。

When the banner runs for $14$14 days with $3$3 featured characters, the cost is $5,700$5,700 diamonds.

當活動開放 $14$14 天且有 $3$3 位限定角色時，總消耗為 $5,700$5,700 水晶。

(a) Find an equation connecting $D$D, $t$t, and $n$n.

(a) 試求連接 $D$D、$t$t 與 $n$n 的方程式。

(b) Find the cost of diamonds if the game developer runs a major anniversary event lasting $20$20 days featuring $4$4 rate-up characters.

(b) 若營運團隊舉辦長達 $20$20 天且包含 $4$4 位限定角色的週年盛典，需要消耗多少水晶？

Solution / 解答:

(a)

Let $D=k_{1}t+k_2{n}^2$D=k1​t+k2​n2, where $k_1$k1​ and $k_2$k2​ are variation constants.

設 $D=k_{1}t+k_2{n}^2$D=k1​t+k2​n2，其中 $k_1$k1​ 與 $k_2$k2​ 為常數。

When $t=7$t=7 and $n=2$n=2, $D=2700$D=2700:

當 $t=7$t=7 及 $n=2$n=2 時，$D=2700$D=2700：$$2700=7k_1+k_2(2{)}^2$$2700=7k1​+k2​(2)2$$7k_1+4k_2=2700\text{— (1)}$$7k1​+4k2​=2700— (1)

When $t=14$t=14 and $n=3$n=3, $D=5700$D=5700:

當 $t=14$t=14 及 $n=3$n=3 時，$D=5700$D=5700：$$5700=14k_1+k_2(3{)}^2$$5700=14k1​+k2​(3)2$$14k_1+9k_2=5700\text{— (2)}$$14k1​+9k2​=5700— (2)

Multiply equation (1) by $2$2:

將方程式 (1) 乘以 $2$2：$$14k_1+8k_2=5400\text{— (3)}$$14k1​+8k2​=5400— (3)

Subtracting equation (3) from equation (2):

以方程式 (2) 減去方程式 (3)：$$(14k_1+9k_2)-(14k_1+8k_2)=5700-5400$$(14k1​+9k2​)−(14k1​+8k2​)=5700−5400$$k_2=300$$k2​=300

Substitute $k_2=300$k2​=300 into equation (1):

把 $k_2=300$k2​=300 代入方程式 (1)：$$7k_1+4(300)=2700$$7k1​+4(300)=2700$$7k_1+1200=2700$$7k1​+1200=2700$$7k_1=1500⟹k_1=\frac{1500}{7}$$7k1​=1500⟹k1​=71500​

The equation connecting $D$D, $t$t, and $n$n is:

連接 $D$D、$t$t 與 $n$n 的方程式為：$$D=\frac{1500}{7}t+300n^2$$D=71500​t+300n2

(b)

Substitute $t=20$t=20 and $n=4$n=4:

代入 $t=20$t=20 及 $n=4$n=4：$$D=\frac{1500}{7}(20)+300(4{)}^2$$D=71500​(20)+300(4)2$$D=\frac{30000}{7}+300(16)$$D=730000​+300(16)$$D\approx 4285.71+4800=9085.71$$D≈4285.71+4800=9085.71

The event will require approximately $9,086$9,086 diamonds (or $\frac{63600}{7}$763600​ exact diamonds).

舉辦該週年活動大約需要消耗 $9,086$9,086 顆魔法水晶（準確值為 $\frac{63600}{7}$763600​ 水晶）。

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Teacher’s Checkpoint & Encouragement / 導師互動專區

Great work on completing all four sections of variation problems!

恭喜各位同學順利完成這四種變數關係的學習！

Notice how every complex problem breaks down once you correctly write the initial algebraic statement: $y=kx$y=kx, $y=\frac{k}{x}$y=xk​, $z=kxy$z=kxy, or $y=k_1+k_{2}x$y=k1​+k2​x!

大家可以看到，只要先準確寫出正變、反變、聯變或部分變的基礎代數結構，再複雜的題目也能迎刃而解！

Quick Check Question for You / 即堂思考小問題:

If a gaming café in Mong Kok charges an entrance fee that is partly fixed and partly varies directly as the hours played, how would you set up the initial equation?

如果一間位於旺角的電競館收費機制為：一部分為固定入場費，另一部分與遊玩時數成正變，你會如何列出這條基本代數式呢？

Feel free to share your thoughts, and keep up the fantastic work in mathematics!

隨時告訴老師你的答案，繼續加油，掌握好基礎數學的每一步！