Porisma

Worked solution

Combine a capped geometric plan with arithmetic increments

A student with a knee problem plans swimming and jogging to prepare for a 2.4 km NAPFA run. On day 1 he jogs 200200 m and swims 200200 m. From day 2, jogging rises by 20%20\% of the previous day’s distance but is capped at 400400 m per day; swimming increases by 5050 m each day. Find the combined distance after 20 days, and the first day his combined daily distance reaches 2.42.4 km.

[2012/PJC/Promo/8]

CJC · Discussion · Question 17

Answer

Combined 20-day distance 20973.620973.6 m; first 2.42.4 km day is day 37.

Apply the cap at the correct day

The printed daily pattern is:

DayJog (m)Swim (m)
1200200
2240250
3288300
4345.6350
5400 (cap)400

The uncapped fifth jog would exceed 400, so only days 1–4 use the GP; days 5–20 give sixteen capped jogs.

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Sum each activity without rounding exact distances

Swimming totals 202[2(200)+19(50)]=13500\tfrac{20}2[2(200)+19(50)]=13500 m. Jogging totals 200(1.24−1)/(1.2−1)+16(400)=7473.6200(1.2^4-1)/(1.2-1)+16(400)=7473.6 m. The exact combined total is 20973.620973.6 m; the printed note correctly says not to round this exact value to 3 significant figures.

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Find the first qualifying daily distance

After day 4 the daily jog is 400400, so the daily swim must be at least 20002000 m. Solve 200+(n−1)50≥2000200+(n-1)50\ge2000 to get n≥37n\ge37. The combined daily distance first reaches 24002400 m on day 37 (day 36 gives 23502350 m).

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Common mistakes

This asks for a daily threshold, not the day the cumulative distance passes 2.4 km.

Read the hint