Distance

If Danka increases her walking speed by 1km/h, she will cover a distance greater than 20 km in 4 hours. If she decreases her speed by 1km/h, she will not cover more than 20 km in 5 hours. At what speed did Danka walk? (s=v. t;s- distance, v- speed, t- time). Solve both inequalities and check for correctness for both; write down the resulting intervals

Correct answer:

v1 =  4 km/h
v2 =  5 km/h

Step-by-step explanation:

v1<v<v2 t1=4 h t2=5 h s=20 km Δ=1 km/h  s=v t (v+Δ) t1>s (vΔ) t2<s  t1 v > sΔ t1 t2 v < s+Δ t2  v > (sΔ t1)/t1 v < (s+Δ t2)/t2  v1<v<v2  v1=(sΔ t1)/t1=(201 4)/4=4 km/h
v2=(s+Δ t2)/t2=(20+1 5)/5=5 km/h



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You need to know the following knowledge to solve this word math problem:

algebraarithmeticUnits of physical quantitiesthemes, topicsGrade of the word problem

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