Two workers

Two workers should fulfill certain task together for 5 days. If the first worker increased their performance twice and second twice fell, it took them just four days. For how many days would handle the entire task first worker himself?

Correct result:

a =  10 d

Solution:

1/a+1/b=1/5 2/a+1/(2b)=1/4  1/b=1/5−1/a 2/a+0.5⋅ (1/5−1/a)=1/4 2+0.5⋅ (1/5a−1)=1/4a 1.5+1/10a=1/4a a=1.5/(1/4−1/10)=10=10 d b=1/(1/5−1/a)=1/(1/5−1/10)=10 d

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