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?

Result

a =  10 d

Solution:

1/a+1/b=1/5 2/a+1/(2b)=1/4  1/b=1/51/a 2/a+0.5 (1/51/a)=1/4 2+0.5 (1/5a1)=1/4a 1.5+1/10a=1/4a a=1.5/(1/41/10)=10 d b=1/(1/51/a)=1/(1/51/10)=10 d1/a+1/b=1/5 \ \\ 2/a+1/(2b)=1/4 \ \\ \ \\ 1/b=1/5 - 1/a \ \\ 2/a+0.5 \cdot \ (1/5-1/a)=1/4 \ \\ 2 + 0.5 \cdot \ (1/5a-1)=1/4a \ \\ 1.5 + 1/10a=1/4a \ \\ a=1.5 / (1/4-1/10)=10 \ \text{d} \ \\ b=1/(1/5-1/a)=1/(1/5-1/10)=10 \ \text{d}



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