Two workers finished a job in 12 days. How long would it take each worker to do the job by himself if one of the workers needs 10 more days to finish the job than the other worker?

20 days

Step-by-step explanation:

t = time required by one worker to complete the job alone

(t+8) = time required by the other worker

So:

7.5/ t + 7.5/ (t + 8) = 1

Solve for t.

...

t = 12 days, the first guy working alone

The other worker would finish the job in 20 days.

7 days and 17 days

Step-by-step explanation:

We can write an equation to solve this. An equation could be: ½x + ½y = 12. x - y = 10. Variables that solve this equation are x=7 and y=17. 7 + 17 = 24÷2 (since there are 2 workers) =12. Also, ½(7) + ½17 = 3.5 + 8.5 = 12. So, we know that the faster worker will take 7 days and the slower worker will take 17 days.

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step-by-step explanation:

the original amount doesn't ''re just looking for the time it takes for 1/2 of the substance to we have

(1/2) =)     take the ln   of both sides

ln (1/2)   = )   and we can write

ln(1/2) = ) lne     and lne   = 1     so we have

ln(1/2) = -.1092t     divide both sides by -.1092

t = ln(1/2)/-.1092 = about 6.347 days = about 6.3 days (rounded)

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

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