How many unit-tests the second programmer wrote per hour?

Question

Two programmers have to write 27 unit-tests each. The second programmer had started his work 27 minutes earlier than the first programmer. They finished 2/3 of their work at the same time. In order to finish the work at the same time and go to a cafe, second programmer wrote 1 unit-test of the first programmer. How many unit-tests the second programmer wrote per hour?

by rough calculation, my answer is 10. Is that true?

Answer 1

Best way to do this is with a computer or a scientific calculator with matrix algebra,

taking 3 equations and solving for 3 unknowns. 10 doesn't look right. For what it may be worth here's calculations by hand, using simple algebra, with a result of 0.30 units per hour for the 2nd programmer. It's a little tedious.

This is like distance = velocity x time, only instead of distance, it's number of test units completed

instead of d=vt we have n=vt The first completed 28 test units, the 2nd 2 less or 26

If 10 is correct, then 28=(10/60)t or t=28/6= 4.6666... hours

and 26=v₁(t-27/60) Units have to be the same. if you have 10 mph time must be in hours not minutes, so divide by 60

or 26(60)=v₁t-27v₁

1560=v₁(280)-27v₁

v₁=1587/253=per hour which does not equal 10 per hour

they finished 2/3 of the total 2x27=54 at the same time. That's 36 total in (2/3)t time

N at (2/3)t/60 =v₁t/60 + v₂(t-27)/60 = 36 or v₁t + v₂t -27v₂ = (2/3)t

t=-27v₂/(-v₁-v₂+2/3)

v₁t/60=28 v₂(t-27/60)=26

we have 3 equations, 3 unknowns. time and two rates of completion. Solve the system of equations by substitution or matrix algebra

v₁=(60/28)t=(15/7)t substitute t(15/7) into the first equation, getting 15/7+v₂t-27v₂=(2/3)t

v₂=26/(t-9/20) substitute 26/(t-9/20) into above last equation getting 15/7+26t/(t-9/20)+27(26)/(t-9/20)=2t/3

expand, simplify and solve for t

Multiply by t-9/20

15(t-9/20)/7 +26t +27(26) =2t(t-9/20)/3

Multiply by 21

45(t-9/20) + 26(21)t +27(26)(21) = 14t(t-9/20)

expand

456t - 81/4 + 546t + 14742 = 14t² - (63/10)t

Rewrite as a standard quadratic equation

14t² - 1008.3t -14742 = 0

Complete the square, ue the quadratic formula or if possible factor, to solve for t

a= 14, b= -1008.3 c=-14742

t = ((1008.3 + or - ((1008.3)² - 4(14)(-14742))/2(14)

= 1008.3/28 + or - (a number bigger than 1008.3/28) so ignore the negative square root

=1008.3/28 + (square root of 1871704.89)/28

=(1008.3 + 1368.1)/28 = 84.87 hours or about 85 hours

No substitute into the equation relating v₂ to t

v₂ = 26/(t-9/20) = 26/(84.87-9/20) = 26/84.42 = 0.31 unit tests per hour = about 3/10 per hour

vt for programmer 2 = about 8,5x3= about 26 total units completed

programmer 1 completed 28 total units in 84.4 hours, completing them at a rate of 0.33 per hour

When 2/3 were completed, they had together completed 2/3 of 2 times 27 or 36 units

2/3 of about 85 hours = about 57 hours.

at 1/3 per hour, programmer 1 completed 1/3 x 57 = 19 units

at 3/10 per hour programmer 2 completed .3 x 56 = 17 units together they completed about 36 units

programmer 1 completed 19+9=28

and programmer 2 completed 2 less or 26, so they together completed 54 or equivalent to the 27 each required.

Given rounding errors, and approximations, and whether they both completed 2/3 at the same time involved partial units or integers, they may be some errors in the above calculations, but it seems to be close to what is required.

They complete the total required in a bout 85 hours, each working at 1/3 or 3/10 per hour.