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19 Dec 2025, 00:13
Kudos
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We could write an equation like this:
t*r + (14-t)(r-8) = 240, where t is the time taken on the first day, and r is the rate of work on the first day.
Note, 0 < t < 14 and r > 8.
Taking each of the options,
I. 9: this is r-8=9, r = 17
Equation becomes 17t + (14-t)9 = 240, gives t = 14.25 > 14, not possible.
II. 14, r becomes 22,
22t + (14-t)14 = 240, gives t = 5.5 < 14, possible.
III. 19, r becomes 27
27t + (14-t)19 = 240, t becomes negative, not possible.
Only II is possible, option B.
t*r + (14-t)(r-8) = 240, where t is the time taken on the first day, and r is the rate of work on the first day.
Note, 0 < t < 14 and r > 8.
Taking each of the options,
I. 9: this is r-8=9, r = 17
Equation becomes 17t + (14-t)9 = 240, gives t = 14.25 > 14, not possible.
II. 14, r becomes 22,
22t + (14-t)14 = 240, gives t = 5.5 < 14, possible.
III. 19, r becomes 27
27t + (14-t)19 = 240, t becomes negative, not possible.
Only II is possible, option B.
Bunuel
19 Dec 2025, 00:26
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Let the rate be r1 and r2 for first and second days respectively.
Let the time worked on first and second days be t1 and t2.
Given r2 = r1 - 8
t1 + t2 = 14
Then t1(r1) + t2(r1-8) = 240
r1(t1 + t2) - 8t2 = 240
r1(14) - 8t2 = 240
t2 = (14r1 - 240)/ 8 = ( 7r1 - 120)/4
Analysing options
I)
r2 = 9
r1 = 17
then t2 = (7*17) - 120 /4 = -0.25
negative time is not possible so I is incorrect
II)
r2 = 14
r1 = 22
then t2 = (7*22) - 120 /4 = 34/4 = 8.5 hrs, t1 = 14 - 8.5 = 5.5
this is possible
III)
r2 = 19
r1 = 27
then t2 = (7*27) - 120 / 4 = 17.25. This is not possible as total number of hours is 14
Hence II only is the answer
Let the time worked on first and second days be t1 and t2.
Given r2 = r1 - 8
t1 + t2 = 14
Then t1(r1) + t2(r1-8) = 240
r1(t1 + t2) - 8t2 = 240
r1(14) - 8t2 = 240
t2 = (14r1 - 240)/ 8 = ( 7r1 - 120)/4
Analysing options
I)
r2 = 9
r1 = 17
then t2 = (7*17) - 120 /4 = -0.25
negative time is not possible so I is incorrect
II)
r2 = 14
r1 = 22
then t2 = (7*22) - 120 /4 = 34/4 = 8.5 hrs, t1 = 14 - 8.5 = 5.5
this is possible
III)
r2 = 19
r1 = 27
then t2 = (7*27) - 120 / 4 = 17.25. This is not possible as total number of hours is 14
Hence II only is the answer
Bunuel
19 Dec 2025, 00:36
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If the second day rate = r.
Then the first day rate = r+8
Lets say time taken on day 1: t1.
and time taken on day 2: t2.
t1+t2 = 14.
Total area = 240.
(r+8)t1 + (t2*r) = 240.
Replacing t2 = 14-t1 we should get:
t1 = (240 - 14r)/8.
So since time cannot be negative, t1>14 and 14-t2>0.
From 1:
If r = 9:
t1 becomes: (240-126)/8 = 14.25.
but t2 = 14 - 14.25.
Which cannot be possible.
From 2; R=14
t1 becomes = 5.5
t2 becomes = 8.5 which is possible.
From 3:
R = 19.
t1 = -26/8 which is negative and cannot be possible.
Hence 2 only.
B.
Then the first day rate = r+8
Lets say time taken on day 1: t1.
and time taken on day 2: t2.
t1+t2 = 14.
Total area = 240.
(r+8)t1 + (t2*r) = 240.
Replacing t2 = 14-t1 we should get:
t1 = (240 - 14r)/8.
So since time cannot be negative, t1>14 and 14-t2>0.
From 1:
If r = 9:
t1 becomes: (240-126)/8 = 14.25.
but t2 = 14 - 14.25.
Which cannot be possible.
From 2; R=14
t1 becomes = 5.5
t2 becomes = 8.5 which is possible.
From 3:
R = 19.
t1 = -26/8 which is negative and cannot be possible.
Hence 2 only.
B.
Bunuel
