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18 Dec 2025, 11:10
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Bunuel
Let x = day 1 rate, y= day 1 hours, and z=day 2 hours.
\(y+z=14\)
Total work done:
\(xy+(x-8)z=240\)
\(xy+(x-8)(14-y)=240\)
Simplifies to:
\(7x-4y=181\)
Now we need to find rates from the options that, when subbed in, make RHS divisible by -4.
I. \(y=\frac{118}{-4}\), not an integer, incorrect.
II. \(y=\frac{142}{-4}\), not an integer, incorrect.
III. \(y=\frac{48}{4}\), integer, correct.
Therefore answer is C.
18 Dec 2025, 11:21
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Answer B
Let x be rate on 2nd day, and x=8 the rate of work on 1st day
Let h1 be hours worked on first day. h2 is hours worked on 2nd day.
First day work is h1 x (x+8) + 2nd day h2 x (x) H2 + h1 = 14 h2 = 14 - h1
Total cleared 240 = h1 (x +8) + h2 x (x) = h1(x +8) + 14 - h1 (x)
240 = h1x + 8 h1 + 14x - h1(x)
240 = 8 h1 + 14x
h1 = 240 - 14x / 8
for 9 240 - 126 / 8 = 14.25 greater than 14 hours incorrect
for 14 240 - 14(14) / 8 = 5.5 hours works
for 19 240 - 14(19) / 8 = 240 - 266 / 8 negative number incorrect
Let x be rate on 2nd day, and x=8 the rate of work on 1st day
Let h1 be hours worked on first day. h2 is hours worked on 2nd day.
First day work is h1 x (x+8) + 2nd day h2 x (x) H2 + h1 = 14 h2 = 14 - h1
Total cleared 240 = h1 (x +8) + h2 x (x) = h1(x +8) + 14 - h1 (x)
240 = h1x + 8 h1 + 14x - h1(x)
240 = 8 h1 + 14x
h1 = 240 - 14x / 8
for 9 240 - 126 / 8 = 14.25 greater than 14 hours incorrect
for 14 240 - 14(14) / 8 = 5.5 hours works
for 19 240 - 14(19) / 8 = 240 - 266 / 8 negative number incorrect
Bunuel
18 Dec 2025, 11:42
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Bunuel
A farmer is working for two days. D1 and D2 respectively.
Given that the rate of day 2 is 8 less than the rate of day 1.
Since the answer is given as rate for day 2, we can assume rate for day 2 as a.
Let the rates of the farmer on day 1 and day be : (a+8) and a respectively.
Given, the total time is 14 hours.
If the time taken on Day 1 is t hours, then time taken for Day 2 = (14-t) hours.
In these 2 days, the total work done is 240.
(a+8)*t + a * (14-t) = 240
at + 8t + 14a - at = 240
8t + 14a = 240
4t + 7a = 120
Lets look into the options,
I) a = 9, then 120 - 7a = 4t , solving we get t = 14.xx
Since t >= 14 , a=9 cannot be the answer.
II) a = 14, then 120 -7a = 4t , solving t = 5.5
a =14 can be an answer, as t =<14.
III) a=19, 120 - 7*(19) = 4t , solving we get t = negative.
Not possible.
Option B (II)
18 Dec 2025, 11:42
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Rate of day 1 = R1
Rate of day 2 = R1 - 8
W = 240
T1+T2 = 14
240 = R1T1 + (R1-8)T2
240 = R1(T1+T2) - 8T2
240 = 14R1 - 8T2
120 = 7R1 - 4T2
R1 - 8 = (64+4T2)/7
T2 = (7(R1-8) - 64)/4
Use the answer choices to substitute into R1-8
1. R1-8 = 9 => T2 = (7*9 - 64)/4 => -ve -- Invalid because T2>0
2. R1-8 = 14 => T2 = (7*14 - 64)/4 = 34/4 -- Valid because 0<T2<14
3. R1-8 = 19 => T2 = (7*19 - 64)/4 = 99/4 -- Invalid because T2>14
Answer B
Rate of day 2 = R1 - 8
W = 240
T1+T2 = 14
240 = R1T1 + (R1-8)T2
240 = R1(T1+T2) - 8T2
240 = 14R1 - 8T2
120 = 7R1 - 4T2
R1 - 8 = (64+4T2)/7
T2 = (7(R1-8) - 64)/4
Use the answer choices to substitute into R1-8
1. R1-8 = 9 => T2 = (7*9 - 64)/4 => -ve -- Invalid because T2>0
2. R1-8 = 14 => T2 = (7*14 - 64)/4 = 34/4 -- Valid because 0<T2<14
3. R1-8 = 19 => T2 = (7*19 - 64)/4 = 99/4 -- Invalid because T2>14
Answer B
Bunuel
