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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
Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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First day rate= X+8
Second day= X
2 days work= 240m
Time= 14hrs
Rate= 240/14= 17.1
possible number is 14
if X is 14 so
first day= 22
Both are positive primes hece the answer is 14
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Suppose the rate on day 2 is x, then rate on day 1 = x+8
Let time on day 2 be t, then on day it will be 14-t

equating to total work

xt + (x+8) (14-t) = 240
xt +14x - xt + 112 - 8t = 240
14x - 8t = 128

7x - 4t = 64
or

t = 7x - 64/ 4

I tried by putting the options then
1) 9 will give you a negative ans as 63-64 becomes negative
2) 14 works as it gives t = 34/4 = 8.5
3) 19 doesnt work as it gives 69/4 = 17 something, which is more than our total t = 14

Ans B
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Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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let farmer rate on day 1= r
on day 2 = r-8
total time= 14
time on day 1= t
on day 2 time = 14-t
given r t +r(14-t) = 240
on simplifying ,t=(176-7R)/4
using option if day 2 rate = 9,14,16
so r = 9+8, 14+8, 19+8
so r= 17,22,27
and t as per question , t<14
using each value of r in the equation and checking the value of t from the inequality only r=14 satisfies the equation
so second day rate = 14
ans=b
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Given:
First DaySecond DayBoth days
Ratey+8y240/17
Timet14-t14
Area Covered(y+8)ty(14-t)240

Formula:
(y+8)t + y(14-t) = 240

Say y=9,
17t + 9(14-t) = 240
Solving for t, t=14.25 which is more than 14 hours, cannot be a possible value.

Say y=14,
22t + 14(14-t) = 240
=> t=5.5, possible.

Say y=19,
27t + 19(14-t) = 240
=> t=-3.25, cannot be negative, not possible value.

Hence (B) is the answer
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Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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As per the question

Rate of day 2 = x
Rate of day 1 - r1 = 8+x

We need one more person to solve this question - which is easily ignored. at the first go. Time!!

Time worked in day 1 = t1
Time worked in day 2 = 14-t1( given in the question total time worked is 14)

Total work is 240.

240= (8+x)t1 + (14-t1)x

Solving we will get

240= 8t1 + xt1 + 14x - xt1
240=8t1+14x
120=4t1+7x

Plug in the options and you will get the following values t=14.25, t = 5 , t=- 3.25..
hence B is the answer
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since total time for the two days is 14 hrs and total work is 240 sq.mtr.
let x is the rate on the second day and time worked be 14-t hrs
so x+8 is the rate on the first day and let time worked be t hrs
i.e. x*(14-t)+ (x+8)*t=240
14x-8t=240
t=7x/4 - 30
if we take x=9 or 14 value of t comes -ve , non possible
so x can take value of 19 only, so C (III only)
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Let the time taken on Day 1 be T hours

Time taken on Day 2 would be (14 - T) hours

Say, if x is the rate of work done on Day 1, (x - 8) will be the rate of work done on day 2.
Work done on Day 1 = xT
Work done on Day 2 = (x - 8)(14 - T)


Total work done is 240 units
Therefore,
xT + (x - 8)(14 - T) = 240

We can now substitute the values from given options.
Taking the second value as (x - 8) = 14 will give x = 22

Placing these values in the above equation,

22T + 14(14 - T) = 240
22T + 196 - 14 T = 240
8T = 44
T = 5.5 hrs

Checking with the equation

22 x 5.5 + 14(14 - 5.5)
=> 121 + 119
=>240
This is consistent with the given options.

Similarly trying with other options does not give consistent result.
Hence, answer is B - II only
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I set this up in my RxT = W table. Since we have a total work of 240 and the total time for the two days is 14 hours, I divided 240 by 14 to get ~17.2 square meters per hour. Since it's the same party doing the work, the average of day 1 and day 2 = ~17.2. I let Day 1 = x and Day 2 = x-8. I solved and got ~14.
Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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Total work = 240 sq m
Total Time = 14 hours
Average rate over 14 hours
240/14 ≈17.14 sq m/h
Day 2 rate = r
Day 1 rate = r+8
Testing the options
1. 9
Day 1 = 9+8 = 17
Avg must lie between 9 and 17 , but 17.14 too high ❌

2 .14
DAY 1 = 22
Avg must lie between 14 and 22 , 17.14 fitting in here .✅

3. 19
Day 1 = 19+8 = 27
17.14 is below 19 ❌
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Let r denotes rate of farmer on day 1. then on day 2 farmer rate will be r-8. Let t1 be the time taken to complete work on day1. 14-t1 will be the time taken to complete work on day2. so the final equation will be :

rt1 + (r-8)(14-t1) = 240
14r+8t1 =352
Let's take r-8=9 , r=17 , then t1 will be approx. 14.25 , which is not possible as total time taken is itself 14.
Let's take r-8=14 , r= 22, t1 = 5.5 , possible solution
let's take r-8=19 , r=27 , t1 <0 , not a possible solution.

So Answer is B.
Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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We know that r1t1 + r2t2=240
we need to solve for r2

r1= r2-8
t1=t2-14

(r2-8)(t2-14)+r2t2=240
solving we will get 7r2=64+4t2

hence t2=7r2-64/4
we know that t2 >0
(7r2 -64) >0
r2>64/7=9.xx

we also know that t2<14
7r2-64/4<14
7r2-64<56
r2<120/7=17.xx

hence we get the range of r2 i.e. 9<r2<17

answer is 14. II only.
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Hello ,
lets take rate of second day as X
rate of first day will be x+8

total work done is 240 in 14 hours
240=(x+8)t+x*(14-t)
240=8t+14X
120=4t+7X
so here the x can take two value to get T and third value will give negative value for T so
only possible value for rate in second day will be 9 and 14

Hence option D is correct
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Say a is the working rate on the second day, t is the number of hours working on the first day.

(a+8)t + a(14-t) = 240
=> t = (240-14a)/8

Scan the options:
a = 9 => t = 14.25 > 14 (eliminate)
a = 14 => t = 5.5
a = 19 => t < 0 (eliminate)

Answer: B
Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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Bunuel
A farmer worked on clearing a field over two days. On the second day, the farmer’s rate was 8 square meters per hour less than on the first day. Over the two days together, the farmer cleared 240 square meters in 14 hours of work. Which of the following could be the farmer’s rate, in square meters per hour, on the second day?

I. 9
II. 14
III. 19

A. I only
B. II only
C. III only
D. I and II only
E. II and III only

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2nd day rate = f2 = f1 - 8
1st day rate = f1
h1 + h2 = 14
total work done. = 240

f2 = ?

f1 * h1 + (f1 - 8) * h2 = 240
f1 * (14 - h2) + f1h2 - 8h2 = 240
14f1 - f1h2 + f1h2 - 8h2 = 240
(14f1 - 240) / 8 = h2

substituting f1 from the options (f2 + 8) give time which is valid (under 14h) only for option B

option 1 gives a -ve value and option 3 gives a value greater than 14
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First day:
Rate = r , Time taken = t1
Second Day:
Rate= r-8 , Time taken = t2
t1+t2= 14 (Given)
Total Work = 240
So, rxt1 + (r-8)xt2 = 240
r(14-t2) + (r-8)xt2 = 240 [ substituting the value of t2]
Simplify:
t2 = (14r - 240)/8
0 <(14r-240)/8<14
240 < 14r < 352
17.14 <r <25.14 ---> Rate for day 1
9.14 < r - 8 < 17.14 ---> Rate for day 2

So only (II) 14 is the only option which satisfies the inequality
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Average rate = 240/14 = 17.14
Let's assume:
First day rate as r and
second day rate as r+8 (given)

Thus, average must be between
r< 17.14< r+8
Using inequality ,
9.14 < r < 17.14

Now check options :
Option 1 : 9 (smaller than 9.14) So reject
Option 2 : 14 satisfies
Option 3 : 19 (too large than 17.14)
option 2 is valid.

Hence, answer B.
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