Around the World in 80 Questions (Day 5): Two workers, Larry and Jerry

Question

Two workers, Larry and Jerry, are planning to paint a classroom floor, each at their constant rate. Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor. Consequently, when the task is complete, Larry has painted 8 m^2 more of the floor than Jerry. If Larry could paint the floor by himself in 3.6 hours, what is the area of the floor that Jerry painted?

A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

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A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

pintukr · Accepted Answer

Two workers, Larry and Jerry, are planning to paint a classroom floor, each at their constant rate. Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor. Consequently, when the task is complete, Larry has painted 8 m^2 more of the floor than Jerry. If Larry could paint the floor by himself in 3.6 hours, what is the area of the floor that Jerry painted?

A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

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A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

Comparing the work ratios (and efficiency ratios)

(A) J : L = 9:10
(B) J : L = 5:6
(C) J : L = 9:11
(D) J : L = 4:5
(E) J : L = 3:4

for option (D) . time ratio (J:L) = 5 : 4 = 4.5 : 3.6 ....(for 72m^2)
 time ratio (J:L) = 5 : 4 = 4.5/72 : 3.6/72 ....(for 1 m^2)

J's time - L's time = 1/16 - 1/20 = 1/80  (as given in the stimulus)

(D) is the CORRECT answer

vjbhagat · Answer

If we go by the conventional method of solving time-work problems, we will reach very difficult to solve equations. So consider this approach:-

Lets assume Jerry painted  x   m^2  of wall in " t " hrs. So in the same " t " hrs, Jerry paints  (x+8)   m^2  of wall.

Now we have also been told that, for every 1  m^2  painting, Larry would take  \frac{1}{80}  hrs less than Jerry. So for painting  x   m^2  of wall when Jerry takes " t " hrs, Larry would take " t  -  \frac{x}{80} " hrs. But both should finish work in "t" hrs. So clearly Larry will finish those remaining 8  m^2  of wall in remaining  \frac{x}{80}  hrs.

So Larry's rate of work is:- 8  m^2  in  \frac{x}{80}  hrs =  \frac{640}{x}   \frac{m^2}{hrs} .

Also given that, Larry can finish the complete wall in 3.6 hrs, 
Therefore, equating the total wall paint area we get:-  \frac{640}{x}   \frac{m^2}{hrs}  * 3.6 hrs =  x   m^2  +  (x+8)   m^2

2x^2 + 8x -2304 = 0 
Solving for  x  we get,  x = 32 m^2

Therefor Jerry painted  32   m^2  area, hence option D

Kinshook · Answer

Given: Two workers, Larry and Jerry, are planning to paint a classroom floor, each at their constant rate. Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor. Consequently, when the task is complete, Larry has painted 8 m^2 more of the floor than Jerry.

Asked: If Larry could paint the floor by himself in 3.6 hours, what is the area of the floor that Jerry painted?

Let the area of the floor be x m^2
Larry taken 3.6 hours to paint the floor = x m^2
Larry takes hours to paint 1 m^2 of the floor = 3.6/x hours
Jerry takes hours to paint 1 m^2 of the floor = 3.6/x + 1/80 = (288 + x)/80x hours
Jerry takes hour the paint the floor of x m^2 = (3.6 + x/80) hours

Area painted by Larry in 1 hour = x/3.6 m^2
Area painted by Jerry in 1 hour = 80x/(288+x) m^2
Area painted by Larry / Area painted by Jerry = x/3.6 * (288+x)/80x = (288+x)/288
Area painted by Jerry / Total Area = 288/(576+x)
Area painted by Jerry = 288x/(576+x)
Area painted by Larry = (288+x)x/(576+x)

Area painted by Larry - Area painted by Jerry = x^2/(576+x) = 8
x^2 = 8x + 4608
x^2 - 8x - 4608 = 0
x = 72 m^2

Area painted by Jerry = 288x/(576+x) = 288*72/(576+72) = 288*72/648 = 32 m^2

IMO D

IshanSaini · Answer

Given two painters Larry and Jerry,

Condition 1: 
let t be the time taken by Larry to paint 1 m2 of floor
Larry: t
Jerry: t+1/80

Condition 2:
if Larry and Jerry paint together, then when they finish work, say at time y
let x m2 be painted by Larry, then (Larry painted 8 more than Jerry)
Larry: x
Jerry: x-8

Let total area = d
x + x-8 = d,
or 2x-8=d,
or x = d/2 + 4

Larry:  d/2+4 
Jerry:  d/2-4

Using time, y = area painted * time per m2
as y is same for both Larry and Jerry

we have  (d/2+4)*t = (d/2-4)*(t+1/80) , --(i)

Condition 3:
Larry takes 3.6 hrs for painting complete floor,
 d*t = 3.6 , --(ii)

Solving (i) and (ii) for d and t we get,
 1280t^2+8t-3.6=0 
t = 1/20, d = 72

Final answer,
 painted_area= d/2-4 = 32 m2

jm2k5 · Answer

Let A m^2 be the area of the floor painted by Jerry.
Then Area of Floor painted by Larry = (A+8) m^2

Therefore, total Area of Floor = A+(A+8) = (2A+8) m^2

Larry alone paints (2A+8) m^2 in 3.6 hrs. .........................................................................   (Equation 1)

Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor.

==> Larry requires (A/80) hours less than Jerry to paint A m^2 of the floor.

This means Larry painted the extra 8 m^2 in (A/80) hrs.

==> Larry paints 8 m^2 in (A/80) hrs.
==> Larry paints 8*(80/A) m^2 in 1 hr.
==> Larry paints 8*(80/A)*3.6 m^2 in 3.6 hrs.

But we know from equation 1 that Larry paints (2A+8) m^2 in 3.6 hrs.

==> (2A+8) = 8*(80/A)*3.6
or, A+4 = 32*36/A
or, A^2 + 4A - (32*36) = 0
or, (A-32)(A+36) = 0

or, A = 32 (A can't be negative so we reject -36)

IMO Option D

PK1 · Answer

Two workers, Larry and Jerry, are planning to paint a classroom floor, each at their constant rate. Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor. Consequently, when the task is complete, Larry has painted 8 m^2 more of the floor than Jerry. If Larry could paint the floor by himself in 3.6 hours, what is the area of the floor that Jerry painted?

A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

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A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2
Answer:

This requires equation buildup to solve numerically. Lets test answer choices to see if it matches expectation
A. 72 is the area Jerry could paint
So Larry would have painted 80. So total area = 152
Now Larry could paint himself in 3.6 hours so Larry speed = 152/3.6 not divisible so lets test other choices

B. 40 is the area Jerry could paint
So Larry would have painted 48. So total area = 88
Now Larry could paint himself in 3.6 hours so Larry speed = 48/3.6 not divisible so lets test other choices

C.36 is the area Jerry could paint
So Larry would have painted 44. So total area = 80
Now Larry could paint himself in 3.6 hours so Larry speed = 80/3.6 not divisible so lets test other choices

D.32 is the area Jerry could paint
So Larry would have painted 40. So total area = 72
Now Larry could paint himself in 3.6 hours so Larry speed = 72/3.6 = 20 sqm/hours
and Larry could have painted in 2 hours 
So Jerry speed is 32/2 = 16 sqm/hours
Now lets test one more condition of 1/80
in 1 sqm, time taken by Jerry = 1/16
time taken by Larry = time taken by Jerry - 1/80 = 1/16-1/80 = 1/20
So all question stem comes out to be true
So D is the answer

BaoN. · Answer

Let x be the area Jerry painted
The area Larry painted is x+8
The floor is 2x+8 (m^2) (1)
Larry is faster, hence when Jerry finished painting x (m^2), Larry has extra x/80 hr to paint the extra 8 m^2
-> Rate of work of Larry is 8/(x/80)=640/x
Larry completes work alone in 3.6hr, so the work can be expressed as 3.6*640/x (2)

From (1) and (2) we have:
2x+8 = 3.6*640/x
-> x(x+4)=32*36
-> x=32

Answer D

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Oppenheimer1945 · Answer

Weird approach:
Observing the options closely along with the numbers given in 
32+40=72 m^2, and Lary's along time (3.6hrs, =72/20);
 rate of Jerry=20, rate of Lary=20

Hence D is correct)

julialuu · Answer

Two workers, Larry and Jerry, are planning to paint a classroom floor, each at their constant rate. Larry requires 1/80 hours less than Jerry to paint 1 m^2 of the floor. Consequently, when the task is complete, Larry has painted 8 m^2 more of the floor than Jerry. If Larry could paint the floor by himself in 3.6 hours, what is the area of the floor that Jerry painted?

A. 72 m^2
B. 40 m^2
C. 36 m^2
D. 32 m^2
E. 24 m^2

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for the Around the World in 80 Questions

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Call the area of the floor that Jerry painted is x m^2, x >0.
Larry has painted (x+8) m^2 with the same time as Jerry pained x m^2
The classroom floor is (2x+8) m^2 and Larry painted (2x+8) m^2 alone in 3.6hours

To paint 1 m^2, Larry required 1/80 hours less than Jerry
Hence to paint x m^2 Larry required x/80 hours less than Jerry
So in the extra time (x/80 hours) Larry alone painted 8m^2 more.

Combined we have:
Larry painted (2x+8) m^2 alone in 3.6hours
Larry alone painted 8m^2 in x/80 hours
Then we have equation
(2x+8)/3.6 = 8/(x/80)
i.e (2x+8)x=8*80*3.6=8*8*36
i.e. x(x+4)=4*8*36 =32.36
we can easily see x=32 and pick D
the second value x = -36 does not satisfy x >0

Answer: D. 32 m^2

sujoykrdatta · Answer

If the area be 2A m^2: Larry painted 8 m^2 more than Jerry in the same time
=> Larry painted (A + 4)m^2 and Jerry painted (A - 4) m^2
To paint 1 m^2, Larry takes 1/80 hour less than Jerry
=> For the entire area 2A, Larry will take 2A/80 hours less than Jerry
Time Larry would take alone to paint 2A m^2 = 3.6 hours => Rate of work = 2A/3.6 m^2/hr 
=> Actual time for which Larry worked = (A + 4)/(2A/3.6) = 3.6(A + 4)/2A hours ... (i)
Time Jerry would take alone to paint 2A m^2 = (2A/80 + 3.6) hours => Rate of work = 2A/(2A/80 + 3.6) m^2/hr
=> Actual time for which Jerry worked = (A - 4)/(2A/(2A/80 + 3.6)) = (A/40 + 3.6)(A - 4)/2A hours ... (ii)

The actual times must be same since they work together: 
3.6(A + 4)/2A = (A/40 + 3.6)(A - 4)/2A
=> 3.6(A + 4) = (A/40 + 3.6)(A - 4)
Multiplying 10:
36(A + 4) = (A/4 + 36)(A - 4)
Multiplying 4: 
144(A + 4) = (A + 144)(A - 4)
=> 144A + 576 = A^2 + 144A - 4A - 576
=> A^2 - 4A - 1152 = 0
=> A^2 - 36A + 32A - 36*32 = 0
=> A = 36
=> Area Jerry painted = A - 4 = 32 m^2

Alternatively: 
3.6(A + 4) = (A/40 + 3.6)(A - 4)
In this equation, we plugin different values of A: 
Option A: Jerry painted 72 m^2 => A - 4 = 72 => A = 76 - does not satisfy
Option B: Jerry painted 40 m^2 => A - 4 = 40 => A = 74 - does not satisfy
Option C: Jerry painted 36 m^2 => A - 4 = 36 => A = 40 - does not satisfy
Option D: Jerry painted 32 m^2 => A - 4 = 32 => A = 36 - satisfies

Answer D

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