PS-2

[sparsh.21]

Laura can paint 1/x of a certain room in 20 minutes. What
fraction of the same room can Joseph paint in 20 minutes if the two
of them can paint the room in an hour, working together at their
respective rates?

A. 1/3x
B. 3x/(x – 3)
C. (x – 3)/3x
D. x/(x – 3)
E. (x – 3)/x

OA

C

[amitabhprasad]

L -->1/x in 20 min ==> x/3 hrs to paint the room ==> she can paint 3/x part of room in 1 hrs
Say J can paint the room in y hrs ==> in 1 hrs J can paint 1/y par of room
Together
J+L can paint the room completely.
3/x+1/y => 3y+x = xy ==> y = x/x-3
in 1 hrs J will paint x-3/x thus in 20 min he will paint
(x-3)/3x

[bsandhyav]

Laura can paint 1/x of a certain room in 20 minutes. What
fraction of the same room can Joseph paint in 20 minutes if the two
of them can paint the room in an hour, working together at their
respective rates?

A. 1/3x
B. 3x/(x – 3)
C. (x – 3)/3x
D. x/(x – 3)
E. (x – 3)/x

OA

C

The basic formula that needs to be used here is : rate* time= work

rate(Laura) + rate(Joseph) = rate of both working together

rate of both working together = 1/60 ( i.e work/time)

Therefore (1/x)/20 + rate (Joseph) = 1/60
rate(Joseph) = (x-3)/(60*x)

so work(joseph)= rate*time
=(x-3)*20/(60*x)
= (x-3)/3x


Hence C


This is a purely algebric approach. Any body with a better solution?

[cramya]

This is a purely algebric approach. Any body with a better solution

The one Amit/u did would be the safest, IMO. Plus its not time consuming by any means.. Solvable well within 2 minutes..

Regards,
CR

[Mozartain]

I'd strive to avoid unnecessary calculations.

since both of them working together take an hour to paint the whole room, in 20 minutes they together can paint 1/3 of the room.

since Laura paints 1/x of it, Joseph does the rest, which is 1/3 - 1/x