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by gmatnmein2010 » Fri Feb 19, 2010 7:46 pm
Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
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by thephoenix » Fri Feb 19, 2010 7:51 pm
If both J & B work they will finish the job in J*B/(J+B) hrs.
Also given both J&B are even.

Lets start with statement B
(J+B)^2 = 400
If J & B were equal then J=B=10
They will finish the work in 5 hrs i.e 5 PM, statement A is different hence we know J & B are not equal. Sufficient but before we jump to C lets look at statement A alone

J & B finish the job by 4:48p.m they take 4 48/60 hrs = 4 4/5 = 24/5

So we know J*B/(J+B) = 24/5
since J and B are even they can't be fractions, hence
J*B has to be a multiple of 24
J+B has to be a multiple of 5
since J & B are even J+B can only be multiples of 10.
Lets assume J=B and start plugging in values
when J*B/(J+B) = 48/10
J+B = 10, J=B=5 so J*B = 25 too small
when J*B/(J+B) = 96/20
J+B = 10, J=B=10 so J*B = 100 too big
when J*B/(J+B) = 144/30
J+B = 30, J=B=15 so J*B = 225 too big

J*B values keep increasing and will never equal the value derived by assuming j&B are equal.

Hence answer should be A.
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by shashank.ism » Sun Feb 21, 2010 4:28 am
gmatnmein2010 wrote:Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
both start at noon i.e. 12:00 noon
now both finishes work in (1/J +1/B ) hrs

St.1) 1/4.8 = 1/J + 1/B --> this doesn't show J=B not suff.
St.2) (J+B)^2 = 400 --> J+B = 20 , also we know that 1/J + 1/B = 1/T --> (J+B)JB = 1/T --> 20 /JB = 1/T not suff.
combined: 20/JB = 1/4.8 --> JB= 20 x4.8 = 96
J+B = 20 --> (20 -J ) (J) = 96 surely J =/= 10
so J=/= B

hence suff....Ans C
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by harsh.champ » Sun Feb 21, 2010 4:43 am
gmatnmein2010 wrote:Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)^2=400
Let a unit of work be P paints.(an imaginary unit)
Speed of Jane = P/J
Speed of Bill = P/B
Now,they start at 12:00 noon together.

Statement 1:- Total speed =(P/J + P/B)
They took 4(48/60) = 4.8 hours
=>(P/J + P/B) x (4.8) = P
=>(1/J + 1/B) = 1/(4.8)
=>(J + B)/JB =5/4
Its written that both are even numbers,which is not possible according to the statement.
J and B should be 2 each for product to be 4.
Is the 1st option wrong - gmatnmein2010??
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by neoreaves » Wed Mar 03, 2010 10:38 am
The combined work rate will be JB/ (J+H) ....already discussed by other members here

If J = B then the rate becomes J/2 ...just put J = B in the above equation

We also know that J and B are even numbers

1) JB/ (J+H) = 4.8

Lets see what values we get if J=B

J/2 = 4.8 --> J = 9.6 ...which is not an even number so J is not equal to B ....Sufficient

2) We can get the simplified form J + B = 20 ....this is possible in a number of ways ....e.g J = B = 10 and J=12 and B = 8

Insufficient

Thus answer should be A
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by gmatmachoman » Wed Mar 03, 2010 1:34 pm
From A :

(J*B)/(J+B)=4.8 hrs

Probable values of J & B can be (2.4,2.4), (1.4,3.4),(1.2,3.6),(3.4,1.4),(3.6,1.2). Now we have 5 combinations of J&B values combining to 4.8 hrs.

So we can't surely say Value of J= Value of B.

So A is NOT sufficient.

From B

(J+B)^2 =400
again we can have pair of comnibations of J&Bsumming up to 20.
(10,10),(2,18),(4,16),(6,14)...so on

So B is also Insufficient to say J=B.

So I vote for E...

Plz share ur views
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by kstv » Wed Mar 03, 2010 8:03 pm
Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)²=400

1) Jane and Bill takes 4 hrs 48 min = 24/5 hrs
in one hr 1/J + 1/B = J + B / JB , complete work in JB/(J+B) = 24/5
If J = B then JB/(J+B) = J²/2J = J/2 but since J is even J/2 should be an integer , which 24/5 is not. Sufficient to say J <>B.

2) (J+B)² = 400 = J+B = 20 cannot say J = B , diff possibilities even if J and B are even (4,16), (6, 14) (8, 12)
. Insufficient

1) and 2) together JB/(J+B) = 24/5 , J+B = 20 or JB/20 = 24/5 , JB = 24X4 = 96 if J = B, J² = 96 not a perfect square.
so J <> B , in fact of J and B are 12 and 8 but it is not imp to calculate in DS.

IMO A
Last edited by kstv on Thu Mar 04, 2010 4:02 am, edited 1 time in total.
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by gmatmachoman » Wed Mar 03, 2010 11:34 pm
Seems I have been silly...I shuld blame the timing..it was 3 am & all my people were sleeping...so stupid I was...


C is the right answer...

Find the value of J+B from st 2.

Plug them in st 1 to get values of J*B =96

J=12 & B=8

so Jnot equal to B.
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by cunazza » Thu Mar 04, 2010 2:47 am
kstv wrote:Jane can paint the wall in J hours, and Bill can paint the same wall in B hours. They begin at noon together. If J and B are both even numbers is J=B?

(1) Jane and Bill finish at 4:48 p.m.
(2) (J+B)²=400

1) Jane and Bill takes 4 hrs 48 min = 24/5 hrs
in one hr 1/J + 1/B = J + B / JB , complete work in JB/(J+B) = 24/5
If J = B then JB/(J+B) = J²/2J = J/2 but since J is even J/2 should be an integer , which 24/5 is not. In sufficient.

2) (J+B)² = 400 = J+B = 20 cannot say J = B , diff possibilities even if J and B are even (4,16), (6, 14) (8, 12)
. Insufficient

1) and 2) together JB/(J+B) = 24/5 , J+B = 20 or JB/20 = 24/5 , JB = 24X4 = 96 if J = B, J² = 96 not a perfect square.
so J <> B , in fact of J and B are 12 and 8 but it is not imp to calculate in DS.

IMO C
I agree with the solution above, but from Statement 1 we should conclude that J is different from B, so the statement is SUFFICIENT to answer the question! The correct answer should be A in my opinion.
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by kstv » Thu Mar 04, 2010 3:59 am
cunazza is right. The correct option is A not C. I wll change the post.
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