Problem Solving

anshumishra wrote:

Geva@MasterGMAT wrote:

HPengineer wrote:Do you got another beer problem u can create for me to try regarding this concept?

Sure

This is made up on the spot, so it's probably got ambigous holes in it. But here goes:

Geva sits at the pub and drink beers at a constant rate of x beers and hour. A while later, he is joined by HPengineer who matches him pint for pint (i.e. orders a beer only whenever Geva orders one). If a Beer costs $5 each, and the beer tab when they both pass out simultaneously ends up at $150, how much did Geva pay, assuming they both bought nothing but beers?

(1) If their pub stay is measured from their first respective beer to the time they pass out, Geva's pub stay is 50% longer than HPengineer's.

(2) Geva spent 40 minutes drinking beer before he was joined by HP engineer.

Statement 1 :
Since Geva spent 50% more time and beer consumption rate is constant, Hence Geva will be paying 50% more than HPengineer (90$ Geva, 60$ HPengineer) -> Sufficient

Statement 2 :
We can't determine what % of total drinks Geva has already taken based on the time "40" minutes (Depends how fast Geva drinks ) -> Insufficient.

Hence A.

Is it correct to assume that the rate of beer consumption is constant?

problem stem states matched pint for pint so i would think its ok to assume.

HPengineer wrote:problem stem states matched pint for pint so i would think its ok to assume.

ok....

HPengineer wrote:im stumped...

sadly best i can come up with is a total time which might not even be right


geva = 3/2T

hpengineer = T

total time at pub = 5T/2

really stuck as what to do now... X is our respective rates so could we calculate total time multiplied by rate to = 150 or would we take total time and multiply by 5 dollars to get 150... im not really sure where to proceed..

HPengineer wrote: @anshumira

very logical approach... thanks for the eye opener... would still like to see it algebraically to see where i was going wrong.

OK. Lets test it algebraically.

T-> Time in hr; rate -> X ml/hr
geva stayed for 3/2T -> consumed -> 3XT/2 ml
hpengineer stayed for T -> consumed TX ml
total time at pub = 5T/2 -> both consumed a total of -> 5XT/2 ml

Now; 5XT/2 ml costs 150 dollars
Hence; TX costs -> 150*2/5 = 60 $ (which Hpengineer will pay)
3TX/2 costs -> 150*2/5 * 3/2 = 90 $ (which Geva will pay)[/quote]

HPengineer wrote:@anshumira

very logical approach... thanks for the eye opener... would still like to see it algebraically to see where i was going wrong.

Anshumira's approach was what I was aiming for - good job.

Here comes the algebra to prove stat. (1) sufficient:

Place the information in a rate / time / work table, where rate*time = work.

Let Geva's rate be x beers an hour, and so is Hp Engineer's.
Let the time they spend at the pub together be y, so that Geva's time in the pub before HPengineer's arrival is an additional y/2.

thus, the question is dealt with in two phases:
First, Geva drinks alone for y/2 hours at a rate of x beers/hour, so the "work" he does alone - the number of beers - is xy/2.
Then, HP engineer joins in, and the two work together at a combined rate of 2x. This goes on for y hours so the work done in this phase is 2xy.

We now have an expression of the total work done in the two phases: xy/2 + 2xy. from the question stem, we get that this should amount to 150/5 = 30 beers. We cant find x and y from this single equation with 2 unknowns, but we don't need to: the question asked for Geva's beer tab, which is "number of beers drunk by Geva" times $5.
Geva Drank (xy/2) (first phase) + xy (2nd phase) = 3xy/2.
All we really need to know is the value of the product xy, which can be found from the equation xy/2 + 2xy=30.
From the equation we've formed, it is possible to find the value of the expression: if 2xy + xy/2 = 30, then 5xy/2 = 30.