tkarthi4u wrote:
A boat crossed a lake from North to South at the speed of 4 km/h, entered a river and covered twice as much distance going upstream at 3 km/h. It then turned around and stopped at the south shore of the lake. If it averaged 3.8 km/h that day, what was its approximate downstream speed?
This one is really difficult to do algebraically in 2 minutes! I just did that solution and wouldn't recommend it to anyone. Not only are you solving for some variable R (the unknown downstream rate), but you have some unknown distances: X = distance across the lake and 2X = distance upstream = distance downstream. In addition, the RT = D or T = D/R formula means we will encounter fractions in our algebra--not ideal!
Since X is not specified, and it must drop out of the algebra in order to solve for R, why not pick a number for X? If we pick strategically, we can avoid dealing with too many fractions.
Avg. Speed = Total Distance/Total Time.
3.8km/h = (X + 2X + 2X)/Total Time.
Total Time = Time to cross lake + Time to go upstream + Time to go downstream.
In general, Time = Dist/Rate, so
let's pick distance X such that it is divisible by both given rates: 3km/h and 4km/h. Thus, X = 12km.
If the lake is 12km across, Time to cross lake = 12km/4km/h = 3h.
If the travel upstream is 24km, Time to go upstream = 24km/3km/h = 8h.
If the travel downstream is 24km, Time to go downstream = 24km/R. (Remember, we are solving for this R)
So,
3.8 = 5X/(3 + 8 + 24/R) = 60/(11 + 24/R)
3.8(11 + 24/R) = 60
38(11 + 24/R) = 600 (multiplied both sides by 10, just to get rid of decimals)
418 + (38)(24)/R = 600
(38)(24)/R = 182
R = (38)(24)/182 = (19)(24)/91 = 456/91 = approx 5.
tkarthi4u wrote:
Husband is seven times older than his son ? does this translate to W=7S or W - S = 7S.
I think it could be interpreted either way. To avoid this confusion, I think the actual GMAT would present the constraint with different wording. Something like "A father is seven times as old as his son..." (F = 7S)