Data Sufficiency

How long did it take to travel 400km?
a. The car travelled the first 200km in 2.5hrs
b. If the car's average speed had been 20 km/h faster, it would have travelled
the 400km in 1 hour less time.

OA is B

Akansha wrote:How long did it take to travel 400km?
a. The car travelled the first 200km in 2.5hrs
b. If the car's average speed had been 20 km/h faster, it would have travelled
the 400km in 1 hour less time.

Hi!

This question is a perfect place for our data sufficiency hero, "number of equations vs number of unknowns Man", to make an appearance.

Starting by analyzing the stem:

we know that distance = rate * time. We have the distance and we want to solve for the time. Accordingly, we have 1 equation and 2 unknowns; 1 more distinct linear equation and we can solve.

To the statements:

(1) the first 200km took 2.5 hours. No info about the second half of the trip: insufficient.

(2) if the rate had been 20km more, the trip would have taken 1 hour less.

We can convert that info into the following equation:

400 = (r+20)(t-1)

Since we also know that:

400 = rt

we now have 2 distinct equations for our 2 unknowns. However, since our equation has an "rt" term, when we combine them we'll get a quadratic, something that would lead to two solutions if this were pure algebra.

Fortunately for us, this isn't pure algebra: we're talking about rates and speeds, which must be non-negative. So, we can safely discard the negative solution and know that there will only be one possible value for t. Accordingly, (2) is sufficient alone: choose (B).

The "number of equations/number of unknowns" rule is quite possibly THE most powerful weapon in your DS arsenal - be on the lookout for opportunities to use it to wreak destruction on the test!

My choice B

to find - t (time taken to travel 400km)
A) time taken to travel first 200km = 2.5hrs but no information given about next 200 km and thus insufficient.
B)let initial speed = x
Distance = x*t = 400 => x=400/t ------ eqn1
also with new speed, x+20, time taken = t-1
thus x*t = (x+20)(t-1)
Solving xt = xt + 20t -x -20 => 20t -x -20=0
using eqn1, 20t - 400/t -20 =0
=> t^2 - 20t -1 =0 => t =[ 20+-root(404) ] /2
One value will be -ve and as time can't be -ve, we can reject that, leaving us with a unique +ve value.
Thus B is sufficient.