Data Sufficiency

Driving 1.5 times slower, Bill was late for school today. What is the usual time it takes Bill to drive to school? (Assume that each day Bill takes the same route).

1) It took Bill 15 more minutes to drive to school today than usually
2) The distance between home and school is 15 miles


Please explain using example.

A is ans here.

From 1: 2.5vt=v(t+1/4) solve for t.So sufficient. (assume v=speed in km ; t=time in hrs; distance=v*t)

Also if v=the slower speed then actual speed is 1.5 times more i.e. (v+1.5v)=2.5V (actual usual speed); (1/4hr=15min)

A for me too

Suppose distance is d miles and his usual speed is s miles/hour

So, usual time = d/s hours

Today he drove at a speed of s/1.5 miles/hour

Stmt 1 - d/(s/1.5) = d/s + 15/60 solve for d/s SUFF

Stmt 2 - d = 15 miles... nothign else can be determined NOT SUFF

I had hard time figuring out what exactly 1.5 times slower mean but now I know.

Is A the OA?

decrease in speed =1.5 => 3/2

INCREASE IN TIME = 2/3 ( as vt=k)

hence 2/3 of usual time = 15 minutes

usual time =

PLEASE LET ME KNOW CAN WE USE THIS APPROACH?

Regards,
CBM

OA is A

Can someone help with a generic strategy to interpret such questions.

Driving 1.5 times slower

, the statement is so confusing ...!
That means bill drives 2.5 times the speed every day ..doesn't fits into the common sense purview of question writing.

True , it doesn't fit into the convinient way of saying, but that's not the actual value of speed, it's a way of expression vis'-a-vis' slower speed.

If you take x as the actual speed (which is higher speed) then 1.5 times less would be x-1.5x=-.5x which is not possible.

So, the assuming the slower speed as x then higher speed would be 1.5 times more i.e. 1.5x+x=2.5x

Alwys assume the lower value as x or any other variable in these type of cases, to get a positive value for calculation.

Amit

Yep!. I started out with X-1.5X and was like what? -0.5x??