DS - Driving 1.5 times slower

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??