cmal_s wrote:q) Carlos took 0.5 hours from home to the library was the distance he travelled greater than 6 miles ( 1miles = 5280 feet)


1) His average speed in the trip was greater than 16 feet per second

2) His average speed was less than 18 feet per second




answer is E

dont get it

Lets rephrase the question in terms of Rate (R) , Time (T) and Distance (D).

R x T = D

x 30x60 sec = D


So the question asks IS x(1800) > 6 miles (31680 feet )

In directly we need to find x ? or Rate (R)

statement (1) : x > 16

So lets take x = 17

R x T = 17 x 1800 = 30600 < 31680 .

But if we take x = 18 then

R x T = 32400 > 31680

So not sufficient .

statement (2) :


x < 18

So x could be x = 17

And we already know that when x = 17 distance covered is less than 6 miles .

But dont assume rate to be an integer until and unless clearly stated .

Here the rate could be 17.6 also which is less than 18 .

And that is exactly equal to 31680 .

So clearly not sufficient .

Combine (1) and (2)

so 16 < x < 18

again not sufficient hence E.

cmal_s wrote:q) Carlos took 0.5 hours from home to the library was the distance he travelled greater than 6 miles ( 1miles = 5280 feet)


1) His average speed in the trip was greater than 16 feet per second

2) His average speed was less than 18 feet per second




answer is E

dont get it

easier way would be to find the rate when distance travelled is 6 miles.
6miles at 0.5 seconds is = 17.6 ft/s
for distance to be greater than 6, rate would have to be greater than 17.6.
statement 1) r may be 17.9 or 16.1 INSUFFICIENT
statement 2) r may be 17.9 or 16.1 INSUFFICIENT
comibined ) r still may be 17.9 or 16.1 since 18>r>16 INSUFFICIENT

E