If it took Carlos 30 min to cycle from his home to the library, was the distance that he cycled greater than 6 miles? (1 mile=5280feet)
1) His average speed from h to l was greater 16 feet per second
2) His average speed from h to l was less than 18 feet per second
My reasoning:
1) s*t=d, so 16fps = 16*60 fpm *30 = 28800fpm / 5280 = approx. 5.5 miles ->n.s.
2)18fps=18*60*30= approx. 6.1 miles
Ans E hence value is between 5.5 and 6.1
Is there a faster way to recognise that this is not working - My brain just cannot calculate these values so fast ... means it took me a few minutes to solve...
1) His average speed from h to l was greater 16 feet per second
2) His average speed from h to l was less than 18 feet per second
My reasoning:
1) s*t=d, so 16fps = 16*60 fpm *30 = 28800fpm / 5280 = approx. 5.5 miles ->n.s.
2)18fps=18*60*30= approx. 6.1 miles
Ans E hence value is between 5.5 and 6.1
Is there a faster way to recognise that this is not working - My brain just cannot calculate these values so fast ... means it took me a few minutes to solve...
