SAT MATH PROBLEM

Question

MATH PROBLEM WITH SPEED

The boat was traveling 12 km downstream, and 11 km upstream. If it took totally 1 hour the boat to travel and given that the speed of the river is 2 km/h, what is the speed of the boat travelled downstream?

a) 28

b) 12

c) 26

d) 24

Answer 1

"Pick an answer" strategy

option A:

11 = (28-2)*t ---> t = 11/26

12 = (28+2)(1-t) --> 12/30 = 1 - t

t = 1 - 12/30

= 1 - 2/5 = 3/5

NO they do not agree

option B:

11 = (12-2)*t --> 11/10 = t

which is GREATER than 1 hour,

so this cannot be the answer

option C:

11 = (26-2)*t

11 = 24*t

11/24 = t

12 = (26+2)*(1-t)

12 = 28*(1-t)

12/28 = 1 - t

t = 1 - 12/28

= 16/28 = 8/14 = 4/7

No they do not agree

================================

option D:

11 = (24-2)*t

11/22 = t

t = 1/2

12 = (24+2)*(1-t)

12/26 = 1 - t

t = 1 - 12/26

= 14/26 = 7/13

this is the closest, so far

***************************

R=23

11 = (23-2)*t

11/21 = t

12 = (23+2)*(1-t)

12/25 = 1 - t

t = 1 - 12/25 = 13/25

they are almost equal

algebraic solution

--------------------------------

R is the speed of the boat

downstream with the current:

12 = (r+2)*t ---> 12/(r+2) = t

upstream against the current:

11 = (r-2)*(1-t) --> 11/(r-2) = 1-t

1 = 1-t +t = 11/(r-2) + 12/(r+2)

common denominator is (r+2)(r-2)

(r+2)(r-2) = 11(r+2) + 12(r-2)

r^2 - 4 = 11r + 22 + 12r - 24

r^2 - 4 = 23r - 2

r^2 - 4 - 23r + 2 = 0

r^2 - 23r - 2 = 0

23 + sqrt( 23^2 - 4(1)(-2))/ 2

[23 + sqrt( 537)]/2

24 is the closest answer

Answer 2

r*t=d

Let r = rate in still water.

Let t = time traveling upstream (in hrs)

So 1-t = time traveling downstream

Upstream: rate = (r-2)km/hr So ((r-2)t = 12

Downstream: rate = (r +2)km/hr = (r+2)km/hr. So (r+2)(1-t) = 11

Now solve the two bold equations for r+2

Answer 3

t = d / r

t_down = 12 / (s + 2)

t_up = 11 / (s - 2)

t_down = t_up = 1

[12 / (s + 2)] + [11 / (s - 2)] = 1

Can you solve for t and answer?