Problem Solving - 52

A boat traveled up stream a distance of 90 miles at an average speed of (v-3) mph and then traveled the same distance downstream at an average speed of (v+3) mph. If the trip upstream took half an hour longer than the trip downstream, how many hours did it take the boat to travel downstream?

(A) 2.5
(B) 2.4
(C) 2.3
(D) 2.2
(E) 2.1

Answer: A

Total upstream time taken by boat to travel = 90/ (v-3) hrs
Total downstream time taken by boat to travel = 90/ (v+3) hrs

It is given that upstream took half an hour longer than the trip downstream:

=> 90/(v-3) - 90/(v+3) = 1/2
=> [90*(v+3) - 90(v-3)] / [(v^2) -9)] = 1/2
=> 90*[(v+3)-(v-3)] / [(v^2) -9] = 1/2
=> 90*6/ [(v^2) -9] = 1/2
=> v^2 - 9 = 2*90*6
=> v^2 = 2*90*6 + 9 = 2*9*10*6 + 9 = 9(2*10*6+1) = 9(121) = 9*11*11
=> v = 3*11 = 33

We are suppose to calculate = 90/(v+3) = 90/36 = 30/12 = 5/2 = 2.5 ans

Problem Solving - 33

A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the same direction along the same path at a constant rate of 20 miles per hour. The cyclist stops to wait for the hiker 5 minutes after passing her, while the hiker continue to walk at her constant rate. How many minutes must the cyclist wait until the hiker catches up?

A. 20/3
B. 15
C. 20
D. 25
E. 80/3

Answer: C

Hiker's relative speed = 16 m/h
Hiker traveled in 5 min: 16 * 5/60 = 4/3 miles.
Time taken by hiker to cover 4/3 miles: 4/(4/3) = 1/3 hrs = 20 minutes

Manhattan Challenge Problem of the week! - 05/21/07

Edwin is planning to drive from Boston to New Orleans. By what percent would his travel time be reduced if Edwin decides to split the driving time equally with his friend George, instead of making the trip alone?

(1) The driving distance from Boston to New Orleans is 1500 miles.
(2) George’s driving speed is 1.5 times Edwin’s driving speed.

Answer: B

OE - The question asks for the percent decrease in Edwin’s travel time. To determine this, we need to be able to find the ratio between, T1 (the travel time if Edwin drives alone) and T2 (the travel time if Edwin and George drive together). Note that we do NOT need to determine specific values for T1 and T2; we only need to find the ratio between them.

Percentage change is defined as follows: Difference/Original = (T1 - T2)/ T1 = 1 - (T2/ T1)

Ultimately, we can solve the percentage change equation above by simply determining the value of T2 /T1

Using the formula Rate × Time = Distance, we can write equations for each of the 2 possible trips

T1 = Travel time if Edwin drives alone
T2 = Travel time if Edwin and George drive together
E = Edwin’s Rate
G = George’s Rate
D = Distance of the trip

If Edwin travels alone: ET1 = D
If Edwin and George travel together: .5(E + G)T2 = D

(Since Edwin and George split the driving equally, the rate for the trip is equal to the average of Edwin and George’s individual rates).

Since both trips cover the same distance (D), we can combine the 2 equations as follows:

ET1 = .5(E + G)T2

Then, we can isolate the ratio of the times (T2/T1) as follows:

E/ .5(E + G) = T2/ T1

Now we look at the statements to see if they can help us to solve for the ratio of the times.

Statement (1) gives us a value for D, the distance, which does not help us since D is not a variable in the ratio equation above.

Statement (2) tells us that George’s rate is 1.5 times Edwin’s rate. Thus, G = 1.5E. We can substitute this information into the ratio equation above:

E/ .5(E + G) = T2/ T1 ---> E/ .5(E + 1.5E) = T2/ T1 ---> E/ .5E + .75E = T2/T1
---> E/ 1.25E = T2/ T1 ---> 1/ 1.25 = T2/ T1 ---> .8 = T2/ T1

Thus, using this ratio we can see that Edwin’s travel time for the trip will be reduced as follows:

1 - (T2/T1) = 1 - .8 = .2 ---> 20%

Statement (2) alone is sufficient to answer the question.

The correct answer is B.

Manhattan Challenge Problem of the week! - 09/04/07

Stephanie, Regine, and Brian ran a 20 mile race. Stephanie and Regine's combined times exceeded Brian's time by exactly 2 hours. If nobody ran faster than 8 miles per hour, who could have won the race?

I. Stephanie II. Regine III. Brian

(A) I only
(B) II only
(C) III only
(D) I or II only
(E) I, II, or III

Answer: D

Let the race time of Stephanie = S
Let the race time of Regine = R
Let the race time of Brian = B

It is given that -- Stephanie and Regine's combined times exceed Brian's time by 2 hours.

Hence S + R = B + 2 ------- (1)

To win the race, an individual's time must be less than 1/3 of the the combined times of all the runners.

Therefore for Brian to win the race (=> that Brian would have the lowest time) his time would need to be less than 1/3 of the combined times for all the runners.

=> B is less than 1/3 of (S+R+B)

=>3B<> is less than (S+R+B)

=> 2B is less than (S+R)

Making use of (1) in above

2B<> is less than (S+R)

=> 2B<> is less than B+2

=> B<2


Thus to win the race Brian's time must be less than 2 hours. which is impossible as fastest Brian run is 8 miles/hr => that the least amount of time in which he can complete the 20 mile race is 2.5 hrs.

Hence Stephanie and Regine as possible winners. Since the question gives us same information about Stephanie and Regine, we cannot state either one as a possible winner.

Hence, the correct answer is D

Manhattan Challenge Problem of the week ! - Aug 28

High Functioning

The vertical position of an object can be approximated at any given time by the function: p(t) = rt – 5t2 + b where p(t) is the vertical position in meters, t is the time in seconds, and r and b are constants. After 2 seconds, the position of an object is 41 meters, and after 5 seconds the position is 26 meters. What is the position of the object, in meters, after 4 seconds?

(A) 24

(B) 26

(C) 39

(D) 41

(E) 45

Answer - D , for OE click on link below.
Manhattan challenge problem