No Cell Phone No Scrap Paper long will it take to complete the task with both co

Question

No Cell Phone No Scrap Paper long will it take to complete the task with both computers working? tively. How 21. Rowing with the current, a canoeist paddled 18 miles in 3 hours. Against the current, the canoeist could paddle only 6 miles in the same amount of time. Find the rate of the canoest in calm water and the rate of the current. 22. A motorboat leaves a harbor and travels at an average speed of 10 mph towards a small island. Four hours later, a cabin cruiser leaves the same harbor at an average speed of 30 h toward the same island. In how many hours after the cabin cruiser leaves the harbor will it be alongside the motorboat?

Explanation / Answer

20. Two computers let's say A,B

A will take 20 mins and B will take 60 mins to finish a task

whenver you encounter these kind of problems the solution is LCM(lease common multiple) model.

i.e. take LCM for given 2 numbers

LCM(20,60) ----> 60

now take this number (lcm number) as total work needs to be finished.

So A will take 20 mins to finish 60 units of work ie, per min it will complete 3 units of work

B will take 60 mins to finish 60 units of work i.e. per min it will complete 1 unit of work

when working together in one min they will do sum of their works.....so 4 units of work

so totally to finish 60 units of work 60/4 -----------> 15 days required.

Answer is 15

21..

Along with the current 18 miles in 3 hrs ...Against the current 6 miles in 3 hrs...

whaterever it is the basic formula remains same DISTANCE = SPEED * TIME

When you are travelling with the current your speed  = Speed of current(C) + Canoset speed(B)

When you are travelling against the current your speed  = Canoestspeed(B) - Speed of current(C)

so 1st case along the current ----> distanace = 18 miles ; speed = C+B ; time 3 hrs

18 = (C+B)*3 = 3C+3B -->.EQ1

similiaryly

2nd case against the current ----> 6 = (B-C)*3 = 3B-3C --->EQ2

Do 3*EQ2 so LHS of EQ1 and EQ2 (left hand side) will become equal so then equal corresponding RHS.

By doing so we will get 3(B+C) = 9(B-C) ---> 3B+3C = 9B-9C

----------------> 6B= 12C ----> B/C = 2/1

i.e. Substitute this B/C ratio in EQ1 u will get B= 4 and C= 2

so rate of canoest in calm water = 4 and rate of current = 2

22.

motorboat 4 hrs travelled before cruiser starts so 4*10  = 40 miles travelled

After that cruiser started for same isaland so Relative speed = Cruiser speed - motorboat speed

ie. ----> relative speed = 30-10 = 20 miles per hour

we need find time when they will meet? time = distance/speed

time  = 40/20 = 2hrs

so after 2 hrs cruiser started its journey it will align with the motor boat

Cross check your answer as

motorboat with 10mph travlled toatlly (6 hrs-> 4 before cruiser starts + 2 after cruiser started)---> 6*10 = 60

cruiser 2 hrs 30mph 60 miles

so they both will align now.

Any doubt regarding the solution please ask me.

  

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