Math, asked by visunpatel2010, 8 months ago

the river flow is less than the peed of the boat in still water
Find a number greater than ! such that the sum of the number and its reciprocal is
The dilterence of the speed of a faster dur and a slower car is 200 Ader | the slower war​

Answers

Answered by vikhyat04
0

First, let us explain the meaning of "upstream" and "downstream."

When a boat travels in the same direction as the current, we say that it is traveling downstream.

downstream

Thus if b is the speed of the boat in still water, and c is the speed of the current, then its total speed is

Downstream speed = b + c

When a boat travels against the current, it travels upstream.

upstream

In this case, its total speed is

Upstream speed = b − c

Problem.   The speed of a boat in still water is 30 mph.  It takes the same time for the boat to travel 5 miles upstream as it does to travel 10 miles downstream.  Find the speed of the current.

Solution.  The key to this type of problem is same time.  That will give the equation,

Time upstream = Time downstream

Now, speed, or velocity, is distance divided by time -- so many miles per hour:

v  = d

t

Therefore,

t  = d

v

The equation will be

 Time upstream = Time downstream

 

Distance upstream

Speed upstream  = Distance downstream

 Speed downstream

 

       Let x be the speed of the current.  Then according to the problem:

 

  _5_  

30 − x =    10    

30 + x

 

       Therefore,

5(30 + x) = 10(30 − x)

 

150 + 5x = 300 − 10x

 

5x + 10x = 300 − 150

 

15x = 150

 

x = 10 mph

Problem 7.   The speed of a boat in still water is 15 mi/hr.  If the boat travels 8 miles downstream in the same time it takes to travel 4 miles upstream, what is the speed of the current?

To see the equation, pass your mouse over the colored area.

To cover the answer again, click "Refresh" ("Reload").

But do the problem yourself first!

 Time upstream  = Time downstream

 

Distance upstream

Speed upstream    = Distance downstream

 Speed downstream

 

       Let x be the speed of the current.  Then according to the problem:

 

  _4_  

15 − x  =     8    

15 + x

 

4(15 + x)  = 8(15 − x)

 

60 + 4x  = 120 − 8x

 

4x + 8x  = 120 − 60

 

12x  = 60

 

x  = 5 mph

Problem 8.   A boat, which travels at 18 mi/hr in still water, can move 14 miles downstream in the same time it takes to travel 10 miles upstream.  Find the speed of the current.

 Distance upstream

Speed upstream    =  Distance downstream

 Speed downstream

 

  _10_  

18 − x  =     14    

18 + x

 

10(18 + x)  =  14(18 − x)

 

180 + 10x  =  252 − 14x

 

10x + 14x  =  252 − 180

 

24x  =  72

 

x  =  3 mph

Problem 9.   Train A has a speed 15 mi/hr greater than train B.  If train A travels 150 miles in the same time train B travels 120 miles, what are the speeds of the two trains?

Solution.   Let x be the speed of train A.  Then the speed of train B is x − 15.

 Train A's time = Train B's time

 

A's distance

A's speed   = B's distance

 B's speed

 

150

x   =   120  

x − 15

 

150(x − 15) = 120x

 

150x − 150· 15 = 120x

 

30x = 150· 15

 

x = 150· 15

  30

 

 = 150

 2  

 

x = 75 mph

 

x − 15 = 60 mph

Problem 10.   A train travels 30 mi/hr faster than a car.  If the train covers 120 miles in the same time the car covers 80 miles, what is the speed of each of them?

Solution.   Let x be the speed of the train.  Then the speed of the car is x − 30.

 Train's time = Car's time

 

Train's distance

Train's speed   = Car's distance

 Car's speed

 

120

x   =    80    

x − 30

 

120(x − 30) = 80x

 

120x − 120· 30 = 80x

 

40x = 120· 30

 

x = 120· 30

  40

 

 = 3· 30

 

x = 90 mph

 

x − 30 = 60 mph

Example 5.  Total time problem.   Katrina drove her car to Boston at a speed of 100 kph (kilometers per hour).  She drove back at 75 kph.  The total driving time was 7 hours.  How far away was Boston?

Solution.   Let x be the distance to Boston.  Then

Time going + Time returning = Total time.

Again, time is

time problem      

time problem = 7.  

 

time problem = 7. x is a common factor.

 

word problem = 7. Lesson 24 of Arithmetic

 

x = word problem Lesson 9, Example 6.

 = 300 km.  

Problem 11.   You have exactly h hours at your disposal.  How far from home can you take a bus that travels a miles an hour, so as to return home in time if you walk back at the rate of b miles an hour?

Solution.   Let x = that distance.  Then

Time going + Time returning  = Total time

 

x

a  +  x

b  =  h

 

bx + ax  =  hab

 

(a + b)x  =  hab

 

x  =  _hab_

a + b

Example 6.   Job problem.   Raymond can do a job in 3 hours, while it takes Robert 2 hours.  How long will it take them if they work together?

Solution.   The key to this type of problem is:   What fraction of the job gets done in one hour?

For example, if a job takes 3 hours, then in one hour,  1

3  will get done.

In general, if a job takes x hours, then in one hour,  1

x  will get done.

So, let x answer the question.  Let x be how long will it take them if

 they work together.  Then   1

x  is that fraction of the job that gets done in

one hour.

We have,

 1

x   =   1

3  +   1

2

For, in one hour, Raymond does  1

3  of the job, and Robert,  1

2 .

Since x, or its reciprocal, is already isolated on the left, simply add the fractions on the right:

 1

x   =   2 + 3

  6  =   5

6

Therefore, on taking reciprocals,

 x  =   6

5   =  1 1

5  hours.

Problem 12.   Carlos can do a certain job in three days, while it takes Alec six days.  If they work together, how long will it take them?

Let x be that time.  Here is the equation:

  1

x   =   1

3   +   1

6  

 

     =   6  +  3

  18  

 

     =    9  

18

 

  1

x   =   1

2

Therefore,

 x   =   2 days

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