RD Sharma Solutions Class 10 Pair Of Linear Equations In Two Variables Exercise 3.10

RD Sharma Class 10 Solutions Chapter 3 Ex 3.10 PDF Free Download

Exercise 3.10

 

Q1) Points A and B are 70km. apart on a highway. A car starts from A and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars.

 

Sol:

Consider the car starting from point A as x and its speed as x km/hr.

Also, consider the car starting from point B as y and its speed as y km/hr.

Now, there are two cases in the question:

  • Case 1: Car x and y are moving in the same direction
  • Case 2: Car x and y are moving in the opposite direction

Assume that the meeting point in case 1 as P and in case 2 as Q

Now, if two cars are moving in the same direction, (i.e. case 1)

The distance travelled by car x = AP

And, the distance travelled by car Y = BP

As the cars meet in 7 hours,

The distance travelled by car X in 7 hours = 7x km0

⇒ AP = 7x

Similarly,

The distance travelled by car y in 7 hours = 7y km

⇒ BP = 7Y

As the cars are moving in the same direction (i.e. away from each other),

AP — BP = AB

So,

7x – 7y = 70

Take 7 as common,

x — y = 10 ……………………….(Equation 1)

 

Now, if two cars are moving in the same direction, (i.e. case 2)

The distance travelled by car x = AQ

And,

The distance travelled by car y = BQ,

As the cars meet in 1 hour,

The distance travelled by car x in 1 hour = 1x km

⇒ AQ = 1x

Similarly,

The distance travelled by car y in 1 hour = 1y km

⇒ BQ= 1y

Now, since cars are moving in the opposite direction (i.e. towards each other)

AQ + BQ = AB

⇒ x + y = 70 ……………(Equation 2)

 

Now, solve equation (1) and equation (2)

From equation 1, x = 10 + y.

Substitute this value of x in equation 2.

So, (10 + y) + y = 70

⇒ y = 30

Now, put y = 30 in x = 10 + y

⇒ x = 40

Thus, the speed of both cars are obtained:

  • Speed of car starting from point A = 40km/hr.
  • Speed of car starting from point B = 30 km/hr.

 

2) A sailor goes 8km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.

 

Sol:
Assume:

The speed of the sailor in still water = x km/hr

And,

The speed of the current = y km/hr

We know,

Speed upstream = (x – y)Km/hr

Speed downstream = (x + y)km/hr

So, Time taken to cover 8 km upstream = \(\frac{8}{x-y}hrs\)

Time taken to cover 8 km downstream = \(\frac{8}{x+y}hrs\)

But, time taken to cover 8 km downstream in 40 minutes or 40/60 hours or, 2/3 hrs

\(\frac{8}{x+y}\) = \(\frac{2}{3}\)

8 × 3 = 2(x + y)

24 = 2x + 2y

Take 2 as common and rearrange,

x + y = 12……………………(1)

Time taken to cover 8 km upstream in 1hour ,

\(\frac{8}{x-y}\) = 1

8 = 1(x – y)

Or, x – y = 8 ……………..(2)

Now, add equation 1 and equation 2 to get,

2x = 20

Or, x = 10

Now, put value of x in either 1 or 2 to get,

y = 2.

Thus, the speed of sailor is 10km/hr and the speed of the current is 2km/hr.

 

 

3) The boat goes 30km upstream and 44km downstream in 10 hrs. In 13 hours it can go 40km upstream and 55km downstream. Determine the speed of stream and that of boat in still water.

SOLN:

Assume

The speed of the boat in still water = x km/hr and

The speed of the stream be y km/hr

Now, we know

Upstream Speed = (x -y) km/hr

Downstream Speed  = (x + y) km/hr

Now,

Time taken to cover 30 km upstream = \(\frac{30}{x-y}hrs\)

Time taken to cover 44 km downstream =\(\frac{44}{x+y}hrs\)

It is given that the total time of journey is 10 hours. So,

\(\frac{30}{x-y}hrs\) +  \(\frac{44}{x+y}hrs\) = 10 ……..(i)

Time taken to cover 40 km upstream = \(\frac{55}{x+y}hrs\)

In this case total time of journey is given to be 13 hours

∴ \(\frac{40}{x-y}hrs\) + \(\frac{55}{x+y}hrs\) = 13

Putting \(\frac{1}{x-y}\)and \(\frac{1}{x+y} \) in equation (i) and (ii) we get

30u+44v=10

40u +55v =10

30u + 44v-10 = 0 …….(iii)

40u + 551, -13= 0………(iv)

Solving these equations by cross multiplication we get,

\(\frac{u}{44x-13-55x-10} \) = \(\frac{-v}{30x-13-40x-10} \) = \(\frac{1}{30×55-40×44} \)

u = \(\frac{-22}{-110} \)

v  = \(\frac{10}{110} \)

u = \(\frac{2}{10} \)

v  = \(\frac{1}{11} \)

Now,

\(\frac{1}{x-y} \) = \(\frac{2}{10} \)

1 x 10 = 2( x – y )

10 = 2x – 2y / 2

V = \(\frac{1}{11} \)

Substituting x = 8 in equation we get,

X + y = 11

8+ y = 11

Y = 11 – 8

Y = 3

So, the speed of boat in still water is 8 km/hr.

Speed of the stream is 3 km/hr.

 

4) A boat goes 24km upstream and 28km downstream in 6hrs. It goes 30km upstream and 21km downstream in 6.5 hours. Find the speed of the boat in still water and also the speed of the stream.

SOLN:
We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr the

Speed upstream = (x-y)km/hr

Sped down stream= (x+y)km/ hr

Now, Time taken to cover 28 km downstream = \(\frac{28}{x+y}hrs\)

Time taken to cover 24 km upstream =\(\frac{24}{x-y}hrs\)

Since the total time of journey is 6 hours,

\(\frac{24}{x-y}\) + \(\frac{28}{x+y}\) = 6   ….(i)

Time taken to cover 30 km upstream =\(\frac{30}{x-y}\)

Time taken to cover 21km downstream =\(\frac{21}{x+y}\)

In this case total time of journey is given to 6.5hrs.

\(\frac{30}{x-y}\) + \(\frac{21}{x+y}\) = \(\frac{13}{2}\)

24u + 28v – 6 = 0  ……(iii)

30u + 21v – 13/2 = 0  …….(iv)

Solving these equations by cross multiplication we get,

\(\frac{u}{28x-6.5-21x-6}\)  = \(\frac{-v}{24x-6.5-30x-6}\) = \(\frac{1}{24×21-30×28}\)

u = \(\frac{1}{6}\)

v = \(\frac{1}{14}\)

Now,

u = \(\frac{1}{x-y}\) = \(\frac{1}{6}\)

6 = x-y ….(eq)

v = \(\frac{1}{14}\) = \(\frac{1}{x+y}\)

x + y = 14…….(eq)

By solving he equations,

X= 10

Substituting x = 10 in equation, we get,

Speed of the stream = 4km/hr.

Speed of boat = 10km/hr.

 

5) A man walks a certain distance with a certain speed. If he walks ½ km an hour faster, he takes 1 hour less. But, if he walks 1km an hour slower, he takes 3 more hours. Find the distance

SOLN:

Let “x km/hr” be the actual speed of the man and and “y” be the actual time taken by y hours.

Now, we know:

distance covered = speed × distance

So,

Distance = x × y

= xy ………………………….(1)

Case 1: If the speed increase by ½ km/hr

Now, if the man increase his speed by ½ km/hr , the journey time will reduce by 1 hour

⇒ When speed is (x + ½) km/hr, time of journey = y – 1 hours

Now,

Distance covered = xy km   

Or, -2x + y – 1 = 0 …………………..(2)  

Case 1: If the speed reduces by 1 km/hr

Now, if the the speed reduces by 1 km/hr, the time of journey increases by 3 hours

⇒ When speed is (x-1) km/hr, time of journey is (y+3) hours

As, distance covered = xy

⇒ xy = (x-1)(y+3)

Or, xy = (x-1)(y+3)

⇒ xy = xy – 1y + 3x – 3

⇒ xy = xy + 3x – 1y – 3

⇒ 3x – y – 3 = 0 ………………(3)

From equation 2 and equation 3, the value of x can be calculated as

x = 4

Now, y can be obtained by placing the value of in euther of the equation-

y = 9

Now, put the value of x and y in equation (1)

Distance covered = xy

= 4 × 9

= 36 km

Thus, the distance is 36 km and the speed of walking is 4 km/hr.

 

6) A person rowing at the rate of 5km/hr in still water takes thrice as much as time in going 40km upstream as in going 40km downstream. Find the speed of the stream.

SOLN:

Assume “x” as the rate of flowing water

Now,

Speed of boat in downstream = (5+x)

and,

Speed of boat in upstream = (5-x)

Now, if the one way distance is taken as “d”,

Time = \(\frac{distance}{time}\)

It is given that upstream time = 3 × downstream time

So,

\(\frac{d}{5-x} = 3(\frac{d}{5+x})\))

\(\frac{d}{5-x} = (\frac{3d}{5+x})\)

By cross multiplication,

d(5+x) = 3d(5-x)

Now take d as common-

5 + x = 3(5 – x)

x + 3x = 15 – 5

x = 10/4 = 2.5

So, the rate of the current is 2.5 km/hr.

 

7) Ramesh travels 760km to his home partly by train and partly by car. He takes 8hours if he travels 160km, by train and the rest by car. He takes 12 minutes more if he travels 240km by train and the rest by car. Find the speed of train and the car respectively.

SOLN:

Assume:

The speed of the train be x km/hour

The speed of the car = y km/hr

Two scenarios are given in the question:

  • Scenario 1: When Ramesh travels 760 Km by train and the rest by car
  • Scenario 2: When Ramesh travels 240Km by train and the rest by car

Now, if Ramesh travels 760 Km by train and the rest by car (Scenario 1)

Time taken by Ramesh to travel 160 Km by train = \(\frac{160}{x}hrs\)

Total time taken by Ramesh to cover 760Km =  \(\frac{160}{x}hrs\) +  \(\frac{600}{y}hrs\)

Time taken by Ramesh to travel (760-160) = 600 Km by car =\(\frac{600}{y}hrs\)

Total time taken = 8 hours (given in the question)

So,

\(\frac{160}{x}hrs\) + \(\frac{600}{y}hrs\) = 8

Divide the equation by 8:

(\(\frac{20}{x}hrs\) + \(\frac{75}{y}hrs\)) = 1

\(\frac{20}{x}hrs\) +  \(\frac{75}{y}hrs\) = 1 ……………………(1)

 

Now, if Ramesh travels 240Km by train and the rest by car (Scenario 2)

Time taken by Ramesh to travel 240 Km by train =\(\frac{240}{x}hrs\)

Time taken by Ramesh to travel (760-240) =520Km by car = \(\frac{520}{y}hrs\)

For this scneario, Ramesh will take a total of is 8 hours 12 minutes to finish the journey (given)

So,

\(\frac{240}{x} \) + \(\frac{520}{y} \) = 8hrs 12mins.

\(\frac{240}{x} \) + \(\frac{520}{y} \) = \(\frac{41}{5}hrs\)

\(\frac{6}{x}\) + \(\frac{13}{y} \)= \(\frac{41}{200}hrs\)……….(ii)

120 u  = \(\frac{15}{10}\)

u = \(\frac{15}{10}\) x \(\frac{1}{120}\)

u = \(\frac{1}{80}\)

Now,

u = \(\frac{1}{80}\)

\(\frac{1}{x}\) = \(\frac{1}{80}\)

x = 80.

And,

v = \(\frac{1}{100}\)

\(\frac{1}{y}\) = \(\frac{1}{100}\)

y = 100

So, the speed of the train is 80km/hr and the speed of car = 100km/hr.

 

8) A man travels 600 km partly by train and partly by car. If he covers 400km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200km by train and the rest by car, he takes half an hour longer. Find the speed of the train and the speed of the car.

SOLN:

Let the speed of the train be x km/hr that of the car be y km/hr, we have following cases,

CASE 1:

When a man travels 600km by train and the rest by car,

Time taken by a man to travel 400km by train = \(\frac{400}{x}hrs\)

Time taken by a man to cover 600km = \(\frac{400}{x}hrs\) + \(\frac{200}{y}hrs\)

It is given that the total time taken in 8 hours,

\(\frac{400}{x} \) + \(\frac{200}{y} \) = 6hrs 30 mins

\(\frac{400}{x} \) + \(\frac{200}{y}\) = 6 x \(\frac{1}{2}\)

\(\frac{400}{x} \) + \(\frac{200}{y}\) = \(\frac{13}{2}\)

200 (\(\frac{2}{x}\) + \(\frac{1}{y}\)) = \(\frac{13}{2}\)

\(\frac{2}{x}\) + \(\frac{1}{y}\) = \(\frac{13}{400}\)…(i)

 

CASE (ii)

When a man travels 200km by train and rest by car

TT by man to travel 200 km = \(\frac{200}{x}\)hrs.

TT by man to travel 400 km = \(\frac{400}{y}\)hrs.

In this case, total time of the journey in 6hours 30 minutes + 30 minutes that is 7hrs,

\(\frac{200}{x}\) + \(\frac{400}{y}\) = 7

200 (\(\frac{1}{x}\) + \(\frac{2}{y}\)) = 7

\(\frac{1}{x}\) + \(\frac{2}{y}\) = \(\frac{7}{200}\) …..(ii)

Putting \(\frac{1}{x}\) = u, and, \(\frac{1}{y}\) = u,

Then,

2u + 1v = \(\frac{13}{400}\)  …..(iii)

1u + 2v = \(\frac{7}{200}\)  …….(iv)

By equating,

u = \(\frac{1}{100}\)

By substituting u = \(\frac{1}{100}\)  in the equation,

V =\(\frac{1}{80}\)

Now,

u = \(\frac{1}{100}\)

\(\frac{1}{x}\)   =  u = \(\frac{1}{100}\)

X = 100.

And

V =\(\frac{1}{80}\)

\(\frac{1}{y}\) = \(\frac{1}{80}\)

Y = 80.

Hence, the speed of the train is 100km/hr.

The speed of the car is 80km/hr.

 

 

9) Places A and B are 80km apart from each other on a highway. A car starts from A and other from B at the same time. If they move in the same direction, they meet in 8hours and if they move in opposite direction, they meet in 1hour and 20 minutes. Find the speeds of the cars.

SOLN:

Assume that a car x starting from point A and another car y starting from point B respectively.

Also, assume the speed of x and y to be x km/hr and y km/hr respectively.


Case I: When two cars move in the same directions:

Let the two cars meet at a point Q.

Now,

Distance travelled by car x = AQ

Distance travelled by car y = BQ

In 8 hours, 

Distance travelled by car x in 8 hours = 8x km

AQ = 8x and

Distance travelled by car y in 8 hours = 8y km

BQ = 8y

Here, AQ – BQ = AB

8x – 8y = 80

Now, take 8 as common and divide LHS and RHS, 

x – y = 10 ……………………(1)


Case II: When two cars move in opposite direction

For this case, let two cars meet at point P.

Distance travelled by car x = AP

Distance travelled by car y = BP 

Now, 1 hour 20 minutes is  1 hour 20/60 = 1 (1/3) hours = 4/3 hours.

Then,

Distance travelled by car x in 4/3 hours = (4/3)x km

And,

Distance travelled by car y in 4/3 hours = (4/3)y km

We know,

AP + BP = AB

⇒ (4/3)x + (4/3)y = 80

Taking 4/3 as common and shifting to RHS we get,

x + y = 60 ………………………(2)

Now, solve for x by adding equation 1 and 2:

x = 35

Then get y by substituting x in either equation 1 or 2

y = 25

Thus, the speed of car x and y are 35 km/hr and 25 km/ hr respectively.

 

10) A boat goes 12km upstream and 40km downstream in 8 hours. It can go 16km upstream and 32km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.

 

Soln:

We have to find the speed of the boat in still water and speed of the stream

Let the speed of the boat in still water be 1 km/hr and the speed of the stream be r km/hr then

Speed upstream = (x- y)km/hr

Sped down stream= (x+ y)km/hr

Now, Time taken to cover 12 km upstream = \(\frac{12}{x-y}hrs\)

Time taken to cover 40 km downstream =  \(\frac{40}{x+y}hrs\)

But, total time of journey is 8 hours

\(\frac{12}{x-y}\) + \(\frac{40}{x+y}\) = 8 ……(i)

Time taken to cover 16 km upstream =  \(\frac{16}{x-y}\)hrs

x – y Time taken to cover 32km downstream  =  \(\frac{32}{x+y}\)hrs

In this case total time of journey is given to 8hrs

\(\frac{16}{x-y}\) + \(\frac{32}{x+y}\)  =8 …(ii)

12 u + 40 v = 8

16u + 32 v = 8

12u + 40v – 8 = 0 ….(iii)

16u + 32v – 8 = 0 ….(iv)

Solving these equations by cross multiplication, we get,

u= \(\frac{1}{4}\)  and

v= \(\frac{1}{8}\)

Now,

u = \(\frac{1}{x-y}\)

\(\frac{1}{x-y}\)\(\frac{1}{4}\)

4 = x-y  ……(v)

And

v = \(\frac{1}{x+y}\)

\(\frac{1}{x+y}\) = \(\frac{1}{8}\)

x+y = 8  …..(vi)

By solving equation (v) and (vi) we get,

X = 6.

x+y =8

6 + y = 8

y = 8-6

y = 2

Hence, the speed of boat in still water is 6km/hr.

The speed of the stream is 2km/hr.

 

 

11)  Roohi travels 300km to her home partly by train and partly by bus. She takes 4hours if she travels 60km by train and the remaining by bus. If she travels 100kms by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.

SOLN:

Let the speed of the train be X km/hr that of the bus be y km/hr, we have the following cases,

CASE 1:

When Roohi travels 300km by train and rest by bus,

Time taken by Roohi to travel 60km by train = \(\frac{60}{x}hrs\)

Time taken by Roohi to trave (300-60) = 240 km by bus = \(\frac{240}{y}hrs\)

Total time taken by Rohi to cover 300km = \(\frac{60}{x}hrs\) + \(\frac{240}{y}hrs\)

It is given that total time taken in 4hours

\(\frac{60}{x}\)\(\frac{240}{y}\) = 4

60(\(\frac{1}{x}\) + \(\frac{4}{y}\)) = 4

\(\frac{1}{x}\) + \(\frac{4}{y}\) = \(\frac{4}{60}\)

\(\frac{1}{x}\) + \(\frac{4}{y}\) = \(\frac{1}{15}\)…….(i)

 

CASE (ii)

When Roohi travels 100km by train and the rest by bus,

Time taken by Roohi to travel 100km by train = \(\frac{100}{x}\)

Time taken by Roohi to travel (300-100) = 200km by bus = \(\frac{200}{y}\)

In this case total time of the journey is 4 hours 10 minutes

\(\frac{100}{x}\) + \(\frac{200}{y}\) = 4hrs 10minutes.

\(\frac{100}{x}\) + \(\frac{200}{y}\) = \(\frac{25}{6}\)

\(\frac{1}{x}\) + \(\frac{2}{y}\) = \(\frac{25}{6}\)

\(\frac{1}{x}\) + \(\frac{2}{y}\) = \(\frac{1}{24}\) …(i)

1u + 4v = \(\frac{1}{15}\)  …..(iii)

1u + 2v = \(\frac{1}{24}\)  …..(iv)

Subtracting equation (iv) from (iii) we get,

v = \(\frac{1}{80}\)

1u = \(\frac{20-15}{300}\)

1u = \(\frac{5}{300}\)

u = \(\frac{1}{60}\)

Now,

u = \(\frac{1}{60}\)

\(\frac{1}{x}\)  = \(\frac{1}{60}\)

x = 60.

And,

v = \(\frac{1}{80}\)

\(\frac{1}{y}\)   = \(\frac{1}{80}\)

Hence, the speed of the train is 60km/hr.

The speed of the bus is 80km/hr.

 

 

12) Ritu can row downstream 20km in 2hours, and upstream 4km in 2hours. Find her speed of rowing in still water and the spees of the current.

Soln:

Let the speed of rowing be ‘X’km/hr and the speed of the current be ‘Y’km/hr,

Speed upstream = (x-y) km/hr.

Speed downstream = (x+y) km/hr.

Now,

Time taken to cover 20km downstream = \(\frac{20}{x+y}hrs\)

Time taken to cover 4km upstream = \(\frac{4}{x-y}hrs\)

But, time taken to cover 20km downstream in 2hours,

\(\frac{20}{x+y}\)   = 2

20 = 2(x+y)

20 = 2x+2y ……(i)

Time Taken to cover 20km downstream in 2hours

\(\frac{20}{x+y}\)    = 2

20 = 2(x+y)

20 = 2x+2y …..(ii)

By solving (i) and (ii) we get,

2y  = 8

Y = 4

Hence, speed of rowing in still water is 6km/hr.

Speed of current is 4km/hr.

 

 

13)  A motor boat can travel 30km upstream and 28km downstream in 7hours. It can travel 21km upstream and return in 5 hours. Find the speed of the boat in still water and speed of the stream.

SOLN:

Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
Now,

Upstream Speed of Boat = x − y and
Downstream Speed of Boat = x + y
The boat travels 30 km upstream and 28 km downstream in 7 hours (as given).
So, 30/x-y + 28/x+y = 7

Also, the boat travels 21 km upstream and return in 5 hours (given).
So, 21/x-y + 21/x+y = 5

Let 1/x-y = u ……………………….(1)

and 1/x+y = v. …………………(2)

∴ The equation becomes:
30u + 28v = 7 …………………………….(3)
21u + 21v = 5 ……………………………..(4)

Solve these equations 3 and 4 using cross multiplication method to obtain u and v.
⇒ v = 1/14 and u = 1/6

From equation 1 and 2,

1/x-y = 1/6 and 1/x+y = 1/14
⇒ x – y = 6………………….(5)

and x + y = 14 ……………….(6)
Now by solving these equations 5 and 6, the values of x and y can be obtained.
⇒ x = 10 and y = 4

Hence, the speed of boat in still water is 10 km/h and the speed of stream is 4 km/h.

 

Q 14. Abdul travelled 300 km by train and 200 km by taxi, it took him 5 hours and 30 mins. But if he travels 260 km by train and 240 km by taxi he takes 6 mins longer. Find the speed of the train and that of the taxi.

Ans:    Let the speed of the train be x km/hour that of the taxi be y km/hr, we have the following cases

Case I: When Abdul travels 300 Km by train and the 200 Km by taxi

Time taken by Abdul to travel 300 Km by train – \( \frac{300}{x} hrs\)

Time taken by Abdul to travel 200 Km by taxi – \( \frac{200}{y}\)

Total time taken by Abdul to cover 500 Km – \(\frac{300}{x} + \frac{200}{y}\)

It is given that total time taken in 5 hours 30 minutes

\(\frac{300}{x} + \frac{200}{y} = 5 hours\ 30 minutes\)

\(100( \frac{3}{x} + \frac{2}{y}) = 5 \frac{30}{60}\)

\(100( \frac{3}{x} + \frac{2}{y}) = 5 \frac{1}{2}\)

\(100( \frac{3}{x} + \frac{2}{y}) = \frac{11}{2}\)

\(( \frac{3}{x} + \frac{2}{y}) = \frac{1}{2} \times \frac{1}{100}\)

\(\frac{3}{x} + \frac{2}{y} = \frac{11}{200}\)….(i)

Case II: When Abdul travels 260 km by train and the 240 km by taxi

Time taken by abdul to travel 260 km by train = \(\frac{260}{x} hrs\)

Time taken by Abdul to travel 240 km by taxi = \(\frac{240}{y} hrs\)

In this case, total time of the journey is 5 hours 36 minutes.

\(\frac{260}{x} + \frac{240}{y} = 5hrs\ 36 minutes\)

\(\frac{260}{x} + \frac{240}{y} = 5 \frac{36}{60}\)

\(\frac{260}{x} + \frac{240}{y} = 5 \frac{6}{10}\)

\(\frac{260}{x} + \frac{240}{y} = 5 \frac{3}{5}\)

\(20 (\frac{13}{x} + \frac{12}{y}) = \frac{28}{5}\)

\((\frac{13}{x}) = \frac{28}{5} \times \frac{1}{20}\)

\(\frac{13}{x} + \frac{12}{y} = \frac{7}{25}\) ….(ii)

Putting \(\frac{1}{x} = u\ and\ \frac{1}{y} = v\) , the equations (i) and (ii) reduces to

\(3u + 2v = \frac{11}{200}\) ….(iii)

\(13u+12v = \frac{7}{25}\) …..(iv)

Multiplying equation (iii) by 6, the above equation becomes

\(18u+12v= \frac{33}{100}\) ….(v)

Subtracting equation (iv) from (v) we get

\(5u = \frac{33}{100} – \frac{7}{25}\)

\(5u = \frac{33}{100} – \frac{7 \times 4}{25 \times 4}\)

\(5u = \frac{33}{100} – \frac{28}{100}\)

\(5u = \frac{33-28}{100}\)

\(5u = \frac{5}{100}\)

\(u = \frac{5}{100} \times \frac{1}{5}\)

\(u = \frac{1}{100}\)

Putting \(u = \frac{1}{100}\) in equation (iii), we get

\(3u + 2v = \frac{11}{200}\)

\(3 \times \frac{1}{100} +2v = \frac{11}{200}\)

\(\frac{3}{100} + 2v = \frac{11}{200}\)

\(2v = \frac{11}{200} – \frac{3}{100}\)

\(2v = \frac{11}{200} – \frac{3 \times 2}{100 \times 2}\)

\(2v = \frac{11-6}{200}\)

\(2v = \frac{5}{200}\)

\(v = \frac{1}{2}\)

\(v = \frac{1}{80}\)

Now,

\(u = \frac{1}{100}\)

x = 100

And,

\(v = \frac{1}{80}\)

\(\frac{1}{y} = \frac{1}{80}\)

y = 80

Hence, the speed of the train is 100 km/hr

The speed of the taxi is 80 km/hr.

 

 

Q 15. A train covered a certain distance at a uniform speed. If the train could have been 10 km/hr faster, it would have taken 2 hours less than the scheduled time and if the train were slower by 10 km/hr; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer :

Let the actual speed of the train be x km/hr and the actual time is taken by y hours. Then,

Distance = Speed x Time

Distance covered=(xy) km ….(i)

If the speed is increased by 10Km I hr , then the time of journey is reduced by 2 hours.

when speed is (x+10)km I hr , time of journey is (y-2)hours.

Therefore, Distance covered=(x+ I0)(y- 2)

xy=(x+10)(y-2)

xy = xy +10y — 2x —20

—2x +10y —20 = 0

—2x + 3y —12 = 0…..(ii)

 

When the speed is reduced by I0Km/hr, then the time of journey is increased by 3hours.

when speed is (x —10)Km/hr , time of journey is (y+3)hours.

Hence, Distance covered=(x — I0)(y + 3)

xy=(x-10)(y+3)

0 = —10y+3x-30

3x-10y-30=0 ….(iii)

Thus, we obtain the following system of equations: —x +5y —10 =0

– x +5y – 10 = 0

3x – 10y – 30 = 0

By using cross-multiplication, we have

\(\frac{x}{5x – 30 – (-10) \times -10} = \frac{-y}{(-1 \times -30) – (3x – 10)} = \frac{1}{(-1 \times -10) – (3 \times 5)}\)

\(\frac{x}{-150-100} = \frac{-y}{30 + 30} = \frac{1}{10-15}\)

\(\frac{x}{-250} = \frac{-y}{60} = \frac{1}{-5}\)

x = \(\frac{-250}{-5}\)

x = 50

y = \(\frac{-60}{-5}\)

y = 12

Putting the values of x and y in equation (i), we obtain

Distance = xy km

= 50 x 12

= 600 km

Hence, the length of the journey is 600 km

 

 

16) Places A and B are 100km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at different speeds they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of two cars?

SOLN:

Let x and y be two cars starting from points A and B respectively.

Let the speed of the car X be x km/hr and that of the car Y be y km/hr.

Case I: When two cars move in the same directions: Suppose two cars meet at point Q, then,

Distance travelled by car X=AQ

Distance travelled by car Y=BQ

It is given that two cars meet in 5 hours.

Distance travelled by car X in 5 hours =5x km AQ= 5x

Distance travelled by car Y in 5 hours = 5y km BQ= 5y

Clearly AQ-BQ = AB 5x —Sy = 100

Both sides divided by 5, we get x—y = 20 …….(i)

Case II: When two cars move in opposite direction

Suppose two cars meet at point P, then,

Distance travelled by X car X=AP

Distance travelled by Y car Y=BP

In this case, two cars meet in 1 hour

Therefore, Distance traveled by car y in hours = lx km

Distance traveled by car y in 1 hours = ly km

AP + BP = AB

1x + 1y = 100

X + y = 100

By solving, x = 60,

Substituting x=60 in equation, we get,

X+y = 100

60 + y=100

Y = 40

Hence, the speed of car X is 60km/hr.

Speed of car Y = 40km/hr.

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