# Class 10 RD Sharma Solutions – Chapter 3 Pair of Linear Equations in Two Variables – Exercise 3.10

**Question 1. 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.**

**Solution: **

Let’s assume that the car starting from point A as X and its speed as x km/hr.

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

There are two cases given 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.

Let’s assume that the meeting point in case 1 as P and in case 2 as Q.

Case 1 : The distance travelled by car X = AP and,

the distance travelled by car Y = BP

As the time taken for both the cars to meet is 7 hours,

The distance travelled by car X in 7 hours = 7x km (As we know that distance = speed x time)

therefore AP = 7x

Similarly,

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

therefore BP = 7Y

As the cars are moving in the same direction (i.e. away from each other), we can write it as

AP – BP = AB

therefore, 7x – 7y = 70

x – y = 10 ——————(i)

Case 2 : In this case as it’s clearly seen that,

The distance travelled by car X = AQ and,

The distance travelled by car Y = BQ

As the time taken for both the cars to meet is 1 hour,

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

therefore AQ = 1x

Similarly,

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

therefore BQ = 1y

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

AQ + BQ = AB

Hence x + y = 70 —————–(ii)

Hence, by solving eq. (i) and (ii), we get the required solution

From (i), we have x = 10 + y ——————-(iii)

Substituting this value of x in (ii) and we got,

(10 + y) + y = 70

y = 30

Now, put y = 30 in (iii), and we get

x = 40

Hence, Speed of car X = 40km/hr and Speed of car Y = 30 km/hr.

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

**Solution: **

Let’s assume that the speed of the sailor in still water as x km/hr and,

The speed of the current as y km/hr

As we know that,

Speed of the sailor in upstream = (x – y) km/hr

Speed of the sailor in downstream = (x + y) km/hr

therefore, time taken to cover 8 km upstream = 8/ (x – y) hr (As we know that distance = speed x time)

And, time taken to cover 8 km downstream = 8/ (x + y hr

It’s given that time taken to cover 8 km downstream in 40 minutes or, 40/ 60 hour or 2/3 hr.

8/ (x + y) = 2/3

8 × 3 = 2(x + y)

24 = 2x + 2y

x + y = 12 ——————(i)

Similarly, time taken to cover 8 km upstream in 1hour can be written as,

8/ (x – y) = 1

8 = 1(x – y)

x – y = 8 ——————–(ii)

Hence, by solving eq. (i) and (ii) we get the required solution

On adding (i) and (ii) we get,

2x = 20

x = 10

Now, putting the value of x in (i), we find y

10 + y = 12

y = 2

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

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

**Solution: **

Let’s assume that the speed of the boat in still water as x km/hr and,

The speed of the stream as y km/hr

As we know that,

Speed of the boat in upstream = (x – y) km/hr and

Speed of the boat in downstream = (x + y) km/hr

therefore,

Time taken to cover 30 km upstream = 30/ (x − y) hr (As we know that distance = speed x time)

Time taken to cover 44 km downstream = 44/ (x + y) hr (As we know that distance = speed x time)

It’s given that the total time of journey is 10 hours. So, this can expressed as

30/ (x – y) + 44/ (x + y) = 10 —————-(i)

Similarly,

Time taken to cover 40 km upstream = 40/ (x – y) hr (As we know that distance = speed x time)

Time taken to cover 55 km downstream = 55/ (x + y) hr (As we know that distance = speed x time)

And for this case the total time of the journey is given as 13 hours.

Hence, we can write

40/(x – y) + 55/ (x + y) = 13 —————-(ii)

Hence, by solving (i) and (ii) we get the required solution

Taking, 1/ (x – y) = u and 1/ (x + y) = v in equations (i) and (ii) we have

30u + 44v – 10 = 0 —————–(iii)

40u + 55v – 13 = 0 ——————-(iv)

Solving these equations by cross multiplication we get,

u/(44x-13-55x-10) = -v/(30x-13-40x-10) = 1/(30×55 – 40×44)

u = 2/10

v = 1/11

Now,

1/ (x – y) = 2/10

1 x 10 = 2(x – y)

10 = 2x – 2y

x – y = 5 —————–(v)

And,

1/ (x + y) = 1/11

x + y = 11 —————(vi)

Again, solving (v) and (vi)

Adding (v) and (vi), we get

2x = 16

x = 8

Using x in (v), we find y

8 – y = 5

y = 3

Hence, the speed of the boat in still water is 8 km/hr and the speed of the stream is 3 km/hr.

**Question 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 speed of the stream.**

**Solution: **

Let’s assume that the speed of the boat in still water as x km/hr and,

The speed of the stream as y km/hr

As we know that,

Speed of the boat in upstream = (x – y) km/hr and

Speed of the boat in downstream = (x + y) km/hr

therefore,, time taken to cover 28 km downstream = 28/ (x+y) hr (As we know that distance = speed x time)

Time taken to cover 24 km upstream =24/ (x – y) hr (As we know that distance = speed x time)

It’s given that the total time of journey is 6 hours. So, this can be expressed as

24/ (x – y) + 28/ (x + y) = 6 —————-(i)

Similarly,

Time taken to cover 30 km upstream = 30/ (x − y) (As we know that distance = speed x time)

Time taken to cover 21km downstream =21/ (x + y) (As we know that distance = speed x time)

And for this case the total time of the journey is given as 6.5 i.e 13/2 hours.

Hence, we can write it as

30/ (x – y) + 21/ (x + y) = 13/2 —————-(ii)

Hence, by solving (i) and (ii) we get the required solution

Taking, 1/ (x – y) = u and 1/ (x + y) = v in equations (i) and (ii) and we have,

24u + 28v – 6 = 0 ————–(iii)

30u + 21v – 13/2 = 0 ————–(iv)

Solving these equations by cross multiplication we get,

u/(28x-6.5-21x-6) = -v/(24x-6.5-30x-6) = 1/(24×21 – 30×28)

u = 1/6 and v = 1/14

Now,

u = 1/ (x − y) = 1/ 6

x – y = 6 …. (v)

v = 1/ (x + y) = 1/ 14

x + y = 14 —————(vi)

On Solving (v) and (vi)

Adding (v) and (vi), we get

2x = 20

x = 10

Using x = 10 in (v), we find y

10 + y = 14

y = 4

Hence, Speed of the stream = 4km/hr and Speed of boat = 10km/hr.

**Question 5. A man walks a certain distance with a certain speed. If he walks 1/2 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 covered by the man and his original rate of walking.**

**Solution: **

Let’s assume that the actual speed of the man be x km/hr and y be the actual time taken by him in hours.

As we know that,

Distance covered = speed × distance

Distance = x × y = xy —————-(i)

Case I :

If the speed of the man increase by 1/2 km/hr, the journey time will reduce by 1 hour.

it can be written in the form of equation,

When speed is (x + 1/2) km/hr, time of journey = y – 1 hours

Now,

Distance covered = (x + 1/2) x (y – 1) km

Since the distance is the same i.e xy we can equate it, therefore

xy = (x + 1/2) x (y – 1)

And finally we get,

-2x + y – 1 = 0 ——————–(ii)

Case II : If the speed reduces by 1 km/hr then the time of journey increases by 3 hours.

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

Since, the distance covered = xy

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

xy = xy – 1y + 3x – 3

xy = xy + 3x – 1y – 3

3x – y – 3 = 0 —————–(iii)

Add eq. (ii) and (iii), and we get

x – 4 = 0

x = 4

Now, y can be obtained by using x = 4 in (ii)

-2(4) + y – 1 = 0

y = 1 + 8 = 9

Hence, putting the value of x and y in equation (i), we find the distance

Distance covered = xy

= 4 × 9

= 36 km

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

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

**Solution: **

Let’s assume x to be the speed of the stream.

As we know that,

Speed of boat in downstream = (5 + x) and,

Speed of boat in upstream = (5 – x)

It is given that,

The distance in one way is 40km. and,

Time taken during upstream = 3 × time taken during the downstream

Expressing it by equations, we have

40/ (5 – x) = 3 x 40/ (5 + x) (As we know that distance = speed x time)

By cross multiplication, we get

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

5 + x = 3(5 – x)

x + 3x = 15 – 5

x = 10/4 = 2.5

Hence, the speed of the stream is 2.5 km/hr.

**Question 7. Ramesh travels 760km to his home partly by train and partly by car. He takes 8 hours 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 the train and car respectively.**

**Solution: **

Let’s assume that the speed of the train be x km/hr and,

The speed of the car = y km/hr

From the question, it’s understood that there are two cases

Case 1 : When Ramesh travels 160 Km by train and the rest by car.

Case 2 : When Ramesh travels 240Km by train and the rest by car.

Case 1 : Time taken by Ramesh to travel 160 km by train = 160/x hrs (As we know that distance = speed x time)

Time taken by Ramesh to travel the remaining (760 – 160) km i.e., 600 km by car = 600/y hrs

therefore, the total time taken by Ramesh to cover 760Km = 160/x hrs + 600/y hrs

given that,

Total time taken for this journey = 8 hours

therefore, 160/x + 600/y = 8

20/x + 75/y = 1 —————–(i)

Case 2 : Time taken by Ramesh to travel 240 km by train = 240/x hrs,

Time taken by Ramesh to travel (760 – 240) = 520km by car = 520/y hrs

given that Ramesh will take a total of is 8 hours and 12 minutes to finish.

therefore, 240/x + 520/y = 8hrs 12mins = 8 + (12/60) = 41/5 hr

240/x + 520/y = 41/5

6/x + 13/y = 41/200 —————(ii)

Solving (i) and (ii), we get,

Let’s take 1/x = u and 1/y = v,

therefore, (i) and (ii) becomes,

20u + 75v = 1 —————(iii)

6u + 13v = 41/200 —————-(iv)

On multiplying (iii) by 3 and (iv) by 10, we get

60u + 225v = 3

60u + 130v = 41/20

Subtracting the above two equations, we get

(225 – 130)v = 3 – 41/20

95v = 19/ 20

v = 19/ (20 x 95) = 1/100

y = 1/v = 100

Using v = 1/100 in (iii) to find v,

20u + 75(1/100) = 1

20u = 1 – 75/100

20u = 25/100 = 1/4

u = 1/80

x = 1/u = 80

Therefore, the speed of the train is 80km/hr and the speed of car is100km/hr.

**Question 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.**

**Solution: **

Let’s assume that the speed of the train be x km/hr and,

The speed of the car = y km/hr

From the question, it’s understood that there are two cases

Case 1 : When the man travels 400 km by train and the rest by car.

Case 2 : When Ramesh travels 200 km by train and the rest by car.

Case 1 : Time taken by the man to travel 400km by train = 400/x hrs (As we know that distance = speed x time)

Time taken by the man to travel (600 – 400) = 200km by car = 200/y hrs

Time taken by a man to cover 600km = 400/x hrs + 200/y hrs

Total time taken for this journey = 6 hours + 30 mins = 6 + 1/2 = 13/2

therefore,

400/x + 200/y = 13/2

400/x + 200/y = 13/2

400/x + 200/y = 13/2

200 (2/x + 1/y) = 13/2

2/x + 1/y = 13/400 —————-(i)

Case 2 : Time taken by the man to travel 200 km by train = 200/x hrs. (As we know that distance = speed x time)

Time taken by the man to travel (600 – 200) = 400km by car = 200/y hrs

For the part, the total time of the journey is given as 6hours 30 mins + 30 mins that is 7hrs,

200/x + 400/y = 7

200 (1/x + 2/y) = 7

1/x + 2/y = 7/200 ————-(ii)

Taking 1/x = u, and 1/y = v,

So, the equations (i) and (ii) becomes,

2u + v = 13/400 —————(iii)

u + 2v = 7/200 ————-(iv)

Solving (iii) and (iv), we get

3v = 14/200 – 13/400

3v = 1/400 x (28 – 13)

3v = 15/400

v = 1/80

y = 1/v = 80

Now, using v in (iii) we get u,

2u + (1/80) = 13/400

2u = 13/400 – 1/80

2u = 8/400

u = 1/100

x = 1/u = 100

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

**Question 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 8 hours and if they move in opposite direction, they meet in 1hour and 20 minutes. Find the speeds of the cars.**

**Solution: **

Let’s consider the car starting from point A as X and its speed as x km/hr and,

the car starting from point B as Y and its speed as y km/hr.

From the question, it’s understood that there are two cases

Case 1 : Car X and Y are moving in the same direction

Case 2 : Car X and Y are moving in the opposite direction

Let’s assume that the meeting point in case 1 as P and in case 2 as Q.

Case 1 : The distance travelled by car X = AP and,

the distance travelled by car Y = BP.

As the time taken for both the cars to meet is 8 hours,

The distance travelled by car X in 7 hours = 8x km (As we know that distance = speed x time)

AP = 8x

Similarly, The distance travelled by car Y in 8 hours = 8y km

therefore, BP = 8Y

As the cars are moving in the same direction (i.e. away from each other), we can write it as

AP – BP = AB

therefore, 8x – 8y = 80

x – y = 10 ———————(i)

Case 2 : The distance travelled by car X = AQ and,

The distance travelled by car Y = BQ

As the time taken for both the cars to meet is 1 hour and 20 min, ⇒1 + (20/60) = 4/3 hr

The distance travelled by car x in 4/3 hour = 4x/3 km

AQ = 4x/3

Similarly, The distance travelled by car y in 4/3 hour = 4y/3 km

BQ = 4y/3

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

AQ + BQ = AB

4x/3 + 4y/3 = 80

4x + 4y = 240

x + y = 60 ——————(ii)

Hence, by solving (i) and (ii), we get

From (i), we have x = 10 + y ——————-(iii)

Substituting this value of x in (ii).

(10 + y) + y = 60

2y = 50

y = 25

Now, using y = 30 in (iii), we get

x = 35

Hence, Speed of car X = 35 km/hr and Speed of car Y = 25 km/hr.

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

**Solution: **

Let’s assume that 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 = 12/(x-y) hrs,

Time taken to cover 40 km downstream = 40/(x+y) hrs.

But, total time of journey is 8 hours

12/(x-y) + 40/(x+y) = 8 —————–(i)

Time taken to cover 16 km upstream = 16/(x-y) hrs,

Time taken to cover 32 km downstream = 32/(x+y) hrs

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

16/(x-y) + 32/(x+y) = 8 ————–(ii)

12u + 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 = 1/4 and v = 1/8

Now,

u = x/(x-y)

1/(x-y) = 1/4

4 = x – y —————–(v)

and,

v = 1/(x+y)

1/(x+y) = 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

Therefore, The speed of boat in still water is 6 km/hr. and the speed of the stream is 2 km/hr.

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

**Solution: **

Let’s assume that 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 = 60/x hrs

Time taken by Roohi to travel (300 – 60) = 240 km by bus = 240/y hrs

Total time taken by Roohi to cover 300 km = 60/x + 240/y

It is given that total time taken in 4 hours therefore,

60/x + 240/y = 4

1/x + 4/y = 1/15 ————-(i)

Case 2 : When Roohi travels 100km by train and the rest by bus,

Time taken by Roohi to travel 100km by train = 100/x,

Time taken by Roohi to travel (300 – 100) = 200 km by bus = 200/y,

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

100/x + 200/y = 4 hrs 10 min

100/x + 200/y = 25/6

1/x + 2/y = 1/24 ————–(ii)

1u + 4v = 1/15 —————(iii)

1u + 2v = 1/24 —————–(iv)

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

v = 1/80

1u = (20-15)/300 = 1/60

u = 1/60 therefore,

x = 60

And,

v = 1/80

1/y = 1/80

Hence, The speed of the train is 60 km/hr. and the speed of the bus is 80 km/hr.

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

**Solution: **

Let’s assume that 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 20 km downstream = 20/(x+y) hrs

Time taken to cover 4 km upstream = 4/(x-y) hrs

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

20/(x+y) = 2

20 = 2x + 2y —————(i)

Time Taken to cover 20 km downstream in 2 hours

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 6 km/hr. and speed of current is 4 km/hr.

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

**Solution: **

Let’s assume that the speed of A and B be X km/hr and Y km/hr respectively. Then,

Time taken by A to cover 30km = 30/x hrs,

and, time taken by B to cover 30km = 30/y hrs.

given that,

30/x – 30/y = 3

(10/x – 10/y) = 1 —————(i)

If A doubles his pace, then speed of A is 2x km/hr.

Time taken by A to cover 30 km = 30/2x hrs,

Time taken by B to cover 30 km = 30/y hrs,

According to given condition we have,

30/y – 30/2x = 3/2

10u – 10v = 1

10u – 10v = 0 —————–(iii)

-10u + 20v = 1

-10u + 20v – 1 = 0 ——————(iv)

Adding equations (iii) and (iv), we get,

v = 1/5

Putting v = 1/5 in equation (iii), we get,

10u – 10v – 1 = 0

10u – 3 = 0

u = 3/10

1/x = 3/10

x = 10/3

and, v = 1/5

1/y = 1/5

y = 5

Hence, The boat A’s speed is 10/3 km/hr. and boat B’s speed is 5 km/hr.

**Question 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.**

**Solution: **

Let’s assume that 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 = 300/x hrs,

Time taken by Abdul to travel 200 Km by taxi = 200/y,

Total time taken by Abdul to cover 500 Km = 300/x + 200/y

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

300/x + 200/y = 5 hours 30 minutes

100(3/x + 2/y) = 5 30⁄60 (3/x + 2/y) = 5 1⁄2

100(3/x + 2/y) = 1/2 x 1/100

3/x + 2/y = 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 = 260/x hrs

Time taken by Abdul to travel 240 km by taxi = 240/y hrs

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

260/x + 240/y = 5 hours 36 minutes

260/x + 240/y = 5 36⁄60

260/x + 240/y = 5 6⁄10

260/x + 240/y = 5 3⁄5

20(13/x + 12/y) = 28/5

(13/x) = 28/5 × 1/20

13/x + 12/y = 7/25 —————–(ii)

Putting 1/x = u and 1/y = v, the equations (i) and (ii) reduces to

3u + 2v = 11/200 —————(iii)

13u + 12v = 7/25 —————-(iv)

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

18u + 12v = 33/100 —————-(v)

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

5u = 33/100 – 7/25

5u = 33/100 – 28/100

5u = 5/100

u = 1/100

Putting u = 1/100 in equation (iii), we get

3u + 2v = 11/200

3 x 1/100 + 2v = 11/200

3/100 + 2v = 11/200

2v = 5/200

v = 1/2

v = 1/80

Now,

u = 1/100

x = 100

And,

v = 1/80

1/y = 1/80

y = 80

Hence, the speed of the train is 100 km/hr and the speed of the taxi is 80 km/hr.

**Question 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.**

**Solution: **

Let’s assume that 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 10 Km 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 + 10) (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 10 Km/hr, then the time of journey is increased by 3 hours.

When speed is (x -10) Km/h, time of journey is (y + 3) hours.

Hence, Distance covered = (x – 10) (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 get

x/(5x-30-(-10)x-10) = -y/(-1 x -30)-(3x-10) = 1/(-1 x -10)-(3 x 5)

x/-250 = -y/60 = 1/-5

x = 50

y = 12

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

Distance = xy km

= 50 × 12 = 600 km

Hence, the length of the journey is 600 km.

**Question 16. Places A and B are 100 km 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?**

**Solution: **

Let’s assume that x and y be two cars starting from points A and B.

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

Case 1 : 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,

Given that two cars meet in 5 hours.

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

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

therefore, AQ – BQ = AB

5x – 5y = 100

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

Case 2 : When two cars move in opposite direction

Suppose two cars meet at point P then,

Distance travelled by X car X = AP and

Distance travelled by Y car Y = BP

In this case it is given that two cars meet in 1 hour, therefore

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

Distance traveled by car y in 1 hours = ly km

therefore AP + BP = AB

1x + 1y = 100

x + y = 100 ————–(ii)

By solving eq. (i) and (ii) we get x = 60,

Substituting x = 60 in equation, we get,

x + y = 100

60 + y = 100

Y = 40

Hence, The speed of car X is 60 km/hr. and speed of car Y = 40 km/hr.

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