Chapter 3: Pair of Linear Equations in Two Variables Exercise – 3.10
Question: 1
Points A and B are 70 km. 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:
We have to find the speed of car
Let x and y be two cars starting from points A and B respectively. Let the speed of car x be x km/hr and that of 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 7 hours.
Therefore, Distance travelled by car X in 7 hours = 7x km
AQ = 7x
Distance traveled by car y in 7 hours = 7y km
BQ = 7Y
Clearly AQ - BQ = AB
7x - 7 y = 70 Dividing both sides by common factor 7 we get,
x - y = 10 . …… (I)
Case II :
When two cars move in opposite direction
Suppose two cars meet at point. Then,
Distance travelled by car x = AP,
Distance travelled by car y = BP,
In this case, two cars meet in 1 hour
Therefore Distance travelled by car X in 1 hour =1x km
AP = Ix
Distance travelled by car Y in 1 hour = ly km
BP = ly
AP + BP = AB
AP + BP = AB
x + y = 70 … (ii)
By solving equation (i) and (ii),
x = 40
Substituting x = 40 in equation (ii) we get
x + y = 70
40 + y = 70
y = 70 - 40
y = 30
Hence, the speed of car starting from point A is 40 km/hr.
The speed of car starting from point B is 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 the speed of the sailor in still water 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 8 km downstream
Time taken to cover 8 km upstream
But, time taken to cover 8 km downstream in 40 minutes or 40/60 hours that is 2/3 hrs.
8 × 3 = 2(x + y)
24 = 2x + 2y
Dividing both sides by common factor 2 we get
12 = x + y …… (1)
Time taken to cover 8km upstream in 1hour,
8 = 1(x - y) …….. (2)
By solving these equation (i) and (ii) we get,
x = 10
Substitute x = 10 in equation (i) we get,
x + y = 12
10 + y = 12
y = 12 – 10
y = 2
Hence, the speed of sailor is 10 km/hr.
Speed of Current is 2 km/hr.
Question: 3
The boat goes 30 km upstream and 44 km downstream in 10 hrs. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of stream and that of boat in still water.
Solution:
Let the speed of the boat in still water be x km/hr and the speed of the stream be y km/hr
Speed upstream = (x - y) km/hr
Speed downstream = (x + y) km/hr Now,
Time taken to cover 30 km upstream
Time taken to cover 44 km downstream
But total time of journey is 10 hours
Time taken to cover 40 km upstream
In this case total time of journey is given to be 13 hours
Therefore,
Putting
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,
Now,
1 × 10 = 2(x – y)
10 = 2x – 2y/2
v = 1/11
Substituting x = 8 in equation we get,
x + y = 11
8 + y = 11
y = 11 – 8
y = 3
Hence, speed of boat in still water is 8 km/hr.
Speed of the stream is 3 km/hr.
Question: 4
A boat goes 24 km upstream and 28 km downstream in 6 hrs. It goes 30 km upstream and 21 km downstream in 6.5 hours. Find the speed of the boat in still water and also speed of the stream.
Solution:
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
Time taken to cover 24 km upstream
But, total time of journey is 6 hours,
Time taken to cover 30 km upstream
Time taken to cover 21km downstream
In this case total time of journey is given to 6.5 hrs.
24u + 28v – 6 = 0 … (iii)
30u + 21v – 13/2 = 0 … (iv)
Solving these equations by cross multiplication we get,
Now,
x + y = 14 … (eq)
By solving he equations,
x = 10
Substituting x = 10 in equation, we get,
Speed of the stream = 4 km/hr.
Speed of boat = 10 km/hr.
Question: 5
A man walks a certain distance with 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
Solution:
Let r = the rate the man walked
then (r + 0.5) = 1 hr less rate
and (r - 1) = 3 hr more rate
let t = time he actually walked,
then (t - 1) = time taken at the faster speed
and (t + 3) = time at the slower speed
Distance for all three situations is the same (dist = rate x time):
So, rt = (r + 0.5)(t - 1)
rt = rt – r + 05t – 0.5
rt – rt + r – 0.5t = -0.5
r – 0.5t = -0.5 (1)
Also, for the slower speed case,
rt = (r - 1)(t + 3)
rt = rt + 3r - t – 3
rt – rt – 3r + t = -3
-3r + t = -3(2)
Multiply the 1st equation by 2 and add to the above equation
2r – t = – 1
-3r + t = – 3
- r = – 4
r = 4 km/h is the actual walking rate
Find t using: r – 0.5t = – 0.5
4 – 0.5t = – 0.5
- 0.5t = -0.5 – 4
- 0.5t = – 4.5
t = 9 hrs actual time
Hence, Distance = r × t
4 × 9 = 36 km.
Question: 6
A person rowing at the rate of 5 km/hr in still water takes thrice as much as time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
Solution:
Let x = the rate water is flowing
Then,
(5 + x) = boat speed downstream
and
(5 - x) = boat speed upstream
Let d = distance one way
write a time equation;
Upstream time = 3 times downstream time
Cross multiply
d(5 + x) = 3d(5 - x)
divide both sides d and eliminate it:
5 + x = 3(5 – x)
5 + x = 15 – 3x
x + 3x = 15 – 5
4x = 10
x = 10/4
x = 2.5 km/hr is the rate of the current.
Question: 7
Ramesh travels 760 km to his home partly by train and partly by car. He takes 8hours if he travels 160 km, by train and the rest by car. He takes 12 minutes more if he travels 240 km by train and the rest by car. Find the speed of train and the car respectively.
Solution:
Let the speed of the train be x km/hour that of the car be y km/hr, we have the following cases
Case I:
When Ramesh travels 760 Km by train and the rest by car
Time taken by Ramesh to travel 160 Km by train = 160/x hrs
Total time taken by Ramesh to cover 760 Km
It is given that total time taken in 8 hours
Time taken by Ramesh to travel
It is given that total time taken in 8 hours
Case II:
When Ramesh travels 240 Km by train and the rest by car
Time taken by Ramesh to travel 240 Km by train = 240/x hrs
Time taken by Ramesh to travel (760 - 240) = 520 Km by car = 520/y hrs
In this case total time of the journey is 8 hours 12 minutes
Now,
x = 80.
And
v = 1/100
1/y = 1/100
y = 100
Hence, the speed of the train is 80 km/hr.
Speed of car = 100 km/hr.
Question: 8
A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But, if he travels 200 km 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 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 400 km by train = 400/x hrs
Time taken by a man to cover 600 km
It is given that the total time taken in 8 hours,
Case II
When a man travels 200 km by train and rest by car
TT by man to travel 200 km = 200/x hrs
TT by man to travel 400 km = 400/y hrs
In this case, total time of the journey in 6 hours 30 minutes + 30 minutes that is 7 hrs,
Putting 1/x = u, and, 1/y = u,
Then,
By equating,
u = 1/100
By substituting u = 1/100 in the equation,
v = 180
Now,
x = 100
y = 80
Hence, the speed of the train is 100 km/hr.
The speed of the car is 80 km/hr.
Question: 9
Places A and B are 80 km 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.
Solution:
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 8 hours.
Distance travelled by car X in 8 hours =8x km.
AQ = 8x
Distance travelled by car Yin 8 hours =8y km
BQ = 8y
Clearly AQ – BQ = AB
8x – 8y = 80
Both sides divided by 8, we get
x – y = 10 .–(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 20 minutes, we can write it as 1 hour = 20/60 hours.
Therefore,
Distance travelled by a car Y in
x + y = 60 … (ii)
By asolving the equations, we get,
x = 35.
By substituting x = 35 in equation (ii), we get
x + y = 60
35 + y = 60
y = 60 – 35
y = 25.
Hence, speed of car x = 35 km/hr.
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:
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
Time taken to cover 40 km downstream
But, total time of journey is 8 hours
Time taken to cover 16 km upstream
x – y Time taken to cover 32 km downstream
In this case total time of journey is given to 8 hrs
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,
Now,
4 = x - y … (v)
And
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 6 km/hr.
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 kms by train and the remaining by bus, she takes 10 minutes longer. Find the speed of the train and the bus separately.
Solution:
Let the speed of the train be x km/hr that of the bus be y km/hr, we have the following cases,
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 trave (300 - 60) = 240 km by bus = 240/y hrs
Total time taken by Rohi to cover 300 km
It is given that total time taken in 4 hours
Case II
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
Subtracting equation (iv) from (iii) we get,
Now,
x = 60
And,
Hence, the speed of the train is 60 km/hr.
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 spees of the current.
Solution:
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 20 km downstream
Time taken to cover 4 km upstream
But, time taken to cover 20 km downstream in 2 hours,
20 = 2(x + y)
20 = 2x + 2y … (i)
Time Taken to cover 20 km downstream in 2 hours
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.
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 speed of the stream.
Solution:
Let 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.
By the given conditions, we have,
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,
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
y = 5
Hence, the A’s speed is 10/3 km/hr.
Then 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 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
It is given that total time taken in 5 hours 30 minutes
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.
Putting 1/x = u and 1/y = v, the equations (i) and (ii) reduces to
Multiplying equation (iii) by 6, the above equation becomes
Subtracting equation (iv) from (v) we get
Putting u = 1/100 in equation (iii), we get
v = 1/2
v = 1/80
Now,
u = 1/100
x = 100
And,
y = 80
Hence, the speed of the train is 100 km/hr
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 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 have
y = 12
Putting the values of x and y in equation (i), we obtain
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 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 60 km/hr.
Speed of car Y = 40 km/hr.