CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 24

Question 1. The present age of a father and his son: if twice the son’s age in years is added to the father’s age, the sum is 70. But if twice the father’s age is added to the son’s age, the sum is 95. Find the ages of the father and the son.
Answer:
Let the father’s age be x years and the son’s age be y years.
Twice the son’s age added to the father’s age gives 70, so \( x + 2 y = 70 \).
Twice the father’s age added to the son’s age gives 95, so \( 2 x + y = 95 \).
So we have the pair of equations
\( x + 2 y = 70 \) ...(i)    \( 2 x + y = 95 \) ...(ii)
Multiplying (i) by 2: \( 2 x + 4 y = 140 \) ...(iii)
Subtracting (ii) from (iii), we get \( 3 y = 45 \Rightarrow y = 15 \).
Putting \( y = 15 \) in (i): \( 2 \left(15\right) + x = 70 \Rightarrow x = 40 \).
Thus, \( x = 40,\ y = 15 \).
Hence, the father is 40 years old and the son is 15 years old.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 2. A father is three times as old as his son. After twelve years, his age will be twice that of his son then. Find their present ages.
Answer:
Let the father’s age be x years and the son’s age be y years.
The father is three times as old as the son, so \( x = 3 y \).
After 12 years, x + 12 = 2(y + 12), i.e., \( x = 2 y + 12 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 3 y = 0 \) ...(i)    \( x - 2 y = 12 \) ...(ii)
From equation (i), \( x = 3 y \) ...(iii)
Substituting this in equation (ii): \( \left(3 y\right) - 2 y = 12 \Rightarrow y = 12 \).
Putting \( y = 12 \) in (iii): \( x = 36 \).
Thus, \( x = 36,\ y = 12 \).
Hence, the father is 36 years old and the son is 12 years old.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 3. I am three times as old as my son. Five years later, I shall be two and a half times as old as my son. How old am I and how old is my son?
Answer:
Let my age be x years and my son’s age be y years.
I am three times as old as my son, so \( x = 3 y \).
After 5 years, x + 5 = 2.5(y + 5), i.e., 2x - 5y = 15, so \( 2 x - 5 y = 15 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 3 y = 0 \) ...(i)    \( 2 x - 5 y = 15 \) ...(ii)
From equation (i), \( x = 3 y \) ...(iii)
Substituting this in equation (ii): \( 2 \left(3 y\right) - 5 y = 15 \Rightarrow y = 15 \).
Putting \( y = 15 \) in (iii): \( x = 45 \).
Thus, \( x = 45,\ y = 15 \).
Hence, my age is 45 years and my son’s age is 15 years.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 4. Ten years ago, a father was twelve times as old as his son. Ten years hence, he will be twice as old as his son will be. Find their present ages.
Answer:
Let the father’s present age be x years and the son’s present age be y years.
Ten years ago, x - 10 = 12(y - 10), i.e., \( x - 12 y = -110 \).
Ten years hence, x + 10 = 2(y + 10), i.e., \( x - 2 y = 10 \).
So we have the pair of equations
\( x - 12 y = -110 \) ...(i)    \( x - 2 y = 10 \) ...(ii)
Subtracting (ii) from (i), we get \( - 10 y = -120 \Rightarrow y = 12 \).
Putting \( y = 12 \) in (i): \( - 12 \left(12\right) + x = -110 \Rightarrow x = 34 \).
Thus, \( x = 34,\ y = 12 \).
Hence, the father is 34 years old and the son is 12 years old.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 5. Five years hence, a father’s age will be three times the age of his son. Five years ago, the father was seven times as old as his son. Find their present ages.
Answer:
Let the father’s present age be x years and the son’s present age be y years.
Five years hence, x + 5 = 3(y + 5), i.e., \( x - 3 y = 10 \).
Five years ago, x - 5 = 7(y - 5), i.e., \( x - 7 y = -30 \).
So we have the pair of equations
\( x - 3 y = 10 \) ...(i)    \( x - 7 y = -30 \) ...(ii)
Subtracting (ii) from (i), we get \( 4 y = 40 \Rightarrow y = 10 \).
Putting \( y = 10 \) in (i): \( - 3 \left(10\right) + x = 10 \Rightarrow x = 40 \).
Thus, \( x = 40,\ y = 10 \).
Hence, the father is 40 years old and the son is 10 years old.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 6. The present age of a father is three years more than three times the age of his son. Three years hence, the father’s age will be 10 years more than twice the age of the son. Determine their present ages.
Answer:
Let the father’s present age be x years and the son’s present age be y years.
The father is three years more than three times the son’s age, so \( x = 3 y + 3 \).
Three years hence, x + 3 = 2(y + 3) + 10, i.e., \( x = 2 y + 13 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 3 y = 3 \) ...(i)    \( x - 2 y = 13 \) ...(ii)
From equation (i), \( x = 3 y + 3 \) ...(iii)
Substituting this in equation (ii): \( \left(3 y + 3\right) - 2 y = 13 \Rightarrow y = 10 \).
Putting \( y = 10 \) in (iii): \( x = 33 \).
Thus, \( x = 33,\ y = 10 \).
Hence, the father is 33 years old and the son is 10 years old.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 7. A and B are friends and their ages differ by 2 years. A’s father D is twice as old as A, and B is twice as old as his sister C. The ages of D and C differ by 40 years. Find the ages of A and B.
Answer:
Let the ages of A and B be x years and y years. Then D’s age is 2x years and C’s age is y/2 years.
D and C differ by 40 years, so \( 2x - \frac{y}{2} = 40 \Rightarrow 4x - y = 80 \) ...(i)
The ages of A and B differ by 2 years, so either \( x - y = 2 \) (A is older) or \( y - x = 2 \) (B is older).
Case 1: \( x - y = 2 \) ...(ii). Subtracting (ii) from (i): \( 3x = 78 \Rightarrow x = 26 \), and \( y = 24 \). So A is 26 years and B is 24 years old.
Case 2: \( y - x = 2 \) ...(iii). Adding (i) and (iii): \( 3x = 82 \Rightarrow x = 27\frac{1}{3} \), and \( y = 29\frac{1}{3} \). So A is \( 27\frac{1}{3} \) years and B is \( 29\frac{1}{3} \) years old.
Hence, A is 26 years and B is 24 years old (or, if B is the elder, A is \( 27\frac{1}{3} \) years and B is \( 29\frac{1}{3} \) years old). The first pair is the natural answer for ages in whole years.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 8. A is elder to B by 2 years. A’s father F is twice as old as A, and B is twice as old as his sister S. If the ages of the father and the sister differ by 40 years, find the age of A.
Answer:
Let A’s age be x years and B’s age be y years. Then F is 2x years old and S is y/2 years old.
A is elder to B by 2 years, so \( x - y = 2 \).
The father and the sister differ by 40 years, so 2x - y/2 = 40, i.e., \( 4 x - y = 80 \).
So we have the pair of equations
\( x - y = 2 \) ...(i)    \( 4 x - y = 80 \) ...(ii)
Subtracting (ii) from (i), we get \( - 3 x = -78 \Rightarrow x = 26 \).
Putting \( x = 26 \) in (i): \( \left(26\right) - y = 2 \Rightarrow y = 24 \).
Thus, \( x = 26,\ y = 24 \).
Hence, A is 26 years old (and B is 24 years old).
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 9. A father’s age is three times the sum of the ages of his two children. After 5 years, his age will be twice the sum of the ages of the two children. Find the age of the father.
Answer:
Let the father’s age be x years and the sum of the ages of the two children be y years.
The father is three times the sum of the children’s ages, so \( x = 3 y \).
After 5 years the father is x + 5 and the children’s ages add up to y + 10, so x + 5 = 2(y + 10), i.e., \( x = 2 y + 15 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 3 y = 0 \) ...(i)    \( x - 2 y = 15 \) ...(ii)
From equation (i), \( x = 3 y \) ...(iii)
Substituting this in equation (ii): \( \left(3 y\right) - 2 y = 15 \Rightarrow y = 15 \).
Putting \( y = 15 \) in (iii): \( x = 45 \).
Thus, \( x = 45,\ y = 15 \).
Hence, the father is 45 years old (the children’s ages add up to 15 years).
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 10. Points A and B are 90 km apart on a highway. A car starts from A and another from B at the same time. If they go in the same direction, they meet in 9 hours, and if they go in opposite directions, they meet in 9/7 hours. Find their speeds.
Answer:
Let the speeds of the cars from A and B be x km/h and y km/h, with x > y.
In the same direction, the faster car gains 90 km in 9 hours: (x - y) x 9 = 90, so \( x - y = 10 \).
In opposite directions, they cover 90 km together in 9/7 hours: (x + y) x 9/7 = 90, so \( x + y = 70 \).
So we have the pair of equations
\( x - y = 10 \) ...(i)    \( x + y = 70 \) ...(ii)
Subtracting (ii) from (i), we get \( - 2 y = -60 \Rightarrow y = 30 \).
Putting \( y = 30 \) in (i): \( - \left(30\right) + x = 10 \Rightarrow x = 40 \).
Thus, \( x = 40,\ y = 30 \).
Hence, the speeds are 40 km/h and 30 km/h.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 11. Points A and B are 70 km apart on a highway. A car starts from A and another from B simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other, they meet in 1 hour. Find the speeds of the two cars.
Answer:
Let the speeds of the cars from A and B be x km/h and y km/h, with x > y.
In the same direction, (x - y) x 7 = 70, so \( x - y = 10 \).
Towards each other, (x + y) x 1 = 70, so \( x + y = 70 \).
So we have the pair of equations
\( x - y = 10 \) ...(i)    \( x + y = 70 \) ...(ii)
Subtracting (ii) from (i), we get \( - 2 y = -60 \Rightarrow y = 30 \).
Putting \( y = 30 \) in (i): \( - \left(30\right) + x = 10 \Rightarrow x = 40 \).
Thus, \( x = 40,\ y = 30 \).
Hence, the speeds are 40 km/h and 30 km/h.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 12. Points A and B are 80 km apart on a highway. A car starts from A and another from B at the same time. If they move in the same direction, they meet in 8 hours, and if they move in opposite directions, they meet in 1 hour 20 minutes. Find the speeds of the two cars.
Answer:
Let the speeds of the cars from A and B be x km/h and y km/h, with x > y. Note that 1 hour 20 minutes = 4/3 hours.
In the same direction, (x - y) x 8 = 80, so \( x - y = 10 \).
Towards each other, (x + y) x 4/3 = 80, so \( x + y = 60 \).
So we have the pair of equations
\( x - y = 10 \) ...(i)    \( x + y = 60 \) ...(ii)
Subtracting (ii) from (i), we get \( - 2 y = -50 \Rightarrow y = 25 \).
Putting \( y = 25 \) in (i): \( - \left(25\right) + x = 10 \Rightarrow x = 35 \).
Thus, \( x = 35,\ y = 25 \).
Hence, the speeds are 35 km/h and 25 km/h.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 13. Rahul travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and the rest by car. He takes 20 minutes longer if he travels 200 km by train and the rest by car. Find the speed of the train and the car.
Answer:
Let the speed of the train be x km/h and the speed of the car be y km/h.
120 km by train and 480 km by car take 8 hours, so \( \frac{480}{y} + \frac{120}{x} = 8 \).
200 km by train and 400 km by car take 8 hours 20 minutes = 25/3 hours, so \( \frac{400}{y} + \frac{200}{x} = \frac{25}{3} \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 15 u + 60 v = 1 \) ...(i)    \( 24 u + 48 v = 1 \) ...(ii)
Multiplying (i) by 8: \( 120 u + 480 v = 8 \) ...(iii)
Multiplying (ii) by 5: \( 120 u + 240 v = 5 \) ...(iv)
Subtracting (iv) from (iii), we get \( 240 v = 3 \Rightarrow v = \frac{1}{80} \).
Putting \( v = \frac{1}{80} \) in (i): \( 60 \left(\frac{1}{80}\right) + 15 u = 1 \Rightarrow u = \frac{1}{60} \).
Hence \( u = \frac{1}{60} \) and \( v = \frac{1}{80} \).
Therefore \( x = 60 \) and \( y = 80 \).
Thus, \( x = 60,\ y = 80 \).
Hence, the speed of the train is 60 km/h and the speed of the car is 80 km/h.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 14. A boat covers 32 km upstream and 36 km downstream in 7 hours. Also, it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.
Answer:
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h. Then the upstream speed is (x - y) km/h and the downstream speed is (x + y) km/h.
32 km upstream and 36 km downstream take 7 hours, so \( \frac{36}{x + y} + \frac{32}{x - y} = 7 \).
40 km upstream and 48 km downstream take 9 hours, so \( \frac{48}{x + y} + \frac{40}{x - y} = 9 \).
Let \( \frac{1}{x - y} = u \) and \( \frac{1}{x + y} = v \). Then the given equations become linear in u and v.
\( 32 u + 36 v = 7 \) ...(i)    \( 40 u + 48 v = 9 \) ...(ii)
Multiplying (i) by 4: \( 128 u + 144 v = 28 \) ...(iii)
Multiplying (ii) by 3: \( 120 u + 144 v = 27 \) ...(iv)
Subtracting (iv) from (iii), we get \( 8 u = 1 \Rightarrow u = \frac{1}{8} \).
Putting \( u = \frac{1}{8} \) in (i): \( 32 \left(\frac{1}{8}\right) + 36 v = 7 \Rightarrow v = \frac{1}{12} \).
Hence \( u = \frac{1}{8} \) and \( v = \frac{1}{12} \).
Therefore \( x - y = 8 \) and \( x + y = 12 \).
Solving these two equations for x and y, we get \( x = 10,\ y = 2 \).
Hence, the speed of the boat in still water is 10 km/h and the speed of the stream is 2 km/h.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 15. 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.
Answer:
Let the speed of the sailor in still water be x km/h and the speed of the current be y km/h. 40 minutes = 2/3 hour.
Downstream, (x + y) x 2/3 = 8, so \( x + y = 12 \).
Upstream, (x - y) x 1 = 8, so \( x - y = 8 \).
So we have the pair of equations
\( x + y = 12 \) ...(i)    \( x - y = 8 \) ...(ii)
Subtracting (ii) from (i), we get \( 2 y = 4 \Rightarrow y = 2 \).
Putting \( y = 2 \) in (i): \( \left(2\right) + x = 12 \Rightarrow x = 10 \).
Thus, \( x = 10,\ y = 2 \).
Hence, the speed of the sailor in still water is 10 km/h and the speed of the current is 2 km/h.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 16. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
30 km upstream and 44 km downstream take 10 hours, so \( \frac{44}{x + y} + \frac{30}{x - y} = 10 \).
40 km upstream and 55 km downstream take 13 hours, so \( \frac{55}{x + y} + \frac{40}{x - y} = 13 \).
Let \( \frac{1}{x - y} = u \) and \( \frac{1}{x + y} = v \). Then the given equations become linear in u and v.
\( 15 u + 22 v = 5 \) ...(i)    \( 40 u + 55 v = 13 \) ...(ii)
Multiplying (i) by 5: \( 75 u + 110 v = 25 \) ...(iii)
Multiplying (ii) by 2: \( 80 u + 110 v = 26 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 5 u = -1 \Rightarrow u = \frac{1}{5} \).
Putting \( u = \frac{1}{5} \) in (i): \( 15 \left(\frac{1}{5}\right) + 22 v = 5 \Rightarrow v = \frac{1}{11} \).
Hence \( u = \frac{1}{5} \) and \( v = \frac{1}{11} \).
Therefore \( x - y = 5 \) and \( x + y = 11 \).
Solving these two equations for x and y, we get \( x = 8,\ y = 3 \).
Hence, the speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 17. A boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream in 6 1/2 hours. Find the speed of the boat in still water and also the speed of the stream.
Answer:
Let the speed of the boat in still water be x km/h and the speed of the stream be y km/h.
24 km upstream and 28 km downstream take 6 hours, so \( \frac{28}{x + y} + \frac{24}{x - y} = 6 \).
30 km upstream and 21 km downstream take 13/2 hours, so \( \frac{21}{x + y} + \frac{30}{x - y} = \frac{13}{2} \).
Let \( \frac{1}{x - y} = u \) and \( \frac{1}{x + y} = v \). Then the given equations become linear in u and v.
\( 12 u + 14 v = 3 \) ...(i)    \( 60 u + 42 v = 13 \) ...(ii)
Multiplying (i) by 3: \( 36 u + 42 v = 9 \) ...(iii)
Subtracting (ii) from (iii), we get \( - 24 u = -4 \Rightarrow u = \frac{1}{6} \).
Putting \( u = \frac{1}{6} \) in (i): \( 12 \left(\frac{1}{6}\right) + 14 v = 3 \Rightarrow v = \frac{1}{14} \).
Hence \( u = \frac{1}{6} \) and \( v = \frac{1}{14} \).
Therefore \( x - y = 6 \) and \( x + y = 14 \).
Solving these two equations for x and y, we get \( x = 10,\ y = 4 \).
Hence, the speed of the boat in still water is 10 km/h and the speed of the stream is 4 km/h.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 18. X takes 3 hours more than Y to walk 30 km. But, if X doubles his pace, he is ahead of Y by 1 1/2 hours. Find their speeds of walking.
Answer:
Let the speeds of X and Y be x km/h and y km/h.
X takes 3 hours more than Y: 30/x - 30/y = 3, so \( - \frac{30}{y} + \frac{30}{x} = 3 \).
If X doubles his pace he is 1.5 hours ahead of Y: 30/y - 30/(2x) = 3/2, so \( - \frac{30}{y} + \frac{15}{x} = - \frac{3}{2} \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 10 u - 10 v = 1 \) ...(i)    \( 10 u - 20 v = -1 \) ...(ii)
Subtracting (ii) from (i), we get \( 10 v = 2 \Rightarrow v = \frac{1}{5} \).
Putting \( v = \frac{1}{5} \) in (i): \( - 10 \left(\frac{1}{5}\right) + 10 u = 1 \Rightarrow u = \frac{3}{10} \).
Hence \( u = \frac{3}{10} \) and \( v = \frac{1}{5} \).
Therefore \( x = \frac{10}{3} \) and \( y = 5 \).
Thus, \( x = \frac{10}{3},\ y = 5 \).
Hence, X walks at \( \frac{10}{3} \) km/h and Y walks at 5 km/h.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).

Question 19. While covering a distance of 30 km, Ajeet takes 2 hours more than Amit. If Ajeet doubles his speed, he would take 1 hour less than Amit. Find their speeds of walking.
Answer:
Let the speeds of Ajeet and Amit be x km/h and y km/h.
Ajeet takes 2 hours more than Amit: 30/x - 30/y = 2, so \( - \frac{30}{y} + \frac{30}{x} = 2 \).
If Ajeet doubles his speed he takes 1 hour less: 30/(2x) - 30/y = -1, so \( - \frac{30}{y} + \frac{15}{x} = -1 \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 15 u - 15 v = 1 \) ...(i)    \( 15 u - 30 v = -1 \) ...(ii)
Subtracting (ii) from (i), we get \( 15 v = 2 \Rightarrow v = \frac{2}{15} \).
Putting \( v = \frac{2}{15} \) in (i): \( - 15 \left(\frac{2}{15}\right) + 15 u = 1 \Rightarrow u = \frac{1}{5} \).
Hence \( u = \frac{1}{5} \) and \( v = \frac{2}{15} \).
Therefore \( x = 5 \) and \( y = \frac{15}{2} \).
Thus, \( x = 5,\ y = \frac{15}{2} \).
Hence, Ajeet walks at 5 km/h and Amit walks at 7.5 km/h.
In simple words: Each sentence of the problem gives one relation, and two independent relations are enough to find two unknowns.

Teacher's Note:
1. Choose the unknowns so that the equations stay simple.
2. Write the final answer with its unit.

Question 20. 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 1 km an hour slower, he takes 3 more hours. Find the distance covered by the man and his original rate of walking.
Answer:
Let the original speed be x km/h and the usual time be y hours. The distance is xy km.
If he walks 1/2 km/h faster, the time is 1 hour less: (x + 1/2)(y - 1) = xy, which simplifies to \( - x + \frac{y}{2} = 1 \cdot \frac{1}{2} \).
If he walks 1 km/h slower, the time is 3 hours more: (x - 1)(y + 3) = xy, which simplifies to \( 3 x - y = 3 \).
So we have the pair of equations
Simplifying the given equations, we get
\( - 2 x + y = 1 \) ...(i)    \( 3 x - y = 3 \) ...(ii)
Adding (i) and (ii), we get \( x = 4 \Rightarrow x = 4 \).
Putting \( x = 4 \) in (i): \( - 2 \left(4\right) + y = 1 \Rightarrow y = 9 \).
Thus, \( x = 4,\ y = 9 \).
Hence, the original speed is 4 km/h, the usual time is 9 hours and the distance is 36 km.
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Start by stating clearly what x and y stand for.
2. Always verify the answer against the original words of the problem, not just the equations.

Question 21. A train covered a certain distance at a uniform speed. If the train had been 6 km/h faster, it would have taken 4 hours less than the scheduled time. And if the train had been slower by 6 km/h, it would have taken 6 hours more than the scheduled time. Find the length of the journey.
Answer:
Let the speed of the train be x km/h and the scheduled time be y hours. The distance is xy km.
With 6 km/h more speed the time is 4 hours less: (x + 6)(y - 4) = xy, which simplifies to \( - 4 x + 6 y = 24 \).
With 6 km/h less speed the time is 6 hours more: (x - 6)(y + 6) = xy, which simplifies to \( 6 x - 6 y = 36 \).
So we have the pair of equations
Simplifying the given equations, we get
\( - 2 x + 3 y = 12 \) ...(i)    \( x - y = 6 \) ...(ii)
Multiplying (ii) by 2: \( 2 x - 2 y = 12 \) ...(iv)
Adding (i) and (iv), we get \( y = 24 \Rightarrow y = 24 \).
Putting \( y = 24 \) in (i): \( 3 \left(24\right) - 2 x = 12 \Rightarrow x = 30 \).
Thus, \( x = 30,\ y = 24 \).
Hence, the speed is 30 km/h, the scheduled time is 24 hours, and the length of the journey is 720 km.
In simple words: Give each unknown a name, write one equation for each condition in the problem, and solve the pair.

Teacher's Note:
1. Use two different conditions of the problem for the two equations.
2. After solving, check that the values are sensible (positive, whole numbers where needed).