CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 23

Question 1. 4 chairs and 3 tables cost Rs. 2100, and 5 chairs and 2 tables cost Rs. 1750. Find the cost of a chair and of a table separately.
Answer:
Let the cost of a chair be Rs. x and the cost of a table be Rs. y.
4 chairs and 3 tables cost Rs. 2100, so \( 4 x + 3 y = 2100 \).
5 chairs and 2 tables cost Rs. 1750, so \( 5 x + 2 y = 1750 \).
So we have the pair of equations
\( 4 x + 3 y = 2100 \) ...(i)    \( 5 x + 2 y = 1750 \) ...(ii)
Multiplying (i) by 2: \( 8 x + 6 y = 4200 \) ...(iii)
Multiplying (ii) by 3: \( 15 x + 6 y = 5250 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 7 x = -1050 \Rightarrow x = 150 \).
Putting \( x = 150 \) in (i): \( 4 \left(150\right) + 3 y = 2100 \Rightarrow y = 500 \).
Thus, \( x = 150,\ y = 500 \).
Hence, a chair costs Rs. 150 and a table costs Rs. 500.
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. 37 pens and 53 pencils together cost Rs. 320, while 53 pens and 37 pencils together cost Rs. 400. Find the cost of a pen and that of a pencil.
Answer:
Let the cost of a pen be Rs. x and the cost of a pencil be Rs. y.
37 pens and 53 pencils cost Rs. 320, so \( 37 x + 53 y = 320 \).
53 pens and 37 pencils cost Rs. 400, so \( 53 x + 37 y = 400 \).
So we have the pair of equations
\( 37 x + 53 y = 320 \) ...(i)    \( 53 x + 37 y = 400 \) ...(ii)
Adding (i) and (ii), we get \( 90 x + 90 y = 720 \), i.e., \( x + y = 8 \) ...(iii)
Subtracting (ii) from (i), we get \( - 16 x + 16 y = -80 \), i.e., \( x - y = 5 \) ...(iv)
Solving (iii) and (iv): \( x = \frac{13}{2} \) and \( y = \frac{3}{2} \).
Thus, \( x = \frac{13}{2},\ y = \frac{3}{2} \).
Hence, a pen costs Rs. 6.5 and a pencil costs Rs. 1.5.
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. 4 tables and 3 chairs together cost Rs. 2250, and 3 tables and 4 chairs cost Rs. 1950. Find the cost of 2 chairs and 1 table.
Answer:
Let the cost of a table be Rs. x and the cost of a chair be Rs. y.
4 tables and 3 chairs cost Rs. 2250, so \( 4 x + 3 y = 2250 \).
3 tables and 4 chairs cost Rs. 1950, so \( 3 x + 4 y = 1950 \).
So we have the pair of equations
\( 4 x + 3 y = 2250 \) ...(i)    \( 3 x + 4 y = 1950 \) ...(ii)
Adding (i) and (ii), we get \( 7 x + 7 y = 4200 \), i.e., \( x + y = 600 \) ...(iii)
Subtracting (ii) from (i), we get \( x - y = 300 \) ...(iv)
Solving (iii) and (iv): \( x = 450 \) and \( y = 150 \).
Thus, \( x = 450,\ y = 150 \).
Hence, a table costs Rs. 450 and a chair costs Rs. 150. The cost of 2 chairs and 1 table is 2 x 150 + 450 = Rs. 750.
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. A and B each have a certain number of oranges. A says to B, “If you give me 10 of your oranges, I will have twice the number of oranges left with you.” B replies, “If you give me 10 of your oranges, I will have the same number of oranges as are left with you.” Find the number of oranges with A and B separately.
Answer:
Let A have x oranges and B have y oranges.
If B gives 10 oranges to A, A has x + 10 and B has y - 10, so \( x + 10 = 2 \left(y - 10\right) \).
If A gives 10 oranges to B, B has y + 10 and A has x - 10, so \( y + 10 = x - 10 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 2 y = -30 \) ...(i)    \( - x + y = -20 \) ...(ii)
Adding (i) and (ii), we get \( - y = -50 \Rightarrow y = 50 \).
Putting \( y = 50 \) in (i): \( - 2 \left(50\right) + x = -30 \Rightarrow x = 70 \).
Thus, \( x = 70,\ y = 50 \).
Hence, A has 70 oranges and B has 50 oranges.
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. A and B each have a certain number of mangoes. A says to B, “If you give 30 of your mangoes, I will have twice as many as left with you.” B replies, “If you give me 10, I will have thrice as many as left with you.” How many mangoes does each have?
Answer:
Let A have x mangoes and B have y mangoes.
If B gives 30 mangoes to A, A has x + 30 and B has y - 30, so \( x + 30 = 2 \left(y - 30\right) \).
If A gives 10 mangoes to B, B has y + 10 and A has x - 10, so \( y + 10 = 3 \left(x - 10\right) \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 2 y = -90 \) ...(i)    \( - 3 x + y = -40 \) ...(ii)
Multiplying (ii) by 2: \( - 6 x + 2 y = -80 \) ...(iv)
Adding (i) and (iv), we get \( - 5 x = -170 \Rightarrow x = 34 \).
Putting \( x = 34 \) in (i): \( \left(34\right) - 2 y = -90 \Rightarrow y = 62 \).
Thus, \( x = 34,\ y = 62 \).
Hence, A has 34 mangoes and B has 62 mangoes.
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. One says, “Give me a hundred, friend! I shall then become twice as rich as you.” The other replies, “If you give me ten, I shall be six times as rich as you.” Tell me what is the amount of their respective capital.
Answer:
Let the first person have Rs. x and the second person have Rs. y.
If the second gives Rs. 100 to the first, the first has x + 100 and the second has y - 100, so \( x + 100 = 2 \left(y - 100\right) \).
If the first gives Rs. 10 to the second, the second has y + 10 and the first has x - 10, so \( y + 10 = 6 \left(x - 10\right) \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 2 y = -300 \) ...(i)    \( - 6 x + y = -70 \) ...(ii)
Multiplying (ii) by 2: \( - 12 x + 2 y = -140 \) ...(iv)
Adding (i) and (iv), we get \( - 11 x = -440 \Rightarrow x = 40 \).
Putting \( x = 40 \) in (i): \( \left(40\right) - 2 y = -300 \Rightarrow y = 170 \).
Thus, \( x = 40,\ y = 170 \).
Hence, their capitals are Rs. 40 and Rs. 170.
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. Reena has pens and pencils which together are 40 in number. If she has 5 more pencils and 5 fewer pens, then the number of pencils would become 4 times the number of pens. Find the original number of pens and pencils.
Answer:
Let Reena have x pens and y pencils.
Together they are 40, so \( x + y = 40 \).
With 5 fewer pens and 5 more pencils, y + 5 = 4(x - 5), so \( y + 5 = 4 \left(x - 5\right) \).
So we have the pair of equations
Simplifying the given equations, we get
\( x + y = 40 \) ...(i)    \( - 4 x + y = -25 \) ...(ii)
Subtracting (ii) from (i), we get \( 5 x = 65 \Rightarrow x = 13 \).
Putting \( x = 13 \) in (i): \( \left(13\right) + y = 40 \Rightarrow y = 27 \).
Thus, \( x = 13,\ y = 27 \).
Hence, Reena has 13 pens and 27 pencils.
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 8. A man has only 20 paise coins and 25 paise coins in his purse. If he has 50 coins in all, totalling Rs. 11.25, how many coins of each kind does he have?
Answer:
Let the number of 20 paise coins be x and the number of 25 paise coins be y. Rs. 11.25 = 1125 paise.
The total number of coins is 50, so \( x + y = 50 \).
The total value is 1125 paise, so \( 20 x + 25 y = 1125 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x + y = 50 \) ...(i)    \( 4 x + 5 y = 225 \) ...(ii)
Multiplying (i) by 4: \( 4 x + 4 y = 200 \) ...(iii)
Subtracting (ii) from (iii), we get \( - y = -25 \Rightarrow y = 25 \).
Putting \( y = 25 \) in (i): \( \left(25\right) + x = 50 \Rightarrow x = 25 \).
Thus, \( x = 25,\ y = 25 \).
Hence, he has 25 coins of 20 paise and 25 coins of 25 paise.
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 9. A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs. 27 for a book kept for seven days, while Susy paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer:
Let the fixed charge be Rs. x and the charge for each extra day be Rs. y.
Saritha kept the book for 7 days, i.e., 4 extra days, so \( x + 4 y = 27 \).
Susy kept the book for 5 days, i.e., 2 extra days, so \( x + 2 y = 21 \).
So we have the pair of equations
\( x + 4 y = 27 \) ...(i)    \( x + 2 y = 21 \) ...(ii)
Subtracting (ii) from (i), we get \( 2 y = 6 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (i): \( 4 \left(3\right) + x = 27 \Rightarrow x = 15 \).
Thus, \( x = 15,\ y = 3 \).
Hence, the fixed charge is Rs. 15 and the charge for each extra day is Rs. 3.
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 10. The sum of two numbers is 35 and their difference is 13. Find the numbers.
Answer:
Let the numbers be x and y with x > y.
Their sum is 35, so \( x + y = 35 \).
Their difference is 13, so \( x - y = 13 \).
So we have the pair of equations
\( x + y = 35 \) ...(i)    \( x - y = 13 \) ...(ii)
Subtracting (ii) from (i), we get \( 2 y = 22 \Rightarrow y = 11 \).
Putting \( y = 11 \) in (i): \( \left(11\right) + x = 35 \Rightarrow x = 24 \).
Thus, \( x = 24,\ y = 11 \).
Hence, the numbers are 24 and 11.
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 11. The sum of two numbers is 8. If their sum is 4 times their difference, find the numbers.
Answer:
Let the numbers be x and y with x > y.
Their sum is 8, so \( x + y = 8 \).
Their sum is 4 times their difference, so x + y = 4(x - y), i.e., \( 3 x - 5 y = 0 \).
So we have the pair of equations
\( x + y = 8 \) ...(i)    \( 3 x - 5 y = 0 \) ...(ii)
Multiplying (i) by 3: \( 3 x + 3 y = 24 \) ...(iii)
Subtracting (ii) from (iii), we get \( 8 y = 24 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (i): \( \left(3\right) + x = 8 \Rightarrow x = 5 \).
Thus, \( x = 5,\ y = 3 \).
Hence, the numbers are 5 and 3.
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 12. The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers.
Answer:
Let the numbers be x and y with x > y.
Their sum is 1000, so \( x + y = 1000 \).
The difference of their squares is 256000, so (x + y)(x - y) = 256000. Since x + y = 1000, we get x - y = 256, i.e., \( x - y = 256 \).
So we have the pair of equations
\( x + y = 1000 \) ...(i)    \( x - y = 256 \) ...(ii)
Subtracting (ii) from (i), we get \( 2 y = 744 \Rightarrow y = 372 \).
Putting \( y = 372 \) in (i): \( \left(372\right) + x = 1000 \Rightarrow x = 628 \).
Thus, \( x = 628,\ y = 372 \).
Hence, the numbers are 628 and 372.
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 13. In a two-digit number, the unit’s digit is twice the ten’s digit. If 27 is added to the number, the digits interchange their places. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y. The number is 10x + y.
The unit’s digit is twice the ten’s digit, so \( y = 2 x \).
Adding 27 interchanges the digits: 10x + y + 27 = 10y + x, i.e., \( 9 x - 9 y = -27 \).
So we have the pair of equations
Simplifying the given equations, we get
\( - 2 x + y = 0 \) ...(i)    \( x - y = -3 \) ...(ii)
Adding (i) and (ii), we get \( - x = -3 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (i): \( - 2 \left(3\right) + y = 0 \Rightarrow y = 6 \).
Thus, \( x = 3,\ y = 6 \).
Hence, x = 3 and y = 6, so the number is 36.
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 14. In a two-digit number, the ten’s digit is three times the unit’s digit. When the number is decreased by 54, the digits are reversed. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The ten’s digit is three times the unit’s digit, so \( x = 3 y \).
Decreasing the number by 54 reverses the digits: 10x + y - 54 = 10y + x, i.e., \( 9 x - 9 y = 54 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x - 3 y = 0 \) ...(i)    \( x - y = 6 \) ...(ii)
Subtracting (ii) from (i), we get \( - 2 y = -6 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (i): \( - 3 \left(3\right) + x = 0 \Rightarrow x = 9 \).
Thus, \( x = 9,\ y = 3 \).
Hence, x = 9 and y = 3, so the number is 93.
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 15. The sum of the digits of a two-digit number is 15. The number obtained by reversing the order of the digits exceeds the given number by 9. Find the given number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The sum of the digits is 15, so \( x + y = 15 \).
The reversed number 10y + x exceeds the given number 10x + y by 9, so 10y + x - (10x + y) = 9, i.e., \( - 9 x + 9 y = 9 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x + y = 15 \) ...(i)    \( - x + y = 1 \) ...(ii)
Adding (i) and (ii), we get \( 2 y = 16 \Rightarrow y = 8 \).
Putting \( y = 8 \) in (i): \( \left(8\right) + x = 15 \Rightarrow x = 7 \).
Thus, \( x = 7,\ y = 8 \).
Hence, x = 7 and y = 8, so the number is 78.
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 16. The sum of the digits of a two-digit number is 8 and the difference between the number and that formed by reversing the digits is 18. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y. The sum of the digits gives \( x + y = 8 \) ...(i)
The difference of the number and its reversal is \( (10x + y) - (10y + x) = 9(x - y) \), and this is 18 in magnitude, so \( x - y = 2 \) or \( y - x = 2 \).
Case 1: \( x - y = 2 \) ...(ii).
Subtracting (ii) from (i), we get \( 2 y = 6 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (i): \( \left(3\right) + x = 8 \Rightarrow x = 5 \).
Hence, \( x = 5,\ y = 3 \).
So the number is 53.
Case 2: \( y - x = 2 \) ...(iii).
Adding (i) and (ii), we get \( 2 y = 10 \Rightarrow y = 5 \).
Putting \( y = 5 \) in (i): \( \left(5\right) + x = 8 \Rightarrow x = 3 \).
Hence, \( x = 3,\ y = 5 \).
So the number is 35.
Hence, the number is 53 (the number exceeds its reversal by 18). If the question allows the reversed number to be the larger one, 35 also satisfies the condition.
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 17. The sum of a two-digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits of the first number. Find the first number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The number plus its reversal is 110: 11x + 11y = 110, i.e., \( x + y = 10 \).
Subtracting 10 from the number gives 5(x + y) + 4: 10x + y - 10 = 5(x + y) + 4, i.e., \( 5 x - 4 y = 14 \).
So we have the pair of equations
\( x + y = 10 \) ...(i)    \( 5 x - 4 y = 14 \) ...(ii)
Multiplying (i) by 4: \( 4 x + 4 y = 40 \) ...(iii)
Adding (iii) and (ii), we get \( 9 x = 54 \Rightarrow x = 6 \).
Putting \( x = 6 \) in (i): \( \left(6\right) + y = 10 \Rightarrow y = 4 \).
Thus, \( x = 6,\ y = 4 \).
Hence, x = 6 and y = 4, so the first number is 64.
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. The sum of a two-digit number and the number formed by interchanging its digits is 132. If 12 is added to the number, the new number becomes 5 times the sum of the digits. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The number plus its reversal is 132: 11x + 11y = 132, i.e., \( x + y = 12 \).
Adding 12 to the number gives 5(x + y): 10x + y + 12 = 5(x + y), i.e., \( 5 x - 4 y = -12 \).
So we have the pair of equations
\( x + y = 12 \) ...(i)    \( 5 x - 4 y = -12 \) ...(ii)
Multiplying (i) by 4: \( 4 x + 4 y = 48 \) ...(iii)
Adding (iii) and (ii), we get \( 9 x = 36 \Rightarrow x = 4 \).
Putting \( x = 4 \) in (i): \( \left(4\right) + y = 12 \Rightarrow y = 8 \).
Thus, \( x = 4,\ y = 8 \).
Hence, x = 4 and y = 8, so the number is 48.
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. The sum of a two-digit number and the number obtained by reversing the order of its digits is 121, and the two digits differ by 3. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y. Then \( (10x + y) + (10y + x) = 121 \Rightarrow 11(x + y) = 121 \Rightarrow x + y = 11 \) ...(i)
The digits differ by 3, so \( x - y = 3 \) or \( y - x = 3 \).
Case 1: \( x - y = 3 \) ...(ii).
Subtracting (ii) from (i), we get \( 2 y = 8 \Rightarrow y = 4 \).
Putting \( y = 4 \) in (i): \( \left(4\right) + x = 11 \Rightarrow x = 7 \).
Hence, \( x = 7,\ y = 4 \).
So the number is 74.
Case 2: \( y - x = 3 \) ...(iii).
Adding (i) and (ii), we get \( 2 y = 14 \Rightarrow y = 7 \).
Putting \( y = 7 \) in (i): \( \left(7\right) + x = 11 \Rightarrow x = 4 \).
Hence, \( x = 4,\ y = 7 \).
So the number is 47.
Hence, the number is 74 or 47.
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 two-digit number is 3 more than 4 times the sum of its digits. If 18 is added to the number, the digits are reversed. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The number is 3 more than 4 times the sum of the digits: 10x + y = 4(x + y) + 3, i.e., \( 6 x - 3 y = 3 \).
Adding 18 reverses the digits: 10x + y + 18 = 10y + x, i.e., \( 9 x - 9 y = -18 \).
So we have the pair of equations
Simplifying the given equations, we get
\( 2 x - y = 1 \) ...(i)    \( x - y = -2 \) ...(ii)
Subtracting (ii) from (i), we get \( x = 3 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (i): \( 2 \left(3\right) - y = 1 \Rightarrow y = 5 \).
Thus, \( x = 3,\ y = 5 \).
Hence, x = 3 and y = 5, so the number is 35.
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 21. A fraction becomes 4/5 if 1 is added to both the numerator and the denominator. If, however, 5 is subtracted from both the numerator and the denominator, the fraction becomes 1/2. What is the fraction?
Answer:
Let the fraction be x/y.
Adding 1 to both: (x + 1)/(y + 1) = 4/5, so 5x + 5 = 4y + 4, i.e., \( 5 x - 4 y = -1 \).
Subtracting 5 from both: (x - 5)/(y - 5) = 1/2, so 2x - 10 = y - 5, i.e., \( 2 x - y = 5 \).
So we have the pair of equations
\( 5 x - 4 y = -1 \) ...(i)    \( 2 x - y = 5 \) ...(ii)
From equation (ii), \( y = 2 x - 5 \) ...(iii)
Substituting this in equation (i): \( - 4 \left(2 x - 5\right) + 5 x = -1 \Rightarrow x = 7 \).
Putting \( x = 7 \) in (iii): \( y = 9 \).
Thus, \( x = 7,\ y = 9 \).
Hence, the fraction is \( \frac{7}{9} \).
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 22. A fraction is such that if the numerator is multiplied by 3 and the denominator is reduced by 3, we get 18/11. But if the numerator is increased by 8 and the denominator is doubled, we get 2/5. Find the fraction.
Answer:
Let the fraction be x/y.
3x/(y - 3) = 18/11, so 33x = 18y - 54, i.e., \( 11 x - 6 y = -18 \).
(x + 8)/(2y) = 2/5, so 5x + 40 = 4y, i.e., \( 5 x - 4 y = -40 \).
So we have the pair of equations
\( 11 x - 6 y = -18 \) ...(i)    \( 5 x - 4 y = -40 \) ...(ii)
Multiplying (i) by 2: \( 22 x - 12 y = -36 \) ...(iii)
Multiplying (ii) by 3: \( 15 x - 12 y = -120 \) ...(iv)
Subtracting (iv) from (iii), we get \( 7 x = 84 \Rightarrow x = 12 \).
Putting \( x = 12 \) in (i): \( 11 \left(12\right) - 6 y = -18 \Rightarrow y = 25 \).
Thus, \( x = 12,\ y = 25 \).
Hence, the fraction is \( \frac{12}{25} \).
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 23. The denominator of a fraction is 4 more than twice the numerator. When both the numerator and the denominator are decreased by 6, the denominator becomes 12 times the numerator. Determine the fraction.
Answer:
Let the fraction be x/y.
The denominator is 4 more than twice the numerator, so \( y = 2 x + 4 \).
After decreasing both by 6: y - 6 = 12(x - 6), i.e., \( y = 12 x - 66 \).
So we have the pair of equations
Simplifying the given equations, we get
\( - 2 x + y = 4 \) ...(i)    \( - 12 x + y = -66 \) ...(ii)
From equation (i), \( y = 2 x + 4 \) ...(iii)
Substituting this in equation (ii): \( \left(2 x + 4\right) - 12 x = -66 \Rightarrow x = 7 \).
Putting \( x = 7 \) in (iii): \( y = 18 \).
Thus, \( x = 7,\ y = 18 \).
Hence, the fraction is \( \frac{7}{18} \).
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 24. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Answer:
Let the fraction be x/y.
The numerator is 4 less than the denominator, so \( y = x + 4 \).
After the changes: y + 1 = 8(x - 2), i.e., \( y = 8 x - 17 \).
So we have the pair of equations
Simplifying the given equations, we get
\( - x + y = 4 \) ...(i)    \( - 8 x + y = -17 \) ...(ii)
From equation (i), \( y = x + 4 \) ...(iii)
Substituting this in equation (ii): \( \left(x + 4\right) - 8 x = -17 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (iii): \( y = 7 \).
Thus, \( x = 3,\ y = 7 \).
Hence, the fraction is \( \frac{3}{7} \).
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 25. The sum of the numerator and denominator of a fraction is 4 more than twice the numerator. If the numerator and denominator are increased by 3, they are in the ratio 2 : 3. Determine the fraction.
Answer:
Let the fraction be x/y.
x + y = 2x + 4, i.e., \( x - y = -4 \).
After adding 3 to both: (x + 3)/(y + 3) = 2/3, so 3x + 9 = 2y + 6, i.e., \( 3 x - 2 y = -3 \).
So we have the pair of equations
\( x - y = -4 \) ...(i)    \( 3 x - 2 y = -3 \) ...(ii)
Multiplying (i) by 2: \( 2 x - 2 y = -8 \) ...(iii)
Subtracting (ii) from (iii), we get \( - x = -5 \Rightarrow x = 5 \).
Putting \( x = 5 \) in (i): \( \left(5\right) - y = -4 \Rightarrow y = 9 \).
Thus, \( x = 5,\ y = 9 \).
Hence, the fraction is \( \frac{5}{9} \).
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 26. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are each decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Answer:
Let the fraction be x/y.
x + y = 2y - 3, i.e., \( x - y = -3 \).
After decreasing both by 1: x - 1 = (y - 1)/2, so 2x - 2 = y - 1, i.e., \( 2 x - y = 1 \).
So we have the pair of equations
\( x - y = -3 \) ...(i)    \( 2 x - y = 1 \) ...(ii)
Subtracting (ii) from (i), we get \( - x = -4 \Rightarrow x = 4 \).
Putting \( x = 4 \) in (i): \( \left(4\right) - y = -3 \Rightarrow y = 7 \).
Thus, \( x = 4,\ y = 7 \).
Hence, the fraction is \( \frac{4}{7} \).
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.