CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 25

Question 1. In a triangle ABC, \( \angle C = 3\angle B = 2(\angle A + \angle B) \). Find the three angles.
Answer:
Let \( \angle A = x^\circ \) and \( \angle B = y^\circ \). Then \( \angle C = 3y^\circ \).
Since \( \angle C = 2(\angle A + \angle B) \), we get \( 3y = 2(x + y) \Rightarrow 2x - y = 0 \) ...(i)
By the angle sum property, \( x + y + 3y = 180 \Rightarrow x + 4y = 180 \) ...(ii)
Multiplying (ii) by 2: \( 2 x + 8 y = 360 \) ...(iv)
Subtracting (iv) from (i), we get \( - 9 y = -360 \Rightarrow y = 40 \).
Putting \( y = 40 \) in (i): \( - \left(40\right) + 2 x = 0 \Rightarrow x = 20 \).
Hence, \( x = 20,\ y = 40 \).
Hence, \( \angle A = 20^\circ,\ \angle B = 40^\circ,\ \angle C = 120^\circ \).
In simple words: Turn the story into two equations using two unknowns, solve them, and state the answer in words.

Teacher's Note:
1. Use the angle sum property of a triangle as one of the two equations.
2. Check that the three angles add up to 180 degrees.

Question 2. Find the four angles of a cyclic quadrilateral ABCD in which \( \angle A = (2x - 1)^\circ,\ \angle B = (y + 5)^\circ,\ \angle C = (2y + 15)^\circ \) and \( \angle D = (4x - 7)^\circ \).
Answer:
In a cyclic quadrilateral, opposite angles are supplementary.
\( \angle A + \angle C = 180^\circ \Rightarrow (2x - 1) + (2y + 15) = 180 \Rightarrow x + y = 83 \) ...(i)
\( \angle B + \angle D = 180^\circ \Rightarrow (y + 5) + (4x - 7) = 180 \Rightarrow 4x + y = 182 \) ...(ii)
Subtracting (ii) from (i), we get \( - 3 x = -99 \Rightarrow x = 33 \).
Putting \( x = 33 \) in (i): \( \left(33\right) + y = 83 \Rightarrow y = 50 \).
Hence, \( x = 33,\ y = 50 \).
Hence, \( \angle A = 2(33) - 1 = 65^\circ,\ \angle B = 50 + 5 = 55^\circ,\ \angle C = 2(50) + 15 = 115^\circ,\ \angle D = 4(33) - 7 = 125^\circ \).
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. Opposite angles of a cyclic quadrilateral add up to 180 degrees.
2. Check that all four angles add up to 360 degrees.

Question 3. In a triangle ABC, \( \angle A = x^\circ,\ \angle B = 3x^\circ \) and \( \angle C = y^\circ \). If \( 3y - 5x = 30 \), prove that the triangle is right-angled.
Answer:
By the angle sum property, \( x + 3x + y = 180 \Rightarrow 4x + y = 180 \) ...(i)
Also \( 3y - 5x = 30 \Rightarrow 5x - 3y = -30 \) ...(ii)
Multiplying (i) by 3: \( 12 x + 3 y = 540 \) ...(iii)
Adding (iii) and (ii), we get \( 17 x = 510 \Rightarrow x = 30 \).
Putting \( x = 30 \) in (i): \( 4 \left(30\right) + y = 180 \Rightarrow y = 60 \).
Hence, \( x = 30,\ y = 60 \).
So \( \angle A = 30^\circ,\ \angle B = 3 \times 30^\circ = 90^\circ,\ \angle C = 60^\circ \).
Since \( \angle B = 90^\circ \), the triangle is right-angled at B.
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. Find the angles first and then look for a 90 degree angle.
2. A triangle is right-angled if any one angle is exactly 90 degrees.

Question 4. The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and the breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area is increased by 67 square units. Find the length and breadth of the rectangle.
Answer:
Let the length be x units and the breadth be y units. The area is xy.
(x - 5)(y + 3) = xy - 9, which simplifies to \( 3 x - 5 y = 6 \).
(x + 3)(y + 2) = xy + 67, which simplifies to \( 2 x + 3 y = 61 \).
So we have the pair of equations
\( 3 x - 5 y = 6 \) ...(i)    \( 2 x + 3 y = 61 \) ...(ii)
Multiplying (i) by 2: \( 6 x - 10 y = 12 \) ...(iii)
Multiplying (ii) by 3: \( 6 x + 9 y = 183 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 19 y = -171 \Rightarrow y = 9 \).
Putting \( y = 9 \) in (i): \( - 5 \left(9\right) + 3 x = 6 \Rightarrow x = 17 \).
Thus, \( x = 17,\ y = 9 \).
Hence, the length is 17 units and the breadth is 9 units.
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 5. If in a rectangle the length is increased and the breadth is reduced each by 2 units, the area is reduced by 28 square units. If, however, the length is reduced by 1 unit and the breadth is increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.
Answer:
Let the length be x units and the breadth be y units.
(x + 2)(y - 2) = xy - 28, which simplifies to \( x - y = 12 \).
(x - 1)(y + 2) = xy + 33, which simplifies to \( 2 x - y = 35 \).
So we have the pair of equations
\( x - y = 12 \) ...(i)    \( 2 x - y = 35 \) ...(ii)
Subtracting (ii) from (i), we get \( - x = -23 \Rightarrow x = 23 \).
Putting \( x = 23 \) in (i): \( \left(23\right) - y = 12 \Rightarrow y = 11 \).
Thus, \( x = 23,\ y = 11 \).
Hence, the length is 23 units and the breadth is 11 units, so the area is 253 square units.
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 6. In a rectangle, if the length is increased by 3 metres and the breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If the length is reduced by 1 metre and the breadth is increased by 4 metres, the area is increased by 89 square metres. Find the dimensions of the rectangle.
Answer:
Let the length be x metres and the breadth be y metres.
(x + 3)(y - 4) = xy - 67, which simplifies to \( 4 x - 3 y = 55 \).
(x - 1)(y + 4) = xy + 89, which simplifies to \( 4 x - y = 93 \).
So we have the pair of equations
\( 4 x - 3 y = 55 \) ...(i)    \( 4 x - y = 93 \) ...(ii)
Multiplying (ii) by 3: \( 12 x - 3 y = 279 \) ...(iv)
Subtracting (iv) from (i), we get \( - 8 x = -224 \Rightarrow x = 28 \).
Putting \( x = 28 \) in (i): \( 4 \left(28\right) - 3 y = 55 \Rightarrow y = 19 \).
Thus, \( x = 28,\ y = 19 \).
Hence, the length is 28 m and the breadth is 19 m.
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 7. A man starts his job with a certain monthly salary and earns a fixed increment every year. If his salary was Rs. 1500 after 4 years of service and Rs. 1800 after 10 years of service, what was his starting salary and what is the annual increment?
Answer:
Let the starting salary be Rs. x and the annual increment be Rs. y.
After 4 years the salary is x + 4y = 1500, so \( x + 4 y = 1500 \).
After 10 years the salary is x + 10y = 1800, so \( x + 10 y = 1800 \).
So we have the pair of equations
\( x + 4 y = 1500 \) ...(i)    \( x + 10 y = 1800 \) ...(ii)
Subtracting (ii) from (i), we get \( - 6 y = -300 \Rightarrow y = 50 \).
Putting \( y = 50 \) in (i): \( 4 \left(50\right) + x = 1500 \Rightarrow x = 1300 \).
Thus, \( x = 1300,\ y = 50 \).
Hence, the starting salary is Rs. 1300 and the annual increment is Rs. 50.
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 8. A railway half ticket costs half the full fare, and the reservation charge is the same on a half ticket as on a full ticket. One reserved first class ticket from Mumbai to Ahmedabad costs Rs. 216, and one full and one half reserved first class tickets cost Rs. 327. What is the basic first class full fare and what is the reservation charge?
Answer:
Let the basic full fare be Rs. x and the reservation charge be Rs. y.
One full reserved ticket costs Rs. 216, so \( x + y = 216 \).
A full ticket and a half ticket together cost 327: (x + y) + (x/2 + y) = 327, so \( 3 x + 4 y = 654 \).
So we have the pair of equations
\( x + y = 216 \) ...(i)    \( 3 x + 4 y = 654 \) ...(ii)
Multiplying (i) by 3: \( 3 x + 3 y = 648 \) ...(iii)
Subtracting (ii) from (iii), we get \( - y = -6 \Rightarrow y = 6 \).
Putting \( y = 6 \) in (i): \( \left(6\right) + x = 216 \Rightarrow x = 210 \).
Thus, \( x = 210,\ y = 6 \).
Hence, the basic full fare is Rs. 210 and the reservation charge is Rs. 6.
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. Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give her Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received.
Answer:
Let the number of Rs. 50 notes be x and the number of Rs. 100 notes be y.
She got 25 notes in all, so \( x + y = 25 \).
The total amount is Rs. 2000, so \( 50 x + 100 y = 2000 \).
So we have the pair of equations
Simplifying the given equations, we get
\( x + y = 25 \) ...(i)    \( x + 2 y = 40 \) ...(ii)
Subtracting (ii) from (i), we get \( - y = -15 \Rightarrow y = 15 \).
Putting \( y = 15 \) in (i): \( \left(15\right) + x = 25 \Rightarrow x = 10 \).
Thus, \( x = 10,\ y = 15 \).
Hence, she received 10 notes of Rs. 50 and 15 notes of Rs. 100.
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. Yash scored 40 marks in a test, getting 3 marks for each right answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Answer:
Let the number of right answers be x and the number of wrong answers be y.
With the first scheme, 3x - y = 40, so \( 3 x - y = 40 \).
With the second scheme, 4x - 2y = 50, so \( 4 x - 2 y = 50 \).
So we have the pair of equations
Simplifying the given equations, we get
\( 3 x - y = 40 \) ...(i)    \( 2 x - y = 25 \) ...(ii)
Subtracting (ii) from (i), we get \( x = 15 \Rightarrow x = 15 \).
Putting \( x = 15 \) in (i): \( 3 \left(15\right) - y = 40 \Rightarrow y = 5 \).
Thus, \( x = 15,\ y = 5 \).
Hence, Yash answered 15 questions correctly and 5 incorrectly, so there were 15 + 5 = 20 questions.
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. The incomes of X and Y are in the ratio 8 : 7 and their expenditures are in the ratio 19 : 16. If each saves Rs. 1250, find their incomes.
Answer:
Let the incomes of X and Y be Rs. 8k and Rs. 7k, and their expenditures Rs. 19m and Rs. 16m.
X saves Rs. 1250, so 8k - 19m = 1250, i.e., \( 8 x - 19 y = 1250 \).
Y saves Rs. 1250, so 7k - 16m = 1250, i.e., \( 7 x - 16 y = 1250 \).
So we have the pair of equations
\( 8 x - 19 y = 1250 \) ...(i)    \( 7 x - 16 y = 1250 \) ...(ii)
Multiplying (i) by 7: \( 56 x - 133 y = 8750 \) ...(iii)
Multiplying (ii) by 8: \( 56 x - 128 y = 10000 \) ...(iv)
Subtracting (iv) from (iii), we get \( - 5 y = -1250 \Rightarrow y = 250 \).
Putting \( y = 250 \) in (i): \( - 19 \left(250\right) + 8 x = 1250 \Rightarrow x = 750 \).
Thus, \( x = 750,\ y = 250 \).
Writing x for k and y for m, we get k = 750 and m = 250. Hence, the income of X is 8 x 750 = Rs. 6000 and the income of Y is 7 x 750 = Rs. 5250.
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. The ratio of incomes of two persons is 9 : 7 and the ratio of their expenditures is 4 : 3. If each of them saves Rs. 200 per month, find their monthly incomes.
Answer:
Let the incomes be Rs. 9k and Rs. 7k, and the expenditures Rs. 4m and Rs. 3m.
First person: 9k - 4m = 200, i.e., \( 9 x - 4 y = 200 \).
Second person: 7k - 3m = 200, i.e., \( 7 x - 3 y = 200 \).
So we have the pair of equations
\( 9 x - 4 y = 200 \) ...(i)    \( 7 x - 3 y = 200 \) ...(ii)
Multiplying (i) by 3: \( 27 x - 12 y = 600 \) ...(iii)
Multiplying (ii) by 4: \( 28 x - 12 y = 800 \) ...(iv)
Subtracting (iv) from (iii), we get \( - x = -200 \Rightarrow x = 200 \).
Putting \( x = 200 \) in (i): \( 9 \left(200\right) - 4 y = 200 \Rightarrow y = 400 \).
Thus, \( x = 200,\ y = 400 \).
Writing x for k and y for m, we get k = 200 and m = 400. Hence, the incomes are 9 x 200 = Rs. 1800 and 7 x 200 = Rs. 1400.
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. 2 men and 7 boys can do a piece of work in 4 days. The same work is done in 3 days by 4 men and 4 boys. How long would it take one man and one boy to do it?
Answer:
Let one man alone take x days and one boy alone take y days. In one day a man does 1/x and a boy does 1/y of the work.
2 men and 7 boys take 4 days: 4(2/x + 7/y) = 1, so \( \frac{28}{y} + \frac{8}{x} = 1 \).
4 men and 4 boys take 3 days: 3(4/x + 4/y) = 1, so \( \frac{12}{y} + \frac{12}{x} = 1 \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 8 u + 28 v = 1 \) ...(i)    \( 12 u + 12 v = 1 \) ...(ii)
Multiplying (i) by 3: \( 24 u + 84 v = 3 \) ...(iii)
Multiplying (ii) by 2: \( 24 u + 24 v = 2 \) ...(iv)
Subtracting (iv) from (iii), we get \( 60 v = 1 \Rightarrow v = \frac{1}{60} \).
Putting \( v = \frac{1}{60} \) in (i): \( 28 \left(\frac{1}{60}\right) + 8 u = 1 \Rightarrow u = \frac{1}{15} \).
Hence \( u = \frac{1}{15} \) and \( v = \frac{1}{60} \).
Therefore \( x = 15 \) and \( y = 60 \).
Thus, \( x = 15,\ y = 60 \).
Hence, one man alone can finish the work in 15 days and one boy alone in 60 days.
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. 2 women and 5 men can together finish a piece of embroidery in 4 days, while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone to finish the embroidery, and that taken by 1 man alone.
Answer:
Let one woman alone take x days and one man alone take y days.
2 women and 5 men take 4 days: 4(2/x + 5/y) = 1, so \( \frac{20}{y} + \frac{8}{x} = 1 \).
3 women and 6 men take 3 days: 3(3/x + 6/y) = 1, so \( \frac{18}{y} + \frac{9}{x} = 1 \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 8 u + 20 v = 1 \) ...(i)    \( 9 u + 18 v = 1 \) ...(ii)
Multiplying (i) by 9: \( 72 u + 180 v = 9 \) ...(iii)
Multiplying (ii) by 8: \( 72 u + 144 v = 8 \) ...(iv)
Subtracting (iv) from (iii), we get \( 36 v = 1 \Rightarrow v = \frac{1}{36} \).
Putting \( v = \frac{1}{36} \) in (i): \( 20 \left(\frac{1}{36}\right) + 8 u = 1 \Rightarrow u = \frac{1}{18} \).
Hence \( u = \frac{1}{18} \) and \( v = \frac{1}{36} \).
Therefore \( x = 18 \) and \( y = 36 \).
Thus, \( x = 18,\ y = 36 \).
Hence, one woman alone can finish the embroidery in 18 days and one man alone in 36 days.
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. Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row, there would be 3 rows more. Find the number of students in the class.
Answer:
Let the number of students in each row be x and the number of rows be y. The number of students is xy.
(x + 1)(y - 2) = xy, which simplifies to \( - 2 x + y = 2 \).
(x - 1)(y + 3) = xy, which simplifies to \( 3 x - y = 3 \).
So we have the pair of equations
\( - 2 x + y = 2 \) ...(i)    \( 3 x - y = 3 \) ...(ii)
Adding (i) and (ii), we get \( x = 5 \Rightarrow x = 5 \).
Putting \( x = 5 \) in (i): \( - 2 \left(5\right) + y = 2 \Rightarrow y = 12 \).
Thus, \( x = 5,\ y = 12 \).
Hence, there are 5 students in each row and 12 rows, so the class has 60 students.
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. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer:
Let the number of students in each row be x and the number of rows be y. The number of students is xy.
(x + 3)(y - 1) = xy, which simplifies to \( - x + 3 y = 3 \).
(x - 3)(y + 2) = xy, which simplifies to \( 2 x - 3 y = 6 \).
So we have the pair of equations
\( - x + 3 y = 3 \) ...(i)    \( 2 x - 3 y = 6 \) ...(ii)
Multiplying (i) by 2: \( - 2 x + 6 y = 6 \) ...(iii)
Adding (iii) and (ii), we get \( 3 y = 12 \Rightarrow y = 4 \).
Putting \( y = 4 \) in (i): \( 3 \left(4\right) - x = 3 \Rightarrow x = 9 \).
Thus, \( x = 9,\ y = 4 \).
Hence, there are 9 students in each row and 4 rows, so the class has 36 students.
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. If 1 is added to each of the numerator and denominator of a fraction, it becomes 2/3. However, if 1 is subtracted from each of the numerator and denominator, it becomes 3/5. Find the fraction.
Answer:
Let the fraction be x/y.
Adding 1 to both: (x + 1)/(y + 1) = 2/3, so 3x + 3 = 2y + 2, i.e., \( 3 x - 2 y = -1 \).
Subtracting 1 from both: (x - 1)/(y - 1) = 3/5, so 5x - 5 = 3y - 3, i.e., \( 5 x - 3 y = 2 \).
So we have the pair of equations
\( 3 x - 2 y = -1 \) ...(i)    \( 5 x - 3 y = 2 \) ...(ii)
Multiplying (i) by 3: \( 9 x - 6 y = -3 \) ...(iii)
Multiplying (ii) by 2: \( 10 x - 6 y = 4 \) ...(iv)
Subtracting (iv) from (iii), we get \( - x = -7 \Rightarrow x = 7 \).
Putting \( x = 7 \) in (i): \( 3 \left(7\right) - 2 y = -1 \Rightarrow y = 11 \).
Thus, \( x = 7,\ y = 11 \).
Hence, the fraction is \( \frac{7}{11} \).
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 18. A two-digit number is four times the sum of its digits and twice the product of its digits. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y. The number is \( 10x + y \).
The number is four times the sum of the digits: \( 10x + y = 4(x + y) \Rightarrow 6x = 3y \Rightarrow y = 2x \) ...(i)
The number is twice the product of the digits: \( 10x + y = 2xy \) ...(ii)
Putting y = 2x in (ii): \( 10x + 2x = 2x(2x) \Rightarrow 12x = 4x^2 \Rightarrow x = 3 \) (x cannot be 0, since the number is a two-digit number).
Then from (i), \( y = 2 \times 3 = 6 \).
Check: 36 = 4 x (3 + 6) = 36 and 36 = 2 x (3 x 6) = 36.
Hence, the number is 36.
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. Here one equation is linear and the other is not, so use the linear one to substitute.
2. The case x = 0 is rejected because the number must have two digits.

Question 19. A two-digit number is equal to 7 times the sum of its digits. When 27 is subtracted from the number, the digits interchange places. Find the number.
Answer:
Let the ten’s digit be x and the unit’s digit be y.
The number is 7 times the sum of its digits: 10x + y = 7(x + y), i.e., \( 3 x - 6 y = 0 \).
Subtracting 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
\( x - 2 y = 0 \) ...(i)    \( x - y = 3 \) ...(ii)
Subtracting (ii) from (i), we get \( - y = -3 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (i): \( - 2 \left(3\right) + x = 0 \Rightarrow x = 6 \).
Thus, \( x = 6,\ y = 3 \).
Hence, x = 6 and y = 3, so the number is 63.
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 20. A man sold a table and a chair together for Rs. 850, at a loss of 10% on the table and a gain of 10% on the chair. By selling them together for Rs. 950, he would have made a gain of 10% on the table and a loss of 10% on the chair. Find the cost price of each.
Answer:
Let the cost price of the table be Rs. x and that of the chair be Rs. y.
Selling the table at a 10% loss gives 0.9x and the chair at a 10% gain gives 1.1y, so 0.9x + 1.1y = 850, i.e., \( 9 x + 11 y = 8500 \).
Selling the table at a 10% gain gives 1.1x and the chair at a 10% loss gives 0.9y, so 1.1x + 0.9y = 950, i.e., \( 11 x + 9 y = 9500 \).
So we have the pair of equations
\( 9 x + 11 y = 8500 \) ...(i)    \( 11 x + 9 y = 9500 \) ...(ii)
Adding (i) and (ii), we get \( 20 x + 20 y = 18000 \), i.e., \( x + y = 900 \) ...(iii)
Subtracting (ii) from (i), we get \( - 2 x + 2 y = -1000 \), i.e., \( x - y = 500 \) ...(iv)
Solving (iii) and (iv): \( x = 700 \) and \( y = 200 \).
Thus, \( x = 700,\ y = 200 \).
Hence, the cost price of the table is Rs. 700 and that of the chair is Rs. 200.
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. Abdul travelled 300 km by train and 200 km by taxi, and it took him 5 hours 30 minutes. But if he travels 260 km by train and 240 km by taxi, he takes 6 minutes longer. Find the speed of the train and that of the taxi.
Answer:
Let the speed of the train be x km/h and the speed of the taxi be y km/h.
300 km by train and 200 km by taxi take 5.5 hours, so \( \frac{200}{y} + \frac{300}{x} = \frac{11}{2} \).
260 km by train and 240 km by taxi take 5 hours 36 minutes = 28/5 hours, so \( \frac{240}{y} + \frac{260}{x} = \frac{28}{5} \).
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \). Then the given equations become linear in u and v.
\( 600 u + 400 v = 11 \) ...(i)    \( 325 u + 300 v = 7 \) ...(ii)
Multiplying (i) by 3: \( 1800 u + 1200 v = 33 \) ...(iii)
Multiplying (ii) by 4: \( 1300 u + 1200 v = 28 \) ...(iv)
Subtracting (iv) from (iii), we get \( 500 u = 5 \Rightarrow u = \frac{1}{100} \).
Putting \( u = \frac{1}{100} \) in (i): \( 600 \left(\frac{1}{100}\right) + 400 v = 11 \Rightarrow v = \frac{1}{80} \).
Hence \( u = \frac{1}{100} \) and \( v = \frac{1}{80} \).
Therefore \( x = 100 \) and \( y = 80 \).
Thus, \( x = 100,\ y = 80 \).
Hence, the speed of the train is 100 km/h and the speed of the taxi is 80 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 22. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes 1/2. Find the fraction.
Answer:
Let the fraction be x/y.
The sum of the numerator and denominator is 12, so \( x + y = 12 \).
If the denominator is increased by 3, x/(y + 3) = 1/2, so 2x = y + 3, i.e., \( 2 x - y = 3 \).
So we have the pair of equations
\( x + y = 12 \) ...(i)    \( 2 x - y = 3 \) ...(ii)
Adding (i) and (ii), we get \( 3 x = 15 \Rightarrow x = 5 \).
Putting \( x = 5 \) in (i): \( \left(5\right) + y = 12 \Rightarrow y = 7 \).
Thus, \( x = 5,\ y = 7 \).
Hence, the fraction is \( \frac{5}{7} \).
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 sum of a numerator and denominator of a fraction is 8. If 3 is added to both the numerator and the denominator, the fraction becomes 3/4. Find the fraction.
Answer:
Let the fraction be x/y.
The sum of the numerator and denominator is 8, so \( x + y = 8 \).
After adding 3 to both: (x + 3)/(y + 3) = 3/4, so 4x + 12 = 3y + 9, i.e., \( 4 x - 3 y = -3 \).
So we have the pair of equations
\( x + y = 8 \) ...(i)    \( 4 x - 3 y = -3 \) ...(ii)
Multiplying (i) by 3: \( 3 x + 3 y = 24 \) ...(iii)
Adding (iii) and (ii), we get \( 7 x = 21 \Rightarrow x = 3 \).
Putting \( x = 3 \) in (i): \( \left(3\right) + y = 8 \Rightarrow y = 5 \).
Thus, \( x = 3,\ y = 5 \).
Hence, the fraction is \( \frac{3}{5} \).
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. Five years hence, the age of a father shall be three times the age of his son, while five years earlier the age of the father was 7 times that of 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 earlier, 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: 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 monthly incomes of A and B are in the ratio 9 : 7 and their monthly expenditures are in the ratio 4 : 3. If each saves Rs. 1600 per month, find the monthly incomes of each.
Answer:
Let the incomes be Rs. 9k and Rs. 7k, and the expenditures Rs. 4m and Rs. 3m.
A saves Rs. 1600, so 9k - 4m = 1600, i.e., \( 9 x - 4 y = 1600 \).
B saves Rs. 1600, so 7k - 3m = 1600, i.e., \( 7 x - 3 y = 1600 \).
So we have the pair of equations
\( 9 x - 4 y = 1600 \) ...(i)    \( 7 x - 3 y = 1600 \) ...(ii)
Multiplying (i) by 3: \( 27 x - 12 y = 4800 \) ...(iii)
Multiplying (ii) by 4: \( 28 x - 12 y = 6400 \) ...(iv)
Subtracting (iv) from (iii), we get \( - x = -1600 \Rightarrow x = 1600 \).
Putting \( x = 1600 \) in (i): \( 9 \left(1600\right) - 4 y = 1600 \Rightarrow y = 3200 \).
Thus, \( x = 1600,\ y = 3200 \).
Writing x for k and y for m, we get k = 1600 and m = 3200. Hence, A’s monthly income is Rs. 14400 and B’s monthly income is Rs. 11200.
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. 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 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 5 = 100, so \( x - y = 20 \).
Towards each other, (x + y) x 1 = 100, so \( x + y = 100 \).
So we have the pair of equations
\( x - y = 20 \) ...(i)    \( x + y = 100 \) ...(ii)
Subtracting (ii) from (i), we get \( - 2 y = -80 \Rightarrow y = 40 \).
Putting \( y = 40 \) in (i): \( - \left(40\right) + x = 20 \Rightarrow x = 60 \).
Thus, \( x = 60,\ y = 40 \).
Hence, the speeds are 60 km/h and 40 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.