CBSE Class 10 Mathematics Pair Of Linear Equations In 2 Variables Worksheet Set 02

Question- A told B, "when I was as old as you are now, then your age was four years less than half of my present age". If the sum of the present ages of A and B is 61 years, what is B's present age? (in years).

Question- If 173x + 197y = 149 and 197x + 173y = 221, then find (x,y).
(1) (3, - 2)
(2) (2,1)
(3) (1, - 2)
(4) (2,-1)
Answer : D

Very Short Answer type Questions

Question. Solve for x and y: 0.4x – 1.5y = 6.5, 0.3x – 0.2y = 0.9.
Answer : x = 5 and y = –3.

Question. Find the values of k for which the system of equations kx – y = 2, 6x – 2y = 3 has
(i) a unique solution, (ii) no solution. (iii) Is there a value of k for which the given system has infinitely many solutions?
Answer : (i) k ≠ 3, (ii) k =3, (iii) no real value of k

Question. Find the values of k for which the system of equations x – 2y = 3, 3x + ky =1 has a unique solution.
Answer : All real values of k, other than –6.

Question. Solve graphically the system of linear equations 4x – 5y + 16 = 0 and 2x + y – 6 = 0. Determine the vertices of the triangle formed by these lines and the x-axis.
Answer : Vertices are (1, 4), (–4, 0) and (3, 0)

Question. Find the value of k for which the following pair of linear equations has infinitely many solutions: 2x – 3y = 7, (k + 1)x + (1 – 2k)y = (5k – 4) .
Answer : k = 5

Question. Five years ago, A was thrice as old as B and ten years later A shall be twice as old as B. What are the present ages of A and B?
Answer : A’s present age = 50 years, B’s present age = 20 years

Question. Find the values of k for which the pair of linear equations kx + 3y = k – 2 and 12x + ky = k has no solution.
Answer : k = 6 or k = – 6.

Question. Find the value of k for which the given system of equations has infinitely many solutions: x + (k + 1)y = 5, (k + 1)x + 9y + (1 – 8k) = 0.
Answer : k = 2.

Question. Solve the following system of linear equations graphically: 4x – 5y – 20 = 0 and 3x + 5y – 15 = 0. Determine the vertices of the triangle formed by the lines representing the above equations and the y-axis.
Answer : (0, –4), (0, 3) and (5, 0) .

Question. The sum of the digits of a two-digit number is 12. The number obtained by interchanging its digits exceeds the given number by 18. Find the number
Answer : 57

1. Solve graphically the following pairs of linear equations:
(i) 2x – y = 4
3y – x = 3 Also, find the coordinates of the points where these lines intersect the 2 axes.
(ii) 2x + 3y = 12
x – y = 1 Shade the region (area) between the 2 lines and x axis.

2. Find graphically the coordinates of the vertices of a triangle whose sides have the equations:
(i) y = x, y = 0 and 2x + 3y = 30 
(ii) 2y –x = 8, 5y – x = 14 and y – 2x = 1
(iii) y = x, 3y = x and x + y = 8

3. Plot a graph for each of the following pairs of equations and shade the region bounded by the 2 lines and the x-axis.
(i) x – y + 1 = 0 
2x + y – 10 = 0
(ii) 4x – 3y + 4 = 0
4x + 3y – 20 = 0
(iii) 2x + y = 6 
2x – y + 2 = 0
(iv) x + y = 5
2x – y +2 =0

4. Solve the following pair of linear equations graphically: 3x + y – 11 = 0; x – y – 1 = 0
Shade the region bound by these lines and the axis of y.

5. Solve each of the following pairs of linear equations graphically:
(i) 5x – 6y + 30 = 0 
5x + 4y – 20 = 0
(ii) 3x – 4y + 6 = 0
3x + y – 9 = 0

6. Solve the following pairs of equations:
(i) 5m – 5n = 12; 2m + 9n = 20 
(ii) 4/x + 5y = 7
3/x + 4y = 5
(iii) x – y = 0.9
11 / x + y = 2 
(iv) 8x – 3y = 5xy; 6x – 5y = - 2xy
(v) 99x + 101y = 499; 101x + 99y = 501
(vi) 39 x + 41y = 76; 41x + 39y = 84
(vii) x/a + y/b = a + b
 x/a² + y/b² = 2
(viii) a(x + y) + b(x – y) = a² + b² - ab
a (x + y) – b(x – y) =a² + b² + ab
(ix) x/a - y/b = a – b; x/a² - y/b² = 0 (a ≠ b)

7. If 2x + y = 35 and 3x + 4y = 65, find the value of x/y

8. Find the value of c for which the pair of equations : 2x + cy = 1; 3x + 5y = 7 will have
(i) a unique solution; (ii) no solution. Is there a value of c for which the system has infinite number of solutions?

9. Find the value of k for which the following pairs of equations have unique solutions:
(i) 7x – 2y = 3; 22x – ky = 8 (ii) 2x + ky = 1; 3x – 5y = 7 (iii) 2x + 3y – 5 = 0; kx – 6y – 8 = 0

10. For what value(s) of k will the pair of linear equations: kx + 3y = k – 3; 12x + ky = k have a unique solution?

11. Last year 1 kg of tea and 3 kg of sugar together cost Rs 96. This year, the rates of tea increased by 15% and that of sugar by 10%. So the amount of tea and sugar now cost Rs. 108.60. Find the per kg rates of tea & sugar last year.

12. A boat goes 24 km upstream & 28 km downstream in 6 hours. In 6.5 hours, it can go 30 km upstream & 21 km downstream. Find the speed of stream and the speed of boat in still water.

13. A person invests some amount @ 12% S.I. and some other amount @ 10% S.I..He receives an annual interest of Rs.1300. But if he interchanges the amounts invested, he shall receive Rs.40 more as interest. How much has he invested at each rate?

14. If 1 is added to both the numerator and the denominator of a fraction, it becomes equal to
7/8. If, however, 1 is subtracted from both the numerator & denominator of the same fraction, it becomes equal to 6/7. Find the fraction.

15. The age of a father 8 yrs back was 5 times that of his son. After 8 yrs, his age will be 8 yrs more than double the age of his son. Find their present ages.

16. There are some lotus flowers in a lake. If 1 butterfly sits on each flower, one butterfly is left behind. If 2 butterflies sit on each flower, 1 flower is left behind. What is the no. of flowers? What is the no. of butterflies?

1. Solve for x and y :

5. Find the values of k for which the following system of equations is inconsistent

(3k+1)x+3y=2

13. Solve graphically:

SECTION A: (1 MARK)

1. What is the point of intersection of lines represented by 3x – 2y = 6 and the y-axis? (CBSE 

2. Find the value of k for which the pair of equations 4x – 5y = 5 and kx + 3y = 3 is consistent. 

3. If x = a, y = b is the solution of the equations x – y = 2 and x + y = 4, then find the values of a and b .

SECTION B: (2 MARKS)

4. Solve for x: 99x + 101y = 499 ; 101x + 99y = 501

5. The angles of a cyclic quadrilateral taken in order are (3y – 5), (4y + 20), (7x + 5) and 4x. Find the angles of the cyclic quadrilateral.

6. For what value of k the following system of linear equations has a no solution? x + 2y = 3 (k – 1)x + (k + 1)y = k + 2

7. Solve for x and y: 3𝑥−𝑦 =27 ; 3𝑥+𝑦 =243 (NCERT EXEMPLAR PROBLEM)

SECTION C: (3 MARKS)

8. Solve for x and y: (a + b)x + (a – b)y = a2 + b2 (a – b)x + (a + b)y = a2 + b2 (NCERT 

9. Determine the values of m and n so that the following system of linear equations have infinite number of solutions:
(2m – 1)x + 3y – 5 = 0
3x + (n – 1) y – 2 = 0

10. Solve for x and y: x/a + y/b = a + b
x/a2 + y/b2 = 2

11. Solve graphically the pair of linear equations and write the coordinates of the vertices of the triangle formed by these two lines with x – axis. 3x + y – 3 = 0 ;     2x – y + 8 = 0

SECTION D: (4 MARKS)

12. An honest person invested some amount at the rate of 12% simple interest and some other amount at the rate of 10% simple interest. He received yearly interest of ₹130. But if he had interchanged amounts invested, he would have received ₹4 more as interest. How much amount did he invest at different rates? (

13. If a box containing red and white marbles, half the number of white marbles is equal to one-third the number of red marbles. Thrice the total number of marbles exceeds seven times the number of white marbles by 6. How many marbles of each colour does the box contain? (NCERT 

14. Two candles of equal height but different thickness are lighted. The first burns off in 6 hours and the second in 8 hours. How long, after lighting both, will the first candle be half the height of the second? 

15. The ages of two friends Sunaina and Tanishtha differ by 2 years. Sunaina’s father is twice as old as Sunaina and Tanishtha is twice as old as his brother Shiva. The ages of Sunaina’s father and Shiva differ by 40 years. Find the ages of Sunaina and Tanishtha. 

1. What will be the solution of these equations ax+by=a-b, bx-ay=a+b

Q1. If 1 is added to each of the given two numbers, then their ratio is 1:2. If 5 is subtracted from each of the numbers, then their ratio is 5:11. Find the numbers.

Q2. Out of a 940 km journey, a part of the journey was covered by a motor car at a speed of 72 km/hr.The remaining part of the journey was covered by train at a speed of 84 km/hr. If the total distance was covered in 12 hrs, find the distance travelled by the train and the time taken by it to cover that distance.

Q3. The ratio of the 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.

Q4. A person starts his job with a certain monthly salary and earns a fixed increment every year . If his salary was Rs. 4500 after 4 years of service and Rs. 5700 after 12 years of service , find his initial salary and the annual increment .

Q5. Two years ago , the father was five times as old as his son . Two years later , his age will be 8 more than three times the age of the son . Find their present ages .

Q6. Seven times a 2- digit number is equal to four times the number obtained by reversing the order of the digits . If the sum of both the digits is 9 , find the number.

Q7. Students of a class are made to stand in rows. If 4 students are extra in a row , there would be two rows less . If 4 students are less in a row , there would be 4 more rows . Find the number of students in the class.

Q8. It takes 12 hours to fill a swimming pool using two pipes. If the larger pipe is used for 4 hours and the smaller pipe for 9 hours, only half the pool is filled. How long would it take for each pipe alone to fill the pool?

Q9. A piece of work is done by 6 men and 5 women in 6days or 3 men and 4 women in 10 days . How many days will it take for 9 men and 15 women to finish that work ?

Q10. A man sold a table and 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 loss of 10% on the chair. Find the cost price of each.

Q11. A person invested some amount at the rate of 12% simple interest and some at the rate of 10% simple interest. He received an yearly interest of Rs. 130 . If he had interchanged the amounts invested , he would have received Rs. 4 more as interest . How much amount did he invest at different rates?

Q12. 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:-

1). 35 , 71

2). 532km , 6 hrs. 20 mins.

3) Rs. 1800 , Rs. 1400

4) Rs. 3900 , Rs. 150

5) 10 , 42

6) 36

7) 96

8) 20 , 30

9) 3 days

10) Chair at Rs. 200 and Table at Rs. 700

11) Rs. 500 at 12% and Rs. 700 at 10% p.y.

12) 10,15

Question. On reversing the digit of a two-digit number, number obtained is 9 less than three times the original number. If the difference of these two numbers is 45, find the original number
Answer : let the digit on unit place be x and tens digit be y
Then the number = 10y + x
Number formed by reversing the digits = 10x + y
Then,
10x + y = 3(10y + x) - 9
7x - 29y = -9 ………………(i)
Also, x - y = 5
x = y + 5 ……………(ii)
(ii) in (i)
9(y + 5) - 29y = -9
y = 44/22 = 2
x = 2 + 5 = 7
the number = 10(2) + 7 = 27

Question. Solve for x and y:
𝑥𝑎 + 𝑦𝑏 = 2 𝑎𝑛𝑑
𝑎𝑥 − 𝑏𝑦 = 𝑎² − 𝑏²
Answer : x/a + y/b = 2 ⟹ bx + ay = 2ab…………..(i)
ax − by = a² − b² …………….(ii)
(i) ×a ⟹ abx + a²y = 2a²b……………(iii)
(ii) ×b ⟹ abx − b²y = a²b − b³……….(iv)
Solving y = b and x = a

Question. Solve: 𝑎𝑥 + 𝑏𝑦 = 𝑎 − 𝑏 𝑎𝑛𝑑 𝑏𝑥 − 𝑎𝑦 = 𝑎 + 𝑏
Answer : ax + by = a-b……………………..(1)
bx-ay = a + b ………………………(2)
solve the equation by using substitution / elimination
then x = 1 and y = -1

Question. Solve by elimination method:
3𝑥 + 4𝑦 = 10
2𝑥 − 2𝑦 = 2
Answer : 3x + 4y = 10 ………………..(1)
2x − 2y = 2 ………………….(2)
Multiplying (2) by 2 and adding to (1), we get
7x = 14
x = 2
Putting x = 2 in (1), we get 3 (2) + 4y = 10
y = 1
Hence x = 2, y = 1

Question. The larger of the supplementary angles exceeds the smaller by 180. Find the angles
Answer : Let x be larger angle and y be smear angle
Then, x + y = 180⁰ ------------ (1)
x-y = 18⁰ -------------- (2)
Solving (1) and (2) , we get x = 99⁰ and y = 81⁰

Question. A fraction becomes 1/3 when 2 is subtracted from the numerator and it becomes 1/2 when 1 is subtracte from its denominator. Find the fraction.
Answer : Let the fraction be xy
Then, x − 2/y = 1/3 ⟹ 3x − y……………….(1)
x/y − 1 = 1/2 ⟹ 2x − y = − 1………………….(2)
Solving ,we get x = 7 and y = 15
Required fraction is 7/15

Question. For each of the following system of equations determine the values of k for which the given system has no solution
3𝑥 − 4𝑦 + 7 = 0
𝑘𝑥 + 3𝑦 − 5 = 0
Answer : Here a1 = 3, b₁ = -4, c₁ = 7
a₂ = k, b₂ = 3, c₂ = -5
For no solution, we must have a₁a₂ = b₁b₂≠c₁c₂
We have b₁b₂ = − 43 and c₁c₂ = − 75
Clearly, b₁b₂ ≠ c₁c₂ .So the given system will have no solution. a₁/a₂ = b₁/b₂ ⟹ 3/k = − 4/3
⟹ k = − 9/4

Question. Solve for 𝑥 𝑎𝑛𝑑 𝑦 by method of elimination:
47𝑥 + 31𝑦 = 63
31𝑥 + 47𝑦 = 15
Answer : x = 2 and y = -1

Question. The monthly incomes of A and B are in the ratio 5:4 and their expenditure are in the ratio 7:5. If each save 3000/- per month, find the monthly income of each.
Answer : By the given conditions
5x-7y = 3000 …………….(1)
4x-5y = 3000 ……………(2)
Solving, we get x = 2000 /-
Monthly income of A = 5x = 5×2000 = 10000/-
Monthly income of B = 4x = 4 × 2000 = 8000/-

Question. A man has only 20paisa coins and 25 paisa coins in his purse. If he has 50 coins in all totalling 11.25/-, how many coins of each kind does he have?
Answer : Let no. of 20 paisa coins be x and that of 25 paisa coins be y, then
x + y = 50 ……………(i)
20 x + 25y = 1125 ⟹ 4x + 5y = 225 ……….. (ii)
Solving, we get x = 25 and y = 25
Hence there are 24 points of each kind

LINEAR EQUATIONS IN TWO VARIABLES

Q1. Draw the graph of the equation 2x + y =7. From the graph :
a) Find whether the point (3,4) lies on the graph.
b) Find whether x=3 , y = 1 is a solution of the equation .
c) Find the value of x , when y = 1 .
d) Find the point where the equation meets the x-axis .

Q2. Draw the graph using the followig table :
x 0 1 2 b
y 1 3 a -3
From the graph, find the values of 'a' and 'b' .

Q3. Solve the following system of linear equations graphically
a) 2x + 3y = 12 b) 3x - 4y - 12 = 0
2y - 1 = x x + 2y - 4 = 0

Q4. Draw the graph of the system of equations x+y=5 and 2x -y +2 =0 . Shade the region bounded by these lines and the x- axis . Find the area of the shaded region .

Q5. Solve graphically the system
2x - 3y =1
3x - 4y =1
Does the point (3,2) lie on any of the line ? Write its equation .

Q6. Draw the graphs of 2x - y = 1 and x + 2y = 13 . Find the coordinates of the vertices of the triangle formed by the two lines and the y-axis ?

Q7. By comparing the ratios a1/a² , b1/b2 and c1/c2 , find out for what value (s) of α , the lines representing the following equations have a unique solution , no solution or infinitely many solution 
αx + 3y = α - 3
12x + αy = α

Q8. Determine the value of k so that the following pairs of equations are inconsistent
(3k + 1) x + 3y -2 = 0
(k² + 1)x + (k - 2)y - 5 = 0

Q9. Given below are three linear equations . Two of them have infinitely many solutions and two have a unique solution . State the pairs :
4x -5y =3 , 8x - 10y = 6 , 5x - 4y = 5

Q10. Solve the following pair of linear equations :
a) x/6 + y/4 = 1 , 3x/4 - (x-y)/2 = 7/4
b) (a + 2b)x + (2a - b)y = 2 (a - 2b)x + ( 2a + b)y =
c) (a - b)x + (a + b)y = a² - 2ab - b² (a + b) ( x+y) = a² + b²
d) ax/b - by/a = a + b ax - by = 2ab
e) 5/( x+1) - 2/( y- 1) = 1/2 10/( x+1) + 2/( y-1) = 5/2
f) √7x + √11y = 0 √3x - √5y = 0
g) mx - ny = m² + n² x - y = 2n
h) xy/( x + y) = 6/5 xy/( y - x) = 6 {( x +y) ≠ 0 , (y - x) ≠ 0}
i) x/a - y/b = (a- b) x/a² - y/b² = 0
j) b²x/a - a²y/b = ab( a+ b) b²x - a²y = 2a²b²

ANSWERS:-

Ans. 7. Unique sol. : α≠6 or -6 , No solution : α= -6 , Infinitely : α = 6

Ans. 8. k= -1

Ans. 10. a) x = 3 , y = 2
b) x = (5b-2a)/ 10ab , y = (a+10b)/ 10ab
c) x = a + b , y = -2ab/ (a + b)
d) x = b , y = -a
e) x = 4 , y = 5
f) x = 0 , y = 0
g) x = m + n , y = m - n
h) x = 2 , y = 3
i) x = a² , y = b²
j) x = a² , y = -b²

1. Solve the following pair of linear equations by the substitution method or by elimination method.
(i) x + y = 14, x-y = 4
(ii) 3x – y = 3, 9x – 3y = 9

2. Solve the following questions by cross-multiplication method
0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3

3. Solve:
67 x + 112y = - 89
112x + 67 y = -269

4. Five years hence, the age of Jacob will be three times that of his son. \ Five years ago, Jacob’s age was seven times that of his son. What aretheir present ages?

5. 5. A fraction becomes 9/11, if 2 is added to both numerator and denominator. If, 3 is added to the numerator and denominator, then itbecomes 5/6. Find the fraction

6. Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx + 3.

7. 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz. Form the pair of linear equations for the above problem, and find its solution graphically

8. Comparing a1/a2, b1/b2, find out if the lines representing the following pair of linear equations, are intersecting, parallel or coincident.
5x – 4y + 8 = 0, 7x +6y – 9 = 0

9. Given the linear equation 2x + 3y – 8 = 0, write another linear equation in two variables such that the geometrical representation of the pair so formed is:

1. Solve graphically the system of linear equations: x + 3y = 11, 3x + 2y = 12 (3, 5)

2. Draw the graph of the equation x – y + 1 = 0 and 3x + 2y – 12 = 0. Determine the coordinates of the vertices of the triangle Formed by these lines and the x – axis, and shade the triangular region (-1, 0), (2, 3) and (4, 0) 3. Solve:
a) x y = 1 , x y = 1 x + y 5 x – y 7 (x = -1, y = 1/6)
b) 37x + 43y = 123, 43x + 37y = 117 (x = 1, y = 2)
c) (a – b) x + (a + b) y = a² – 2ab – b² , (a + b) x + (a + b) y = a² + b² (x = a +b, y = - 2ab/a +b)

4. Find the value(s) of k for which the pair of linear equations k x + 3y = k – 2 and 12x + k y = k has no solution (k = ± 6)

5. Find the value of k, for which the pair of equations 3x + 5y = 0, k x + 10y = 0, has a non zero solution (k = 6)

6. Find the value of a and b for which the system of equation has infinitely many solutions:
a) 2x + 3y = 7, (a – b) x + (a + b) y = 3a +b – 2 (5, 1)

7. Find the value of k, for which the given linear pair has a unique solution: 2x + 3y – 5 = 0, k x – 6y -8 = 0 (k ≠ -4)

8. 10 students of class x took part in mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and number of girls who took part in the quiz. (3, 7)

9. The larger of the two supplementary angles exceeds the smaller by 18 degrees. Find the angles (99˚, 81˚)

10. In a two digit number, the sum of the digits is 9. If the digits are reversed, the number is increased by 9. Find the number (45)

11. A fraction becomes 4/5 if 1 is added to both the numerator and the denominator. However, if 5 is subtracted from both the numerator and the denominator the fraction becomes ½. Find the fraction (7/9)

12. Two years ago, a father was five times as old as his son. Two years later, his age will be 8 more than three times the age of the

son. Find the present ages of father and son (10, 42)

13 90% and 97% pure acid solutions and mixed to obtain 21 litres of 95% pure acid solution. Find the amount of each Type of acid to be mixed to form the mixture (x=6, y=15)

14. 2 women and 5 men can together finished a piece of work in 4 days, while 3 women and 6 men can finis h it in 3 days. Find the time taken by 1 woman alone to finish the work, and that taken by 1 man alone. (18, 36)

15. A boat goes 16km upstream and 24km downstream in 6hrs. It can go 12km up and 36km down in the same time. Find the speed of the boat in still water and the speed of the stream. (8, 4)

16. Students of a class are made to stand in rows. If 4 students are extra in a row , their would be 2 rows less. If 4 students are less in a row, there would be 4 more rows. Find the number of students

17. The perimeter of a rectangle is 44 cm. If its length is increased by 4 cm and its breadth is increased by 2cm, its area is increased by 72 sqcm. Find the dimensions of the rectangle.

18. The sum of two numbers is 1000 and the difference between their squares is 256000. Find the numbers (266,744)

19. If (x + 2) is a factor of x³ + ax² + 4bx + 12 and a + b = - 4, find the values of a and b (-3, -1)

20. Two numbers are in the ratio 3: 4 and if 4 are added to each, the ratio becomes 4:5. Find the numbers (12, 16)

21. Solve by the method of cross multiplication:
(a – b) x + (a + b) y = a² – 2ab – b2, (a + b)(x+y) = a² + b² (a + b, -2ab/a +b)

22. 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 expenditures. (Rs1800, Rs1400)

23. Sum of the areas of two squares is 468m².If the difference of their perimeter is 24m, find the sides of two square

24. A boy travels for x hrs at 8km/hr and then for y hrs at 7km/hr. If he goes 37km altogether in 5hrs, find x and y (2, 3)

25. Places A and B are 100km 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 hrs. If they travel towards each other they meet in 1 hour. What are the Speeds of the two cars (60km/hr, 40km/hr)

Question 01. Draw the graphs of \(2x + y = 6\) and \(2x - y + 2 = 0\). Shade the triangles bounded by these lines and the a} \(x\)-axis, b} \(y\)-axis. Also find the area of these triangles.
Answer: Let us first find the intercepts and intersection points of the two lines:
Line 1: \( 2x + y = 6 \)
- For \( x = 0 \), \( y = 6 \implies (0,6) \)
- For \( y = 0 \), \( 2x = 6 \implies x = 3 \implies (3,0) \)
Line 2: \( 2x - y + 2 = 0 \)
- For \( x = 0 \), \( -y = -2 \implies y = 2 \implies (0,2) \)
- For \( y = 0 \), \( 2x = -2 \implies x = -1 \implies (-1,0) \)
Solving the two equations together:
\( (2x + y) + (2x - y) = 6 - 2 \implies 4x = 4 \implies x = 1 \).
Substituting \( x = 1 \) in \( 2x + y = 6 \), we get \( y = 4 \).
The lines intersect at the point \( P(1, 4) \).

(a) For the triangle bounded by the lines and the \(x\)-axis:
The vertices of the base are \( A(-1, 0) \) and \( B(3, 0) \), so the base length is \( 3 - (-1) = 4 \) units.
The height is the \(y\)-coordinate of the intersection point \( P \), which is 4 units.
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 4 = 8 \text{ sq. units} \).

(b) For the triangle bounded by the lines and the \(y\)-axis:
The vertices of the base are \( C(0, 2) \) and \( D(0, 6) \), so the base length is \( 6 - 2 = 4 \) units.
The height is the \(x\)-coordinate of the intersection point \( P \), which is 1 unit.
\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 1 = 2 \text{ sq. units} \).
In simple words: Find where the lines cross the coordinate axes and each other. Use these boundary points as base coordinates and the intersection point \( (1,4) \) to calculate the heights of the triangles.

Exam Tip: Be sure to sketch the graph clearly and shade the exact region defined by the question constraints to secure full marks.

Question 02. Solve the following pair of equations:-
a) \( \frac{5}{x+y} + \frac{2}{x-y} = 3 \), \( \frac{15}{x+y} - \frac{4}{x-y} = -1 \)
b) \( 99x + 101y = 499 \), \( 101x + 99y = 501 \)
c) \( \frac{2}{3x+2y} + \frac{3}{3x-2y} = \frac{17}{5} \), \( \frac{5}{3x+2y} + \frac{1}{3x-2y} = 2 \)
d) \( \frac{2}{\sqrt{x}} + \frac{3}{\sqrt{y}} = 2 \), \( \frac{4}{\sqrt{x}} - \frac{9}{\sqrt{y}} = -1 \)
e) \( 152x - 378y = -74 \), \( -378x + 152y = -604 \)
Answer: Let us solve each part step-by-step:
a) Substituting \( \frac{1}{x+y} = u \) and \( \frac{1}{x-y} = v \):
\( 5u + 2v = 3 \) and \( 15u - 4v = -1 \).
Multiply the first equation by 2 and add: \( 25u = 5 \implies u = \frac{1}{5} \).
Substituting \( u \), we find \( v = 1 \).
This gives: \( x + y = 5 \) and \( x - y = 1 \).
Solving these gives \( x = 3, y = 2 \).

b) Adding the equations gives: \( 200x + 200y = 1000 \implies x + y = 5 \).
Subtracting the equations gives: \( 2x - 2y = 2 \implies x - y = 1 \).
Solving \( x + y = 5 \) and \( x - y = 1 \) yields \( x = 3, y = 2 \).

c) Substituting \( \frac{1}{3x+2y} = u \) and \( \frac{1}{3x-2y} = v \):
\( 2u + 3v = \frac{17}{5} \) and \( 5u + v = 2 \implies v = 2 - 5u \).
Substituting \( v \) into the first equation: \( 2u + 3(2 - 5u) = \frac{17}{5} \implies u = \frac{1}{5} \).
This gives \( v = 1 \).
So, \( 3x + 2y = 5 \) and \( 3x - 2y = 1 \).
Solving these linear equations yields \( x = 1, y = 1 \).

d) Substituting \( \frac{1}{\sqrt{x}} = u \) and \( \frac{1}{\sqrt{y}} = v \):
\( 2u + 3v = 2 \) and \( 4u - 9v = -1 \).
Multiplying the first equation by 3 and adding: \( 10u = 5 \implies u = \frac{1}{2} \).
This gives \( v = \frac{1}{3} \).
So, \( \sqrt{x} = 2 \implies x = 4 \) and \( \sqrt{y} = 3 \implies y = 9 \).

e) Adding the symmetric equations: \( -226x - 226y = -678 \implies x + y = 3 \).
Subtracting the second equation from the first: \( 530x - 530y = 530 \implies x - y = 1 \).
Solving \( x + y = 3 \) and \( x - y = 1 \) yields \( x = 2, y = 1 \).
In simple words: For equations with variables in the denominators or square roots, use temporary variables like \( u \) and \( v \) to simplify. For equations with large symmetric coefficients, add and subtract them first to reduce the numbers.

Exam Tip: Always make sure to perform final substitutions back to the original variables \( x \) and \( y \) to complete the problem.

Question 03. Solve for x and y:-
a) \( x + y = a + b \), \( ax - by = a^2 - b^2 \)
b) \( \frac{x}{a} + \frac{y}{b} = 2 \), \( ax - by = a^2 - b^2 \)
Answer: Let us solve each equation systematically:
a) Multiply the first equation \( x + y = a + b \) by \( b \):
\( bx + by = ab + b^2 \).
Add this to the second equation \( ax - by = a^2 - b^2 \):
\( (a + b)x = a^2 + ab \implies (a + b)x = a(a + b) \implies x = a \) (since \( a+b \neq 0 \)).
Substituting \( x = a \) into \( x + y = a + b \), we find \( y = b \).

b) Simplify the first fractional equation by taking LCM:
\( bx + ay = 2ab \implies y = \frac{2ab - bx}{a} \).
Substitute this \( y \) into \( ax - by = a^2 - b^2 \):
\( ax - b\left(\frac{2ab - bx}{a}\right) = a^2 - b^2 \implies a^2x - 2ab^2 + b^2x = a^3 - ab^2 \) (multiplying by \( a \)).
\( \implies (a^2 + b^2)x = a^3 + ab^2 \implies (a^2 + b^2)x = a(a^2 + b^2) \implies x = a \).
Substituting \( x = a \) into \( bx + ay = 2ab \) gives \( ab + ay = 2ab \implies ay = ab \implies y = b \).
In simple words: Use algebraic elimination to cancel out the \( y \)-terms, which simplifies the calculations to reveal the values \( x = a \) and \( y = b \).

Exam Tip: Be careful with brackets and negative signs when distributing variables like \( a \) and \( b \) during substitution.

Question 04. A boat can go 20 km upstream and 30 km downstream in 3 hours. It can go 20 km downstream and 10 km upstream in \(1 \frac{2}{3}\) hours. Find the speed of the boat in still water and the speed of the stream.
Answer: Let the speed of the boat in still water be \( u \) km/h and the speed of the stream be \( v \) km/h.
Upstream speed \( = u - v \) and downstream speed \( = u + v \).
Let \( \frac{1}{u-v} = x \) and \( \frac{1}{u+v} = y \).
First condition:
\( 20x + 30y = 3 \) - (Equation 1)
Second condition:
\( 10x + 20y = \frac{5}{3} \) - (Equation 2)
Multiplying Equation 2 by 2:
\( 20x + 40y = \frac{10}{3} \) - (Equation 3)
Subtracting Equation 1 from Equation 3:
\( 10y = \frac{10}{3} - 3 = \frac{1}{3} \implies y = \frac{1}{30} \).
Substituting \( y = \frac{1}{30} \) in Equation 1:
\( 20x + 30\left(\frac{1}{30}\right) = 3 \implies 20x + 1 = 3 \implies 20x = 2 \implies x = \frac{1}{10} \).
This gives:
\( u - v = 10 \) and \( u + v = 30 \).
Solving these:
\( 2u = 40 \implies u = 20 \) km/h, and \( 2v = 20 \implies v = 10 \) km/h.
Thus, the speed of the boat in still water is 20 km/h, and the speed of the stream is 10 km/h.
In simple words: Write equations for the total time in both cases based on the upstream and downstream speeds. Solving these reveals the boat travels at 20 km/h and the stream flows at 10 km/h.

Exam Tip: Upstream speed is always less than downstream speed, so \( u - v \) must always be smaller than \( u + v \).

Question 05. A man sold a chair and a table together for Rs. 12000 gaining 10% on the chair and 15% on the table. Had he sold them for Rs. 11625, the profit % would have interchanged. Find the C.P. of each of them.
Answer: Let the cost price of the chair be Rs. \( x \) and the table be Rs. \( y \).
In the first case (10% profit on chair and 15% profit on table):
\( 1.10x + 1.15y = 12000 \implies 110x + 115y = 1200000 \implies 22x + 23y = 240000 \) - (Equation 1)
In the second case (15% profit on chair and 10% profit on table):
\( 1.15x + 1.10y = 11625 \implies 115x + 110y = 1162500 \implies 23x + 22y = 232500 \) - (Equation 2)
Adding Equation 1 and Equation 2:
\( 45x + 45y = 472500 \implies x + y = 10500 \) - (Equation 3)
Subtracting Equation 2 from Equation 1:
\( -x + y = 7500 \implies y - x = 7500 \) - (Equation 4)
Adding Equation 3 and Equation 4:
\( 2y = 18000 \implies y = 9000 \) (Cost price of the table).
Substituting \( y \) into Equation 3:
\( x = 10500 - 9000 = 1500 \) (Cost price of the chair).
Therefore, the cost price of the chair is Rs. 1500 and that of the table is Rs. 9000.
In simple words: Write equations for the total selling prices in both cases. Solving these shows that the chair costs Rs. 1500 and the table costs Rs. 9000.

Exam Tip: Interchange questions are highly symmetric. Always add and subtract the equations to easily find the values of \( x+y \) and \( y-x \).

Question 06. In cyclic quadrilateral ABCD, \( \angle A = (2x + 4)^\circ \), \( \angle B = (y + 3)^\circ \), \( \angle C = (2y + 10)^\circ \) and \( \angle D = (4x - 5)^\circ \). Find the angles.
Answer: In a cyclic quadrilateral, opposite angles are supplementary (their sum is \( 180^\circ \)):
\( \angle A + \angle C = 180^\circ \implies 2x + 4 + 2y + 10 = 180 \implies 2x + 2y = 166 \implies x + y = 83 \) - (Equation 1)
\( \angle B + \angle D = 180^\circ \implies y + 3 + 4x - 5 = 180 \implies 4x + y = 182 \) - (Equation 2)
Subtracting Equation 1 from Equation 2:
\( 3x = 99 \implies x = 33 \).
Substituting \( x = 33 \) in Equation 1:
\( 33 + y = 83 \implies y = 50 \).
Now we can find the individual angles:
\( \angle A = 2(33) + 4 = 70^\circ \)
\( \angle B = 50 + 3 = 53^\circ \)
\( \angle C = 2(50) + 10 = 110^\circ \)
\( \angle D = 4(33) - 5 = 127^\circ \)
The angles are \( \angle A = 70^\circ \), \( \angle B = 53^\circ \), \( \angle C = 110^\circ \), and \( \angle D = 127^\circ \).
In simple words: Since opposite corners in a cyclic quadrilateral sum up to \( 180^\circ \), form two linear equations. Solving them yields \( x = 33 \) and \( y = 50 \), which gives the corner angles.

Exam Tip: Be sure to write the final values with the degree symbol \( (^\circ) \) as they represent angle measurements.

Question 07. Students of a class are made to stand in rows. If one student extra in a row, there would have been two rows less. If one student less in a row, there would have been 3 rows more. Find the number of students in the class.
Answer: Let the number of rows be \( r \) and the number of students in each row be \( s \).
Total number of students \( = r \times s \).
According to the first condition:
\( (s + 1)(r - 2) = rs \implies rs - 2s + r - 2 = rs \implies r - 2s = 2 \) - (Equation 1)
According to the second condition:
\( (s - 1)(r + 3) = rs \implies rs + 3s - r - 3 = rs \implies 3s - r = 3 \implies -r + 3s = 3 \) - (Equation 2)
Adding Equation 1 and Equation 2:
\( s = 5 \) students per row.
Substituting \( s = 5 \) into Equation 1:
\( r - 2(5) = 2 \implies r = 12 \) rows.
Total number of students \( = r \times s = 12 \times 5 = 60 \).
In simple words: Create equations based on the variations in rows and columns. Solving these gives 12 rows with 5 students each, making 60 students in total.

Exam Tip: The total number of students \( r \times s \) remains constant, so setting up the equations in terms of \( rs \) is crucial for cancelling it out.

Question 08. A says to B "if you give me 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.
According to A's condition:
\( x + 30 = 2(y - 30) \implies x + 30 = 2y - 60 \implies x - 2y = -90 \) - (Equation 1)
According to B's condition:
\( y + 10 = 3(x - 10) \implies y + 10 = 3x - 30 \implies 3x - y = 40 \implies y = 3x - 40 \) - (Equation 2)
Substituting Equation 2 into Equation 1:
\( x - 2(3x - 40) = -90 \implies x - 6x + 80 = -90 \implies -5x = -170 \implies x = 34 \).
Substituting \( x = 34 \) into Equation 2:
\( y = 3(34) - 40 = 102 - 40 = 62 \).
So, A has 34 mangoes and B has 62 mangoes.
In simple words: Write equations for the mango transfers between A and B. Solving them shows that A starts with 34 mangoes and B with 62.

Exam Tip: Always read the "left with you" part carefully to write the terms like \( y-30 \) and \( x-10 \) accurately.

Question 09. A man when asked how many hens and cows he has, he told that his animals have 120 eyes and 180 legs. How many animals of each kind has he?
Answer: Let the number of hens be \( x \) and the number of cows be \( y \).
Since each animal has 2 eyes, the total number of eyes is:
\( 2x + 2y = 120 \implies x + y = 60 \implies y = 60 - x \) - (Equation 1)
Hens have 2 legs and cows have 4 legs, so the total number of legs is:
\( 2x + 4y = 180 \implies x + 2y = 90 \) - (Equation 2)
Substituting Equation 1 into Equation 2:
\( x + 2(60 - x) = 90 \implies x + 120 - 2x = 90 \implies -x = -30 \implies x = 30 \).
This gives:
\( y = 60 - 30 = 30 \).
Thus, the man has 30 hens and 30 cows.
In simple words: Formulate linear equations based on the count of eyes and legs. Solving this shows there are 30 hens and 30 cows.

Exam Tip: Simplify the eye-count equation immediately by dividing both sides by 2 to keep calculations simple.

Question 10. 90% and 97% pure acid solution are mixed to obtain 21 liters 95% pure solution. How many liters of each solution are needed?
Answer: Let \( x \) liters of the 90% solution and \( y \) liters of the 97% solution be mixed.
The total volume of the mixture is:
\( x + y = 21 \implies y = 21 - x \) - (Equation 1)
The total amount of pure acid in the mixture is:
\( 0.90x + 0.97y = 0.95(21) \implies 90x + 97y = 1995 \) - (Equation 2) (multiplying by 100).
Substituting Equation 1 into Equation 2:
\( 90x + 97(21 - x) = 1995 \)
\( \implies 90x + 2037 - 97x = 1995 \)
\( \implies -7x = -42 \implies x = 6 \).
Substituting \( x = 6 \) in Equation 1:
\( y = 21 - 6 = 15 \).
So, 6 liters of the 90% solution and 15 liters of the 97% solution are required.
In simple words: Write equations for the total volume and total pure acid. Solving the system shows we need 6 liters of the 90% acid and 15 liters of the 97% acid.

Exam Tip: Multiplying chemical percentage equations by 100 clears the decimals and prevents arithmetic errors.