CBSE Class 10 Pair of Linear Equations in Two Variables Sure Shot Questions Set 03

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Solve each of the following system of linear equations graphically:

1. x+2y=3;         4x+3y=2.

2. 2x+3y= 8;       x-2y+3=0. 

3. x +2y+2=0;     3x+2y- 2 = 0.

4. 4x+3y=5;        2y-x=7.

5. 2x-3y=1;         3x-4y=1.

6. 2x+3y=4;        3x-y=-5.

7. x-y+1=0;         3x+2y-12=0.

8. 3x+2y=4;         2x-3y=7.

9. 2x+3y=2;         x-27=8.

10. 2x-5y+4=0;    2x+y-8=0.

11. 3x+y+1=0;     2x-3y+8=0.

12. Solve the following system of linear equations graphically:

2x- 3y- 17 =0;  4x+y-13=0 . Shade the region bounded by the above lines and x-axis.

13. Solve the following system of linear equations graphically: 2x+ 3y= 4;   3x-y=-5. Shade the region bounded by the above lines and y-axis.

14. Solve the following system of linear equations graphically: 4x-y= 4;    3x+ 2y=14.  Shade the region bounded by the above lines and y-axis.

15. Solve the following system of linear equations graphically: x+2=5;   2x-3y=-4. Shade the region bounded by the above lines and y-axis.

16. Draw the graphs of the equations 4x-y-8= 0;    2x-3y+6=0 . Also determine the vertices of the triangle formed by the lines and x-axis.

17. Solve the following system of linear equations graphically: 2x-y= 1;    x- y =-1 . Shade the region bounded by the above lines and y-axis.

18. Solve the following system of linear equations graphically: 5x-y= 7;     x- y+1=0 . Calculate the area bounded by these lines and y-axis. 

Question. Graphically, the pair of equations 6x – 3y + 10 = 0 and 2x – y + 9 = 0 represents two lines which are:
(A) Intersecting at exactly one point.
(B) Intersecting at exactly two points.
(C) Coincident.
(D) Parallel
Answer: (D)
Explanation: Here
\[ \frac{a_1}{a_2} = \frac{6}{2} = \frac{3}{1} \]
\[ \frac{b_1}{b_2} = \frac{-3}{-1} = \frac{3}{1} \]
\[ \frac{c_1}{c_2} = \frac{10}{9} \]
This implies
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
Therefore, lines are parallel.

Question. The pair of equations x + 2y – 5 = 0 and −3x – 6y + 15 = 0 have:
(A) A unique solution
(B) Exactly two solutions
(C) Infinitely many solutions
(D) No solution
Answer: (C)
Explanation: Here,
\[ \frac{a_1}{a_2} = \frac{1}{-3} \]
\[ \frac{b_1}{b_2} = \frac{2}{-6} = \frac{1}{-3} \]
\[ \frac{c_1}{c_2} = \frac{-5}{15} = \frac{-1}{3} \]
This implies
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Therefore, the pair of equations has infinitely many solutions.

Question. If a pair of linear equations is consistent, then the lines will be:
(A) Parallel
(B) Always coincident
(C) Intersecting or coincident
(D) Always intersecting
Answer: (C)
Explanation: If a pair of linear equations is consistent the two lines represented by these equations definitely have a solution, this implies that either lines are intersecting or coincident.

Question. The pair of equations y = 0 and y = –7 has
(A) One solution
(B) Two solutions
(C) Infinitely many solutions
(D) No solution
Answer: (D)
Explanation: The graph of equations will be parallel lines. So the equations have no solution.

Question. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then the value of k is
(A) 5/4
(B) 2/5
(C) 15/4
(D) 3/2
Answer: (C)
Explanation: For parallel lines
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
\[ \Rightarrow \frac{3}{2} = \frac{2k}{5} \]
\[ \Rightarrow k = \frac{15}{4} \]

Question. The value of c for which the pair of equations cx – y = 2 and 6x – 2y = 3 will have infinitely many solutions is
(A) 3
(B) – 3
(C) –12
(D) no value
Answer: (A)
Explanation: For infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
\[ \Rightarrow \frac{c}{6} = \frac{-1}{-2} \]
\[ \Rightarrow c = 3 \]

Question. One equation of a pair of dependent linear equations is –5x + 7y – 2 = 0. The second equation can be
(A) 10x + 14y + 4 = 0
(B) –10x – 14y + 4 = 0
(C) –10x + 14y + 4 = 0
(D) 10x – 14y = –4
Answer: (D)
Explanation: For dependent pair, the two lines must have.
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
For option (D)
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} = \frac{-1}{2} \]

Question. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Then the numbers are:
(A) 40, 42
(B) 42, 48
(C) 40, 48
(D) 44, 50
Answer: (C)
Explanation: According to given information.
\[ \frac{x}{y} = \frac{5}{6} \]
\[ \Rightarrow 6x - 5y = 0 \text{ .....(i)} \]
\[ \frac{x - 8}{y - 8} = \frac{4}{5} \]
\[ \Rightarrow 5x - 40 = 4y - 32 \]
\[ \Rightarrow 5x - 4y = 8 \text{ .....(ii)} \]
From equation (i)
\( 6x - 5y = 0 \)
\( \Rightarrow x = \frac{5y}{6} \)
Substituting in ...(2)
\( 5x - 4y = 8 \)
\( \Rightarrow 5 \left( \frac{5y}{6} \right) - 4y = 8 \)
\( \Rightarrow \frac{y}{6} = 8 \)
\( \Rightarrow y = 48 \)
Therefore
\( x = \frac{5y}{6} \)
\( \Rightarrow x = \frac{5 \times 48}{6} \)
\( \Rightarrow x = 40 \)

Question. The solution of the equations x – y = 2 and x + y = 4 is:
(A) 3 and 5
(B) 5 and 3
(C) 3 and 1
(D) –1 and –3
Answer: (C)
Explanation: Adding both equations, we have:
\( x - y = 2 \)
\( x + y = 4 \)
\( 2x = 6 \)
\( \Rightarrow x = 3 \)
This implies
\( x - y = 2 \)
\( \Rightarrow 3 - y = 2 \)
\( \Rightarrow y = 1 \)

Question. For which values of a and b, will the following pair of linear equations have infinitely many solutions?
x + 2y = 1
(a – b)x + (a + b)y = a + b – 2
(A) a = 2 and b = 1
(B) a = 2 and b = 2
(C) a = -3 and b = 1
(D) a = 3 and b = 1
Answer: (D)
Explanation: For infinitely many solutions:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
\[ \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} = \frac{1}{a + b - 2} \]
\[ \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} \]
\[ \Rightarrow a + b = 2a - 2b \]
\[ \Rightarrow a - 3b = 0 \text{ .....(i)} \]
And
\[ \frac{2}{a + b} = \frac{1}{a + b - 2} \]
\[ \Rightarrow 2a + 2b - 4 = a + b \]
\[ \Rightarrow a + b = 4 \text{ .....(ii)} \]
Solving equation (i) and (ii), we get a = 3 and b = 1.

Question. The father’s age is six times his son’s age. Four years hence, the age of the father will be four times his son’s age. The present ages, in years, of the son and the father are, respectively.
(A) 4 and 24
(B) 5 and 30
(C) 6 and 36
(D) 3 and 24
Answer: (C)
Explanation: Let the age of father be x and of son is y.
Then according to question,
x = 6y …..(i)
Four years hence age of son will be y + 4 and age of father will be x + 4
Then according to question,
x + 4 = 4 (y + 4)
x – 4y = 12 …..(ii)
Solving equations (i) and (ii) we get:
y = 6 and x = 36

Question. Rakshita has only Rs. 1 and Rs. 2 coins with her. If the total number of coins that she has is 50 and the amount of money with her is Rs 75, then the number of Rs.1 and Rs.2 coins is, respectively
(A) 35 and 15
(B) 35 and 20
(C) 15 and 35
(D) 25 and 25
Answer: (D)
Explanation: Let her number of Rs.1 coins are x
Let the number of Rs.2 coins are y
Then
By the given conditions
x + y = 50 …..(i)
1 × x + 2 × y = 75
⇒ x + 2y = 75 …..(ii)
Solving equations (i) and (ii) we get:
(x + 2y) – (x + y) = 75 – 50
⇒ y = 25
Therefore, x = 50 – 25 = 25
So the number of coins are 25, 25 each.

Question. In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
(A) 100
(B) 95
(C) 90
(D) 60
Answer: (A)
Explanation: Let x be the number of correct answers of the questions in a competitive exam.
Then, 120 − x be the number of wrong answers
Then by given condition
\[ 1 \times x - (120 - x) \times \frac{1}{2} = 90 \]
\[ \Rightarrow x - 60 + \frac{x}{2} = 90 \]
\[ \Rightarrow \frac{3x}{2} = 150 \]
\[ \Rightarrow x = \frac{150 \times 2}{3} \]
\[ \Rightarrow x = 100 \]

Question. The angles of a cyclic quadrilateral ABCD are:
\( \angle A = (6x+10)^\circ, \angle B = (5x)^\circ \)
\( \angle C = (x+y)^\circ, \angle D = (3y-10)^\circ \)
Then value of x and y are:
(A) x = 20° and y = 30°
(B) x = 40° and y = 10°
(C) x = 44° and y = 15°
(D) x = 15° and y = 15°
Answer: (A)
Explanation: In cyclic quadrilateral, sum of opposite angles is 180°
Therefore
6x + 10 + x + y = 180
⇒ 7x + y = 170 …..(i)
5x + 3y – 10 = 180
⇒ 5x + 3y = 190 …..(ii)
Multiplying equations (i) and (ii), we get:
x = 20° and y = 30°

Question. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days, and an additional charge for each day thereafter. Reema paid Rs. 22 for a book kept for six days, while Ruchika paid Rs 16 for the book kept for four days, then the charge for each extra day is:
(A) Rs 5
(B) Rs 4
(C) Rs 3
(D) Rs.2
Answer: (C)
Explanation: Let Rs. x be the fixed charge and Rs. y be the charge for each extra day.
Then by the given conditions
x + 4y = 22 …..(i)
x + 2y = 16 …..(ii)
Subtracting equation (ii) from (i), we get:
y = Rs. 3

Very Short Answer Type Questions

Question. Do the \( x = 2y \) and \( y = 2x \) pair of linear equations have no solution? Justify your answer.
Answer: Sol. Here, \( x = 2y \) and \( y = 2x \)
\( \Rightarrow x - 2y = 0 \) and \( 2x - y = 0 \)
Here, \( a_1 = 1, b_1 = -2, c_1 = 0 \) and \( a_2 = 2, b_2 = -1, c_2 = 0 \)
Now, \( \frac{a_1}{a_2} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{-2}{-1} = \frac{2}{1} \)
Here, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
So, the given pair of linear equations has unique solution.

Question. Do the following equations represent a pair of coincident lines? Justify your answer. \( \frac{x}{2} + y + \frac{2}{5} = 0 \) and \( 4x + 8y + \frac{5}{16} = 0 \).
Answer: Sol. Given: \( \frac{x}{2} + y + \frac{2}{5} = 0 \) and \( 4x + 8y + \frac{5}{16} = 0 \)
On comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \),
We have:
\( a_1 = \frac{1}{2}, b_1 = 1, c_1 = \frac{2}{5} \) and \( a_2 = 4, b_2 = 8, c_2 = \frac{5}{16} \).
Here, \( \frac{a_1}{a_2} = \frac{1/2}{4} = \frac{1}{8}, \frac{b_1}{b_2} = \frac{1}{8}, \frac{c_1}{c_2} = \frac{2/5}{5/16} = \frac{32}{25} \)
\( \Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Therefore, the given pair of linear equation has no solution and lines are parallel.

Question. Are the following pair of linear equations consistent? Justify your answer.
(i) \( \frac{3}{5}x - y = \frac{1}{2} \) and \( \frac{1}{5}x - 3y = \frac{1}{6} \)
(ii) \( 2ax + by = a \) and \( 4ax + 2by - 2a = 0; a, b \neq 0 \)
Answer: Sol. (i) Yes, \( \frac{3}{5}x - y - \frac{1}{2} = 0 \) and \( \frac{1}{5}x - 3y - \frac{1}{6} = 0 \) ...[Given
Here, \( a_1 = \frac{3}{5}, b_1 = -1, c_1 = -\frac{1}{2} \) and \( a_2 = \frac{1}{5}, b_2 = -3, c_2 = -\frac{1}{6} \)
Now, \( \frac{a_1}{a_2} = \frac{3/5}{1/5} = 3, \frac{b_1}{b_2} = \frac{-1}{-3} = \frac{1}{3}, \frac{c_1}{c_2} = \frac{-1/2}{-1/6} = 3 \)
\( \Rightarrow \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Therefore, the given pair of linear equations has unique solution or is consistent.
(ii) Yes, \( 2ax + by - a = 0 \) and \( 4ax + 2by - 2a = 0; a, b \neq 0 \)
Here, \( a_1 = 2a, b_1 = b, c_1 = -a \) and \( a_2 = 4a, b_2 = 2b, c_2 = -2a \)
Now, \( \frac{a_1}{a_2} = \frac{2a}{4a} = \frac{1}{2}, \frac{b_1}{b_2} = \frac{b}{2b} = \frac{1}{2}, \frac{c_1}{c_2} = \frac{-a}{-2a} = \frac{1}{2} \)
\( \Rightarrow \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \)
Therefore, the given pair of linear equations has infinitely many solutions or is consistent.

Short Answer Type Questions

Question. For which value(s) of \( \lambda \), do the pair of linear equations \( \lambda x + y = \lambda^2 \) and \( x + \lambda y = 1 \) have
(i) no solution?
(ii) infinitely many solutions?
(iii) a unique solution?
Answer: Sol. Given: \( \lambda x + y - \lambda^2 = 0 \) and \( x + \lambda y - 1 = 0 \) ...(i)
Here, \( a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2 \) and \( a_2 = 1, b_2 = \lambda, c_2 = -1 \)
(i) For no solution:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \Rightarrow \frac{\lambda}{1} = \frac{1}{\lambda} \neq \frac{-\lambda^2}{-1} \)
\( \lambda^2 - 1 = 0 \Rightarrow \lambda^2 = 1 \Rightarrow \lambda = \pm 1 \)
Ignore \( \lambda = 1 \), because for which system of equations will have infinitely many solutions. Therefore, \( \lambda = -1 \).
(ii) For infinitely many solutions:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{\lambda}{1} = \frac{1}{\lambda} = \frac{\lambda^2}{1} \Rightarrow \frac{\lambda}{1} = \frac{\lambda^2}{1} \Rightarrow \lambda(\lambda - 1) = 0 \)
When \( \lambda \neq 0 \), then \( \lambda = 1 \).
(iii) For a unique solution:
\( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \Rightarrow \frac{\lambda}{1} \neq \frac{1}{\lambda} \Rightarrow \lambda^2 \neq 1 \Rightarrow \lambda \neq \pm 1 \)
Therefore, for all real values of \( \lambda \) except \( \pm 1 \), equation will have unique solution.

Question. For which value (s) of \( k \) will the pair of equations \( kx + 3y = k - 3, 12x + ky = k \) have no solution?
Answer: Sol. Given: \( kx + 3y - (k - 3) = 0 \) and \( 12x + ky - k = 0 \)
On comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) we have
Here, \( a_1 = k, b_1 = 3 \) and \( c_1 = -(k - 3) \)
\( a_2 = 12, b_2 = k \) and \( c_2 = -k \) ...(i)
Condition for a pair of linear equations to have no solution is:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) ...(ii)
Using (i) and (ii), we get:
\( \frac{k}{12} = \frac{3}{k} \neq \frac{-(k - 3)}{-k} \Rightarrow \frac{k}{12} = \frac{3}{k} \Rightarrow k^2 = 36 \) or \( k = \pm 6 \) ...(iii)
Again, from (ii) we have: \( \frac{3}{k} \neq \frac{k - 3}{k} \Rightarrow 3k \neq k(k - 3) \Rightarrow 3k - k(k - 3) \neq 0 \Rightarrow k(3 - k + 3) \neq 0 \Rightarrow k(6 - k) \neq 0 \)
\( \Rightarrow k \neq 0 \) and \( k \neq 6 \) ...(iv)
Thus, from (iii) and (iv), it’s clear that at \( k = -6 \) the given pair of linear equations will have no solution.

Question. For which values of \( a \) and \( b \) will the following pair of linear equations has infinitely many solutions?
\( x + 2y = 1 \) and \( (a - b)x + (a + b)y = a + b - 2 \)
Answer: Sol. Given equation are: \( x + 2y - 1 = 0 \) and \( (a - b)x + (a + b)y - (a + b - 2) = 0 \) ...(i)
On comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \) we have:
\( a_1 = 1, b_1 = 2, c_1 = -1 \) and \( a_2 = (a - b), b_2 = (a + b), c_2 = -(a + b - 2) \)
Condition for infinitely many solutions is:
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) ...(ii)
\( \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} = \frac{-1}{-(a + b - 2)} \Rightarrow \frac{1}{a - b} = \frac{2}{a + b} \)
\( \Rightarrow a + b = 2a - 2b \Rightarrow a = 3b \) ...(iii)
And \( \frac{2}{a + b} = \frac{1}{a + b - 2} \Rightarrow 2a + 2b - 4 = a + b \Rightarrow a + b = 4 \) ...(iv)
Using (iii) in (iv): \( 3b + b = 4 \Rightarrow 4b = 4 \Rightarrow b = 1 \)
Putting the value of \( b \) in (iii) we get \( a = 3(1) = 3 \).
Hence, \( a = 3, b = 1 \).

Question. If \( 2x + y = 23 \) and \( 4x - y = 19 \), then find the values of \( 5y - 2x \) and \( y/x - 2 \).
Answer: Sol. Given pair of linear equations is:
\( 2x + y = 23 \) ...(i)
and \( 4x - y = 19 \) ...(ii)
On adding (i) and (ii), we get: \( 6x = 42 \Rightarrow x = 7 \)
Putting the value of \( x \) in equation (i), we get: \( 2(7) + y = 23 \Rightarrow 14 + y = 23 \Rightarrow y = 9 \)
Thus, \( 5y - 2x = 5 \times 9 - 2 \times 7 = 45 - 14 = 31 \)
and \( \frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = -\frac{5}{7} \)
Therefore, the value of \( (5y - 2x) \) and \( (\frac{y}{x} - 2) \) are \( 31 \) and \( -\frac{5}{7} \) respectively.

Question. Solve the following pairs of equations:
(i) \( x + y = 3.3, \frac{0.6}{3x - 2y} = -1, 3x - 2y \neq 0 \)
(ii) \( 43x + 67y = -24, 67x + 43y = 24 \)
(iii) \( \frac{x}{a} + \frac{y}{b} = a + b, \frac{x}{a^2} + \frac{y}{b^2} = 2, a, b \neq 0 \)
(iv) \( \frac{2xy}{x + y} = \frac{3}{2}, \frac{xy}{2x - y} = \frac{-3}{10}, x + y \neq 0, 2x - y \neq 0 \)
Answer: Sol. (i) \( x + y = 3.3 \) ...(i)
\( \frac{0.6}{3x - 2y} = -1 \Rightarrow 0.6 = -3x + 2y \) or \( 3x - 2y = -0.6 \) ...(ii)
Multiplying (i) by 2, and (ii) by 1, we get
\( 2x + 2y = 6.6 \)
\( 3x - 2y = -0.6 \)
Adding: \( 5x = 6 \Rightarrow x = \frac{6}{5} = 1.2 \)
Putting the value of \( x \) in equation (i), we get \( 1.2 + y = 3.3 \Rightarrow y = 3.3 - 1.2 \Rightarrow y = 2.1 \)
Hence, \( x = 1.2 \) and \( y = 2.1 \).
(ii) \( 43x + 67y = -24 \) ...(i)
\( 67x + 43y = 24 \) ...(ii)
Multiplying (ii) by 67 and (i) by 43, we get
\( (67)^2x + 43 \times 67y = 24 \times 67 \)
\( (43)^2x + 43 \times 67y = -24 \times 43 \)
Subtracting:
\( \{(67)^2 - (43)^2\}x = 24(67 + 43) \Rightarrow [(67 + 43)(67 - 43)]x = 24 \times 110 \)
\( \Rightarrow 110 \times 24x = 24 \times 110 \Rightarrow x = 1 \)
Putting the value of \( x \) in equation (i), we get \( 43 \times 1 + 67y = -24 \Rightarrow 67y = -24 - 43 \Rightarrow 67y = -67 \Rightarrow y = -1 \)
Hence, \( x = 1, y = -1 \).
(iii) \( \frac{x}{a} + \frac{y}{b} = a + b \) ...(i) and \( \frac{x}{a^2} + \frac{y}{b^2} = 2 \) ...(ii)
Multiplying (i) by \( \frac{1}{a} \) and (ii) by 1, we get
\( \frac{x}{a^2} + \frac{y}{ab} = 1 + \frac{b}{a} \) and \( \frac{x}{a^2} + \frac{y}{b^2} = 2 \)
Subtracting: \( y(\frac{1}{b^2} - \frac{1}{ab}) = 2 - (1 + \frac{b}{a}) \Rightarrow y(\frac{a - b}{ab^2}) = 1 - \frac{b}{a} = \frac{a - b}{a} \)
\( \Rightarrow y = \frac{ab^2}{a} = b^2 \)
Putting \( y = b^2 \) in (ii), we get \( \frac{x}{a^2} + \frac{b^2}{b^2} = 2 \Rightarrow \frac{x}{a^2} + 1 = 2 \Rightarrow \frac{x}{a^2} = 1 \Rightarrow x = a^2 \).
(iv) \( \frac{2xy}{x + y} = \frac{3}{2} \Rightarrow \frac{x + y}{xy} = \frac{4}{3} \Rightarrow \frac{1}{y} + \frac{1}{x} = \frac{4}{3} \) ...(A)
\( \frac{xy}{2x - y} = -\frac{3}{10} \Rightarrow \frac{2x - y}{xy} = -\frac{10}{3} \Rightarrow \frac{2}{y} - \frac{1}{x} = -\frac{10}{3} \) ...(B)
Let \( \frac{1}{x} = u, \frac{1}{y} = v \). Then \( v + u = \frac{4}{3} \) and \( 2v - u = -\frac{10}{3} \).
Adding: \( 3v = \frac{4 - 10}{3} = -2 \Rightarrow v = -\frac{2}{3} \).
Then \( -\frac{2}{3} + u = \frac{4}{3} \Rightarrow u = 2 \).
So \( x = \frac{1}{2}, y = -\frac{3}{2} \).

Question. Find the solution of the pair of equations \( \frac{x}{10} + \frac{y}{5} - 1 = 0 \) and \( \frac{x}{8} + \frac{y}{6} = 15 \) and find \( \lambda \), if \( y = \lambda x + 5 \).
Answer: Sol. Given: \( \frac{x + 2y}{10} = 1 \Rightarrow x + 2y = 10 \) ...(i)
And \( \frac{6x + 8y}{48} = 15 \Rightarrow 6x + 8y = 720 \Rightarrow 3x + 4y = 360 \) ...(ii)
Multiplying (i) by 2, we get \( 2x + 4y = 20 \). Subtracting from (ii):
\( (3x + 4y) - (2x + 4y) = 360 - 20 \Rightarrow x = 340 \).
Putting \( x = 340 \) in (i): \( 340 + 2y = 10 \Rightarrow 2y = -330 \Rightarrow y = -165 \).
Now, \( y = \lambda x + 5 \Rightarrow -165 = \lambda(340) + 5 \Rightarrow 340\lambda = -170 \Rightarrow \lambda = -1/2 \).
Hence, \( x = 340, y = -165 \) and \( \lambda = -1/2 \).

Question. If \( (x + 1) \) is a factor of \( 2x^3 + ax^2 + 2bx + 1 \), then find the value of \( a \) and \( b \) given that \( 2a - 3b = 4 \).
Answer: Sol. Let \( f(x) = 2x^3 + ax^2 + 2bx + 1 \). Since \( (x + 1) \) is a factor, \( f(-1) = 0 \).
\( 2(-1)^3 + a(-1)^2 + 2b(-1) + 1 = 0 \Rightarrow -2 + a - 2b + 1 = 0 \Rightarrow a - 2b - 1 = 0 \) ...(i)
Given: \( 2a - 3b = 4 \Rightarrow b = \frac{2a - 4}{3} \).
Putting value of \( b \) in (i): \( a - 2(\frac{2a - 4}{3}) - 1 = 0 \Rightarrow 3a - 4a + 8 - 3 = 0 \Rightarrow -a + 5 = 0 \Rightarrow a = 5 \).
Then \( 5 - 2b - 1 = 0 \Rightarrow 2b = 4 \Rightarrow b = 2 \).
Hence, \( a = 5, b = 2 \).

Question. The age of the father is twice the sum of the ages of his two children. After 20 year, his age will be equal to the sum of the ages of his children. Find the age of the father.
Answer: Sol. Let the present age of father \( = x \) years and sum of the ages of his two children \( = y \) years.
Case I: \( x = 2y \) ...(i)
After 20 years: Father's age \( = x + 20 \), Sum of children's ages \( = y + 40 \).
Case II: \( x + 20 = y + 40 \Rightarrow y = x - 20 \) ...(ii)
Putting (ii) in (i): \( x = 2(x - 20) \Rightarrow x = 2x - 40 \Rightarrow x = 40 \).
Hence, father's age is 40 years.

Question. A shopkeeper gives books on rent for reading. She takes a fixed charge for the first two days and an additional charge for each day thereafter. Latika paid Rs 22 for a book kept for six days, while Anand paid Rs 16 for the book kept for four days. Find the fixed charges and the charge for each extra day.
Answer: Sol. Let fixed charge for first two days \( = Rs x \) and additional charge for each extra day \( = Rs y \).
Case I: \( x + 4y = 22 \) ...(i)
Case II: \( x + 2y = 16 \) ...(ii)
Subtracting (ii) from (i): \( 2y = 6 \Rightarrow y = 3 \).
Putting \( y = 3 \) in (ii): \( x + 2(3) = 16 \Rightarrow x = 10 \).
So, fixed charges \( = Rs 10 \) and additional charges \( = Rs 3 \) per day.

Question. In a competitive examination, 1 mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Answer: Sol. Let the no. of questions attempted correctly \( = x \). No. of questions answered \( = 120 \).
So, wrong answer attempted \( = (120 - x) \).
Given: \( x \times 1 - (120 - x) \times \frac{1}{2} = 90 \)
\( x - 60 + \frac{x}{2} = 90 \Rightarrow \frac{3x}{2} = 150 \Rightarrow x = \frac{150 \times 2}{3} = 100 \).
Hence, Jayanti answered 100 questions correctly.

Long Answer Type Questions

Question. Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour, if she travels 2 km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 min longer. Find the speed of the rickshaw and of the bus.
Answer: Sol. Let speed of rickshaw \( = x \) km/h and bus \( = y \) km/h.
Case I: \( \frac{2}{x} + \frac{12}{y} = \frac{1}{2} \) ...(i)
Case II: \( \frac{4}{x} + \frac{10}{y} = \frac{1}{2} + \frac{9}{60} = \frac{1}{2} + \frac{3}{20} = \frac{13}{20} \) ...(ii)
Let \( \frac{1}{x} = u, \frac{1}{y} = v \).
\( 2u + 12v = \frac{1}{2} \Rightarrow 4u + 24v = 1 \)
\( 4u + 10v = \frac{13}{20} \)
Subtracting: \( 14v = 1 - \frac{13}{20} = \frac{7}{20} \Rightarrow v = \frac{7}{20 \times 14} = \frac{1}{40} \).
Then \( 2u + 12(\frac{1}{40}) = \frac{1}{2} \Rightarrow 2u = \frac{1}{2} - \frac{3}{10} = \frac{2}{10} \Rightarrow u = \frac{1}{10} \).
So \( x = 10, y = 40 \). Rickshaw speed \( = 10 \) km/h, Bus speed \( = 40 \) km/h.

Question. A person, rowing at the rate of 5 km/h in still water, takes thrice as much time in going 40 km upstream as in going 40 km downstream. Find the speed of the stream.
Answer: Sol. Let speed of stream \( = x \) km/h. Person's speed \( = 5 \) km/h.
Downstream speed \( = (5 + x) \) km/h, Upstream speed \( = (5 - x) \) km/h.
\( t_1 = \frac{40}{5 + x} \), \( t_2 = \frac{40}{5 - x} \).
Given \( t_2 = 3t_1 \Rightarrow \frac{40}{5 - x} = 3(\frac{40}{5 + x}) \Rightarrow 5 + x = 15 - 3x \Rightarrow 4x = 10 \Rightarrow x = 2.5 \) km/h.

Question. A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.
Answer: Sol. Let boat speed \( = x \), stream speed \( = y \).
Case I: \( \frac{30}{x - y} + \frac{28}{x + y} = 7 \). Case II: \( \frac{21}{x - y} + \frac{21}{x + y} = 5 \Rightarrow \frac{1}{x - y} + \frac{1}{x + y} = \frac{5}{21} \).
Let \( \frac{1}{x - y} = p, \frac{1}{x + y} = q \).
\( 30p + 28q = 7 \) and \( p + q = \frac{5}{21} \Rightarrow 28p + 28q = \frac{140}{21} = \frac{20}{3} \).
Subtracting: \( 2p = 7 - \frac{20}{3} = \frac{1}{3} \Rightarrow p = \frac{1}{6} \).
Then \( \frac{1}{6} + q = \frac{5}{21} \Rightarrow q = \frac{5}{21} - \frac{1}{6} = \frac{10 - 7}{42} = \frac{3}{42} = \frac{1}{14} \).
\( x - y = 6, x + y = 14 \Rightarrow 2x = 20 \Rightarrow x = 10 \). Then \( y = 4 \).
Boat speed \( = 10 \) km/hr, Stream speed \( = 4 \) km/hr.

Question. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.
Answer: Sol. Let the number be \( 10x + y \).
Case I: \( 8(x + y) - 5 = 10x + y \Rightarrow 8x + 8y - 5 = 10x + y \Rightarrow 2x - 7y = -5 \) ...(i)
Case II: \( 16(x - y) + 3 = 10x + y \Rightarrow 16x - 16y + 3 = 10x + y \Rightarrow 6x - 17y = -3 \) ...(ii)
Multiplying (i) by 3: \( 6x - 21y = -15 \). Subtracting (ii): \( 4y = 12 \Rightarrow y = 3 \).
Then \( 2x - 7(3) = -5 \Rightarrow 2x = 16 \Rightarrow x = 8 \). Number \( = 83 \).

Question. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from the stations A to B costs Rs 2530. Also, one reserved first class ticket and one reserved first class half ticket from stations A to B costs Rs 3810. Find the full first class fare from station A to B and also the reservation charges for a ticket.
Answer: Sol. Let full fare \( = Rs x \), reservation charge \( = Rs y \). Half fare \( = Rs x/2 \).
Case I: \( x + y = 2530 \) ...(i)
Case II: \( (x + y) + (x/2 + y) = 3810 \Rightarrow \frac{3x}{2} + 2y = 3810 \Rightarrow 3x + 4y = 7620 \) ...(ii)
Multiplying (i) by 4: \( 4x + 4y = 10120 \). Subtracting (ii): \( x = 2500 \).
Then \( 2500 + y = 2530 \Rightarrow y = 30 \).
Full fare \( = Rs 2500 \), Reservation charge \( = Rs 30 \).

Question. A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby, getting a sum Rs 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs 1028 then find the cost of the saree and the list price (price before discount) of the sweater.
Answer: Sol. Let CP of saree \( = Rs x \), list price of sweater \( = Rs y \).
Case I: \( 1.08x + 0.9y = 1008 \Rightarrow y = \frac{1008 - 1.08x}{0.9} \) ...(i)
Case II: \( 1.1x + 0.92y = 1028 \) ...(ii)
Substituting \( y \) in (ii): \( 1.1x + 0.92(\frac{1008 - 1.08x}{0.9}) = 1028 \).
Solving: \( 0.99x + 927.36 - 0.9936x = 925.2 \Rightarrow -0.0036x = -2.16 \Rightarrow x = 600 \).
Then \( y = \frac{1008 - 1.08(600)}{0.9} = 400 \).
CP of saree \( = Rs 600 \), List price of sweater \( = Rs 400 \).

Question. Susan invested certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs 20 more as annual interest. How much money did she invest in each scheme?
Answer: Sol. Let investment in Scheme A \( = Rs x \) and Scheme B \( = Rs y \).
Case I: \( \frac{8x}{100} + \frac{9y}{100} = 1860 \Rightarrow 8x + 9y = 186000 \) ...(i)
Case II: \( \frac{9x}{100} + \frac{8y}{100} = 1860 + 20 = 1880 \Rightarrow 9x + 8y = 188000 \) ...(ii)
Multiplying (i) by 9 and (ii) by 8: \( 72x + 81y = 1674000 \), \( 72x + 64y = 1504000 \).
Subtracting: \( 17y = 170000 \Rightarrow y = 10000 \).
Then \( 8x + 90000 = 186000 \Rightarrow 8x = 96000 \Rightarrow x = 12000 \).
Susan invested Rs 12000 in Scheme A and Rs 10000 in Scheme B.

Question. Vijay had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Rs 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Rs 1 per banana and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.
Answer: Sol. Let bananas in Lot A \( = x \) and Lot B \( = y \).
Case I: \( \frac{2}{3}x + y = 400 \Rightarrow 2x + 3y = 1200 \) ...(i)
Case II: \( x + \frac{4}{5}y = 460 \Rightarrow 5x + 4y = 2300 \) ...(ii)
Multiplying (i) by 4 and (ii) by 3: \( 8x + 12y = 4800 \), \( 15x + 12y = 6900 \).
Subtracting: \( -7x = -2100 \Rightarrow x = 300 \).
Then \( 2(300) + 3y = 1200 \Rightarrow 3y = 600 \Rightarrow y = 200 \).
Total bananas \( = x + y = 300 + 200 = 500 \).

Please click the link below to download CBSE Class 10 Pair of Linear Equations Sure Shot Questions Set C.