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

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PRACTICE QUESTIONS

CLASS X : CHAPTER - 3

PAIR OF LINEAR EQUATIONS IN TWO VARIABLES

GRAPHICAL QUESTIONS

Solve each of the following system of linear equations graphically: [3 Marks]

1. x + 2y = 3; 4x + 3y = 2

Answer:
1. For equation x + 2y = 3, when x = 1, y = 1 and when x = -1, y = 2.
2. For equation 4x + 3y = 2, when x = 2, y = -2 and when x = -1, y = 2.
3. The two lines intersect at the point (-1, 2).
4. Hence, the solution is x = -1 and y = 2.

Teacher's Note:
a) Plot at least three coordinate points for each linear equation to ensure accuracy.
b) A common student error is interchanging x and y coordinates when plotting.
c) The intersection point of the two lines represents the unique solution of the system.

2. 2x + 3y = 8; x - 2y + 3 = 0 [3 Marks]

Answer:
1. For equation 2x + 3y = 8, when x = 1, y = 2 and when x = 4, y = 0.
2. For equation x - 2y + 3 = 0, when x = 1, y = 2 and when x = -3, y = 0.
3. The two lines intersect at the point (1, 2).
4. Hence, the solution is x = 1 and y = 2.

Teacher's Note:
a) Rearrange equations into standard form before finding coordinate points.
b) Students often make sign errors while transferring terms across the equals sign.
c) Scale selection on graph paper must be uniform and clearly labeled.

3. x + y + 2 = 0; 3x + 2y - 2 = 0 [3 Marks]

Answer:
1. For equation x + y + 2 = 0, when x = 0, y = -2 and when x = -2, y = 0.
2. For equation 3x + 2y - 2 = 0, when x = 2, y = -2 and when x = 0, y = 1.
3. The two lines intersect at the point (-6, 10).
4. Hence, the solution is x = -6 and y = 10.

Teacher'sNote:
a) Consistent equations yield intersecting lines with a unique solution.
b) Watch out for fractional coordinate values during calculation.
c) Always verify the intersection point by substituting back into both equations.

4. 4x + 3y = 5; 2y - x = 7 [3 Marks]

Answer:
1. For equation 4x + 3y = 5, when x = 2, y = -1 and when x = -1, y = 3.
2. For equation 2y - x = 7, when x = 1, y = 4 and when x = -3, y = 2.
3. The two lines intersect at the point (-1, 3).
4. Hence, the solution is x = -1 and y = 3.

Teacher's Note:
a) Rewrite the second equation as x - 2y + 7 = 0 to avoid plotting confusion.
b) Ensure proper arrowheads are drawn on the coordinate axes.
c) Label both intersecting lines clearly with their respective equations.

5. 2x - 3y = 1; 3x - 4y = 1 [3 Marks]

Answer:
1. For equation 2x - 3y = 1, when x = 2, y = 1 and when x = 5, y = 3.
2. For equation 3x - 4y = 1, when x = 3, y = 2 and when x = 7, y = 5.
3. The two lines intersect at the point (1, 1).
4. Hence, the solution is x = 1 and y = 1.

Teacher's Note:
a) Check parallel or intersecting nature using coefficient ratios before plotting.
b) A common mistake is misplotting negative intercepts on axes.
c) Write the final coordinates of the solution clearly below the graph.

6. 2x + 3y = 4; 3x - y = -5 [3 Marks]

Answer:
1. For equation 2x + 3y = 4, when x = 2, y = 0 and when x = -1, y = 2.
2. For equation 3x - y = -5, when x = -1, y = 2 and when x = 0, y = 5.
3. The two lines intersect at the point (-1, 2).
4. Hence, the solution is x = -1 and y = 2.

Teacher's Note:
a) Finding common points directly from table values saves plotting time.
b) Pay close attention to negative signs in coefficients.
c) Use a sharp pencil and ruler for accurate graphical representation.

7. x - y + 1 = 0; 3x + 2y - 12 = 0 [3 Marks]

Answer:
1. For equation x - y + 1 = 0, when x = 0, y = 1 and when x = 2, y = 3.
2. For equation 3x + 2y - 12 = 0, when x = 0, y = 6 and when x = 4, y = 0.
3. The two lines intersect at the point (2, 3).
4. Hence, the solution is x = 2 and y = 3.

Teacher's Note:
a) Intercept form makes finding axis-crossing points very quick.
b) Students frequently confuse x-intercepts with y-intercepts.
c) The graphical method visually reinforces algebraic substitution methods.

8. 3x + 2y = 4; 2x - 3y = 7 [3 Marks]

Answer:
1. For equation 3x + 2y = 4, when x = 0, y = 2 and when x = 2, y = -1.
2. For equation 2x - 3y = 7, when x = 2, y = -1 and when x = 5, y = 1.
3. The two lines intersect at the point (2, -1).
4. Hence, the solution is x = 2 and y = -1.

Teacher's Note:
a) Notice how a common point (2, -1) appears in both tables.
b) Care must be taken when plotting points in the fourth quadrant.
c) Always shade or mark the intersection clearly to highlight the answer.

9. 2x + 3y = 2; x - 2y = 8 [3 Marks]

Answer:
1. For equation 2x + 3y = 2, when x = 1, y = 0 and when x = -2, y = 2.
2. For equation x - 2y = 8, when x = 8, y = 0 and when x = 4, y = -2.
3. The two lines intersect at the point (4, -2).
4. Hence, the solution is x = 4 and y = -2.

Teacher's Note:
a) Choose convenient integer values for x to calculate clean y-coordinates.
b) Avoid choosing points too close together as it reduces drawing accuracy.
c) Extend lines sufficiently far across the grid to locate the intersection.

10. 2x - 5y + 4 = 0; 2x + y - 8 = 0 [3 Marks]

Answer:
1. For equation 2x - 5y + 4 = 0, when x = 0, y = \(\frac{4}{5}\) and when x = 3, y = 2.
2. For equation 2x + y - 8 = 0, when x = 0, y = 8 and when x = 3, y = 2.
3. The two lines intersect at the point (3, 2).
4. Hence, the solution is x = 3 and y = 2.

Teacher's Note:
a) Select points that yield integer values to simplify plotting.
b) Watch out for decimal approximations when fractions cannot be avoided.
c) Both lines sharing (3, 2) confirms it as the exact solution.

11. 3x + y + 1 = 0; 2x - 3y + 8 = 0 [3 Marks]

Answer:
1. For equation 3x + y + 1 = 0, when x = 0, y = -1 and when x = -1, y = 2.
2. For equation 2x - 3y + 8 = 0, when x = -1, y = 2 and when x = 2, y = 4.
3. The two lines intersect at the point (-1, 2).
4. Hence, the solution is x = -1 and y = 2.

Teacher's Note:
a) Rearranging terms carefully prevents arithmetic errors.
b) Verify the coordinates by plugging x = -1 and y = 2 into both equations.
c) Consistent labeling of axes is essential for scoring full marks.

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. [4 Marks]

Answer:
1. For equation 2x - 3y - 17 = 0, when x = 4, y = -3 and when x = 7, y = -1.
2. For equation 4x + y - 13 = 0, when x = 3, y = 1 and when x = 4, y = -3.
3. The two lines intersect at the point (4, -3).
4. The lines meet the x-axis at (8.5, 0) and (3.25, 0); the region between these points and (4, -3) is shaded.

Teacher's Note:
a) Identifying the correct boundary limits (x-axis and intersection lines) is crucial.
b) Students often shade the wrong region by including the y-axis instead of the x-axis.
c) Clearly mark the vertices of the bounded region on the graph.
d) Cross-check intersection coordinates using algebraic elimination.

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. [4 Marks]

Answer:
1. For equation 2x + 3y = 4, when x = 2, y = 0 and when x = -1, y = 2.
2. For equation 3x - y = -5, when x = -1, y = 2 and when x = 0, y = 5.
3. The two lines intersect at the point (-1, 2).
4. The lines meet the y-axis at (0, \(\frac{4}{3}\)) and (0, 5); the region between these intercepts and (-1, 2) is shaded.

Teacher's Note:
a) Bounded regions involving axes require accurate intercept calculations.
b) Ensure shading is neat and restricted strictly to the specified boundaries.
c) Label all corner points of the shaded polygon clearly.
d) Double-check that the y-axis forms one of the boundary sides.

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. [4 Marks]

Answer:
1. For equation 4x - y = 4, when x = 1, y = 0 and when x = 0, y = -4.
2. For equation 3x + 2y = 14, when x = 2, y = 4 and when x = 4, y = 1.
3. The two lines intersect at the point (2, 4).
4. The lines meet the y-axis at (0, -4) and (0, 7); the region bounded by these intercepts and (2, 4) along the y-axis is shaded.

Teacher's Note:
a) Plotting intercepts carefully ensures the enclosed polygon shape is correct.
b) A common error is extending the shaded region beyond the axis boundaries.
c) State the coordinates of all vertices forming the shaded region.
d) Check the solution point (2, 4) by substituting into both given equations.

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

Answer:
1. For equation x + 2y = 5, when x = 1, y = 2 and when x = 5, y = 0.
2. For equation 2x - 3y = -4, when x = 1, y = 2 and when x = 4, y = 4.
3. The two lines intersect at the point (1, 2).
4. The lines meet the y-axis at (0, 2.5) and (0, \(\frac{4}{3}\)); the region bounded by these points and (1, 2) along the y-axis is shaded.

Teacher's Note:
a) Intersection point (1, 2) directly serves as one vertex of the shaded region.
b) Pay attention to whether the bounding axis is the x-axis or y-axis.
c) Use shading pencils lightly so that graph grid lines remain visible.
d) Verify all table coordinate points before drawing lines.

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. [4 Marks]

Answer:
1. For equation 4x - y - 8 = 0, when x = 2, y = 0 and when x = 0, y = -8.
2. For equation 2x - 3y + 6 = 0, when x = 0, y = 2 and when x = -3, y = 0.
3. The two lines intersect at the point (3, 4).
4. The lines meet the x-axis at (2, 0) and (-3, 0); hence the vertices of the triangle are (3, 4), (2, 0), and (-3, 0).

Teacher's Note:
a) Identifying triangle vertices requires finding both line intersection and x-intercepts.
b) Students often forget to list the vertex coordinates explicitly.
c) Ensure the triangle is fully enclosed by the two lines and the x-axis.
d) Double-check calculations for the intersection point (3, 4).

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. [4 Marks]

Answer:
1. For equation 2x - y = 1, when x = 1, y = 1 and when x = 0, y = -1.
2. For equation x - y = -1, when x = 0, y = 1 and when x = 2, y = 3.
3. The two lines intersect at the point (2, 3).
4. The lines meet the y-axis at (0, -1) and (0, 1); the region bounded by these y-intercepts and the intersection point (2, 3) is shaded.

Teacher's Note:
a) Simple integer coordinates make plotting swift and reliable.
b) Confirm that the y-axis forms the third enclosing side of the shaded figure.
c) Clearly write down the vertices enclosing the shaded area.
d) Verify the solution (2, 3) algebraically to ensure graphical accuracy.

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. [5 Marks]

Answer:
1. For equation 5x - y = 7, when x = 1, y = -2 and when x = 2, y = 3.
2. For equation x - y + 1 = 0, when x = 0, y = 1 and when x = 2, y = 3.
3. The two lines intersect at the point (2, 3).
4. The lines meet the y-axis at (0, -7) and (0, 1), forming a triangle with base along the y-axis of length 8 units.
5. Area = \(\frac{1}{2}\) × base × height = \(\frac{1}{2}\) × 8 × 2 = 8 square units.

Teacher's Note:
a) Area calculation requires identifying the base length along the axis and the perpendicular height (x-coordinate of intersection).
b) A common mistake is using incorrect units or forgetting square units in the final answer.
c) Base is the distance between y-intercepts (-7 and 1), which equals 8.
d) Height corresponds to the x-coordinate of the intersection point (2, 3), which is 2.

19. Solve the following system of linear equations graphically: 4x - 3y + 4 = 0; 4x + 3y - 20 = 0. Calculate the area bounded by these lines and the x-axis.

Answer:
1. Line 4x - 3y + 4 = 0 meets the x-axis at (-1, 0).
2. Line 4x + 3y - 20 = 0 meets the x-axis at (5, 0).
3. The two lines intersect at (2, 4).
4. The triangle has vertices (-1, 0), (5, 0) and (2, 4).
5. Base = 5 - (-1) = 6 units; height = 4 units.
6. Area = ½ × 6 × 4 = 12 square units.

Teacher's Note:
a) The intersection point is found by adding the two equations: 8x - 16 = 0, so x = 2 and y = 4.

20. Solve the following system of linear equations graphically: 4x - 5y - 20 = 0; 3x + 5y - 15 = 0. Find the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Answer:
1. Line 4x - 5y - 20 = 0 meets the y-axis at (0, -4).
2. Line 3x + 5y - 15 = 0 meets the y-axis at (0, 3).
3. The two lines intersect at (5, 0).
4. The vertices of the triangle are (5, 0), (0, -4) and (0, 3).

Teacher's Note:
a) Check: (5, 0) satisfies both equations, since 4(5) - 20 = 0 and 3(5) - 15 = 0.

21. Solve the following system of linear equations graphically: 2x - 5y + 4 = 0; 2x + y - 8 = 0. Find the points where these lines meet the y-axis.

Answer:
1. Line 2x - 5y + 4 = 0 meets the y-axis at (0, 4/5), that is (0, 0.8).
2. Line 2x + y - 8 = 0 meets the y-axis at (0, 8).

Teacher's Note:
a) For the y-axis, put x = 0 in each equation and solve for y.

22. Solve the following system of linear equations graphically: 2x + y - 5 = 0; x + y - 3 = 0. Find the points where these lines meet the y-axis.

Answer:
1. Line 2x + y - 5 = 0 meets the y-axis at (0, 5).
2. Line x + y - 3 = 0 meets the y-axis at (0, 3).

Teacher's Note:
a) Put x = 0 in each equation: y = 5 and y = 3.

23. Solve the following system of linear equations graphically: 4x - 5y + 16 = 0; 2x + y - 6 = 0. Find the coordinates of the vertices of the triangle formed by these lines and the y-axis.

Answer:
1. Line 4x - 5y + 16 = 0 meets the y-axis at (0, 16/5), that is (0, 3.2).
2. Line 2x + y - 6 = 0 meets the y-axis at (0, 6).
3. The two lines intersect at (1, 4).
4. The vertices of the triangle are (0, 3.2), (0, 6) and (1, 4).

Teacher's Note:
a) The intersection is found by substituting y = 6 - 2x into 4x - 5y + 16 = 0, which gives x = 1 and y = 4.

24. Draw the graphs of the equations 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.

Answer:
1. Line x - y + 1 = 0 meets the x-axis at (-1, 0).
2. Line 3x + 2y - 12 = 0 meets the x-axis at (4, 0).
3. The two lines intersect at (2, 3).
4. The vertices of the triangle are (-1, 0), (4, 0) and (2, 3). Shade the region enclosed by these three points.

Teacher's Note:
a) For the graph, plot the x-intercepts and the intersection point, then join them with straight lines.

25. Solve the following system of linear equations graphically: 3x + y - 11 = 0; x - y - 1 = 0. Shade the region bounded by these lines and the y-axis. Find the points where these lines cut the y-axis.

Answer:
1. Line 3x + y - 11 = 0 cuts the y-axis at (0, 11).
2. Line x - y - 1 = 0 cuts the y-axis at (0, -1).
3. The two lines intersect at (3, 2).
4. The region bounded by these lines and the y-axis is the triangle with vertices (0, 11), (0, -1) and (3, 2).

Teacher's Note:
a) Check: (3, 2) satisfies both equations, since 3(3) + 2 - 11 = 0 and 3 - 2 - 1 = 0.

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 \( = x + y = 300 + 200 = 500 \).