CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Worksheet Set 13

Short Answer Questions-II

Question. In figure, ABCDE is a pentagon with \( BE \parallel CD \) and \( BC \parallel DE \). BC is perpendicular to CD. \( AB = 5 \) cm, \( AE = 5 \) cm, \( BE = 7 \) cm, \( BC = x - y \) and \( CD = x + y \). If the perimeter of ABCDE is 27 cm. Find the value of x and y, given \( x, y \neq 0 \).
Answer: \( x + y = 7 \) and \( 2(x - y) + x + y + 5 + 5 = 27 \)
\(\therefore\) \( x + y = 7 \) and \( 3x - y = 17 \)
Solving, we get, \( x = 6 \) and \( y = 1 \).  

Question. If \( 2x + y = 23 \) and \( 4x - y = 19 \), find the value of \( (5y - 2x) \) and \( \left( \frac{y}{x} - 2 \right) \). 
Answer: Given equations are
\( 2x + y = 23 \) ... (i)
\( 4x - y = 19 \) ... (ii)
Adding given pair of linear equations, we get
\( 2x + y = 23 \)
\( 4x - y = 19 \)
-----------------
\( 6x = 42 \)
\( \implies \) \( x = \frac{42}{6} = 7 \)
\(\therefore\) \( x = 7 \)
Putting the value of \( x = 7 \) in eq. (i), we get
\( 2 \times 7 + y = 23 \)
\( \implies \) \( 14 + y = 23 \)
\( \implies \) \( y = 23 - 14 = 9 \)
\( \implies \) \( y = 9 \)
\(\therefore\) \( x = 7, y = 9 \)
Now, \( 5y - 2x = 5 \times 9 - 2 \times 7 = 45 - 14 = 31 \)
and, \( \frac{y}{x} - 2 = \frac{9}{7} - 2 = \frac{9 - 14}{7} = \frac{-5}{7} \)

Question. The present age of a father is three years more than three times the age of his son. Three years hence the father's age will be 10 years more than twice the age of the son. Determine their present ages.  
Answer: Let father's present age be x years and his son's present age be y years.
According to question,
\( x = 3y + 3 \)
\( \implies \) \( x - 3y = 3 \) ... (i)
After three years,
Father's age will be \( (x + 3) \) years and
his son's age will be \( (y + 3) \) years.
Again, according to question
\( \implies \) \( x + 3 = 2(y + 3) + 10 \)
\( \implies \) \( x + 3 = 2y + 6 + 10 \)
\( \implies \) \( x - 2y = 13 \) ... (ii)
Subtracting (ii) from (i), we get
\( x - 3y = 3 \)
\( x - 2y = 13 \)
- + -
-----------------
\( -y = -10 \)
\( \implies \) \( y = 10 \)
Putting the value of \( y = 10 \) in equation (i), we get
\( x - 3 \times 10 = 3 \)
\( \implies \) \( x = 33 \)
\(\therefore\) Present age of father is 33 years and his son's present age is 10 years.

Question. Taxi charges in a city consist of fixed charges and the remaining charges depend upon the distance travelled. For a journey of 10 km, the charge paid is Rs. 75 and for a journey of 15 km, the charge paid is Rs. 110. Find the fixed charge and charges per km. Hence, find the charge of covering a distance of 35 km. 
Answer: Let Rs. x be the fixed charge and Rs. y be the charge per km.
Therefore, charge paid for 10 km \( = x + 10y \)
\( \implies \) \( 75 = x + 10y \)
\( \implies \) \( x + 10y = 75 \) ... (i)
Also, charge paid for 15 km \( = x + 15y \)
\( \implies \) \( x + 15y = 110 \) ... (ii)
Subtracting equation (ii) from (i), we have
\( x + 10y = 75 \)
\( x + 15y = 110 \)
- - -
-----------------
\( -5y = -35 \)
\( \implies \) \( y = 7 \)
Putting the value of \( y = 7 \) in equation (i), we get
\( x + 10 \times 7 = 75 \)
\( \implies \) \( x = 75 - 70 = 5 \)
\( \implies \) \( x = 5 \)
\(\therefore\) Fixed charge \( = \) Rs. 5 and charge per km is Rs. 7
Now, charge for covering a distance of 35 km
\( = x + 35 y \)
\( = 5 + 35 \times 7 = 5 + 245 = \) Rs. 250

Long Answer Questions

Each of the following questions are of 5 marks.

Question. A train covered a certain distance at a uniform speed. If the train would have been 6 km/h faster, it would have taken 4 hours less than the scheduled time and if the train were slower by 6 km/h, it would have taken 6 hours more than the scheduled time. Find the length of the journey. 
Answer: Let original speed of the train be x km/h and scheduled time of journey be y hours.
\(\therefore\) Distance covered \( = xy \)
Now, When speed is 6 km/h faster and time taken is 4 hour less
\( \implies \) \( xy = (x + 6)(y - 4) \)
\( \implies \) \( xy = xy - 4x + 6y - 24 \)
\( \implies \) \( 4x - 6y = -24 \)
\( \implies \) \( 2x - 3y = -12 \) ... (i)
Again, when speed be 6 km/h slower and time taken 6 hours more
\(\therefore\) Distance \( = (x - 6)(y + 6) \)
\( \implies \) \( xy = xy + 6x - 6y - 36 \)
\( \implies \) \( 6x - 6y = 36 \)
\( \implies \) \( x - y = 6 \) ... (ii)
Multiply (ii) by 3 and subtract from (i), we get
\( 2x - 3y = -12 \)
\( 3x - 3y = 18 \)
- + -
-----------------
\( -x = -30 \)
\( \implies \) \( x = 30 \) km/h
Putting the value of \( x = 30 \) in equation (ii), we have
\( 30 - y = 6 \)
\( \implies \) \( y = 30 - 6 = 24 \)
\(\therefore\) \( y = 24 \) hours
\(\therefore\) Total distance covered \( = xy = 30 \times 24 = 720 \) km.
Length of journey \( = 720 \) km.

Question. A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours, it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer: Let the speed of the boat be x km/h and speed of the stream be y km/h.
\(\therefore\) Speed of boat in upstream \( = (x - y) \) km/h
and speed of boat in downstream \( = (x + y) \) km/h
\(\therefore\) According to question,
\( \frac{30}{x - y} + \frac{44}{x + y} = 10 \) ... (i)
and \( \frac{40}{x - y} + \frac{55}{x + y} = 13 \) ... (ii)
(i) x 4 - (ii) x 3 gives
\( \frac{1}{x + y}(176 - 165) = 1 \)
\( \implies \) \( 11 = x + y \)
\( \implies \) \( x + y = 11 \) ... (iii)
Putting the value of \( x + y = 11 \) in equation (i), we have
\( \frac{30}{x - y} + \frac{44}{11} = 10 \)
\( \frac{30}{x - y} + 4 = 10 \)
\( \frac{30}{x - y} = 10 - 4 = 6 \)
\( x - y = 5 \) ... (iv)
From equation (iii) and (iv), we have
\( x + y = 11 \)
\( x - y = 5 \)
-----------------
\( 2x = 16 \)
(on adding)
\( \implies \) \( x = 8 \)
Putting \( x = 8 \) in equation (iii), we have
\( 8 + y = 11 \)
\( \implies \) \( y = 11 - 8 = 3 \)
\( \implies \) \( y = 3 \)
\(\therefore\) Speed of boat is 8 km/h and speed of stream is 3 km/h.

Question. Students of a class are made to stand in rows. If one student is extra in each row, there would be 2 rows less. If one student is less in each row, there would be 3 rows more. Find the number of students in the class. 
Answer: Let total number of rows be y and total number of students in each row be x.
\(\therefore\) Total number of students \( = xy \)
Case I: If one student is extra in each row, there would be two rows less.
Now, number of rows \( = (y - 2) \)
Number of students in each row \( = (x + 1) \)
Total number of students \( = \) number of rows \( \times \) number of students in each row
\( xy = (y - 2)(x + 1) \)
\( \implies \) \( xy = xy + y - 2x - 2 \)
\( \implies \) \( xy - xy + y + 2x = -2 \)
\( \implies \) \( 2x - y = -2 \) ... (i)
Case II: If one student is less in each row, there would be 3 rows more.
Now, number of rows \( = (y + 3) \)
and number of students in each row \( = (x - 1) \)
Total number of students \( = \) number of rows \( \times \) number of students in each row
\(\therefore\) \( xy = (y + 3)(x - 1) \)
\( xy = xy - y + 3x - 3 \)
\( xy - xy + y - 3x = -3 \)
\( \implies \) \( -3x + y = -3 \) ... (ii)
On adding equations (i) and (ii), we have
\( 2x - y = -2 \)
\( -3x + y = -3 \)
-----------------
\( -x = -5 \)
or \( x = 5 \)
Putting the value of x in equation (i), we get
\( 2(5) - y = -2 \)
\( 10 - y = -2 \)
\( -y = -2 - 10 \)
\( -y = -12 \) or \( y = 12 \)
\(\therefore\) Total number of students in the class \( = 5 \times 12 = 60 \).

Question. Draw the graph of \( 2x + y = 6 \) and \( 2x - y + 2 = 0 \). Shade the region bounded by these lines and x-axis. Find the area of the shaded region.
Answer: We have, \( 2x + y = 6 \)
\( \implies \) \( y = 6 - 2x \)
When \( x = 0 \), we have \( y = 6 - 2 \times 0 = 6 \)
When \( x = 3 \), we have \( y = 6 - 2 \times 3 = 0 \)
When \( x = 2 \), we have \( y = 6 - 2 \times 2 = 2 \)
When \( x = 1 \), we have \( y = 6 - 2 \times 1 = 4 \)
Thus, we get the following table:
[Table: Points (0, 6), (3, 0), (2, 2), (1, 4)]
Now, we plot the points \( A(0, 6) \), \( F(1, 4) \), \( C(2, 2) \) and \( B(3, 0) \) on the graph paper as shown. We join A, B and C and extend it on both sides to obtain the graph of the equation \( 2x + y = 6 \).
We have, \( 2x - y + 2 = 0 \)
\( \implies \) \( y = 2x + 2 \)
When \( x = 0 \), we have \( y = 2 \times 0 + 2 = 2 \)
When \( x = -1 \), we have \( y = 2 \times (-1) + 2 = 0 \)
When \( x = 1 \), we have \( y = 2 \times 1 + 2 = 4 \)
Thus, we have the following table:
[Table: Points (0, 2), (-1, 0), (1, 4)]
Now, we plot the points \( D(0, 2) \), \( E(-1, 0) \) and \( F(1, 4) \) on the same graph paper. We join D, E and F and extend it on both sides to obtain the graph of the equation \( 2x - y + 2 = 0 \).
[Graph showing intersection at point F(1, 4)]
It is evident from the graph that the two lines intersect at point \( F(1, 4) \). The area enclosed by the given lines and x-axis is shown in figure by the shaded region.
Thus, \( x = 1 \), \( y = 4 \) is the solution of the given system of equations. Draw FM perpendicular from F on x-axis.
Clearly, we have \( FM = \) y-coordinate of point \( F(1, 4) = 4 \) and \( BE = 4 \)
\(\therefore\) Area of the shaded region \( = \) Area of \( \triangle FBE \)
\( \implies \) Area of the shaded region \( = \frac{1}{2}(\text{base} \times \text{height}) = \frac{1}{2}(BE \times FM) \)
\( = \left( \frac{1}{2} \times 4 \times 4 \right) = 8 \) sq. units.

Case Study-based Questions

Each of the following questions are of 4 marks.

Read the following and answer any four questions from (i) to (v).
Amit is planning to buy a house and the layout is given below figure. The design and the measurement has been made such that areas of two bedrooms and kitchen together is 95 sq.m. [CBSE Question Bank]
[Layout showing: Bedroom 1 (5m x x), Bedroom 2 (5m x x), Kitchen (5m x y), Bathroom/Living room details]

Question. (i) The pair of linear equations in two variables from this situation are
(a) \( x + y = 19 \)
(b) \( 2x + y = 19 \)
(c) \( 2x + y = 19 \)
(d) none of the options
\( x + y = 13 \)
\( x + 2y = 13 \)
\( x + y = 13 \)
Answer: (c) \( 2x + y = 19 \); \( x + y = 13 \)

Question. (ii) The length of the outer boundary of the layout is
(a) 50 m
(b) 52 m
(c) 54 m
(d) 56 m
Answer: (c) 54 m

Question. (iii) The area of each bedroom and kitchen in the layout is
(a) 30 \( m^2, 40 m^2 \)
(b) 30 \( m^2, 35 m^2 \)
(c) 30 \( m^2, 45 m^2 \)
(d) 35 \( m^2, 45 m^2 \)
Answer: (b) 30 \( m^2, 35 m^2 \)

Question. (iv) The area of living room in the layout is
(a) 60 \( m^2 \)
(b) 75 \( m^2 \)
(c) 80 \( m^2 \)
(d) 100 \( m^2 \)
Answer: (b) 75 \( m^2 \)

Question. (v) The cost of laying tiles in kitchen at the rate of Rs. 50 per sq.m is
(a) Rs. 1700
(b) Rs. 1800
(c) Rs. 1900
(d) Rs. 1750
Answer: (d) Rs. 1750

Answer: Sol. We have length of each bedroom be x m and length of kitchen be y m.
(i) Areas of two bedrooms and kitchen together
\( = 2(x \times 5) + y \times 5 \)
\( \implies \) \( 95 = 10x + 5y \)
\( \implies \) \( 2x + y = 19 \) ...(a)
Also, total length \( = x + 2 + y \)
\( \implies \) \( 15 = x + 2 + y \)
\( \implies \) \( x + y = 13 \) ...(b)
\(\therefore\) Option (c) is correct.
(ii) Length of outer boundary of the layout \( = 15 + 12 + 15 + 12 \)
\( = 54 \) metre
\(\therefore\) Option (c) is correct.
(iii) Subtracting (b) from (a), we get
\( x = 6 \)
Putting \( x = 6 \) in (b), we get \( y = 7 \)
\(\therefore\) \( x = 6 \) and \( y = 7 \)
Area of each bedroom \( = \text{length} \times \text{breadth} \)
\( = x \times 5 = 6 \times 5 = 30 \) \( m^2 \)
and area of kitchen \( = \text{length} \times \text{breadth} \)
\( = y \times 5 = 7 \times 5 = 35 \) \( m^2 \)
\(\therefore\) Option (b) is correct.
(iv) Area of living room \( = 15 \times 7 - \text{area of bedroom 2} \)
\( = 15 \times 7 - x \times 5 \)
\( = 15 \times 7 - 6 \times 5 = 105 - 30 = 75 \) \( m^2 \)
\(\therefore\) Option (b) is correct.
(v) We have area of kitchen \( = 35 \) \( m^2 \)
\(\therefore\) Total cost of laying tiles in the kitchen at the rate of Rs. 50 per \( m^2 \)
\( = 35 \times 50 \)
\( = \) Rs. 1750
\(\therefore\) Option (d) is correct.

Read the following and answer any four questions from (i) to (v).
It is common that governments revise travel fares from time to time based on various factors such as inflation (a general increase in prices and fall in the purchasing value of money) on different types of vehicles like auto, rickshaws, taxis, radio cab etc. The auto charges in a city comprise of a fixed charge together with the charge for the distance covered. Study the following situations:
[Table: City A - Distance 10km, Amount 75; Distance 15km, Amount 110. City B - Distance 8km, Amount 91; Distance 14km, Amount 145]
Situation 1: In city A, for a journey of 10 km, the charge paid is Rs. 75 and for a journey of 15 km, the charge paid is Rs. 110.
Situation 2: In city B, for a journey of 8 km, the charge paid is Rs. 91 and for a journey of 14 km, the charge paid is Rs. 145.
Refer situation 1

Question. (i) If the fixed charges of auto rickshaw be Rs. x and the running charges be Rs. y per km, the pair of linear equations representing the situation is
(a) \( x + 10y = 110 \), \( x + 15y = 75 \)
(b) \( x + 10y = 75 \), \( x + 15y = 110 \)
(c) \( 10x + y = 110 \), \( 15x + y = 75 \)
(d) \( 10x + y = 75 \), \( 15x + y = 110 \)
Answer: (b) \( x + 10y = 75 \), \( x + 15y = 110 \)

Question. (ii) A person travels a distance of 50 km. The amount he has to pay is
(a) Rs. 155
(b) Rs. 255
(c) Rs. 355
(d) Rs. 455
Answer: (c) Rs. 355

Refer situation 2
Question. (iii) What will a person have to pay for travelling a distance of 30 km?
(a) Rs. 185
(b) Rs. 289
(c) Rs. 275
(d) Rs. 305
Answer: (b) Rs. 289

Question. (v) Out of both the city, which one has cheaper fare?
(a) City A
(b) City B
(c) Both are same
(d) cannot be decided
Answer: (a) City A

Answer: Sol. (i) In city A, for journey of 10 km, the charge paid is Rs. 75.
\(\therefore\) \( x + 10y = 75 \) ...(i)
Where x be the fixed charge and y be the running charge per km.
Also, for journey of 15 km, the charge paid is Rs. 110.
\(\therefore\) \( x + 15y = 110 \) ...(ii)
\(\therefore\) Option (b) is correct.
(ii) When a person travels a distance of 50 km.
\(\therefore\) Amount he has to pay \( = x + 50 y \) ...(iii)
On solving equation (i) and (ii), we get \( x = 5, y = 7 \)
Putting in (iii), we have
Total payment \( = x + 50y = 5 + 50 \times 7 = Rs. 355 \)
\(\therefore\) Option (c) is correct.
(iii) Referring Situation 2
We have, In a city B, for a journey of 8 km, the charge paid is Rs. 91 and for a journey of 14 km, the charge paid is Rs. 145.
\(\therefore\) \( x + 8y = 91 \) ...(i)
\( x + 14y = 145 \) ...(ii)
be the required pair of linear equations.
Subtracting (i) from (ii), we have
\( 6y = 54 \) \( \implies \) \( y = 9 \)
from (i), we have
\( x + 8 \times 9 = 91 \)
\( \implies \) \( x = 91 - 72 = 19 \)
\(\therefore\) \( x = 19 \)
Total payment for travelling a distance of 30 km
\( = x + 30y = 19 + 30 \times 9 = 19 + 270 = Rs. 289 \)
\(\therefore\) Option (b) is correct.
(iv) From situation 2, we have pair of linear equations
\( x + 8y = 91 \)
\( x + 14y = 145 \)
and point of intersection of these lines is (19, 9).
\(\therefore\) Option (c) is correct.
(v) From the table given, we can easily find out that city A is more cheaper than city B as per the fare charge.
\(\therefore\) Option (a) is correct.

Read the following and answer any four questions from (i) to (v).
A test consists of 'True' or 'False' questions. One mark is awarded for every correct answer while \( \frac{1}{4} \) mark is deducted for every wrong answer. A student knew correct answers of some of the questions. Rest of the questions he attempted by guessing. He answered 120 questions and got 90 marks.
[Table: Type of Question - Marks given for correct answer (1), Marks deducted for wrong answer (0.25)]
Based on the above information answer the following questions.

Question. (i) (a) How many number of questions did he guess?
(b) If answer to all questions he attempted by guessing were wrong and answered 80 correctly, then how many marks will he get?
(ii) (a) If answer to all questions he attempted by guessing were wrong, then how many questions were answered correctly to score 95 marks?
(b) How many maximum marks can a student score?
Answer: Sol. Let the student answered x question correctly and y question incorrectly (wrong).
\(\therefore\) Total number of questions \( = 120 \)
\( x + y = 120 \) ...(i)
Also, one mark is awarded for each correct answer and \( \frac{1}{4} \) mark is deducted for every wrong answer.
\(\therefore\) Total marks student got \( = 90 \)
\( x - \frac{1}{4}y = 90 \) ...(ii)
Subtracting (ii) from (i), we get
\( y + \frac{1}{4}y = 120 - 90 = 30 \)
\( \frac{5y}{4} = 30 \) \( \implies \) \( y = 24 \)
From (ii), \( x - \frac{1}{4} \times 24 = 90 \)
\( \implies \) \( x = 96 \)
(i) (a) The number of questions student guess (do incorrect) \( = 120 - 96 = 24 \).
(b) As the student answered all 120 questions in which 80 questions answered correctly i.e. rest 40 questions do incorrectly.
\(\therefore\) Student got the marks \( = 80 - \frac{1}{4} \times 40 = 70 \) marks
(ii) (a) Let student answered correctly x questions.
\(\therefore\) \( x - \frac{1}{4} \times (120 - x) = 95 \)
\( \implies \) \( x - 30 + \frac{x}{4} = 95 \)
\( \implies \) \( \frac{5x}{4} = 125 \)
\( \implies \) \( x = 100 \)
(b) Since the total questions are 120.
If a student answered all questions correctly then he can score maximum marks i.e; 120.

Objective Type Questions: 

Choose and write the correct option in each of the following questions.

Question. (i) If \( x^{2n-1} + y^{m-4} = 0 \) is a linear equation, which of these is also a linear equation?
(a) \( x^n + y^m = 0 \)
(b) \( x^{1/2n} + y^{m/5} = 0 \)
(c) \( x^{n+1/2} + y^{m+4} = 0 \)
(d) \( x^{\frac{n}{5}} + y^{\frac{m}{5}} = 0 \)
Answer: (d) \( x^{\frac{n}{5}} + y^{\frac{m}{5}} = 0 \)

Question. (ii) Raghav earned Rs. 3550 by selling some bags each for Rs. 500 and some baskets each for Rs. 150. Aarav earned Rs. 3400 by selling the same number of bags each for Rs. 400 and the same number of baskets each for Rs. 200 as Raghav sold. Which of these equations relate the number of bags x, and the number of baskets, y?
(a) \( 500x + 150y = 3400 \) and \( 400x + 200y = 3550 \)
(b) \( 400x + 150y = 3550 \) and \( 500x + 200y = 3400 \)
(c) \( 500x + 150y = 3550 \) and \( 400x + 200y = 3400 \)
(d) \( 500x + 200y = 3550 \) and \( 400x + 150y = 3400 \)
Answer: (c) \( 500x + 150y = 3550 \) and \( 400x + 200y = 3400 \)

Question. (iii) 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) no solution

Question. For what value of \( k \), do the equations \( 3x - y + 8 = 0 \) and \( 6x - ky = - 16 \) represent coincident lines?
(a) \( \frac{1}{2} \)
(b) \( - \frac{1}{2} \)
(c) 2
(d) -2
Answer: (c) 2

Question. For which value(s) of \( p \) will the lines represented by the following pair of linear equations be parallel : \( 3x - y - 5 = 0 \) and \( 6x - 2y - p = 0 \)?
(a) all real values except 10
(b) 10
(c) \( \frac{5}{2} \)
(d) \( \frac{1}{2} \)
Answer: (a) all real values except 10

Very Short Answer Questions: 

Question. Do the following equations represent a pair of coincident lines?
\( -2x - 3y = 1 \), \( 6y + 4x = - 2 \)
Answer: Yes

Question. Write the number of solutions of the following pair of linear equations:
\( 3x - 7y = 1 \) and \( 6x - 14y - 3 = 0 \)
Answer: No solution

Question. Find the value of \( k \) for which the system of equations \( kx - y = 2, 6x - 2y = 3 \), has a unique solution.
Answer: \( k \neq 3 \)

Question. Find the value of \( k \) for which the system of equations \( 2x + 3y = 7 \) and \( 8x + (k + 4)y - 28 = 0 \) has infinitely many solutions.
Answer: \( k = 8 \)

Question. Find the value of \( k \) for which the system of equations \( 2x + y - 3 = 0 \) and \( 5x + ky + 7 = 0 \) has no solution.
Answer: \( k = \frac{5}{2} \)

Question. If \( x = a, y = b \) is the solution of the pair of equations \( x - y = 2 \) and \( x + y = 4 \), find the value of \( a \) and \( b \).
Answer: \( a = 3, b = 1 \)

Short Answer Questions-I:

Question. Find the value of \( k \) for which the lines \( (k + 1)x + 3ky + 15 = 0 \) and \( 5x + ky + 5 = 0 \) are coincident.
Answer: 14

Question. On comparing the ratios \( \frac{a_1}{a_2}, \frac{b_1}{b_2} \) and \( \frac{c_1}{c_2} \), find out whether the linear equations \( 4x - 5y = 8 \) and \( 3x - \frac{15}{4}y = 6 \) are consistent or inconsistent.
Answer: consistent

Question. Find the value of \( k \) for which the system of equations \( x + 3y - 4 = 0 \) and \( 2x + ky = 7 \) is inconsistent.
Answer: 6

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

Question. Find the value(s) of \( k \) for which the pair of equations
\[ \begin{cases} kx + 2y = 3 \\ 3x + 6y = 10 \end{cases} \]
has unique solution.
Answer: \( k \neq 1 \) (the pair of equations have unique solution for all real values of \( k \) except 1)

Question. If the system of equations \( 2x + 3y = 7 \) and \( (a + b)x + (2a - b)y = 21 \) has infinitely many solutions, then find \( a \) and \( b \).
Answer: \( a = 5, b = 1 \)

Question. Solve the following pair of linear equations as :
\( 3x - 5y = 4 \)
\( 2y + 7 = 9x \)
Answer: \( x = \frac{9}{13}, y = \frac{-5}{13} \)

Question. Sumit is 3 times as old as his son. Five years later, he shall be two and a half time as old as his son. How old is Sumit at present?
Answer: 45 years

Short Answer Questions-II: 

Question. Find the solution of the pair of equations \( \frac{x}{10} + \frac{y}{5} - 1 = 0 \) and \( \frac{x}{8} + \frac{y}{6} = 15 \). Hence, find \( \lambda \), if \( y = \lambda x + 5 \).
Answer: \( x = 340, y = -165, \lambda = \frac{-1}{2} \)

Question. If \( 3x + 7y = -1 \) and \( 4y - 5x + 14 = 0 \), then find the values of \( 3x - 8y \) and \( \frac{y}{x} - 2 \).
Answer: \( 14, -\frac{5}{2} \)

Question. A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has taken food in the mess. When a student \( A \) takes food for 25 days, he has to pay Rs. 4,500, whereas a student \( B \) who takes food for 30 days, has to pay Rs. 5,200. Find the fixed charges per month and the cost of food per day.
Answer: Fixed charge = Rs. 1000; cost of food per day = Rs. 140

Question. There are some students in the two examination halls \( A \) and \( B \). To make the number of students equal in each hall, 10 students are sent from \( A \) to \( B \). But if 20 students are sent from \( B \) to \( A \), the number of students in \( A \) becomes double the number of students in \( B \). Find the number of students in the two halls.
Answer: 100 students in Hall \( A \), 80 students in Hall \( B \)

Question. The age of the father is twice the sum of the ages of his two children. After 20 years, his age will be equal to the sum of the ages of his children. Find the age of the father.
Answer: 40 years

Question. The angles of a cyclic quadrilateral \( ABCD \) are \( \angle A = (2x + 4)^\circ \), \( \angle B = (y + 3)^\circ \), \( \angle C = (2y + 10)^\circ \), \( \angle D = (4x - 5)^\circ \). Find \( x \) and \( y \) and hence the values of the four angles.
Answer: \( x = 33, y = 50, \angle A = 70^\circ, \angle B = 53^\circ, \angle C = 110^\circ, \angle D = 127^\circ \)

Question. Two numbers are in the ratio of \( 1 : 3 \). If 5 is added to both the numbers, the ratio becomes \( 1 : 2 \). Find the numbers.
Answer: 5 and 15

Long Answer Questions: 

Question. Draw the graphs of the equations \( y = -1, y = 3 \) and \( 4x - y = 5 \). Also, find the area of the quadrilateral formed by the lines and the y-axis.
Answer: 6 square units

Question. Solve the following system of linear equations graphically and shade the region between the two lines and x-axis.
(i) \( 3x + 2y - 4 = 0 \) and \( 2x - 3y - 7 = 0 \)
(ii) \( 3x + 2y - 11 = 0 \) and \( 2x - 3y + 10 = 0 \)
Answer: (i) \( x = 2, y = -1 \) (ii) \( x = 1, y = 4 \)

Question. The sum of a two digit number and the number formed by interchanging its digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of the digits in the first number. Find the first number.
Answer: 64

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 minutes longer. Find the speed of the rickshaw and of the bus.
Answer: 10 km/h, 40 km/h

Question. The sum of the numerator and denominator of a fraction is 3 less than twice the denominator. If the numerator and denominator are decreased by 1, the numerator becomes half the denominator. Determine the fraction.
Answer: \( \frac{4}{7} \)

Question. Points \( A \) and \( B \) are 70 km apart on a highway. A car starts from \( A \) and another car starts from \( B \) simultaneously. If they travel in the same direction, they meet in 7 hours, but if they travel towards each other, they meet in one hour. Find the speed of the two cars.
Answer: Speed of car from point \( A = 40 \) km/h and from point \( B = 30 \) km/h

Question. \( A \) takes 6 days less than \( B \) to do a work. If both \( A \) and \( B \) working together can do it in 4 days, how many days will \( B \) take to finish it?
Answer: 12 days