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

Question. A fraction becomes \( \frac{1}{3} \) when 2 is subtracted from the numerator and it becomes \( \frac{1}{2} \) when 1 is subtracted from the denominator. Find the fraction.  
Answer: Let the fraction be \( \frac{x}{y} \).
According to question
\( \frac{x - 2}{y} = \frac{1}{3} \)
or \( 3(x – 2) = y \)
or \( 3x – y = 6 \) ...(i)
again, According to question
\( \frac{x}{y - 1} = \frac{1}{2} \)
or \( 2x = y – 1 \)
or \( 2x – y = – 1 \) ...(ii)
On solving equation (i) and (ii), we get
\( x = 7, y = 15 \)
\(\therefore\) The required fraction is \( \frac{7}{15} \).

Question. Ritu can row downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find her speed of rowing in still water and the speed of current.  
Answer: Let the speed of rowing in still water be \( x \) km/hr and speed of current be \( y \) km/hr.
We know, Speed = \( \frac{Distance}{Time} \)
\(\therefore\) According to the question,
\( x + y = \frac{20}{2} = 10 \) ...(i)
and \( x – y = \frac{4}{2} = 2 \) ...(ii)
Adding equations (i) and (ii), we get
\( 2x = 12 \)
\(\Rightarrow x = 6 \)
Putting \( x = 6 \) in equation (i), we get
\( 6 + y = 10 \)
\(\Rightarrow y = 4 \)
Thus, the speed of rowing in still water is 6 km/hr and speed of current is 4 km/hr.

Question. ABCD is a cyclic quadrilateral. Find the angles of the quadrilateral. 
Answer: We know, opposite angles of a cyclic quadrilateral are supplementary.
\( \therefore \angle A + \angle C = 180^\circ \)
\(\Rightarrow 4y + 20 – 4x = 180^\circ \)
\(\Rightarrow 4y – 4x = 160^\circ \)
\(\Rightarrow y – x = 40^\circ \) ...(i)
Also, \( \angle B + \angle D = 180^\circ \)
\(\Rightarrow 3y – 5 + (– 7x + 5) = 180^\circ \)
\(\Rightarrow 3y – 7x = 180^\circ \) ...(ii)
Multiplying equation (i) by 3 and then subtracting it from equation (ii), we get
\( 3y – 7x = 180^\circ \)
\( 3y – 3x = 120^\circ \)
— + —
\( – 4x = 60^\circ \)
\(\Rightarrow x = – 15^\circ \)
Putting \( x = – 15^\circ \) in equation (i), we get
\( y – (– 15^\circ) = 40^\circ \)
\(\Rightarrow y = 40^\circ – 15^\circ = 25^\circ \)
\( \therefore \angle A = 4y + 20 = 4 \times 25 + 20 = 120^\circ \)
\( \angle B = 3y – 5 = 3 \times 25 – 5 = 70^\circ \)
\( \angle C = – 4x = – 4 \times (– 15) = 60^\circ \)
\( \angle D = – 7x + 5 = – 7 \times (– 15) + 5 = 110^\circ \).

Passage Based Questions

Read the following passage and answer the questions that follows :
A student write four pair of linear equations in two variables on his copy as given below :
(a) \( 3x – y = 9, x – \frac{y}{3} = 3 \)
(b) \( 4x + 3y = 24, – 2x + 3y = 6 \)
(c) \( 5x – y = 10, 10x – 2y = 20 \)
(d) \( – 2x + y = 3, – 4x + 2y = 10 \)

Question. How many pair of linear equations are consistent?
Answer: On comparing the above equations with standard form of pair of linear equations \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we get
(a) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) as \( \frac{3}{1} = \frac{3}{1} = \frac{-9}{-3} \), consistent
(b) \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \) as \( \frac{4}{-2} \neq 1 \), consistent
(c) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \) as \( \frac{5}{10} = \frac{1}{2} = \frac{10}{20} \), consistent
So three pair of linear equations are consistent.

Question. Which pair of linear equations are inconsistents ?
Answer: (d) \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \) as \( \frac{-2}{-4} = \frac{1}{2} \neq \frac{3}{10} \), inconsistent. The pair (d) of linear equations is inconsistent.

Question. What is the condition for unique solution ?
Answer: Condition for unique solution is \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \).

Read the following passage and answer the questions :
Akhil went to a fair in his village. He wanted to enjoy rides on the Giant Wheel and play Hoopla. The number of times he played Hoopla is half the number of rides he had on the Giant Wheel. He played Hoopla 3 times. Each ride costs ₹ 10, and a game of Hoopla costs ₹ 5.

Question. Find out the number of rides he had in the fair.
Answer: The number of rides is twice the number of times he played Hoopla. So, he had 6 rides in the fair.

Question. How much did he spend on rides?
Answer: Money spent on the rides = ₹ \( (6 \times 10) \) = ₹ 60.

Question. If he had ₹ 150 in his wallet, how much money did he save?
Answer: Money spent on playing Hoopla = ₹ \( (3 \times 5) \) = ₹ 15. His savings = ₹ \( (150 – 60 – 15) \) = ₹ 75.

Long Answer Type Questions

Question. Solve for \( x \) and \( y \) :
\( \frac{10}{x + y} + \frac{2}{x - y} = 4 \) and \( \frac{15}{x + y} - \frac{9}{x - y} = -2 \).  
Answer: Given,
\( \frac{10}{x + y} + \frac{2}{x - y} = 4 \)
and \( \frac{15}{x + y} - \frac{9}{x - y} = -2 \)
Let \( \frac{1}{x + y} = u \) and \( \frac{1}{x - y} = v \)
Thus, \( 10u + 2v = 4 \Rightarrow 5u + v = 2 \) ...(i)
And \( 15u – 9v = – 2 \) ...(ii)
Multiplying equation (i) by 9, we get \( 45u + 9v = 18 \) ...(iii)
Adding equations (ii) and (iii), we get
\( 60u = 16 \Rightarrow u = \frac{16}{60} = \frac{4}{15} \)
Thus, \( \frac{1}{x + y} = \frac{4}{15} \Rightarrow 4x + 4y = 15 \) ...(iv)
Substituting \( u = \frac{4}{15} \) in equation (iii), we get
\( 45 \left(\frac{4}{15}\right) + 9v = 18 \Rightarrow 12 + 9v = 18 \Rightarrow 9v = 6 \Rightarrow v = \frac{2}{3} \)
Thus \( \frac{1}{x - y} = \frac{2}{3} \Rightarrow 2x – 2y = 3 \) or \( 4x – 4y = 6 \) ...(v)
Adding equations (iv) and (v), we get
\( 8x = 21 \Rightarrow x = \frac{21}{8} = 2 \frac{5}{8} \)
Substituting \( x = \frac{21}{8} \) in equation (iv), we get
\( 4 \left(\frac{21}{8}\right) + 4y = 15 \Rightarrow \frac{21}{2} + 4y = 15 \Rightarrow 4y = 15 – \frac{21}{2} = \frac{9}{2} \Rightarrow y = \frac{9}{8} = 1 \frac{1}{8} \)
Thus \( x = 2 \frac{5}{8} \) and \( y = 1 \frac{1}{8} \).

Question. Solve for \( x \) and \( y \) :
\( \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \) and \( \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8} \). 
Answer: Given,
\( \frac{1}{3x + y} + \frac{1}{3x - y} = \frac{3}{4} \)
and \( \frac{1}{2(3x + y)} - \frac{1}{2(3x - y)} = -\frac{1}{8} \)
Let \( \frac{1}{3x + y} = u \) and \( \frac{1}{3x - y} = v \)
Thus, \( u + v = \frac{3}{4} \Rightarrow 4u + 4v = 3 \) ...(i)
And \( \frac{u}{2} - \frac{v}{2} = -\frac{1}{8} \Rightarrow 4u – 4v = – 1 \) ...(ii)
Adding equations (i) and (ii), we get
\( 8u = 2 \Rightarrow u = \frac{1}{4} \)
Thus, \( \frac{1}{3x + y} = \frac{1}{4} \Rightarrow 3x + y = 4 \) ...(iii)
Subtracting equation (ii) from equation (i), we get
\( 8v = 4 \Rightarrow v = \frac{1}{2} \)
Thus, \( \frac{1}{3x - y} = \frac{1}{2} \Rightarrow 3x – y = 2 \) ...(iv)
Adding equations (iii) and (iv), we get \( 6x = 6 \Rightarrow x = 1 \). Subtracting equation (iv) from equation (iii), we get \( 2y = 2 \Rightarrow y = 1 \). Thus \( x = y = 1 \).

Question. Five years ago, A was thrice as old as B and ten years later, A shall be twice as old as B. What are the present ages of A and B ?
Answer: Let the present ages of A and B be \( x \) and \( y \) years respectively.
Five years ago,
\( x - 5 = 3(y - 5) \)
\( \Rightarrow x - 5 = 3y - 15 \)
\( \Rightarrow x - 3y = -10 \dots(i) \)
Ten years later,
\( x + 10 = 2(y + 10) \)
\( \Rightarrow x + 10 = 2y + 20 \)
\( \Rightarrow x - 2y = 10 \dots(ii) \)
Subtracting equation (ii) from equation (i), we get
\( -3y - (-2y) = -10 - 10 \)
\( \Rightarrow -y = -20 \)
\( \Rightarrow y = 20 \)
Substituting \( y = 20 \) in equation (i), we get
\( x - 3(20) = -10 \)
\( \Rightarrow x - 60 = -10 \)
\( \Rightarrow x = 60 - 10 \)
\( \Rightarrow x = 50 \)
Thus, the ages of A and B are 50 years and 20 years respectively.

Question. Places A and B are 160 km apart on the highway. One car starts from A and another from B at the same time. If they travel in the same direction, they meet in 8 hours. If they travel in opposite directions, they meet in 2 hours. Find the speed of each car.
Answer: Let the speed of car A be \( x \) km/hr and of B be \( y \) km/hr.
We know, \( \text{Speed (s)} = \frac{\text{Distance (d)}}{\text{Time (t)}} \)
\( d = s \times t \)
In 8 hours time, A travels \( 8x \) km and B travels \( 8y \) km.
Thus, when they travel in the same direction,
\( 8x - 8y = 160 \)
\( \Rightarrow x - y = 20 \dots(i) \)
In 2 hours, A travels \( 2x \) km and B travels \( 2y \) km.
Thus, when they travel in opposite directions,
\( 2x + 2y = 160 \)
\( \Rightarrow x + y = 80 \dots(ii) \)
Adding equations (i) and (ii), we get
\( 2x = 100 \)
\( \Rightarrow x = 50 \)
Substituting \( x = 50 \) in equation (ii), we get
\( 50 + y = 80 \)
\( \Rightarrow y = 30 \)
Thus, the speeds of car A and B are 50 km/hr and 30 km/hr respectively.

Question. Abdul travelled 300 km by train and 200 km by taxi, taking 5 hours and 30 minutes. If he had travelled 260 km by train and 240 km by taxi, he would have taken 6 minutes longer. Find the speed of the train and the taxi.
Answer: Let the speed of the train be \( x \) km/hr and that of the taxi be \( y \) km/hr.
Time taken by the train to travel 300 km = \( \frac{300}{x} \)
Time taken by the train to travel 260 km = \( \frac{260}{x} \)
Time taken by the taxi to travel 200 km = \( \frac{200}{y} \)
Time taken by the taxi to travel 240 km = \( \frac{240}{y} \)
According to the question,
\( \frac{300}{x} + \frac{200}{y} = 5\frac{30}{60} = 5\frac{1}{2} = \frac{11}{2} \)
\( \Rightarrow 600y + 400x = 11xy \dots(i) \)
and
\( \frac{260}{x} + \frac{240}{y} = 5\frac{36}{60} = 5\frac{6}{10} = \frac{56}{10} \)
\( \Rightarrow 2600y + 2400x = 56xy \)
\( \Rightarrow 325y + 300x = 7xy \dots(ii) \)
Multiplying equation (i) with 3 and equation (ii) with 4, we get
\( 1200x + 1800y = 33xy \dots(iii) \)
\( 1200x + 1300y = 28xy \dots(iv) \)
Subtracting equation (iv) from equation (iii),
\( 500y = 5xy \)
\( \Rightarrow x = 100 \)
Substituting \( x = 100 \) in \( \frac{300}{x} + \frac{200}{y} = \frac{11}{2} \), we have
\( \frac{300}{100} + \frac{200}{y} = \frac{11}{2} \)
\( \Rightarrow 3 + \frac{200}{y} = \frac{11}{2} \)
\( \Rightarrow \frac{200}{y} = \frac{11}{2} - 3 \)
\( \Rightarrow \frac{200}{y} = \frac{5}{2} \)
\( \Rightarrow y = 80 \)
Thus, the speed of the train and taxi are 100 km/hr and 80 km/hr respectively.

Question. A boat goes 30 km upstream and 44 km downstream in 10 hours. The same boat goes 40 km upstream and 55 km downstream in 13 hours. Determine the speed of stream and that of boat in still water.
Answer: Let the speed of boat in still water be \( x \) km/hr and the speed of stream be \( y \) km/hr.
\(\therefore\) Speed of boat in upstream = \( (x - y) \) km/hr
Speed of boat in downstream = \( (x + y) \) km/hr
Time taken to cover 30 km upstream = \( \frac{30}{x - y} \) hrs.
Time taken to cover 44 km downstream = \( \frac{44}{x + y} \) hrs.
Total time taken = 10 hrs.
\( \frac{30}{x - y} + \frac{44}{x + y} = 10 \dots(i) \)
Now, time taken to cover 55 km downstream = \( \frac{55}{x + y} \) hrs.
Time taken to cover 40 km upstream = \( \frac{40}{x - y} \) hrs.
Total time taken = 13 hrs.
\( \frac{40}{x - y} + \frac{55}{x + y} = 13 \dots(ii) \)
Let \( \frac{1}{x - y} = u, \frac{1}{x + y} = v \).
\(\therefore 30u + 44v = 10 \)
and \( 40u + 55v = 13 \)
or \( 15u + 22v = 5 \dots(iii) \)
and \( 8u + 11v = \frac{13}{5} \dots(iv) \)
Multiplying equation (iii) by 8 and equation (iv) by 15, we get
\( 1200u + 176v = 40 \) [Note: There's a typo in the book's intermediate multiplication step, but following the logic leading to correct subtraction]
\( 120u + 176v = 40 \)
\( 120u + 165v = 39 \)
On subtracting:
\( 11v = 1 \)
\( v = \frac{1}{11} \)
Putting the value of \( v \) in equation (iii), we get
\( 15u + 22 \times \frac{1}{11} = 5 \)
\( \Rightarrow 15u + 2 = 5 \)
\( \Rightarrow 15u = 3 \)
\( \Rightarrow u = \frac{3}{15} = \frac{1}{5} \)
Now, \( v = \frac{1}{11} \Rightarrow \frac{1}{x + y} = \frac{1}{11} \Rightarrow x + y = 11 \dots(v) \)
and \( u = \frac{1}{5} \Rightarrow \frac{1}{x - y} = \frac{1}{5} \Rightarrow x - y = 5 \dots(vi) \)
On adding equations (v) and (vi), we get
\( 2x = 16 \Rightarrow x = 8 \).
Putting the value of \( x \) in equation (v), we get
\( 8 + y = 11 \Rightarrow y = 3 \)
Thus, the speed of boat in still water = 8 km/hr. The speed of stream = 3 km/hr.

Question. A man sold a chair and a table together for ₹ 760, thereby making a profit of 25% on the chair and 10% on the table. Had they been sold together for ₹ 767.50, he would have made a profit of 10% on the chair and 25% on the table. Find the cost price of each.
Answer: Let the cost of each chair be \( x \) and the cost of each table be \( y \).
Thus, S.P. of chair with 10% profit = \( x + \frac{10}{100}x = x + \frac{1}{10}x = \frac{11}{10}x \)
and S.P. of chair with 25% profit = \( x + \frac{25}{100}x = x + \frac{x}{4} = \frac{5}{4}x \)
Also, S.P. of table with 10% profit = \( y + \frac{10}{100}y = y + \frac{1}{10}y = \frac{11}{10}y \)
and S.P. of table with 25% profit = \( y + \frac{25}{100}y = y + \frac{y}{4} = \frac{5}{4}y \)
Now, \( \frac{5}{4}x + \frac{11}{10}y = 760 \)
\( \Rightarrow \frac{50x + 44y}{40} = 760 \)
\( \Rightarrow 25x + 22y = 760 \times 20 \)
\( \Rightarrow 25x + 22y = 15200 \dots(i) \)
and \( \frac{11}{10}x + \frac{5}{4}y = 767.50 \)
\( \Rightarrow \frac{44x + 50y}{40} = 767.50 \)
\( \Rightarrow 22x + 25y = 7675 \times 2 \)
\( \Rightarrow 22x + 25y = 15350 \dots(ii) \)
Adding equations (i) and (ii), we get
\( 47x + 47y = 30550 \)
\( \Rightarrow x + y = 650 \dots(iii) \)
Subtracting equation (ii) from equation (i), we get
\( 3x - 3y = -150 \Rightarrow x - y = -50 \dots(iv) \)
Adding equations (iii) and (iv), we get
\( 2x = 600 \Rightarrow x = 300 \)
Substituting \( x = 300 \) in equation (iii), we get
\( 300 + y = 650 \Rightarrow y = 350 \)
Thus, the cost of each chair is ₹ 300 and that of each table is ₹ 350.

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately?
Answer: Let the time taken by pipe of larger diameter be \( x \) hours and the time taken by the pipe of smaller diameter by \( y \) hours.
So, amount of water filled by pipe of larger diameter in 1 hr = \( \frac{1}{x} \)
and amount of water filled by pipe of smaller diameter in 1 hr = \( \frac{1}{y} \)
According to the question,
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \dots(i) \)
Now, amount of water filled by pipe of larger diameter in 4 hr = \( \frac{4}{x} \)
and amount of water filled by pipe of smaller diameter in 9 hr = \( \frac{9}{y} \)
According to the question,
\( \frac{4}{x} + \frac{9}{y} = \frac{1}{2} \dots(ii) \)
Consider \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)
So, by equation (i)
\( a + b = \frac{1}{12} \Rightarrow 12a + 12b = 1 \dots(iii) \)
and, by equation (ii)
\( 4a + 9b = \frac{1}{2} \Rightarrow 8a + 18b = 1 \dots(iv) \)
Multiplying equation (iii) by 2 and equation (iv) by 3 and then subtracting them, we get
\( 24a + 24b = 2 \)
\( 24a + 54b = 3 \)
\( -30b = -1 \Rightarrow b = \frac{1}{30} \)
Putting the value of \( b \) in equation (iii), we get
\( 12a + 12(\frac{1}{30}) = 1 \)
\( \Rightarrow 12a + \frac{2}{5} = 1 \)
\( \Rightarrow 12a = 1 - \frac{2}{5} \)
\( \Rightarrow 12a = \frac{3}{5} \Rightarrow a = \frac{1}{20} \)
\(\therefore a = \frac{1}{x} \Rightarrow x = 20 \text{ hr} \)
and \( b = \frac{1}{y} \Rightarrow y = 30 \text{ hr} \)
Hence, the larger diameter pipe would take 20 hours and the smaller diameter pipe would take 30 hours to fill the pool separately.

Case Study

A company manufactures two types of sanitizers Alpha and Beta. The cost of the small bottle of sanitizer is ₹ 10 and for beta sanitizer is ₹ 12. In the month of June, the company sold total 1000 bottles and makes a total sale of ₹ 10,820 seeing the great demand and short of supply, company decided to increase the price of both the sanitizer by ₹ 1. In the next month i.e., July, the company sold 2,500 bottles and total sales of ₹ 29,200.

Question. How many sanitizers of each type was sold in June?
(a) 460, 510
(b) 540, 460
(c) 410, 590
(d) 590, 410
Answer: d

Question. If the store sold 500 bottles of each type sanitizer in June, what would be their sales?
(a) ₹ 5500
(b) ₹ 5600
(c) ₹ 10,500
(d) ₹ 11,000
Answer: d

Question. How many bottles of each type were sold in the next month when rate was increased?
(a) 1200, 1300
(b) 1300, 1200
(c) 1550, 950
(d) 1650, 800
Answer: d

Question. What percent of increased was found in sanitizer in July as compared to June?
(a) 160.55%
(b) 183.35%
(c) 179.66%
(d) 100%
Answer: (c) In July sale of \( \alpha = 1650 \). In June sale of \( \alpha = 590 \). Then % increase = \( \frac{1650 - 590}{590} \times 100 = \frac{1060 \times 10}{59} = 179.66\% \).

Question. In July, if total of 1050 sanitizers of each type were sold, what would be the sale?
(a) ₹ 25,000
(b) ₹ 25,200
(c) ₹ 27,000
(d) ₹ 28,500
Answer: (b) \( 11 \times 1050 + 13 \times 1050 = 11550 + 13650 = 25,200 \).

Akhil went to a fair in her village. He wanted to enjoy rides on the Giant Wheel and play Hoopla. The number of times he played Hoopla is half the number of rides he had on the Giant Wheel. He played Hoopla 3 times. Each ride cost Rs. 10, and a game of Hoopla costs Rs. 5.

Question. Find out the number of rides he had in the fair.
(a) 3
(b) 4
(c) 6
(d) 8
Answer: (c) The number of rides is twice is number of times he played Hoopla. So, he had 6 rides in the fair.

Question. How much money did he spent on rides?
(a) ₹ 10
(b) ₹ 30
(c) ₹ 80
(d) ₹ 60
Answer: (d) Money spent on the rides = ₹ \( (6 \times 10) \) = ₹ 60.

Question. How much money did he spent on Hoopla?
(a) ₹ 15
(b) ₹ 20
(c) ₹ 25
(d) ₹ 35
Answer: (a)  

Question. If he had ₹ 150 in his wallet, how much money did he save?
(a) ₹ 60
(b) ₹ 80
(c) ₹ 75
(d) ₹ 85
Answer: (c) Money spent on playing Hoopla = ₹ \( (3 \times 5) \) = ₹ 15. His savings = ₹ \( (150 - 60 - 15) \) = ₹ 75.

Question. If he spend full ₹ 150 then how many maximum number of both the things he can do.
(a) 9 Rides, 12 Hoopla
(b) 12 Rides, 9 Hoopla
(c) 6 Rides, 10 Hoopla
(d) None of these
Answer: (a) 9 Rides = \( 9 \times 10 = 90 \); 12 Hoopla = \( 12 \times 5 = 60 \). Total = ₹ 150. Thus, maximum ₹ 150 can the man attain by having 9 rides and 12 Hoopla.

Assertion and Reasoning Based Questions

DIRECTIONS : In the following questions, a statement 1 is followed by statement 2. Mark the correct choice as :
(A) If both statement 1 and statement 2 are true and statement 2 is the correct explanation of statement 1.
(B) If both statement 1 and statement 2 are true, but statement 2 is not the correct explanation of statement 1.
(C) If statement 1 is true, but statement 2 is false.
(D) If statement 1 is false, but statement 2 is true.

Question. Statement 1 : A pair of linear equations with two variables are, \( a_1 x + b_1 y = c_1 \), \( a_2 x + b_2 y = c_2 \) where, \( a_1, b_1, c_1, a_2, b_2, c_2 \) real numbers. Statement 2 : The given equations \( 2x + 3y = 9 \) and \( 4x + 6y = 17 \) intersect at two points.
Answer: (C) An equation is said to be linear equation only when the constants are real numbers. The equations given in the statement 1 form linear equations with two variables and each linear equation represent a line. In the graphical representation of the pair of linear equations, the given lines \( 2x + 3y = 9 \) and \( 4x + 6y = 17 \) does not intersect at two points at any case. So, the statement 1 is true but the statement 2 is false.

Question. Statement 1 : The lines forming the pair of linear equations are parallel in graphical representation. Statement 2 : The ratio of the two equations is : \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \)
Answer: (A) When the ratio of the two equations is, \( \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \). On solving the equations, we get no solutions. If there is no solution, then the lines are not intersecting at any point which means the lines are parallel lines. If the ratio are similar to statement 2, the lines formed are parallel lines. Thus, both the statements are true and statement 2 is the reason for the occurrence of statement 1.

Question. Statement 1 : The constants of the two linear equations form the below given condition and on solving this condition, we get the unique solution. \( a_1 b_2 - a_2 b_1 = 0 \). Statement 2 : A pair of linear equation form a line always.
Answer: (D) If a pair of linear equation has unique solution, then the constants are different and the condition will be, \( a_1 b_2 - a_2 b_1 \neq 0 \). If the condition is similar to statement 1, then the solution formed can be no solution or infinite solution. A pair of linear equation always forms a line in graph. It does not form any other shape because the linear equations are line equations. Thus, statement 1 is false and statement 2 is true.

Question. Statement 1 : A pair of linear equations given below is a consistent pair of linear equation. \( 4x + 2y = 9 \), \( x + 3y = 10 \). Statement 2 : There is no solution when there are only consistent pairs of linear equations.
Answer: (C) The equations are,
\( 4x + 2y = 9 \dots(i) \)
\( x + 3y = 10 \dots(ii) \)
On rearranging equation (ii), we get \( x = 10 - 3y \)
On substituting the value of \( x \) in equation (i), we get
\( 4(10 - 3y) + 2y = 9 \)
\( \Rightarrow 40 - 12y + 2y = 9 \)
\( \Rightarrow 40 - 10y = 9 \)
\( \Rightarrow 40 - 9 = 10y \)
\( \Rightarrow 31 = 10y \)
\( \Rightarrow y = \frac{31}{10} \)
On substituting the value of \( y \) in value of \( x \), we get,
\( x = 10 - (3 \times \frac{31}{10}) \)
\( x = 10 - \frac{93}{10} \)
\( x = \frac{100 - 93}{10} = \frac{7}{10} \)
Thus, the solution of the given equation is, \( (x, y) = (\frac{7}{10}, \frac{31}{10}) \).
Since, the given equations form a unique solution, the equations are consistent. So, a consistent pair of equation always has a unique solution. If the equations has no solution, then the equations are inconsistent and dependent as they form parallel lines. Thus, statement 1 is true but the statement 2 is false.

Archives

Question. For what value of \( k \), will the following pair of equations have infinitely many soutions : \( 2x + 3y = 7 \) and \( (k + 2)x - 3 (1 - k)y = 5k + 1 \)
Answer: 4

Question. Find whether the following pair of linear equations is consistent or inconsistent : \( 3x + 2y = 8 \), \( 6x - 4y = 9 \)
Answer: Consistent

Question. Solve by elimination : \( 3x - y = 7 \), \( 2x + 5y + 1 = 0 \)
Answer: \( x = 2, y = -1 \)

Question. Given the linear equation \( x - 2y - 6 = 0 \), write another linear equation in these two varibles, such that the geometrical representation of the pair so formed is : (i) coincident lines, (ii) intersection lines
Answer: [Verification required for specific equations, general form provided in text as Archive section summary list.]

Question. The sum of the digits of a two digit number is 8 and the difference between the number and that formed by reversing the digit is 18. Find the number.
Answer: 53.