CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables VBQs Set 07

Question. Rs. 9,000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons.
Answer: Let original number of persons \( = x \) The increased number of persons \( = y \) According to the given condition, \( y = x + 20 \) ... (1) Total amount \( = \) Rs. 9000 According to the question, original share of each person \( - \) share of each of the increased persons \( = \) Rs. 160
\( \implies \frac{9000}{x} - \frac{9000}{y} = 160 \) [Each got \( = \frac{\text{Total Amount}}{\text{No. of persons}} \)] Dividing both sides by 40, we get \( \frac{225}{x} - \frac{225}{y} = 4 \) ... (2) Putting \( y = x + 20 \) from equation (1) in equation (2), we get \( \frac{225}{x} - \frac{225}{x + 20} = 4 \)
\( \implies \frac{225x + 4500 - 225x}{x(x + 20)} = 4 \)
\( \implies \frac{4500}{x(x + 20)} = 4 \)
\( \implies 4x^2 + 80x = 4500 \)
\( \implies x^2 + 20x = 1125 \)
\( \implies x^2 + 20x - 1125 = 0 \)
\( \implies x^2 + 45x - 25x - 1125 = 0 \)
\( \implies x (x + 45) - 25 (x + 45) = 0 \)
\( \implies (x - 25) (x + 45) = 0 \)
\( \implies x = 25 \) or \( x = -45 \) But, number of persons can't be negative. Hence, the original number of persons are is 25.

Question. Some students planned a picnic. The budget for food was Rs. 500. But 5 of these failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic?
Answer: Let the number of students who attended picnic \( = x \) Number of students who planned picnic \( = y \) According to the question, \( y - x = 5 \)
\( \implies y = 5 + x \) ... (1) Again as given in the question, Cost of food for each member attended \( - \) Cost of food each member planned \( = \) Rs. 5
\( \implies \frac{500}{x} - \frac{500}{y} = 5 \) ... (2) [\( \because \) Cost of food per student \( = \frac{\text{Total Budget}}{\text{No. of students}} \)] Putting \( y = 5 + x \) from equation (1) in equation (2), we get
\( \implies \frac{500}{x} - \frac{500}{5 + x} = 5 \)
\( \implies \frac{100}{x} - \frac{100}{5 + x} = 1 \)
\( \implies \frac{500 + 100x - 100x}{x(5 + x)} = 1 \)
\( \implies x(5 + x) = 500 \)
\( \implies x^2 + 5x = 500 \)
\( \implies x^2 + 5x - 500 = 0 \)
\( \implies x^2 + 25x - 20x - 500 = 0 \)
\( \implies x (x + 25) - 20 (x + 25) = 0 \)
\( \implies (x + 25) (x - 20) = 0 \)
\( \implies x = -25 \) or \( x = 20 \)
\( \implies x = 20 \) and \( x \neq -25 \) (\( \because \) No. of students can't be negative). Hence, the numbers of students attending picnic are 20.

Question. Students of a class are made to stand in rows. If 4 students are extra in a row, there would be 2 rows less. If 4 students are less in a row, there would be 4 more rows. Find the number of students in the class.
Answer: Let the number of students \( = x \) And the number of rows \( = n \). Number of students in each row \( = y = \frac{\text{Total number of students}}{\text{Number of rows}} = \frac{x}{n} \)
\( \implies x = ny \) ... (1) As per the question Case I. \( \frac{x}{y + 4} = n - 2 \)
\( \implies x = (n - 2) (y + 4) \)
\( \implies x = ny - 2y + 4n - 8 \) ... (2) Putting \( x = ny \) in equation (2), we get \( ny = ny - 2y + 4n - 8 \)
\( \implies ny - ny + 2y - 4n + 8 = 0 \)
\( \implies 2y - 4n + 8 = 0 \)
\( \implies y - 2n + 4 = 0 \) ... (3) Case II. \( \frac{x}{y - 4} = n + 4 \)
\( \implies x = (n + 4) (y - 4) \)
\( \implies x = (n + 4) (y - 4) \) ... (4) Putting \( x = ny \) in equation (4), we have \( ny = (n + 4) (y - 4) \)
\( \implies ny = ny - 4n + 4y - 16 \)
\( \implies ny - ny + 4n - 4y + 16 = 0 \)
\( \implies 4n - 4y + 16 = 0 \) [Dividing by 4]
\( \implies n - y + 4 = 0 \) ... (5) Adding equation (5) and equation (3), we get \( n = 8 \) Putting \( n = 8 \) in (3), we get \( y - 2 \times 8 + 4 = 0 \)
\( \implies y - 16 + 4 = 0 \)
\( \implies y - 12 = 0 \)
\( \implies y = 12 \) Putting \( n = 8 \) and \( y = 12 \) in eq. (1), we get \( x = 12 \times 8 = 96 \) Hence, number of students in the class are 96.

Form the pair of linear equations for the following problems and find their solutions by substitution method.

Question. The difference between two numbers is 26 and one number is three times the other. Find them.
Answer: Let the numbers be \( x \) and \( y \). Difference of two numbers is 26. i.e., \( x - y = 26 \) ... (1) One number is three times the other. i.e., \( x = 3y \) ... (2) Putting \( x = 3y \) in (1), we get \( 3y - y = 26 \)
\( \implies 2y = 26 \)
\( \implies y = 13 \) Putting \( y = 13 \) in (2), we get \( x = 3 \times 13 = 39 \) Hence, the numbers are \( x = 39 \) and \( y = 13 \).

Question. The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Answer: Let the angles be \( x \) and \( y \). Then according to the question, \( x + y = 180 \) ... (1) and \( x = y + 18 \) ... (2) Putting \( x = y + 18 \) in (1), we get \( y + 18 + y = 180 \)
\( \implies 2y = 180 - 18 \)
\( \implies 2y = 162 \)
\( \implies y = 81 \) Putting \( y = 81 \) in (2), we get \( x = 81 + 18 = 99 \) Hence, angles are \( x = 99^\circ \) and \( y = 81^\circ \).

Question. The coach of a cricket team buys 7 bats and 6 ball for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Answer: Let the cost of one bat and one ball be Rs. \( x \) and Rs. \( y \) respectively. Then, \( 7x + 6y = 3800 \) ... (1) \( 3x + 5y = 1750 \) ... (2) From (2), \( 5y = 1750 - 3x \) or \( y = \frac{1750 - 3x}{5} \) Substituting \( y = \frac{1750 - 3x}{5} \) in (1), we get \( 7x + 6\left(\frac{1750 - 3x}{5}\right) = 3800 \)
\( \implies 35x + 10500 - 18x = 19000 \)
\( \implies 17x = 19000 - 10500 \)
\( \implies 17x = 8500 \) or \( x = 500 \) Putting \( x = 500 \) in (2), we get \( 3(500) + 5y = 1750 \)
\( \implies 5y = 1750 - 1500 \)
\( \implies 5y = 250 \)
\( \implies y = \frac{250}{5} = 50 \) Hence, the cost of one bat is Rs. 500 and the cost of one ball is Rs. 50.

Question. The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charges per kilometer? How much does a person have to pay for travelling a distance of 25 km?
Answer: Let fixed charges of taxi \( = \) Rs. \( x \) And running charges of taxi \( = \) Rs. \( y \) per km. According to the question, Expenses of travelling 10 km \( = \) Rs. 105
\( x + 10y = 105 \) ... (1) Again expenses of travelling 15 km \( = \) Rs. 155
\( x + 15y = 155 \) ... (2)
\( \implies x = 155 - 15y \) Putting \( x = 155 - 15y \) in (1), we get \( 155 - 15y + 10y = 105 \)
\( 155 - 5y = 105 \)
\( \implies -5y = 105 - 155 \)
\( \implies -5y = -50 \)
\( \implies y = 10 \) Putting \( y = 10 \) in (2), we get \( x + 15 \times 10 = 155 \)
\( x + 150 = 155 \)
\( \implies x = 155 - 150 = 5 \) Hence, fixed charges of taxi is Rs. 5 and running charges per km is Rs. 10 A person should pay for travelling 25 km \( = 5 + 25 \times 10 = 5 + 250 = \) Rs. 255

Question. A fraction becomes \( \frac{9}{11} \), if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \( \frac{5}{6} \). Find the fraction.
Answer: Let the numerator be \( x \) and denominator be \( y \). Then, according to the question, Case I. \( \frac{x + 2}{y + 2} = \frac{9}{11} \)
\( \implies 11(x + 2) = 9(y + 2) \)
\( \implies 11x + 22 = 9y + 18 \)
\( \implies 11x - 9y = -4 \) ... (1) Case II. \( \frac{x + 3}{y + 3} = \frac{5}{6} \)
\( \implies 6(x + 3) = 5(y + 3) \)
\( \implies 6x + 18 = 5y + 15 \)
\( \implies 6x - 5y = -3 \) ... (2)
\( \implies x = \frac{5y - 3}{6} \) Putting \( x = \frac{5y - 3}{6} \) in (1), we get \( 11\left(\frac{5y - 3}{6}\right) - 9y = -4 \)
\( \implies 11(5y - 3) - 54y = -24 \)
\( \implies 55y - 33 - 54y = -24 \)
\( \implies y = 33 - 24 = 9 \) Putting \( y = 9 \) in (1), we get \( 11x - 9 \times 9 = -4 \)
\( \implies 11x = -4 + 81 = 77 \)
\( \implies x = 7 \) Hence, the required fraction is \( \frac{7}{9} \).

Question. Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob's age was seven times that of his son. What are their present ages.
Answer: Let the present age of Jacob and his son be \( x \) and \( y \) respectively. Case I. After five years age of Jacob \( = (x + 5) \), After five years the age of his son \( = (y + 5) \). According to question, \( x + 5 = 3(y + 5) \)
\( \implies x - 3y = 10 \) ... (1) Case II. Five years ago Jacob's age \( = x - 5 \), and his son's age \( = y - 5 \). Then, according to question, \( x - 5 = 7(y - 5) \)
\( \implies x = 7y - 30 \) ... (2) Putting \( x = 7y - 30 \) from (2) in (1), we get \( 7y - 30 - 3y = 10 \)
\( \implies 4y = 40 \)
\( \implies y = 10 \) Putting \( y = 10 \) in (1), we get \( x - 3 \times 10 = 10 \)
\( \implies x = 10 + 30 \)
\( \implies x = 40 \) Hence, age of Jacob is 40 years, and age of his son is 10 years.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

Question. If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes \( \frac{1}{2} \) if we only add 1 to the denominator. What is the fraction?
Answer: Let \( x \) be the numerator and \( y \) be the denominator of the fraction. So, the fraction is \( \frac{x}{y} \). By given conditions: \( \frac{x + 1}{y - 1} = 1 \)
\( \implies x + 1 = y - 1 \)
\( \implies x - y = -2 \) ... (1) and \( \frac{x}{y + 1} = \frac{1}{2} \)
\( \implies 2x = y + 1 \)
\( \implies 2x - y = 1 \) Subtracting (2) from (1), we get \( (x - y) - (2x - y) = -2 - 1 \)
\( \implies x - y - 2x + y = -3 \)
\( \implies -x = -3 \)
\( \implies x = 3 \) Substituting \( x = 3 \) in (1), we get \( 3 - y = -2 \)
\( \implies y = 5 \) Hence, the required fraction is \( \frac{3}{5} \).

Question. Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
Answer: Let the present age of Nuri be \( x \) years and the present age of Sonu by \( y \) years. Five years ago, Nuri's age \( = (x - 5) \) years Sonu's age \( = (y - 5) \) years As per conditions, \( x - 5 = 3(y - 5) \)
\( \implies x - 5 = 3y - 15 \)
\( \implies x - 3y = -15 + 5 \)
\( \implies x - 3y = -10 \) ... (1) Ten years later, Nuri's age \( = (x + 10) \) years Sonu's age \( = (y + 10) \) years As per conditions, \( x + 10 = 2(y + 10) \)
\( \implies x + 10 = 2y + 20 \)
\( \implies x - 2y = 20 - 10 \)
\( \implies x - 2y = 10 \) ... (2) Subtracting (2) from (1), we get \( -y = -20 \)
\( \implies y = 20 \) Putting \( y = 20 \) in (2), we get \( x - 2(20) = 10 \)
\( \implies x = 10 + 40 = 50 \) \( \therefore \) Nuri's present age is 50 years and Sonu's present age is 20 years.

Question. The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the number. Find the number.
Answer: Let the ten's and the unit's digits in the number be \( x \) and \( y \), respectively. So, the number may be written as \( 10x + y \). When the digits are reversed, \( x \) becomes the unit's digit and \( y \) becomes the ten's digit. The number can be written as \( 10y + x \). According to the given condition, \( x + y = 9 \) ... (1) We are also given that nine times the number, i.e., \( 9(10x + y) \) is twice the number obtained by reversing the order of the number, i.e., \( 2(10y + x) \). \( \therefore 9(10x + y) = 2(10y + x) \)
\( \implies 90x + 9y = 20y + 2x \)
\( \implies 90x - 2x + 9y - 20y = 0 \)
\( \implies 88x - 11y = 0 \)
\( \implies 8x - y = 0 \) ... (2) Adding (1) and (2), we get \( 9x = 9 \)
\( \implies x = 1 \) Putting \( x = 1 \) in (1), we get \( y = 9 - 1 = 8 \) Thus, the number is \( 10 \times 1 + 8 = 10 + 8 = 18 \).

Question. Meena went to a bank to withdraw Rs. 2000. She asked the cashier to give Rs. 50 and Rs. 100 notes only. Meena got 25 notes in all. Find how many notes of Rs. 50 and Rs. 100 she received?
Answer: Let the number of notes of Rs. 50 be \( x \), and the number of notes of Rs. 100 be \( y \). Then according to the question, \( x + y = 25 \) ... (1) \( 50x + 100y = 2000 \) ... (2) Multiplying (1) by 50, we get \( 50x + 50y = 1250 \) ... (3) Subtracting (3) from (2), we have \( 50y = 750 \)
\( \implies y = 15 \) Putting \( y = 15 \) in (1), we get \( x + 15 = 25 \)
\( x = 25 - 15 = 10 \) Hence, the number of notes of Rs. 50 was 10 and that of Rs. 100 was 15.

Question. A lending library has fixed charge for the first three days and an additional charge for each day thereafter. Sarita paid Rs. 27 for a book she kept for seven days. While Kavita paid Rs. 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer: Let the fixed charges for 3 days be Rs. \( x \) and charges per day be Rs. \( y \). \( \therefore \) By the given conditions, \( x + 4y = 27 \) ... (1) and \( x + 2y = 21 \) ... (2) Subtracting (2) from (1), we get \( x + 4 \times 3 = 27 \)
\( \implies x = 27 - 12 = 15 \) \( \therefore \) Fixed charges \( = \) Rs. 15 and charges per day \( = \) Rs. 3.

Form the pair of linear equations in the following problems and find their solutions (if they exist) by any algebraic method:

Question. A part of monthly hostel charges is fixed and the remaining depends on the number of days one has taken food in the mess. When a student, A takes food for 20 days, A has to pay Rs. 1000 as hostel charges, whereas a student B, who takes food for 26 days pays Rs. 1180 as hostel charges. Find the fixed charge and the cost of food per day.
Answer: Let the fixed hostel charge be Rs. \( x \) And the cost of food per day be Rs. \( y \) According to the question, A's hostel charges are Fixed Hostel charges + Cost of food for 20 days \( = \) Rs. 1000 \( x + 20y = 1000 \) ... (1) B's hostel charges are Fixed hostel charge + Cost of food for 26 days \( = \) Rs. 1180 \( x + 26y = 1180 \) ... (2) Subtracting equation (1) from equation (2), we get \( 6y = 180 \)
\( \implies y = 30 \) Putting \( y = 30 \) in equation (1), we have \( x + 20 (30) = 1000 \)
\( \implies x + 600 = 1000 \)
\( x = 1000 - 600 = 400 \) Hence, fixed charge \( = \) Rs. 400 and cost of food per day \( = \) Rs. 30.

Question. A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction.
Answer: Let the fraction be \( \frac{x}{y} \). According to the given conditions: \( \frac{x - 1}{y} = \frac{1}{3} \)
\( \implies 3x - 3 = y \)
\( \implies 3x - y - 3 = 0 \) ... (1) and \( \frac{x}{y + 8} = \frac{1}{4} \)
\( \implies 4x = y + 8 \)
\( \implies 4x - y - 8 = 0 \) ... (2) Solving (1) and (2) by cross-multiplication method, we have: \( \frac{x}{8 - 3} = \frac{y}{-12 + 24} = \frac{1}{-3 + 4} \)
\( \implies \frac{x}{5} = \frac{y}{12} = \frac{1}{1} \)
\( \implies x = 5, y = 12 \) Hence, the required fraction is \( \frac{5}{12} \).

Question. Yash scored 40 marks in a test, receiving 3 marks for each correct answer and losing 1 mark for each wrong answer. Had 4 marks been awarded for each correct answer and 2 marks been deducted for each incorrect answer, then Yash would have scored 50 marks. How many questions were there in the test?
Answer: Let the number of correct answers of Yash be \( x \) and number of wrong answers be \( y \). Then according to question: Case I. He gets 40 marks if 3 marks are given for correct answer and 1 mark is deducted for incorrect answers. \( 3x - y = 40 \) ... (1) Case II. He gets 50 marks if 4 marks are given for correct answer and 2 marks are deducted for incorrect answers. \( 4x - 2y = 50 \) ... (2) Multiplying (1) by 2, we get \( 6x - 2y = 80 \) ... (3) Subtracting (2) from (3), we get \( 2x = 30 \)
\( \implies x = \frac{30}{2} = 15 \) Putting \( x = 15 \) in (1), we get \( 3 \times 15 - y = 40 \)
\( \implies 45 - y = 40 \)
\( \implies y = 5 \) Total number of questions \( = \) number of correct answers \( + \) number of incorrect answers \( = 15 + 5 = 20 \).

Question. Places A and B are 100 km apart on the highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at a different speed, they meet in 5 hours. If they travel towards each other, they meet in 1 hour. What are the speeds of the two cars?
Answer: Let the speed of first car, starting from A \( = x \) km/hr. And the speed of second car, starting from B \( = y \) km/hr. Distance travelled by first car in 5 hours \( = \) AC \( = 5x \) km Distance travelled by second car in 5 hours \( = \) BC \( = 5y \) km According to the question, Let they meet at C, when moving in the same direction. AC \( = \) AB \( + \) BC \( 5x = 100 + 5y \)
\( \implies x = 20 + y \) ... (1) When moving in the opposite direction, let they meet at D. Distance travelled by first car in 1 hour \( = \) AD \( = x \) Distance travelled by second car in 1 hour \( = \) BD \( = y \) AD \( + \) BD \( = \) AB
\( \implies x + y = 100 \) ... (2) Substituting \( x = 20 + y \) in (2), we have \( (20 + y) + y = 100 \)
\( \implies 20 + 2y = 100 \)
\( \implies 2y = 100 - 20 = 80 \)
\( \implies y = 40 \) km/hour On putting \( y = 40 \) in (1), we get \( x = 20 + 40 = 60 \) km/hour Hence, the speed of first car \( = 60 \) km/hour, and the speed of the second car \( = 40 \) km/hour.

Question. The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.
Answer: Let the length of rectangle be \( x \) units and the breadth of the rectangle be \( y \) units Area of the rectangle \( = xy \) Case I. According to the first condition, Reduced length \( = (x - 5) \) units Increased breadth \( = (y + 3) \) units Reduced area \( = (x - 5) (y + 3) \) units Reduction in area \( = 9 \) unit\(^2\) Original area \( - \) Reduced area \( = 9 \) \( xy - [(x - 5) (y + 3)] = 9 \)
\( \implies xy - [xy + 3x - 5y - 15] = 9 \)
\( \implies xy - xy - 3x + 5y + 15 = 9 \)
\( \implies 3x - 5y = 6 \) ... (1) Case II. According to the second condition, Increased length \( = (x + 3) \) units Increased breadth \( = (y + 2) \) units Increased area \( = (x + 3) (y + 2) \) unit\(^2\) Increase in area \( = 67 \) unit\(^2\) Increased area \( - \) Original area \( = 67 \) unit\(^2\)
\( \implies (x + 3) (y + 2) - xy = 67 \)
\( \implies xy + 2x + 3y + 6 - xy = 67 \)
\( \implies 2x + 3y = 61 \) ... (2) On solving (1) and (2), we get \( x = 17 \) units and \( y = 9 \) units Hence, length of rectangle \( = 17 \) units, and breadth of rectangle \( = 9 \) units.

Question. The ages of two friends Ani and Biju differ by 3 years. Ani's father Dharam is twice as old as Ani and Biju is twice as old as his sister Cathy. The age of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Answer: Let the ages of Ani and Biju be \( x \) years and \( y \) years respectively. Then, \( x - y = \pm 3 \) Dharam's age \( = 2x \), and Cathy's age \( = \frac{y}{2} \) Clearly, Dharam is older than Cathy. So, \( 2x - \frac{y}{2} = 30 \) or \( 4x - y = 60 \) Thus, we have the following two systems of linear equations: \( x - y = 3 \) ... (1) and \( 4x - y = 60 \) ... (2) or \( x - y = -3 \) ... (3) and \( 4x - y = 60 \) ... (4) Subtracting (1) from (2), we get \( 3x = 57 \) or \( x = 19 \) Putting \( x = 19 \) in (1), we get \( 19 - y = 3 \) or \( y = 19 - 3 = 16 \) Subtracting (4) from (3), we get \( 19 - y = 3 \) or \( x = \frac{63}{3} = 21 \) Putting \( x = 21 \) in (3), we get \( 21 - y = -3 \) or \( y = 21 + 3 = 24 \) Hence, Ani's age \( = 19 \) years and Biju's age \( = 16 \) years or Ani's age \( = 21 \) years and Biju's age \( = 24 \) years.

Question. One says, "Give me a hundred, friend ! I shall then become twice as rich as you." The other replies, "If you give me ten, I shall be six times as rich as you." Tell me what is the amount of their (respectively) capital? [From the Bijaganita of Bhaskara II]
Answer: Let the friends be named as A and B. Let A has Rs. \( x \) and B has Rs. \( y \). As per question, we have: \( x + 100 = 2(y - 100) \)
\( \implies x - 2y + 300 = 0 \) ... (1) and \( y + 10 = 6(x - 10) \)
\( \implies 6x - y - 70 = 0 \) ... (2) Multiplying (2) by 2, we get \( 12x - 2y - 140 = 0 \) ... (3) Subtracting (3) from (1), we get \( -11x + 440 = 0 \)
\( \implies -11x = -440 \) or \( x = 40 \) Putting \( x = 40 \) in (1), we get \( 40 - 2y + 300 = 0 \) or \( -2y = -340 \) or \( y = 170 \).

Question. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h, it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Answer: Let the original speed of the train be \( x \) km/h and let the time taken to complete the journey be \( y \) hours. Then, the distance covered \( = xy \) km Case I: When speed \( = (x + 10) \) km/h time taken \( = (y - 2) \) hour distance \( = (x + 10) (y - 2) \)
\( \implies xy = (x + 10) (y - 2) \)
\( \implies xy = xy - 2x + 10y - 20 \)
\( \implies 2x - 10y + 20 = 0 \) ... (1) Case II: When speed \( = (x - 10) \) km/h time taken \( = (y + 3) \) hour distance \( = (x - 10)(y + 3) \)
\( \implies xy = (x - 10)(y + 3) \)
\( \implies xy = xy + 3x - 10y - 30 \)
\( \implies 3x - 10y - 30 = 0 \) ... (2) Subtracting (1) from (2), we get \( x - 50 = 0 \implies x = 50 \) Putting \( x = 50 \) in (1), we get \( 100 - 10y + 20 = 0 \implies -10y = -120 \)
\( \implies y = 12 \) \( \therefore \) The original speed of the train \( = 50 \) km/h The time taken to complete the journey \( = 12 \) hours Hence, the length of the journey \( = \text{Speed} \times \text{Time} = (50 \times 12) \) km \( = 600 \) km.

Question. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Answer: Let originally there be \( x \) students in each row. Let there be \( y \) rows in total. Therefore, total number of students is \( xy \). Case I: When 3 students are taken extra in each row, then the number of rows in this case becomes \( (y - 1) \).
\( \implies xy = (x + 3)(y - 1) \)
\( \implies xy = xy - x + 3y - 3 \)
\( \implies x - 3y + 3 = 0 \) ... (1) Case II: When 3 students are taken less in each row, then the number of rows becomes \( (y + 2) \).
\( \implies xy = (x - 3)(y + 2) \)
\( \implies xy = xy + 2x - 3y - 6 \)
\( \implies 2x - 3y - 6 = 0 \) ... (2) Solving (1) and (2), we get \( x = 9, y = 4 \) Hence, the number of students \( = xy = 9 \times 4 = 36 \).