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

Word Problems on Pairs of Linear Equations

To solve daily-life problems with the help of simultaneous linear equations in two variables (or equations reducible to them):

1. Represent the unknown quantities by variables x and y.
2. Find the conditions given in the problem and translate them into a pair of linear equations.
3. Solve the equations and obtain the required quantities with appropriate units.

Types of problems: (i) determining two numbers when a relation between them is given; (ii) fractions, digits of a number and ages of persons; (iii) current of a river, time and distance; (iv) mensuration and geometry; (v) time and work; (vi) mixtures, cost of articles, profit and loss, discount and so on.

Solved Examples on Word Problems

Example 11. Find two numbers such that the sum of twice the first and thrice the second is 89 and four times the first exceeds five times the second by 13.
Answer:
Let the numbers be x and y. \( 2x + 3y = 89 \) ... (i) and \( 4x - 5y = 13 \) ... (ii).
Solving, x = 22 and y = 15. The numbers are 22 and 15.

Example 12. The numerator of a fraction is 4 less than the denominator. If the numerator is decreased by 2 and the denominator is increased by 1, then the denominator is eight times the numerator. Find the fraction.
Answer:
Let the numerator be x and the denominator y. \( y - x = 4 \) ... (i) and \( y + 1 = 8(x - 2) \) ... (ii).
Solving, x = 3 and y = 7. The fraction is \( \dfrac{3}{7} \).

Example 13. A number consists of two digits, the sum of the digits being 12. If 18 is subtracted from the number, the digits are reversed. Find the number.
Answer:
Let the units digit be x and the tens digit y; the number is 10y + x. \( 10y + x - 18 = 10x + y \) gives \( y - x = 2 \) ... (i), and \( x + y = 12 \) ... (ii).
Solving, x = 5 and y = 7. The number is 75.

Example 14. The sum of a two-digit number and the number obtained by reversing the order of its digits is 165. If the digits differ by 3, find the number.
Answer:
Let the units digit be x and the tens digit y. \( (10y + x) + (10x + y) = 165 \) gives \( x + y = 15 \) ... (i). The digits differ by 3: \( x - y = 3 \) or \( y - x = 3 \).
If x − y = 3: x = 9, y = 6, the number is 69. If y − x = 3: x = 6, y = 9, the number is 96. The number is 69 or 96.

Example 15. Six years hence a man's age will be three times the age of his son, and three years ago he was nine times as old as his son. Find their present ages.
Answer:
Let the man's age be x and the son's age y. \( x + 6 = 3(y + 6) \Rightarrow x - 3y = 12 \) ... (i) and \( x - 3 = 9(y - 3) \Rightarrow x - 9y = -24 \) ... (ii).
Solving, x = 30 and y = 6. The man is 30 years old and the son is 6 years old.

Example 16. A boat goes 12 km upstream and 40 km downstream in 8 hours. It can go 16 km upstream and 32 km downstream in the same time. Find the speed of the boat in still water and the speed of the stream.
Answer:
Let the speed in still water be x km/h and that of the stream y km/h. Downstream speed = x + y and upstream speed = x − y.
\( \dfrac{12}{x - y} + \dfrac{40}{x + y} = 8 \) ... (i) and \( \dfrac{16}{x - y} + \dfrac{32}{x + y} = 8 \) ... (ii).
Put \( \dfrac{1}{x - y} = u \) and \( \dfrac{1}{x + y} = v \): \( 12u + 40v = 8 \) and \( 16u + 32v = 8 \), giving \( u = \dfrac{1}{4} \), \( v = \dfrac{1}{8} \). So x − y = 4 and x + y = 8.
Speed of the boat in still water = 6 km/h; speed of the stream = 2 km/h.

Example 17. Ramesh travels 760 km to his home partly by train and partly by car. He takes 8 hours if he travels 160 km by train and the rest by car. He takes 12 minutes more if he travels 240 km by train and the rest by car. Find the speed of the train and of the car.
Answer:
Let the speed of the train be x km/h and of the car y km/h.
\( \dfrac{160}{x} + \dfrac{600}{y} = 8 \) ... (i) and \( \dfrac{240}{x} + \dfrac{520}{y} = 8\tfrac{1}{5} = \dfrac{41}{5} \) ... (ii).
Solving, x = 80 and y = 100 (check: 160/80 + 600/100 = 8 and 240/80 + 520/100 = 8.2).
Speed of the train = 80 km/h, speed of the car = 100 km/h.

Example 18. Points A and B are 90 km apart on a highway. A car starts from A and another from B at the same time. If they go in the same direction they meet in 9 hours, and if they go in opposite directions they meet in \( \dfrac{9}{7} \) hours. Find their speeds.
Answer:
Let the speeds of the cars from A and B be x km/h and y km/h.
Same direction: the car from A gains 90 km in 9 hours, so \( 9x - 9y = 90 \) ... (i). Opposite directions: \( \dfrac{9}{7}(x + y) = 90 \) ... (ii).
From (i) x − y = 10 and from (ii) x + y = 70. So x = 40 and y = 30.
The speed of the car from A = 40 km/h and from B = 30 km/h.

Example 19. In a cyclic quadrilateral ABCD, \( \angle A = (2x + 11)^\circ \), \( \angle B = (y + 12)^\circ \), \( \angle C = (3y + 6)^\circ \) and \( \angle D = (5x - 25)^\circ \). Find the angles of the quadrilateral.
Answer:
Opposite angles of a cyclic quadrilateral are supplementary: \( (2x + 11) + (3y + 6) = 180 \Rightarrow 2x + 3y = 163 \) ... (i) and \( (y + 12) + (5x - 25) = 180 \Rightarrow 5x + y = 193 \) ... (ii).
Solving, x = 32 and y = 33. \( \angle A = 75^\circ \), \( \angle B = 45^\circ \), \( \angle C = 105^\circ \), \( \angle D = 135^\circ \) (sum 360°).

Example 20. A vessel contains a mixture of 24 litres of milk and 6 litres of water, and a second vessel contains a mixture of 15 litres of milk and 10 litres of water. How much of the mixture should be taken from the first and the second vessel separately and kept in a third vessel so that the third vessel may contain a mixture of 25 litres of milk and 10 litres of water?
Answer:
Let x litres be taken from the first vessel and y litres from the second. The first mixture has \( \tfrac{4}{5} \) milk and \( \tfrac{1}{5} \) water; the second has \( \tfrac{3}{5} \) milk and \( \tfrac{2}{5} \) water.
\( \dfrac{4}{5}x + \dfrac{3}{5}y = 25 \) ... (i) and \( \dfrac{1}{5}x + \dfrac{2}{5}y = 10 \) ... (ii).
Solving, x = 20 and y = 15. Take 20 litres from the first vessel and 15 litres from the second.

Example 21. A lady has 25 p and 50 p coins in her purse. If in all she has 40 coins totalling Rs. 12.50, find the number of coins of each type.
Answer:
Let x coins be of 25 p and y coins of 50 p. \( x + y = 40 \) ... (i) and \( 25x + 50y = 1250 \) ... (ii).
Solving, x = 30 and y = 10. She has 30 coins of 25 p and 10 coins of 50 p.

Example 22. Students of a class are made to stand in rows. If one student is extra in a row, there would be 2 rows less. If one student is less in a row, there would be 3 rows more. Find the total number of students in the class.
Answer:
Let the original number of rows be x and the number of students in each row y; the total is xy.
\( (y + 1)(x - 2) = xy \Rightarrow x - 2y = 2 \) ... (i) and \( (y - 1)(x + 3) = xy \Rightarrow 3y - x = 3 \) ... (ii).
Solving, x = 12 and y = 5. Total number of students = 12 × 5 = 60.

Example 23. A man started his job with a certain monthly salary and earned a fixed increment every year. If his salary was Rs. 4500 after 5 years of service and Rs. 5550 after 12 years of service, what was his starting salary and what was his annual increment?
Answer:
Let the initial monthly salary be Rs. x and the annual increment Rs. y. \( x + 5y = 4500 \) ... (i) and \( x + 12y = 5550 \) ... (ii).
Subtracting, 7y = 1050, so y = 150 and x = 3750. Starting salary = Rs. 3750 and annual increment = Rs. 150.

Example 24. A dealer sold a VCR and a TV for Rs. 38560, making a profit of 12% on the VCR and 15% on the TV. By selling them for Rs. 38620, he would have realised a profit of 15% on the VCR and 12% on the TV. Find the cost price of each.
Answer:
Let the cost price of the VCR be Rs. x and of the TV Rs. y. \( \dfrac{112}{100}x + \dfrac{115}{100}y = 38560 \) ... (i) and \( \dfrac{115}{100}x + \dfrac{112}{100}y = 38620 \) ... (ii).
Solving, x = 18000 and y = 16000. Cost price of the VCR = Rs. 18000 and of the TV = Rs. 16000.

Check Your Progress: Word Problems

Check Your Progress 1. The cost of 5 pencils and 4 pens together is Rs. 35, while the cost of 2 pencils and 3 pens together is Rs. 21. What is the price of a pencil and of a pen?
Answer:
\( 5p + 4n = 35 \) and \( 2p + 3n = 21 \). Solving, p = 3 and n = 5. A pencil costs Rs. 3 and a pen Rs. 5.

Check Your Progress 2. In a farm there are some cows and hens. If the total number of heads is 35 and the total number of legs is 98, how many cows and hens are there?
Answer:
\( c + h = 35 \) and \( 4c + 2h = 98 \). Solving, c = 14 and h = 21: 14 cows and 21 hens.

Check Your Progress 3. Find the fraction which becomes 2/3 when the numerator and denominator are increased by 1 and becomes 1/2 when they are diminished by 1.
Answer:
\( \dfrac{x + 1}{y + 1} = \dfrac{2}{3} \Rightarrow 3x - 2y = -1 \) and \( \dfrac{x - 1}{y - 1} = \dfrac{1}{2} \Rightarrow 2x - y = 1 \). Solving, x = 3 and y = 5. The fraction is 3/5.

Check Your Progress 4. A certain number of two digits is four times the sum of its digits. If 9 is added to the number, the digits in the number are reversed. Find the number.
Answer:
Let the number be 10a + b. \( 10a + b = 4(a + b) \Rightarrow b = 2a \), and \( 10a + b + 9 = 10b + a \Rightarrow b = a + 1 \). So a = 1, b = 2. The number is 12.

Check Your Progress 5. Six years ago a man was three times as old as his son. In six years he will be twice as old as his son. Find their present ages.
Answer:
\( x - 6 = 3(y - 6) \Rightarrow x - 3y = -12 \) and \( x + 6 = 2(y + 6) \Rightarrow x - 2y = 6 \). Solving, y = 18 and x = 42. Ages: 42 years and 18 years.

Check Your Progress 6. If the length of a rectangle is increased by 8 metres and the breadth by 3 metres, its area will be increased by 200 square metres. If its length is increased by 3 metres and breadth by 8 metres, its area will be increased by 255 square metres. Find the length and breadth of the rectangle.
Answer:
\( (l + 8)(b + 3) = lb + 200 \Rightarrow 3l + 8b = 176 \) and \( (l + 3)(b + 8) = lb + 255 \Rightarrow 8l + 3b = 231 \). Solving, l = 24 and b = 13. Length 24 m, breadth 13 m.

Check Your Progress 7. In a pen (a small enclosure) there are rabbits and pheasants (game birds). They have between them 35 heads and 98 feet. How many rabbits are there?
Answer:
\( r + p = 35 \) and \( 4r + 2p = 98 \). Solving, r = 14 and p = 21. There are 14 rabbits (and 21 pheasants).