DAV Class 8 Maths Chapter 9 Worksheet 2

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DAV Class 8 Maths Chapter 9 Worksheet 2

Linear Equations in One Variable Worksheet 2


1. The present ages of A and B are in the ratio 7 : 5. Ten years later, their ages will be in the ratio 9 : 7. Find their present ages.

Solution

\[ \begin{aligned} \text{Let the present age of A } &= 7x \ years \\ \text{and the present age of B } &= 5x \ years \\[6pt] \text{Ten years } & \text{later,} \\[6pt] \text{A's age} &= (7x + 10) \ years \\ \text{B's age} &= (5x + 10) \ years \\[6pt] \text{According to the } & \text{question,} \\[6pt] (7x + 10):(5x + 10) &= 9:7 \\[6pt] \frac{7x + 10}{5x + 10} &= \frac{9}{7} \\[6pt] 7 \times (7x + 10) &= 9 \times (5x + 10) \\[6pt] 49x + 70 &= 45x + 90 \\[6pt] 49x - 45x &= 90 - 70 \\[6pt] 4x &= 20 \\[6pt] x &= \frac{20}{4} \\[6pt] \color{green} x &= \color{green} 5 \\ \\ \text{Present age of A } &= 7x \\ &= 7(5) \\ &= 35 \text{ years} \\[6pt] \text{Present age of B } &= 5x \\ &= 5(5) \\ &= 25 \text{ years} \end{aligned} \]

Answer \( \text{A's age} = {\color{red} 35 \text{ years }} , \text{ B's age} = {\color{red} 25 \text{ years}} \)

2. Two years ago, father was three times as old as his son and two years hence, twice his age will be equal to five times that of his son. Find their present ages.

Solution

\[ \begin{aligned} \text{Two years } & \text{ago,} \\[6pt] \text{Let age of the son be} &= x \ years \\ \text{Father's age} &= 3x \ years \\[6pt] \text{Present age of son} &= (x + 2) \ years \\ \text{Present age of father} &= (3x + 2) \ years \\[6pt] \text{Two years } & \text{later,} \\[6pt] \text{Age of son } = (x + 2) + 2 &\implies (x + 4) \ years \\ \text{Age of father} = (3x + 2) + 2 &\implies (3x + 4) \ years \\[6pt]\text{According to the } & \text{question,} \\[6pt]2 \times (\text{Father's age}) &= 5 \times (\text{Son's age}) \\[6pt] 2 \times (3x + 4) &= 5 \times (x + 4) \\[6pt] 6x + 8 &= 5x + 20 \\[6pt] 6x - 5x &= 20 - 8 \\[6pt] \color{green} x &= \color{green} 12 \\ \\ \text{Present age of son } &= x + 2 \\ & = 12 + 2 \\ &= 14 \text{ years} \\[6pt] \text{Present age of father } &= 3x + 2 \\ & = 3(12) + 2 \\ & = 36 + 2 \\ &= 38 \text{ years} \end{aligned} \]

Answer \( \text{Father's age} = {\color{red} 38 \text{ years } },\text{ Son's age} = {\color{red} 14 \text{ years}} \)

3. The sum of the digits of a 2-digit number is 8. The number obtained by interchanging the digits exceeds the given number by 18. Find the given numbers.

Solution

\[ \begin{aligned} \text{Sum of digits } &= 8 \\ \text{Let the digit at ones place} &= x \\ \text{Let the digit at tens place} &= 8 - x \\[6pt] \text{Original Number } &= 10 \times (8 - x) + x \\ &= 80 - 10x + x \\ &= 80 - 9x \\[6pt] \text{Interchanging} & \text{ the digits}\\ \text{Ones place} &= 8 - x \\ \text{Tens place} &= x \\[6pt] \text{New Number} &= 10x + 8 - x \\ &= 9x + 8 \\ \text{According to the question,} \\ (\text{New Number}) - (\text{Original Number}) &= 18 \\[6pt] (9x + 8) - (80 - 9x) &= 18 \\[6pt] 9x + 8 - 80 + 9x &= 18 \\[6pt] 18x - 72 &= 18 \\[6pt] 18x &= 18 + 72 \\[6pt] 18x &= 90 \\[6pt] x &= \frac{90}{18} \\[6pt] \color{green} x &= \color{green} 5 \\ \\ \text{Ones digit } & = 5 \\ \text{Tens digit } &= 8 - 5 \\ &= 3 \\ \therefore \ \color{green} \text{Required Number} &= \color{green} 35 \end{aligned} \]

Answer The number is \( \color{red} 35 \)

4. The ones digit of a 2-digit number is twice the tens digit. When the number formed by reversing the digits is added to the original number, the sum is 99. Find the original number.

Solution

\[ \begin{aligned} \text{Let the digit at tens place} &= x \\ \text{Let the digit at ones place} &= 2x \\[6pt] \text{Original Number } &= 10(x) + 2x \\ &= 12x \\[6pt]\text{Reversed Number } &= 10(2x) + x \\ &= 20x + x \\ &= 21x \\[6pt]\text{According to the question,} \\ \text{Original Number + Reversed Number } &= 99 \\[6pt] 12x + 21x &= 99 \\[6pt] 33x &= 99 \\[6pt] x &= \frac{99}{33} \\[6pt] \color{green} x &= \color{green} 3 \\ \\ \text{Tens digit } &= 3 \\ \text{Ones digit } &= 2(3) \\ &= 6 \\ \therefore \ \color{green} \text{Required Number} &= \color{green} 36 \end{aligned} \]

Answer The number is \( \color{red} 36 \)

5. The sum of the digits of a 2-digit number is 11. The number obtained by interchanging the digits exceeds the original number by 27. Find the number.

Solution

\[ \begin{aligned} \text{Sum of digits } &= 11 \\ \text{Let the digit at ones place} &= x \\ \text{Let the digit at tens place} &= 11 - x \\[6pt] \text{Original Number } &= 10(11 - x) + x \\ &= 110 - 10x + x \\ &= 110 - 9x \\[6pt] \text{Interchanging the digits,} \\ \text{New tens digit} &= x \\ \text{New ones digit} &= 11 - x \\[6pt] \text{New Number} &= 10x + (11 - x) \\ &= 10x + 11 - x \\ &= 9x + 11 \\[6pt] \text{According to the question,} \\ \text{New Number} - \text{Original Number} &= 27 \\ (9x + 11) - (110 - 9x) &= 27 \\ 9x + 11 - 110 + 9x &= 27 \\ 18x - 99 &= 27 \\ 18x &= 27 + 99 \\ 18x &= 126 \\[6pt] x &= \frac{126}{18} \\[6pt] \color{green} x &= \color{green} 7 \\ \\ \text{Ones digit} &= 7 \\ \text{Tens digit} &= 11 - 7 \\ &= 4 \\[6pt] \therefore \ \color{green} \text{Required Number} &= \color{green} 47 \end{aligned} \]

Answer The number is \( \color{red} 47 \)

6. The sum of three consecutive multiples of 7 is 777. Find these multiples.

Solution

\[ \begin{aligned} \text{Let the three consecutive multiples of 7 be } & x, x+7, x+14 \\[6pt] \text{Sum of these multiples } &= 777 \\[6pt] x + (x + 7) + (x + 14) &= 777 \\ x + x + 7 + x + 14 &= 777 \\ 3x + 21 &= 777 \\ 3x &= 777 - 21 \\ 3x &= 756 \\ x &= \frac{756}{3} \\[6pt] \color{green} x &= \color{green} 252 \\ \\ \text{First multiple} &= 252 \\[6pt] \text{Second multiple} &= x + 7 \\ &= 252 + 7 \\ &= 259 \\[6pt] \text{Third multiple} &= x + 14 \\ & = 252 + 14 \\ & = 266 \end{aligned} \]

Answer The multiples are \( \color{red} 252, 259, 266 \)

7. The sum of three consecutive multiples of 9 is 999. Find these multiples.

Solution

\[ \begin{aligned} \text{Let the three consecutive multiples of 9 be } & x, x + 9, x + 18 \\[6pt] \text{Sum of these multiples } &= 999 \\[6pt] x + (x+9) + (x+18) &= 999 \\ x + x + 9 + x + 18 &= 999 \\ 3x + 27 &= 999 \\ 3x &= 999 - 27 \\ 3x &= 972 \\[6pt] x &= \frac{972}{3} \\[6pt] \color{green} x &= \color{green} 324 \\[6pt] \text{First multiple} &= 324 \\[6pt] \text{Second multiple} &= 324 + 9 \\ & = 333 \\[6pt] \text{Third multiple} &= 333 + 9 \\ &= 342 \end{aligned} \]

Answer The multiples are \( \color{red} 324, 333, 342 \)

8. The denominator of a rational number is greater than its numerator by 7. If 3 is subtracted from the numerator and 2 is added to its denominator, the new number becomes \( \dfrac{1}{5} \). Find the rational number.

Solution

\[ \begin{aligned} \text{Let the numerator be} &= x \\ \text{Then the denominator} &= x + 7 \\[6pt] \text{Original Rational Number} &= \frac{x}{x+7} \\[6pt] \text{According to the question,} \\ \frac{x \ {\color{magenta} - \ 3}}{(x + 7) \ {\color{magenta} + \ 2}} &= \frac{1}{5} \\[6pt] \frac{x - 3}{x + 9} &= \frac{1}{5} \\[6pt] 5(x - 3) &= 1(x + 9) \\ 5x - 15 &= x + 9 \\ 5x - x &= 9 + 15 \\ 4x &= 24 \\[6pt] x &= \frac{24}{4} \\[6pt] \color{green} x &= \color{green} 6 \\ \\ \text{Numerator} &= 6 \\ \text{Denominator} &= 6 + 7 \\ & = 13 \\[6pt] \therefore \ \color{green} \text{Rational Number} &= \color{green} \frac{6}{13} \end{aligned} \]

Answer The rational number is \( \color{red} \dfrac{6}{13} \)

9. The numerator and denominator of a rational number are in the ratio 3 : 4. If the denominator is increased by 3, the ratio becomes 3 : 5. Find the rational number.

Solution

\[ \begin{aligned} \text{Numerator : Denominator } &= 3 : 4 \\[6pt] \text{Let the numerator be } &= 3x \\ \text{Let the denominator be } &= 4x \\[6pt] \text{Rational Number } &= \frac{3x}{4x} \\[6pt] \text{According to the question,} \\ \frac{3x}{4x \ {\color{magenta} + \ 3}} &= \frac{3}{5} \\[6pt] 5(3x) &= 3(4x + 3) \\ 15x &= 12x + 9 \\ 15x - 12x &= 9 \\ 3x &= 9 \\ \color{green} x &= \color{green} 3 \\ \\ \text{Numerator } &= 3(3) \\ &= 9 \\[6pt] \text{Denominator } &= 4(3) \\ &= 12 \\[6pt] \therefore \ \color{green} \text{Rational Number} &= \color{green} \frac{9}{12} \end{aligned} \]

Answer The rational number is \( \color{red} \dfrac{9}{12} \)

10. A motor boat goes downstream and covers a distance in 4 hours while it covers the same distance upstream in 5 hours. If the speed of the stream is 3 km/hr, find the speed of the motor boat in still water.

Solution

\[ \begin{aligned} \text{Let speed of boat in still water} &= x \text{ km/hr} \\ \text{Speed of stream} &= 3 \text{ km/hr} \\[6pt] \text{Speed of boat downstream} &= (x + 3) \text{ km/hr} \\ \text{Distance covered in 4 hrs} &= \text{Speed} \times \text{Time} \\ & = 4(x + 3) \text{ km} \\[6pt] \text{Speed of boat upstream} &= (x - 3) \text{ km/hr} \\ \text{Distance covered in 5 hrs} &= \text{Speed} \times \text{Time} \\ &= 5(x - 3) \text{ km} \\[6pt] \text{Since the } & \text{distance is same} \\[6pt] 4(x + 3) &= 5(x - 3) \\ 4x + 12 &= 5x - 15 \\ 4x - 5x &= -15 - 12 \\ -x &= -27 \\ \color{green} x &= \color{green} 27 \text{ km/hr} \\[6pt] \therefore \ \color{green} \text{Speed of motor boat} &= \color{green} 27 \text{ km/hr} \end{aligned} \]

Answer Speed of motor boat in still water \( = \color{red} 27 \text{ km/hr} \)

11. A steamer, going downstream in a river, covers the distance between two towns in 15 hours. Coming back upstream, it covers this distance in 20 hours. The speed of the water is 3 km/hr. Find the distance between two towns.

Solution

\[ \begin{aligned} \text{Let speed of steamer in still water} &= x \text{ km/hr} \\ \text{Speed of water} &= 3 \text{ km/hr} \\[6pt] \text{Speed downstream} &= (x + 3) \text{ km/hr} \\ \text{Distance covered in 15 hrs} &= 15(x + 3) \text{ km} \\[6pt] \text{Speed upstream} &= (x - 3) \text{ km/hr} \\ \text{Distance covered in 20 hrs} &= 20(x - 3) \text{ km} \\[6pt] \text{Since the } & \text{distance is same} \\[6pt] 15(x + 3) &= 20(x - 3) \\ 15x + 45 &= 20x - 60 \\ 15x - 20x &= -60 - 45 \\ -5x &= -105 \\[6pt] x &= \frac{105}{5} \\[6pt] \color{green} x &= \color{green} 21 \text{ km/hr} \\[6pt] \text{Distance} &= 15(x + 3) \\ &= 15(21 + 3) \\ &= 15(24) \\ &= 360 \text{ km} \end{aligned} \]

Answer Distance between two towns is \( = \color{red} 360 \text{ km} \)

12. The distance between two towns is 300 km. Two cars start simultaneously from these towns and move towards each other. The speed of one car is more than the other by 7 km/hr. If the distance between the cars after 2 hours is 34 km, find the speed of the cars.

Solution

\[ \begin{aligned} \text{Let the speed of 1st car} &= x \text{ km/hr} \\ \text{Speed of 2nd car} &= (x + 7) \text{ km/hr} \\[6pt] \text{Distance covered by 1st car in 2 hrs} &= 2x \\ \text{Distance covered by 2nd car in 2 hrs} &= 2(x + 7) \\[6pt] \text{Remaining Distance} &= 34 \text{ km} \\[6pt] \text{According to the} & \text{ question,} \\ \text{Total Distance} &= 300 \text{ km} \\ 2x + 2(x + 7) + 34 &= 300 \\ 2x + 2x + 14 + 34 &= 300 \\ 4x + 48 &= 300 \\ 4x &= 300 - 48 \\ 4x &= 252 \\[6pt] x &= \frac{252}{4} \\[6pt] \color{green} x &= \color{green} 63 \\[6pt] \text{Speed of 1st car} &= 63 \text{ km/hr} \\ \text{Speed of 2nd car} &= 63 + 7 \\ &= 70 \text{ km/hr} \end{aligned} \]

Answer Speeds of 1st car \( = \color{red} 63 \text{ km/hr} \) , Speeds of 2nd car \( = \color{red} 70 \text{ km/hr} \)

13. The length of a rectangle is greater than the breadth by 18 cm. If both length and breadth are increased by 6 cm, then area increases by 168 cm². Find the length and breadth of the rectangle.

Solution

\[ \begin{aligned} \text{Let the breadth} &= x \text{ cm} \\ \text{Length} &= (x + 18) \text{ cm} \\[6pt] \text{Original Area} &= x(x + 18) \\ &= (x^2 + 18x) \text{ cm}^2 \\[6pt] \text{New breadth} &= (x + 6) \text{ cm} \\ \text{New length} &= (x + 18) + 6 \\ &= (x + 24) \text{ cm} \\[6pt] \text{New Area} &= (x + 6)(x + 24) \\ &= x(x+24) + 6(x+24) \\ &= x^2 + 24x + 6x + 144 \\ &= (x^2 + 30x + 144) \text{ cm}^2 \\[6pt] \end{aligned} \] \[ \begin{aligned} \text{According to the} & \text{ question,} \\ \text{New Area} - \text{Original Area} &= 168 \\ (x^2 + 30x + 144) - (x^2 + 18x) &= 168 \\ x^2 + 30x + 144 - x^2 - 18x &= 168 \\ 12x + 144 &= 168 \\ 12x &= 168 - 144 \\ 12x &= 24 \\[6pt] x &= \frac{24}{12} \\[6pt] \color{green} x &= \color{green} 2 \\[6pt] \text{Breadth} &= 2 \text{ cm} \\ \text{Length} &= 2 + 18 \\ &= 20 \text{ cm} \end{aligned} \]

Answer Length = \( \color{red} 20 \text{ cm} \), Breadth = \( \color{red} 2 \text{ cm} \)

14. The length of a rectangle is greater than the breadth by 3 cm. If the length is increased by 9 cm and the breadth is reduced by 5 cm, the area remains the same. Find the dimensions of the rectangle.

Solution

\[ \begin{aligned} \text{Let the breadth} &= x \text{ cm} \\ \text{Length} &= (x + 3) \text{ cm} \\[6pt] \text{Original Area} &= x(x + 3) \\ &= (x^2 + 3x) \text{ cm}^2 \\[6pt] \text{New Length} &= (x + 3) + 9 \\ &= (x + 12) \text{ cm} \\ \text{New Breadth} &= (x - 5) \text{ cm} \\[6pt] \text{New Area} &= (x + 12)(x - 5) \\ &= x(x-5) + 12(x-5) \\ &= x^2 - 5x + 12x - 60 \\ &= (x^2 + 7x - 60) \text{ cm}^2 \\[6pt] \text{According to} & \text{ the question,} \\ \text{New Area} &= \text{Original Area} \\ x^2 + 7x - 60 &= x^2 + 3x \\ x^2 + 7x - x^2 - 3x &= 60 \\ 4x &= 60 \\[6pt] x &= \frac{60}{4} \\[6pt] \color{green} x &= \color{green} 15 \\[6pt] \text{Breadth} &= 15 \text{ cm} \\ \text{Length} &= 15 + 3 \\ &= 18 \text{ cm} \end{aligned} \]

Answer Length = \( \color{red} 18 \text{ cm} \), Breadth = \( \color{red} 15 \text{ cm} \)

15. The difference between two positive integers is 30. The ratio of these integers is 2 : 5. Find the integers.

Solution

\[ \begin{aligned} \text{Let the integers be } & 2x \text{ and } 5x \\[6pt] \text{Difference} &= 30 \\ 5x - 2x &= 30 \\ 3x &= 30 \\[6pt] x &= \frac{30}{3} \\[6pt] \color{green} x &= \color{green} 10 \\[6pt] \text{First integer} &= 2(10) \\ &= 20 \\[6pt] \text{Second integer} &= 5(10) \\ &= 50 \end{aligned} \]

Answer The integers are \( \color{red} 20 \) and \( \color{red} 50 \)

16. The sum of two positive integers is 105. The integers are in the ratio 2 : 3. Find the integers.

Solution

\[ \begin{aligned} \text{Ratio of integers} &= 2 : 3 \\ \text{Let the integers be } & 2x \text{ and } 3x \\[6pt] \text{Sum} &= 105 \\ 2x + 3x &= 105 \\ 5x &= 105 \\[6pt] x &= \frac{105}{5} \\[6pt] \color{green} x &= \color{green} 21 \\[6pt] \text{First integer} &= 2(21) \\ &= 42 \\[6pt] \text{Second integer} &= 3(21) \\ &= 63 \end{aligned} \]

Answer The integers are \( \color{red} 42 \) and \( \color{red} 63 \)

17. A money box contains one-rupee and two-rupee coins in the ratio 5 : 6. If the total value of the coins in the money box is ₹ 85, find the number of two-rupee coins.

Solution

\[ \begin{aligned} \text{Let no. of 1-rupee coins} &= 5x \\ \text{Let no. of 2-rupee coins} &= 6x \\[6pt] \text{Value of 1-rupee coins} &= 5x \times 1 \implies 5x \\ \text{Value of 2-rupee coins} &= 6x \times 2 \implies 12x \\[6pt] \text{Total Value} &= 85 \\ 5x + 12x &= 85 \\ 17x &= 85 \\[6pt] x &= \frac{85}{17} \\[6pt] \color{green} x &= \color{green} 5 \\[6pt] \text{Number of two-rupee coins} &= 6x \\ &= 6(5) \\ &= 30 \end{aligned} \]

Answer Number of two-rupee coins is \( \color{red} 30 \)

18. A purse has only one-rupee and two-rupee coins in it. The number of two-rupee coins is one-third the number of one-rupee coins. If the purse has ₹ 115, find the number of two-rupee coins.

Solution

\[ \begin{aligned} \text{Let the no. of one-rupee coins} &= 3x \\ \text{Let the no. of two-rupee coins} &= x \\[6pt] \text{Value of 1-rupee coins} &= 3x \times 1 \implies 3x \\ \text{Value of 2-rupee coins} &= x \times 2 \implies 2x \\[6pt] \text{Total Value} &= 115 \\ 3x + 2x &= 115 \\ 5x &= 115 \\[6pt] x &= \frac{115}{5} \\[6pt] \color{green} x &= \color{green} 23 \\[6pt] \text{Number of two-rupee coins} &= 23 \end{aligned} \]

Answer Number of two-rupee coins is \( \color{red} 23 \)