Selina Concise Solutions for ICSE Class 10 Mathematics Chapter 6 Solving Problems Based On Quadratic Equations

Question 1. The product of two consecutive integers is 56. Find the integers.
Answer:
Solution:
Let the two consecutive integers be x and x + 1.
From the given information,
x(x + 1) = 56
\( x^2 + x - 56 = 0 \)
(x + 8)(x - 7) = 0
x = -8 or 7
Thus, the required integers are -8 and -7; 7 and 8.
In simple words: We need two numbers next to each other. When we multiply them, we get 56. We solve this using algebra and get two pairs: -8,-7 and 7,8.

📝 Teacher's Note: Show students that consecutive integers are like 3,4 or 10,11. They differ by 1. Use simple examples first before solving.

🎯 Exam Tip: Always write "Let the consecutive integers be x and x+1". This gets you method marks. Check both answers by multiplication.

Question 2. The sum of the squares of two consecutive natural numbers is 41. Find the numbers.
Answer:
Solution:
Let the numbers be x and x + 1.
From the given information,
\( x^2 + (x + 1)^2 = 41 \)
\( 2x^2 + 2x + 1 - 41 = 0 \)
\( x^2 + x - 20 = 0 \)
(x + 5)(x - 4) = 0
x = -5, 4
But, -5 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 5.
In simple words: Natural numbers are counting numbers like 1,2,3,4. We square both numbers and add them to get 41. Only 4 and 5 work.

📝 Teacher's Note: Remind students that natural numbers start from 1. Negative numbers are not natural numbers. Always check which solution is valid.

🎯 Exam Tip: Write "natural numbers are positive integers" and reject negative solutions. This shows you understand the question properly.

Question 3. Find the two natural numbers which differ by 5 and the sum of whose squares is 97.
Answer:
Solution:
Let the two numbers be x and x + 5.
From the given information,
\( x^2 + (x + 5)^2 = 97 \)
\( 2x^2 + 10x + 25 - 97 = 0 \)
\( 2x^2 + 10x - 72 = 0 \)
\( x^2 + 5x - 36 = 0 \)
(x + 9)(x - 4) = 0
x = -9 or 4
Since, -9 is not a natural number. So, x = 4.
Thus, the numbers are 4 and 9.
In simple words: We need two natural numbers where one is 5 more than the other. Their squares add up to 97. The answer is 4 and 9.

📝 Teacher's Note: "Differ by 5" means one number is 5 bigger than the other. Use examples like 3,8 or 10,15 to show this concept.

🎯 Exam Tip: Always check your answer. \( 4^2 + 9^2 = 16 + 81 = 97 \). This verification step can save you marks.

Question 4. The sum of a number and its reciprocal is 4.25. Find the number.
Answer:
Solution:
Let the number be x and \( \frac{1}{x} \).
From the given information,
\( x + \frac{1}{x} = 4.25 \)
\( \frac{x^2 + 1}{x} = \frac{425}{100} = \frac{17}{4} \)
\( 4x^2 - 17x + 4 = 0 \)
\( 4x^2 - 16x - x + 4 = 0 \)
\( 4x(x - 4) - 1(x - 4) = 0 \)
(x - 4)(4x - 1) = 0
\( x = 4, \frac{1}{4} \)
\( x = 4 \Rightarrow \frac{1}{x} = \frac{1}{4} \)
\( x = \frac{1}{4} \Rightarrow x = 4 \)
Thus, the numbers are 4 and \( \frac{1}{4} \).
In simple words: A reciprocal is like flipping a fraction. The reciprocal of 4 is 1/4. When we add 4 + 1/4, we get 4.25.

📝 Teacher's Note: Show students that reciprocal of 2 is 1/2, reciprocal of 5 is 1/5. Use simple fractions first before solving the problem.

🎯 Exam Tip: Convert 4.25 to fraction form 17/4 first. This makes the calculation easier and avoids decimal errors.

Question 5. Two natural numbers differ by 3. Find the numbers, if the sum of their reciprocals is \( \frac{7}{10} \).
Answer:
Solution:
Let the numbers be x and x + 3.
From the given information,
\( \frac{1}{x} + \frac{1}{x + 3} = \frac{7}{10} \)
\( \frac{x + 3 + x}{x(x + 3)} = \frac{7}{10} \)
\( \frac{2x + 3}{x^2 + 3x} = \frac{7}{10} \)
\( 20x + 30 = 7x^2 + 21x \)
\( 7x^2 + x - 30 = 0 \)
\( 7x^2 - 14x + 15x - 30 = 0 \)
\( 7x(x - 2) + 15(x - 2) = 0 \)
(x - 2)(7x + 15) = 0
\( x = 2, -\frac{15}{7} \)
Since, x is a natural number, so x = 2.
Thus, the numbers are 2 and 5.
In simple words: We need two natural numbers where one is 3 more than the other. When we add their reciprocals (flipped fractions), we get 7/10.

📝 Teacher's Note: When adding fractions, find common denominator first. Show students step by step how to combine 1/x + 1/(x+3).

🎯 Exam Tip: Cross multiply carefully when dealing with fractions. Write each step clearly to avoid calculation mistakes.

Question 6. Divide 15 into two parts such that the sum of their reciprocals is \( \frac{3}{10} \).
Answer:
Solution:
Let the two parts be x and x - 15.
\( \frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \)
\( \frac{15 - x + x}{x(15 - x)} = \frac{3}{10} \)
\( \frac{15}{15x - x^2} = \frac{3}{10} \)
\( 150 = 45x - 3x^2 \)
\( 3x^2 - 45x + 150 = 0 \)
\( x^2 - 15x + 50 = 0 \)
(x - 5)(x - 10) = 0
x = 5, 10
x = 5 \( \Rightarrow \) One part = 5 and other part = 10
x = 10 \( \Rightarrow \) One part = 10 and other part = 5
Thus, the required two parts are 5 and 10.
In simple words: We split 15 into two parts. When we add the reciprocals of these parts, we get 3/10. The parts are 5 and 10.

📝 Teacher's Note: Show students that if one part is x, the other part is (15-x) since they add up to 15. Use simple examples like splitting 10 into 3 and 7.

🎯 Exam Tip: Always verify: 5+10=15 and 1/5+1/10=2/10+1/10=3/10. This check confirms your answer is correct.

Question 7. The sum of the square of two positive integers is 208. If the square of larger number is 18 times the smaller number, find the numbers.
Answer:
Solution:
Let the two numbers be x and y, y being the bigger number. From the given information,
\( x^2 + y^2 = 208 \) ..... (i)
\( y^2 = 18x \) ..... (ii)
From (i), we get \( y^2 = 208 - x^2 \). Putting this in (ii), we get,
\( 208 - x^2 = 18x \)
\( \Rightarrow x^2 + 18x - 208 = 0 \)
\( \Rightarrow x^2 + 26x - 8x - 208 = 0 \)
\( \Rightarrow x(x + 26) - 8(x + 26) = 0 \)
\( \Rightarrow (x - 8)(x + 26) = 0 \)
\( \Rightarrow \) x can't be a negative number, hence x = 8
\( \Rightarrow \) Putting x = 8 in (ii), we get \( y^2 = 18 \times 8 = 144 \)
\( \Rightarrow \) y = 12, since y is a positive integer
Hence, the two numbers are 8 and 12.
In simple words: We have two positive numbers. Their squares add to 208. The bigger number squared equals 18 times the smaller number. The answer is 8 and 12.

📝 Teacher's Note: Set up two equations from the given conditions. Substitute one equation into the other to solve. This is a common technique in algebra.

🎯 Exam Tip: Always label which number is bigger. Write "Let y be the larger number" clearly. This helps avoid confusion in the solution.

Question 8. The sum of the squares of two consecutive positive even numbers is 52. Find the numbers.
Answer:
Solution:
Let the consecutive positive even numbers be x and x + 2.
From the given information,
\( x^2 + (x + 2)^2 = 52 \)
\( 2x^2 + 4x + 4 = 52 \)
\( 2x^2 + 4x - 48 = 0 \)
\( x^2 + 2x - 24 = 0 \)
(x + 6)(x - 4) = 0
x = -6, 4
Since, the numbers are positive, so x = 4.
Thus, the numbers are 4 and 6.
In simple words: Consecutive even numbers are like 2,4 or 6,8. We need two such numbers whose squares add to 52. The answer is 4 and 6.

📝 Teacher's Note: Consecutive even numbers differ by 2. Examples: 2,4,6,8. Consecutive odd numbers also differ by 2: 1,3,5,7.

🎯 Exam Tip: For consecutive even numbers, use x and x+2. For consecutive odd numbers, also use x and x+2. Don't use x+1.

Question 9. Find two consecutive positive odd numbers, the sum of whose squares is 74.
Answer:
Solution:
Let the consecutive positive odd numbers be x and x + 2.
From the given information,
\( x^2 + (x + 2)^2 = 74 \)
\( 2x^2 + 4x + 4 = 74 \)
\( 2x^2 + 4x - 70 = 0 \)
\( x^2 + 2x - 35 = 0 \)
(x + 7)(x - 5) = 0
x = -7, 5
Since, the numbers are positive, so, x = 5.
Thus, the numbers are 5 and 7.
In simple words: Consecutive odd numbers are like 1,3 or 5,7. We need two such numbers whose squares add to 74. The answer is 5 and 7.

📝 Teacher's Note: Show students the pattern: odd numbers are 1,3,5,7,9. They always differ by 2, not 1. This is important for setting up the equation.

🎯 Exam Tip: Check your answer: \( 5^2 + 7^2 = 25 + 49 = 74 \). Always verify to make sure you solved correctly.

Question 10. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is 2.9; find the fraction.
Answer:
Solution:
Let the required fraction be \( \frac{x}{2x + 1} \).
From the given information,
\( \frac{x}{2x + 1} + \frac{2x + 1}{x} = 2.9 \)
\( \frac{x^2 + 4x^2 + 1 + 4x}{x(2x + 1)} = \frac{29}{10} \)
\( \frac{5x^2 + 1 + 4x}{2x^2 + x} = \frac{29}{10} \)
\( 50x^2 + 10 + 40x = 58x^2 + 29x \)
\( 8x^2 - 11x - 10 = 0 \)
\( x = \frac{11 \pm \sqrt{121 + 320}}{16} \)
\( x = \frac{11 \pm \sqrt{441}}{16} \)
\( x = \frac{11 \pm 21}{16} \)
\( x = 2, -\frac{5}{8} \)
Thus, the required fraction is \( \frac{2}{5} \).
In simple words: We need a fraction where the bottom number is 1 more than twice the top number. The fraction plus its reciprocal equals 2.9. The answer is 2/5.

📝 Teacher's Note: "Denominator is one more than twice the numerator" means if numerator is x, denominator is 2x+1. Break down word problems step by step.

🎯 Exam Tip: Convert 2.9 to 29/10 for easier calculation. Always work with fractions instead of decimals in these problems.

Question 11. Three positive numbers are in the ratio 1/2 : 1/3 : 1/4. Find the numbers if the sum of their squares is 244.
Answer:
Solution:
Given, three positive numbers are in the ratio 1/2 : 1/3 : 1/4 = 6 : 4 : 3
Let the numbers be 6x, 4x and 3x.
From the given information,
\( (6x)^2 + (4x)^2 + (3x)^2 = 244 \)
\( 36x^2 + 16x^2 + 9x^2 = 244 \)
\( 61x^2 = 244 \)
\( x^2 = 4 \)
x = 2 (since numbers are positive)
Thus, the numbers are 12, 8 and 6.
In simple words: The three numbers are in the ratio 6:4:3. When we square them and add, we get 244. The numbers are 12, 8, and 6.

📝 Teacher's Note: To convert ratio 1/2:1/3:1/4, find LCM of 2,3,4 which is 12. Then multiply: 6:4:3. This makes calculation easier.

🎯 Exam Tip: Always simplify ratios first. Write "Let the numbers be 6x, 4x, 3x" clearly. This standard approach gets you method marks.

Question 12. Divide 20 into two parts such that three times the square of one part exceeds the other part by 10.
Answer:
Given: We need to divide 20 into two parts.
Let the two parts be x and y.

Step 1: Set up equations from given information.
x + y = 20
\( \implies \) y = 20 – x

Step 2: Use the condition that three times the square of one part exceeds the other part by 10.
\( 3x^2 = (20 - x) + 10 \)
\( 3x^2 = 30 - x \)
\( 3x^2 + x - 30 = 0 \)

Step 3: Factorize the quadratic equation.
\( 3x^2 - 9x + 10x - 30 = 0 \)
\( 3x(x - 3) + 10(x - 3) = 0 \)
\( (x - 3)(3x + 10) = 0 \)

Step 4: Solve for x.
\( x = 3 \) or \( x = -\frac{10}{3} \)
Since x cannot be negative, x = 3.

Step 5: Find the other part.
y = 20 – 3 = 17

Therefore, the two parts are 3 and 17.
In simple words: We split 20 into two numbers. One number squared and multiplied by 3 is bigger than the other number by 10. The two numbers are 3 and 17.

📝 Teacher's Note: Show students how to set up two equations. One for the sum and one for the special condition. Practice converting word problems into math equations.

🎯 Exam Tip: Always check your answer by putting the values back into the original problem. Write "Therefore" before your final answer to get full marks.

Question 13. Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by 60. Assume the middle number to be x and form a quadratic equation satisfying the above statement. Hence; find the three numbers.
Answer:
Given: Three consecutive natural numbers with middle number as x.
So the numbers are: x – 1, x, x + 1

Step 1: Set up the equation from given condition.
Square of middle number = Difference of squares of other two + 60
\( x^2 = (x + 1)^2 - (x - 1)^2 + 60 \)

Step 2: Expand the squares.
\( x^2 = x^2 + 1 + 2x - x^2 - 1 + 2x + 60 \)
\( x^2 = 4x + 60 \)

Step 3: Form the quadratic equation.
\( x^2 - 4x - 60 = 0 \)

Step 4: Factorize the equation.
\( (x - 10)(x + 6) = 0 \)
\( x = 10 \) or \( x = -6 \)

Step 5: Choose the valid solution.
Since x is a natural number, x = 10.

Therefore, the three consecutive numbers are 9, 10, and 11.
In simple words: We have three numbers in a row. The middle number squared is 60 more than a special calculation with the other two. The three numbers are 9, 10, and 11.

📝 Teacher's Note: Explain that consecutive means "one after another" like 5, 6, 7. Show how to expand (a+b)² and (a-b)² step by step.

🎯 Exam Tip: Always write the quadratic equation clearly. Show factorization steps. Reject negative answers when dealing with natural numbers.

Question 14. Out of three consecutive positive integers, the middle number is p. If three times the square of the largest is greater than the sum of the squares of the other two numbers by 67; calculate the value of p.
Answer:
Given: Three consecutive positive integers with middle number as p.
So the numbers are: p – 1, p, p + 1

Step 1: Set up the equation from given condition.
Three times square of largest = Sum of squares of other two + 67
\( 3(p + 1)^2 = (p - 1)^2 + p^2 + 67 \)

Step 2: Expand all terms.
\( 3(p^2 + 2p + 1) = p^2 - 2p + 1 + p^2 + 67 \)
\( 3p^2 + 6p + 3 = 2p^2 - 2p + 68 \)

Step 3: Simplify to get quadratic equation.
\( 3p^2 + 6p + 3 - 2p^2 + 2p - 68 = 0 \)
\( p^2 + 8p - 65 = 0 \)

Step 4: Factorize the equation.
\( (p + 13)(p - 5) = 0 \)
\( p = -13 \) or \( p = 5 \)

Step 5: Choose the valid solution.
Since the numbers are positive, p cannot be -13.

Therefore, p = 5.
In simple words: We have three numbers in a row with p in the middle. A special calculation with squares gives us p = 5. So the three numbers are 4, 5, and 6.

📝 Teacher's Note: Remind students that "positive integers" means no negative numbers or zero. Practice expanding (a+1)² and (a-1)² carefully.

🎯 Exam Tip: Write "Since numbers are positive" when rejecting negative solutions. This shows you understand the problem conditions.

Question 15. A can do a piece of work in 'x' days and B can do the same work in (x + 16) days. If both working together can do it in 15 days; calculate 'x'.
Answer:
Given:
A completes work in x days
B completes work in (x + 16) days
Together they complete work in 15 days

Step 1: Find work rates.
Work done by A in one day = \( \frac{1}{x} \)
Work done by B in one day = \( \frac{1}{x + 16} \)

Step 2: Set up equation for combined work.
Together A and B can do the work in 15 days
\( \frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15} \)

Step 3: Solve the equation.
\( \frac{x + 16 + x}{x(x + 16)} = \frac{1}{15} \)
\( \frac{2x + 16}{x^2 + 16x} = \frac{1}{15} \)
\( 15(2x + 16) = x^2 + 16x \)
\( 30x + 240 = x^2 + 16x \)
\( x^2 - 14x - 240 = 0 \)

Step 4: Factorize the quadratic equation.
\( (x - 24)(x + 10) = 0 \)
\( x = 24 \) or \( x = -10 \)

Step 5: Choose the valid solution.
Since x cannot be negative, x = 24.

Therefore, x = 24.
In simple words: A finishes work in 24 days. B finishes the same work in 40 days. Together they finish it in 15 days. Think of it like two people helping to clean a house.

📝 Teacher's Note: Explain work rate as "portion of work done per day". Use real examples like painting a wall or filling a tank. Draw fractions to show the concept.

🎯 Exam Tip: Always write work rates as fractions first. Show the equation setup clearly. Check that your answer makes sense - faster worker should take less time.

Question 16. One pipe can fill a cistern in 3 hours less than the other. The two pipes together can fill the cistern in 6 hours 40 minutes. Find the time that each pipe will take to fill the cistern.
Answer:
Given:
One pipe fills cistern in x hours
Other pipe fills cistern in (x - 3) hours
Together they fill in 6 hours 40 minutes = \( 6\frac{40}{60} = 6\frac{2}{3} = \frac{20}{3} \) hours

Step 1: Set up the equation using work rates.
\( \frac{1}{x} + \frac{1}{x - 3} = \frac{3}{20} \)

Step 2: Solve the equation.
\( \frac{x - 3 + x}{x(x - 3)} = \frac{3}{20} \)
\( \frac{2x - 3}{x^2 - 3x} = \frac{3}{20} \)
\( 20(2x - 3) = 3(x^2 - 3x) \)
\( 40x - 60 = 3x^2 - 9x \)
\( 3x^2 - 49x + 60 = 0 \)

Step 3: Factorize the quadratic equation.
\( 3x^2 - 45x - 4x + 60 = 0 \)
\( 3x(x - 15) - 4(x - 15) = 0 \)
\( (x - 15)(3x - 4) = 0 \)
\( x = 15 \) or \( x = \frac{4}{3} \)

Step 4: Check validity of solutions.
If \( x = \frac{4}{3} \), then \( x - 3 = \frac{4}{3} - 3 = -\frac{5}{3} \) (negative, not possible)
So, x = 15

Therefore, one pipe fills the cistern in 15 hours and the other fills in 12 hours.
In simple words: One pipe is slower and takes 15 hours to fill the tank. The other pipe is faster and takes 12 hours. Together they fill it in 6 hours 40 minutes.

📝 Teacher's Note: Convert mixed time to improper fraction first. Show students how to convert 40 minutes to fraction of an hour (40/60 = 2/3).

🎯 Exam Tip: Always convert time to same units first. Check that both pipe times are positive. Write the final answer clearly stating both times.

Question 17. A positive number is divided into two parts such that the sum of the squares of the two parts is 20. The square of the larger part is 8 times the smaller part. Taking x as the smaller part of the two parts, find the number.
Answer:
Given:
Let the smaller part = x
Then, (larger part)² = 8x
So larger part = \( \sqrt{8x} \)

Step 1: Use the condition about sum of squares.
Sum of squares of both parts = 20
\( x^2 + (\sqrt{8x})^2 = 20 \)
\( x^2 + 8x = 20 \)
\( x^2 + 8x - 20 = 0 \)

Step 2: Factorize the quadratic equation.
\( x^2 - 2x + 10x - 20 = 0 \)
\( x(x - 2) + 10(x - 2) = 0 \)
\( (x - 2)(x + 10) = 0 \)
\( x = 2 \) or \( x = -10 \)

Step 3: Choose the valid solution.
Since x = -10 is negative, we reject it.
So x = 2

Step 4: Find the larger part.
Smaller part = 2
Larger part = \( \sqrt{8 \times 2} = \sqrt{16} = 4 \)

Step 5: Find the original number.
The required number = 2 + 4 = 6

Therefore, the number is 6.
In simple words: We split a number into two parts: 2 and 4. When we square them and add (4 + 16 = 20), we get 20. The bigger part squared (16) is 8 times the smaller part (2). The original number is 6.

📝 Teacher's Note: Help students understand that "square of larger part is 8 times smaller part" means (larger)² = 8 × (smaller). Draw diagrams to show the relationship.

🎯 Exam Tip: Clearly define what each variable represents. Check your answer by substituting back into all given conditions. Show that 2² + 4² = 20 and 4² = 8 × 2.

Exercise 6B

Question 1. The sides of a right-angled triangle containing the right angle are 4x cm and (2x – 1) cm. If the area of the triangle is 30 cm²; calculate the lengths of its sides.
Answer:
Given:
Sides containing right angle: 4x cm and (2x - 1) cm
Area of triangle = 30 cm²

Step 1: Use area formula for right triangle.
Area = \( \frac{1}{2} \times \text{base} \times \text{height} \)
\( 30 = \frac{1}{2} \times 4x \times (2x - 1) \)
\( 30 = 2x(2x - 1) \)
\( 30 = 4x^2 - 2x \)
\( 4x^2 - 2x - 30 = 0 \)
\( 2x^2 - x - 15 = 0 \)

Step 2: Factorize the quadratic equation.
\( 2x^2 - 6x + 5x - 15 = 0 \)
\( 2x(x - 3) + 5(x - 3) = 0 \)
\( (x - 3)(2x + 5) = 0 \)
\( x = 3 \) or \( x = -\frac{5}{2} \)

Step 3: Choose the valid solution.
Since x cannot be negative, x = 3.

Step 4: Find the lengths of sides.
Side 1 = 4x = 4 × 3 = 12 cm
Side 2 = 2x - 1 = 2 × 3 - 1 = 5 cm
Hypotenuse = \( \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \) cm

Therefore, the lengths of the three sides are 12 cm, 5 cm, and 13 cm.
In simple words: The two sides that meet at the right angle are 12 cm and 5 cm. The longest side (hypotenuse) is 13 cm. This makes the famous 5-12-13 right triangle.

📝 Teacher's Note: Show students the 5-12-13 Pythagorean triple. Remind them that area of right triangle uses the two perpendicular sides, not the hypotenuse.

🎯 Exam Tip: Always find all three sides including hypotenuse using Pythagoras theorem. Check your area calculation: ½ × 12 × 5 = 30. This confirms your answer.

Question 2. The hypotenuse of a right-angled triangle is 26 cm and the sum of other two sides is 34 cm. Find the lengths of its sides.
Answer:
Given:
Hypotenuse = 26 cm
Sum of other two sides = 34 cm

Step 1: Set up variables.
Let the other two sides be x cm and (34 - x) cm.

Step 2: Apply Pythagoras theorem.
\( 26^2 = x^2 + (34 - x)^2 \)
\( 676 = x^2 + x^2 + 1156 - 68x \)
\( 676 = 2x^2 + 1156 - 68x \)
\( 2x^2 - 68x + 480 = 0 \)
\( x^2 - 34x + 240 = 0 \)

Step 3: Factorize the quadratic equation.
\( x^2 - 10x - 24x + 240 = 0 \)
\( x(x - 10) - 24(x - 10) = 0 \)
\( (x - 10)(x - 24) = 0 \)
\( x = 10 \) or \( x = 24 \)

Step 4: Find both sides.
When x = 10, other side = 34 - 10 = 24 cm
When x = 24, other side = 34 - 24 = 10 cm

Therefore, the lengths of the three sides are 10 cm, 24 cm, and 26 cm.
In simple words: The triangle has sides of 10 cm, 24 cm, and 26 cm. The longest side (26 cm) is the hypotenuse. The other two sides add up to 34 cm.

📝 Teacher's Note: Explain that both x = 10 and x = 24 give the same triangle, just with different labeling of sides. Show the 10-24-26 relationship.

🎯 Exam Tip: Verify your answer using Pythagoras: 10² + 24² = 100 + 576 = 676 = 26². Also check: 10 + 24 = 34. Both conditions are satisfied.

Question 3. The sides of a right-angled triangle are (x – 1) cm, 3x cm and (3x + 1) cm. Find: (i) the value of x, (ii) the lengths of its sides.
Answer:
Given: Sides of right triangle are (x - 1) cm, 3x cm, and (3x + 1) cm

(i) Finding the value of x:

Step 1: Identify the hypotenuse.
The largest side is the hypotenuse = (3x + 1) cm

Step 2: Apply Pythagoras theorem.
\( (3x + 1)^2 = (x - 1)^2 + (3x)^2 \)
\( 9x^2 + 6x + 1 = x^2 - 2x + 1 + 9x^2 \)
\( 9x^2 + 6x + 1 = 10x^2 - 2x + 1 \)
\( -x^2 + 8x = 0 \)
\( x(8 - x) = 0 \)
\( x = 0 \) or \( x = 8 \)

Step 3: Choose the valid solution.
If x = 0, then (x - 1) = -1, which is not possible for a side length.
Therefore, x = 8.

(ii) Finding the lengths of sides:

Step 4: Calculate all sides.
Side 1 = x - 1 = 8 - 1 = 7 cm
Side 2 = 3x = 3 × 8 = 24 cm
Side 3 = 3x + 1 = 3 × 8 + 1 = 25 cm

Therefore, (i) x = 8, and (ii) the lengths of the three sides are 7 cm, 24 cm, and 25 cm.
In simple words: By solving the equation, we find x = 8. This gives us the famous 7-24-25 right triangle. We can check: 7² + 24² = 49 + 576 = 625 = 25².

📝 Teacher's Note: This is another Pythagorean triple: 7-24-25. Show students how to identify which side is the hypotenuse (always the longest). Verify the calculation.

🎯 Exam Tip: Always check which side should be the hypotenuse before applying Pythagoras theorem. Reject solutions that make any side negative or zero.

Question. (iii) its area.
Answer:
Solution:
Longer side = Hypotenuse = (3x + 1) cm
Lengths of other two sides are (x – 1) cm and 3x cm.
Using Pythagoras theorem,
\((3x + 1)^2 = (x – 1)^2 + (3x)^2\)
\(9x^2 + 1 + 6x = x^2 + 1 – 2x + 9x^2\)
\(x^2 – 8x = 0\)
\(x(x – 8) = 0\)
\(x = 0, 8\)
But, if x = 0, then one side = 3x = 0, which is not possible.
So, x = 8
Thus, the lengths of the sides of the triangle are (x – 1) cm = 7 cm, 3x cm = 24 cm and (3x + 1) cm = 25 cm.
Area of the triangle = ½ × 7 cm × 24 cm = 84 cm²
In simple words: We use the Pythagoras theorem to find x. Then we calculate each side length and finally the area using the formula for right triangle area.

📝 Teacher's Note: Always check if the answer makes sense. If x = 0, one side becomes zero which cannot happen in a real triangle. This helps students avoid silly mistakes.

🎯 Exam Tip: Show all steps clearly. Write "Using Pythagoras theorem" and then substitute values. Always check your answer at the end.

Question 4. The hypotenuse of a right-angled triangle exceeds one side by 1 cm and the other side by 18 cm; find the lengths of the sides of the triangle.
Answer:
Solution:
Let the hypotenuse of the triangle be x cm.
From the given information,
Length of one side = (x – 1) cm
Length of other side = (x – 18) cm
Using Pythagoras theorem,
\(x^2 = (x – 1)^2 + (x – 18)^2\)
\(x^2 = x^2 + 1 – 2x + x^2 + 324 – 36x\)
\(x^2 – 38x + 325 = 0\)
\(x^2 – 13x – 25x + 325 = 0\)
\(x(x – 13) – 25(x – 13) = 0\)
\((x – 13)(x – 25) = 0\)
\(x = 13, 25\)
When x = 13, x – 18 = 13 – 18 = -5, which being negative, is not possible.
So, x = 25
Thus, the lengths of the sides of the triangle are x = 25 cm, (x – 1) = 24 cm and (x – 18) = 7 cm.
In simple words: We set up the problem using variables. The hypotenuse is bigger than both sides by given amounts. We solve and check which answer makes sense.

📝 Teacher's Note: Help students understand that side lengths cannot be negative. When we get two values of x, we must check both to see which one gives positive side lengths.

🎯 Exam Tip: Always check your final answer. If any side length comes out negative, that solution is wrong. Write the final answer clearly with units.

Question 5. The diagonal of a rectangle is 60 m more than its shorter side and the larger side is 30 m more than the shorter side. Find the sides of the rectangle.
Answer:
Solution:
Let the shorter side be x m.
Length of the other side = (x + 30) m
Length of diagonal = (x + 60) m
Using Pythagoras theorem,
\((x + 60)^2 = x^2 + (x + 30)^2\)
\(x^2 + 3600 + 120x = x^2 + x^2 + 900 + 60x\)
\(x^2 – 60x – 2700 = 0\)
\(x^2 – 90x + 30x – 2700 = 0\)
\(x(x – 90) + 30(x – 90) = 0\)
\((x – 90)(x + 30) = 0\)
\(x = 90, -30\)
But, x cannot be negative. So, x = 90.
Thus, the sides of the rectangle are 90 m and (90 + 30) m = 120 m.
In simple words: We use the fact that in a rectangle, the diagonal forms a right triangle with the sides. We solve to find the shorter side first.

[Diagram: This shows a rectangle with shorter side x, longer side (x + 30), and diagonal (x + 60).]

📝 Teacher's Note: Draw a rectangle on the board and mark the diagonal. Show students how the diagonal divides the rectangle into two right triangles. This makes the problem visual.

🎯 Exam Tip: Label your diagram clearly with all the given information. Write "Let the shorter side be x" to show your variable clearly.

Question 6. The perimeter of a rectangle is 104 m and its area is 640 m². Find its length and breadth.
Answer:
Solution:
Let the length and the breadth of the rectangle be x m and y m.
Perimeter = 2(x + y) m
∴ 104 = 2(x + y)
x + y = 52
y = 52 – x
Area = 640 m²
∴ xy = 640
x(52 – x) = 640
\(x^2 – 52x + 640 = 0\)
\(x^2 – 32x – 20x + 640 = 0\)
x(x – 32) – 20(x – 32) = 0
(x – 32)(x – 20) = 0
x = 32, 20
When x = 32, y = 52 – 32 = 20
When x = 20, y = 52 – 20 = 32
Thus, the length and breadth of the rectangle are 32 m and 20 m.
In simple words: We use two conditions - perimeter and area. We substitute one variable in terms of the other and solve the quadratic equation.

📝 Teacher's Note: Explain that length and breadth can be interchanged. So 32m × 20m and 20m × 32m represent the same rectangle. Both answers are correct.

🎯 Exam Tip: Use the formulas: Perimeter = 2(l + b) and Area = l × b. Show both substitutions clearly. The final answer can be written either way.

Question 7. A footpath of uniform width runs round the inside of a rectangular field 32 m long and 24 m wide. If the path occupies 208 m², find the width of the footpath.
Answer:
Solution:
Let w be the width of the footpath.
Area of the path = Area of outer rectangle – Area of inner rectangle
∴ 208 = (32)(24) – (32 – 2w)(24 – 2w)
208 = 768 – 768 + 64w + 48w – 4w²
4w² – 112w + 208 = 0
w² – 28w + 52 = 0
w² – 26w – 2w + 52 = 0
w(w – 26) – 2(w – 26) = 0
(w – 26)(w – 2) = 0
w = 26, 2
If w = 26, then breadth of inner rectangle = (24 – 52) m = -28 m, which is not possible.
Hence, the width of the footpath is 2 m.
In simple words: The path area equals the total field area minus the inner grass area. We subtract 2w from each side because the path takes width from both sides.

[Diagram: This shows a rectangular field with outer dimensions 32m × 24m and inner dimensions (32-2w) × (24-2w), with path width w marked.]

📝 Teacher's Note: Draw the rectangle and show that the path width w is removed from both sides. So inner length becomes (32-2w) and inner breadth becomes (24-2w).

🎯 Exam Tip: Remember that path width is subtracted from BOTH sides. So if path width is w, then each inner dimension reduces by 2w. Always check if your answer makes physical sense.

Question 8. Two squares have sides x cm and (x + 4) cm. The sum of their area is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Answer:
Solution:
Given that, two squares have sides x cm and (x + 4) cm.
Sum of their area = 656 cm²
∴ x² + (x + 4)² = 656
x² + x² + 16 + 8x = 656
2x² + 8x – 640 = 0
x² + 4x – 320 = 0
x² + 20x – 16x – 320 = 0
x(x + 20) – 16(x + 20) = 0
(x + 20)(x – 16) = 0
x = -20, 16
But, x being side, cannot be negative.
So, x = 16
Thus, the sides of the two squares are 16 cm and 20 cm.
In simple words: We add the areas of both squares and set it equal to 656. Area of a square is side × side. We solve to find the side length.

📝 Teacher's Note: Remind students that area of a square is side². When we have (x + 4)², we must expand it properly: x² + 8x + 16.

🎯 Exam Tip: Write the algebraic equation clearly: x² + (x + 4)² = 656. Expand step by step. Side length cannot be negative, so reject negative values.

Question 9. The dimensions of a rectangular field are 50 m and 40 m. A flower bed is prepared inside this field leaving a gravel path of uniform width all around the flower bed. The total cost of laying the flower bed and gravelling the path at Rs 30 and Rs 20 per square metre, respectively, is Rs 52,000. Find the width of the gravel path.
Answer:
Solution:
Let the width of the gravel path be w m.
Length of the rectangular field = 50 m
Breadth of the rectangular field = 40 m
Let the length and breadth of the flower bed be x m and y m respectively.
Therefore, we have:
x + 2w = 50 ... (1)
y + 2w = 40 ... (2)
Also, area of rectangular field = 50 × 40 = 2000 m²
Area of the flower bed = xy m²
Area of gravel path = Area of rectangular field – Area of flower bed = (2000 – xy) m²
Cost of laying flower bed + Gravel path = Area × cost of laying per sq. m
52000 = 30xy + 20(2000 – xy)
52000 = 10xy + 40000
xy = 1200
Using (1) and (2), we have:
(50 – 2w)(40 – 2w) = 1200
2000 – 180w + 4w² = 1200
4w² – 180w + 800 = 0
w² – 45w + 200 = 0
w² – 5w – 40w + 200 = 0
w(w – 5) – 40(w – 5) = 0
(w – 5)(w – 40) = 0
w = 5, 40
If w = 40, then x = 50 – 2w = -30, which is not possible.
Thus, the width of the gravel path is 5 m.
In simple words: We set up cost equation using different rates for flower bed and path. Then we use the area relationship to find the path width.

📝 Teacher's Note: This is a complex word problem. Break it into steps: find the cost equation first, then use area relationships. Show that path width affects both length and breadth.

🎯 Exam Tip: Set up the cost equation carefully: flower bed area × Rs 30 + path area × Rs 20 = total cost. Use the fact that flower bed dimensions are (50-2w) and (40-2w).

Question 10. An area is paved with square tiles of a certain size and the number required is 128. If the tiles had been 2 cm smaller each way, 200 tiles would have been needed to pave the same area. Find the size of the larger tiles.
Answer:
Solution:
Let the size of the larger tiles be x cm.
Area of larger tiles = x² cm²
Number of larger tiles required to pave an area is 128.
So, the area needed to be paved = 128x² cm² ... (1)
Size of smaller tiles = (x – 2) cm
Area of smaller tiles = (x – 2)² cm²
Number of smaller tiles required to pave an area is 200.
So, the area needed to be paved = 200(x – 2)² cm² ... (2)
Therefore, from (1) and (2), we have:
128x² = 200(x – 2)²
128x² = 200x² + 800 – 800x
72x² – 800x + 800 = 0
9x² – 100x + 100 = 0
9x² – 90x – 10x + 100 = 0
9x(x – 10) – 10(x – 10) = 0
(x – 10)(9x – 10) = 0
x = 10 or x = 10/9
If x = 10/9, then x – 2 = 10/9 – 2 = -8/9, which is not possible.
Hence, the size of the larger tiles is 10 cm.
In simple words: The total area to be paved stays the same. We use the fact that number of tiles × area of each tile = total area for both cases.

📝 Teacher's Note: Explain that when tiles are smaller, you need more tiles to cover the same area. This is an inverse relationship that students should understand.

🎯 Exam Tip: Set up the equation: 128 × (large tile area) = 200 × (small tile area). The total area paved remains constant. Check that both tile sizes are positive.

Question 11. A farmer has 70 m of fencing, with which he encloses three sides of a rectangular sheep pen; the fourth side being a wall. If the area of the pen is 600 sq. m, find the length of its shorter side.
Answer:
Given:
Total fencing = 70 m
Area of pen = 600 sq. m
Three sides are fenced (fourth side is a wall)

Step 1: Let length and breadth be x and y respectively.
From the given information:
\( x + y + x = 70 \)
\( 2x + y = 70 \) ... (1)

Step 2: Using area formula.
Area = \( xy = 600 \)

Step 3: Substitute from equation (1).
From (1): \( y = 70 - 2x \)
\( x(70 - 2x) = 600 \)
\( 70x - 2x^2 = 600 \)
\( 2x^2 - 70x + 600 = 0 \)
\( x^2 - 35x + 300 = 0 \)

Step 4: Factorize the quadratic equation.
\( x^2 - 15x - 20x + 300 = 0 \)
\( x(x - 15) - 20(x - 15) = 0 \)
\( (x - 15)(x - 20) = 0 \)

Step 5: Find the values of x.
\( x = 15 \) or \( x = 20 \)

Step 6: Find corresponding y values.
If \( x = 15 \), then \( y = 70 - 2(15) = 40 \)
If \( x = 20 \), then \( y = 70 - 2(20) = 30 \)

Therefore: The length of the shorter side is 15 m when the longer side is 40 m, or 20 m when the longer side is 30 m.
In simple words: We made an equation using the fencing and area. Then we solved it to find two possible answers. The shorter side is either 15 m or 20 m.

📝 Teacher's Note: Draw a rectangle on the board. Show students that only three sides need fencing because one side is already a wall. This helps them understand why we write x + y + x = 70, not 2x + 2y = 70.

🎯 Exam Tip: Always write "Let x = length, y = breadth" first. Set up both equations clearly: one for fencing, one for area. Check both answers to see which gives the shorter side.

Question 12. A square lawn is bounded on three sides by a path 4 m wide. If the area of the path is \( \frac{7}{8} \) that of the lawn, find the dimensions of the lawn.
Answer:
Given:
Square lawn with side x m
Path width = 4 m on three sides
Area of path = \( \frac{7}{8} \) × Area of lawn

[Diagram: This diagram shows a square lawn with a 4m wide path on three sides forming a larger rectangle]

Step 1: Calculate areas.
Area of square lawn = \( x^2 \) m²
Area of outer rectangle = \( (x + 4)(x + 8) = x^2 + 12x + 32 \)
Area of path = \( x^2 + 12x + 32 - x^2 = 12x + 32 \)

Step 2: Set up equation using given condition.
\( 12x + 32 = \frac{7}{8} \times x^2 \)
\( 96x + 256 = 7x^2 \)
\( 7x^2 - 96x - 256 = 0 \)

Step 3: Factorize the equation.
\( 7x^2 - 112x + 16x - 256 = 0 \)
\( 7x(x - 16) + 16(x - 16) = 0 \)
\( (x - 16)(7x + 16) = 0 \)

Step 4: Find the solution.
\( x = 16 \) or \( x = -\frac{16}{7} \)
Since x cannot be negative, \( x = 16 \) m

Therefore: Each side of the square lawn is 16 m.
In simple words: The path makes a bigger rectangle around the square lawn. We found the size by making the path area equal to 7/8 of the lawn area.

📝 Teacher's Note: Use graph paper to draw this. Show how the path adds 4 m on three sides, making the outer rectangle (x+4) by (x+8). Students can see why one dimension increases by 4 and the other by 8.

🎯 Exam Tip: Draw the diagram first. Write "Area of path = Area of big rectangle - Area of lawn". This formula is very important for getting marks.

Question 13. The area of a big rectangular room is 300 m². If the length were decreased by 5 m and the breadth increased by 5 m; the area would be unaltered. Find the length of the room.
Answer:
Given:
Original area = 300 m²
New length = (x - 5) m
New breadth = (y + 5) m
New area = 300 m² (unchanged)

Step 1: Set up equations.
Let original length = x m, breadth = y m
Original area: \( xy = 300 \)
\( \implies y = \frac{300}{x} \) ... (1)

Step 2: Use the condition for new dimensions.
New area: \( (x - 5)(y + 5) = 300 \)

Step 3: Substitute equation (1).
\( (x - 5)\left(\frac{300}{x} + 5\right) = 300 \)
\( 300 + 5x - \frac{1500}{x} - 25 = 300 \)
\( 5x - \frac{1500}{x} - 25 = 0 \)
\( 5x^2 - 25x - 1500 = 0 \)
\( x^2 - 5x - 300 = 0 \)

Step 4: Factorize the equation.
\( x^2 - 20x + 15x - 300 = 0 \)
\( x(x - 20) + 15(x - 20) = 0 \)
\( (x - 20)(x + 15) = 0 \)

Step 5: Find the solution.
\( x = 20 \) or \( x = -15 \)
Since x cannot be negative, \( x = 20 \) m

Therefore: The length of the room is 20 m.
In simple words: When we make the room shorter but wider by the same amount, the area stays the same. This only works for one specific length.

📝 Teacher's Note: Show students with rectangles made from graph paper. Cut one and rearrange to show how decreasing length and increasing breadth by the same amount can keep area constant.

🎯 Exam Tip: Write y = 300/x first from the area formula. Then substitute this into the second equation. This substitution method gets you marks even if calculation goes wrong.

Exercise 6C

Question 1. The speed of an ordinary train is x km per hr and that of an express train is (x + 25) km per hr.
(i) Find the time taken by each train to cover 300 km.
(ii) If the ordinary train takes 2 hrs more than the express train; calculate speed of the express train.
Answer:
(i) Speed of ordinary train = x km/hr
Speed of express train = (x + 25) km/hr
Distance = 300 km

Using \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \)

Time taken by ordinary train to cover 300 km = \( \frac{300}{x} \) hrs
Time taken by express train to cover 300 km = \( \frac{300}{x + 25} \) hrs

(ii) Given that ordinary train takes 2 hours more than express train.
\( \frac{300}{x} - \frac{300}{x + 25} = 2 \)
\( \frac{300x + 7500 - 300x}{x(x + 25)} = 2 \)
\( \frac{7500}{x(x + 25)} = 2 \)
\( 7500 = 2x^2 + 50x \)
\( 2x^2 + 50x - 7500 = 0 \)
\( x^2 + 25x - 3750 = 0 \)
\( x^2 + 75x - 50x - 3750 = 0 \)
\( x(x + 75) - 50(x + 75) = 0 \)
\( (x + 75)(x - 50) = 0 \)

Therefore, x = -75 or x = 50
Speed cannot be negative, so x = 50 km/hr
Speed of express train = (x + 25) = 75 km/hr
In simple words: The faster train saves 2 hours for the same distance. We used this fact to find both train speeds.

📝 Teacher's Note: Emphasize that Time = Distance ÷ Speed. Make students practice this formula with simple examples like walking speed before solving train problems.

🎯 Exam Tip: Always write the time formula first. Set up the equation "time difference = 2 hours" clearly. Remember to reject negative speed values.

Question 2. If the speed of a car is increased by 10 km per hr, it takes 18 minutes less to cover a distance of 36 km. Find the speed of the car.
Answer:
Given:
Distance = 36 km
Speed increase = 10 km/hr
Time saved = 18 minutes = \( \frac{18}{60} = \frac{3}{10} \) hours

Step 1: Set up the problem.
Let original speed = x km/hr
New speed = (x + 10) km/hr

Time taken at original speed = \( \frac{36}{x} \) hrs
Time taken at new speed = \( \frac{36}{x + 10} \) hrs

Step 2: Use the given condition.
\( \frac{36}{x} - \frac{36}{x + 10} = \frac{18}{60} \)
\( \frac{36x + 360 - 36x}{x(x + 10)} = \frac{3}{10} \)
\( \frac{360}{x^2 + 10x} = \frac{3}{10} \)
\( \frac{120}{x^2 + 10x} = \frac{1}{10} \)
\( x^2 + 10x = 1200 \)
\( x^2 + 10x - 1200 = 0 \)

Step 3: Factorize the equation.
\( (x + 40)(x - 30) = 0 \)
x = -40 or x = 30

Since speed cannot be negative, x = 30 km/hr
Therefore: The original speed of the car is 30 km/hr.
In simple words: When the car goes 10 km/hr faster, it saves 18 minutes. We used this time difference to find the original speed.

📝 Teacher's Note: Convert minutes to hours carefully: 18 minutes = 18/60 hours = 3/10 hours. Many students make errors in this conversion.

🎯 Exam Tip: Always convert time units to hours when speed is in km/hr. Write the conversion clearly: "18 minutes = 18/60 = 3/10 hours".

Question 3. If the speed of an aeroplane is reduced by 40 km/hr, it takes 20 minutes more to cover 1200 km. Find the speed of the aeroplane.
Answer:
Given:
Distance = 1200 km
Speed reduction = 40 km/hr
Extra time = 20 minutes = \( \frac{20}{60} = \frac{1}{3} \) hours

Step 1: Set up the problem.
Let original speed = x km/hr
New speed = (x - 40) km/hr

Time at original speed = \( \frac{1200}{x} \) hrs
Time at reduced speed = \( \frac{1200}{x - 40} \) hrs

Step 2: Use the given condition.
\( \frac{1200}{x - 40} - \frac{20}{60} = \frac{1200}{x} \)
\( \frac{1200}{x - 40} - \frac{1200}{x} = \frac{20}{60} \)
\( \frac{1200x - 1200x + 48000}{x(x - 40)} = \frac{1}{3} \)
\( \frac{48000}{x(x - 40)} = \frac{1}{3} \)
\( x(x - 40) = 48000 \times 3 \)
\( x^2 - 40x = 144000 \)
\( x^2 - 40x - 144000 = 0 \)

Step 3: Factorize the equation.
\( x^2 - 400x + 360x - 144000 = 0 \)
\( x(x - 400) + 360(x - 400) = 0 \)
\( (x - 400)(x + 360) = 0 \)

Therefore, x = 400 or x = -360
Since speed cannot be negative, x = 400 km/hr
Therefore: The original speed of the aeroplane is 400 km/hr.
In simple words: When the plane flies 40 km/hr slower, it takes 20 minutes extra. We used this to find how fast the plane normally flies.

📝 Teacher's Note: Aeroplanes fly very fast, so 400 km/hr is a reasonable answer. Help students understand that reduced speed means longer time for the same distance.

🎯 Exam Tip: Convert 20 minutes to 1/3 hours. Set up the equation as "slow time - fast time = extra time". Always check that your final answer makes sense for an aeroplane.

Question 4. A car covers a distance of 400 km at a certain speed. Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less. Find the original speed of the car.
Answer:
Given:
Distance = 400 km
Speed increase = 12 km/h
Time saved = 1 hour 40 minutes = \( \frac{5}{3} \) hours

Step 1: Let x km/h be the original speed of the car.
We know that Time = \( \frac{\text{Distance}}{\text{Speed}} \)

Step 2: Set up the time equation.
Time taken at original speed = \( \frac{400}{x} \) hours
Time taken at increased speed = \( \frac{400}{x + 12} \) hours

Step 3: Write the equation based on given condition.
\( \frac{400}{x} - \frac{400}{x + 12} = \frac{5}{3} \)

Step 4: Solve the equation.
\( \frac{400(x + 12) - 400x}{x(x + 12)} = \frac{5}{3} \)

\( \frac{400x + 4800 - 400x}{x(x + 12)} = \frac{5}{3} \)

\( \frac{4800}{x(x + 12)} = \frac{5}{3} \)

\( 3 \times 4800 = 5 \times x \times (x + 12) \)

\( 14400 = 5x^2 + 60x \)

\( 5x^2 + 60x - 14400 = 0 \)

\( x^2 + 12x - 2880 = 0 \)

\( (x + 60)(x - 48) = 0 \)

\( x = -60 \) or \( x = 48 \)

Since speed cannot be negative, x = 48.

The original speed of the car is 48 km/hr.
In simple words: We set up an equation using the fact that faster speed takes less time. We solved the quadratic equation and got the original speed as 48 km/hr.

📝 Teacher's Note: Show students that time = distance ÷ speed. When speed increases, time decreases. This is the key idea behind these word problems.

🎯 Exam Tip: Always write "speed cannot be negative" when rejecting the negative solution. Set up the equation clearly showing time difference equals the given value.

Question 5. A girl goes to her friend's house, which is at a distance of 12 km. She covers half of the distance at a speed of x km/hr and the remaining distance at a speed of (x + 2) km/hr. If she takes 2 hrs 30 minutes to cover the whole distance, find 'x'.
Answer:
Given:
Total distance = 12 km
First half distance = 6 km at speed x km/hr
Second half distance = 6 km at speed (x + 2) km/hr
Total time = 2 hrs 30 mins = \( \frac{5}{2} \) hrs

Step 1: Find time for each part of the journey.
Time taken to cover first 6 km = \( \frac{6}{x} \) hours
Time taken to cover next 6 km = \( \frac{6}{x + 2} \) hours

Step 2: Set up the equation.
\( \frac{6}{x} + \frac{6}{x + 2} = \frac{5}{2} \)

Step 3: Solve the equation.
\( \frac{6(x + 2) + 6x}{x(x + 2)} = \frac{5}{2} \)

\( \frac{6x + 12 + 6x}{x(x + 2)} = \frac{5}{2} \)

\( \frac{12x + 12}{x(x + 2)} = \frac{5}{2} \)

\( 2(12x + 12) = 5x(x + 2) \)

\( 24x + 24 = 5x^2 + 10x \)

\( 5x^2 - 14x - 24 = 0 \)

Using the quadratic formula or factoring:
\( (5x + 6)(x - 4) = 0 \)

\( x = -\frac{6}{5} \) or \( x = 4 \)

Since speed cannot be negative, x = 4.

The value of x is 4 km/hr.
In simple words: The girl travels the first half at 4 km/hr and the second half at 6 km/hr. We added both times to equal the total time given.

📝 Teacher's Note: Explain that when distance is split into parts with different speeds, we add the individual times. Students often forget to convert minutes to hours.

🎯 Exam Tip: Always convert time to the same units (hours or minutes) before solving. Write the final answer clearly with correct units.

Question 6. A car made a run of 390 km in 'x' hours. If the speed had been 4 km/hour more, it would have taken 2 hours less for the journey. Find 'x'.
Answer:
Given:
Distance = 390 km
Original time = x hours
Speed increase = 4 km/hr
Time saved = 2 hours

Step 1: Find the original speed.
Original speed = \( \frac{390}{x} \) km/hr
New speed = \( \frac{390}{x} + 4 = \frac{390 + 4x}{x} \) km/hr

Step 2: Find the new time.
New time = \( \frac{390}{\frac{390 + 4x}{x}} = \frac{390x}{390 + 4x} \) hours

Step 3: Set up the equation based on time difference.
\( x - \frac{390x}{390 + 4x} = 2 \)

Step 4: Solve the equation.
\( \frac{x(390 + 4x) - 390x}{390 + 4x} = 2 \)

\( \frac{390x + 4x^2 - 390x}{390 + 4x} = 2 \)

\( \frac{4x^2}{390 + 4x} = 2 \)

\( 4x^2 = 2(390 + 4x) \)

\( 4x^2 = 780 + 8x \)

\( 4x^2 - 8x - 780 = 0 \)

\( x^2 - 2x - 195 = 0 \)

\( (x - 15)(x + 13) = 0 \)

\( x = 15 \) or \( x = -13 \)

Since time cannot be negative, x = 15.

The car took 15 hours for the original journey.
In simple words: We found that the original journey took 15 hours. At the original speed, it was 26 km/hr, and at the faster speed, it would be 30 km/hr.

📝 Teacher's Note: Help students understand that we are finding time (x), not speed. The original speed can be calculated as 390÷15 = 26 km/hr.

🎯 Exam Tip: Read the question carefully to see what variable you need to find. Here we find time (x hours), not speed.

Question 7. A goods train leaves a station at 6 p.m., followed by an express train which leaves at 8 p.m. and travels 20 km/hour faster than the goods train. The express train arrives at a station, 1040 km away, 36 minutes before the goods train. Assuming that the speeds of both the train remain constant between the two stations; calculate their speeds.
Answer:
Given:
Distance = 1040 km
Express train speed = Goods train speed + 20 km/hr
Express train leaves 2 hours later but arrives 36 minutes earlier
Total time difference = 2 hours + 36 minutes = \( \frac{13}{5} \) hours

Step 1: Let the speed of goods train be x km/hr.
Speed of express train = (x + 20) km/hr

Step 2: Find travel times.
Time taken by goods train = \( \frac{1040}{x} \) hours
Time taken by express train = \( \frac{1040}{x + 20} \) hours

Step 3: Set up the equation.
The goods train travels for 2 hours more + arrives 36 minutes later
\( \frac{1040}{x} - \frac{1040}{x + 20} = \frac{13}{5} \)

Step 4: Solve the equation.
\( \frac{1040(x + 20) - 1040x}{x(x + 20)} = \frac{13}{5} \)

\( \frac{20800}{x(x + 20)} = \frac{13}{5} \)

\( \frac{1600}{x^2 + 20x} = \frac{1}{5} \)

\( x^2 + 20x = 8000 \)

\( x^2 + 20x - 8000 = 0 \)

\( (x - 80)(x + 100) = 0 \)

\( x = 80 \) or \( x = -100 \)

Since speed cannot be negative, x = 80.

Speed of goods train = 80 km/hr
Speed of express train = 100 km/hr
In simple words: The goods train travels at 80 km/hr and the express train at 100 km/hr. Even though the express train starts 2 hours late, it arrives 36 minutes early because it is much faster.

📝 Teacher's Note: Draw a timeline showing when each train starts and arrives. This helps students visualize the time difference problem clearly.

🎯 Exam Tip: Convert all times to the same unit (hours or minutes). Write both speeds in the final answer as asked in the question.

Question 8. A man bought an article for Rs x and sold it for Rs 16. If his loss was x per cent, find the cost price of the article.
Answer:
Given:
Cost Price (C.P.) = Rs x
Selling Price (S.P.) = Rs 16
Loss = Rs (x - 16)
Loss percentage = x%

Step 1: Use the loss percentage formula.
Loss% = \( \frac{\text{Loss}}{\text{C.P.}} \times 100 \)

Step 2: Substitute the values.
\( x = \frac{x - 16}{x} \times 100 \)

Step 3: Solve the equation.
\( x^2 = 100(x - 16) \)

\( x^2 = 100x - 1600 \)

\( x^2 - 100x + 1600 = 0 \)

\( (x - 80)(x - 20) = 0 \)

\( x = 80 \) or \( x = 20 \)

The cost price of the article is Rs 20 or Rs 80.
In simple words: The man could have bought the article for Rs 20 or Rs 80. In both cases, when he sells for Rs 16, his loss percentage equals the cost price number.

📝 Teacher's Note: Check both answers: If C.P. = 20, loss = 4, loss% = 20%. If C.P. = 80, loss = 64, loss% = 80%. Both work!

🎯 Exam Tip: Always verify your answer by substituting back into the original problem. Both solutions are mathematically correct here.

Question 9. A trader bought an article for Rs x and sold it for Rs 52, thereby making a profit of (x - 10) per cent on his outlay. Calculate the cost price.
Answer:
Given:
Cost Price (C.P.) = Rs x
Selling Price (S.P.) = Rs 52
Profit = Rs (52 - x)
Profit percentage = (x - 10)%

Step 1: Use the profit percentage formula.
Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)

Step 2: Substitute the values.
\( x - 10 = \frac{52 - x}{x} \times 100 \)

Step 3: Solve the equation.
\( x^2 - 10x = 100(52 - x) \)

\( x^2 - 10x = 5200 - 100x \)

\( x^2 + 90x - 5200 = 0 \)

\( (x + 130)(x - 40) = 0 \)

\( x = -130 \) or \( x = 40 \)

Since cost price cannot be negative, x = 40.

The cost price is Rs 40.
In simple words: The trader bought the article for Rs 40 and sold it for Rs 52, making a profit of Rs 12. This gives him a profit of 30%, which equals (40-10)%.

📝 Teacher's Note: Verify: C.P. = 40, S.P. = 52, Profit = 12, Profit% = 30%, and (x-10)% = 30%. Everything matches perfectly.

🎯 Exam Tip: When profit percentage is given in terms of x, set up the equation carefully. Always check your final answer by substituting back.

Question 10. By selling a chair for Rs 75, Mohan gained as much per cent as its cost. Calculate the cost of the chair.
Solution:
Let the C.P. of the chair be Rs x
S.P. of chair = Rs 75
Profit = Rs (75 – x)

We know:
Profit% = \( \frac{\text{Profit}}{\text{C.P.}} \times 100 \)
\( \therefore x = \frac{75 - x}{x} \times 100 \)
\( x^2 = 7500 - 100x \)
\( x^2 + 100x - 7500 = 0 \)
\( (x + 150)(x - 50) = 0 \)
\( x = -150, 50 \)

But, C.P. cannot be negative. So, x = 50.
Hence, the cost of the chair is Rs 50.
In simple words: We made an equation where the profit percent equals the cost price. Then we solved to find the cost is Rs 50.

📝 Teacher's Note: Show students that profit percent = cost price is a special condition. Make them write the profit formula first, then substitute values step by step.

🎯 Exam Tip: Always write "C.P. cannot be negative" when rejecting negative answers. Write the profit percentage formula clearly to get full marks.

Exercise 6D

Question 1. The sum S of n successive odd numbers starting from 3 is given by the relation n(n + 2). Determine n, if the sum is 168.
Solution:
From the given information, we have:
n(n + 2) = 168
\( n^2 + 2n - 168 = 0 \)
\( n^2 + 14n - 12n - 168 = 0 \)
\( n(n + 14) - 12(n + 14) = 0 \)
\( (n + 14)(n - 12) = 0 \)
\( n = -14, 12 \)

But, n cannot be negative.
Therefore, n = 12.
In simple words: We need to find how many odd numbers starting from 3 add up to 168. The answer is 12 numbers.

📝 Teacher's Note: Explain that n represents "how many numbers" so it must be positive. Students can verify: 3+5+7+...+(12 numbers) = 168.

🎯 Exam Tip: Always state "n cannot be negative" when rejecting solutions. Write the given formula n(n+2) = 168 as your first step.

Question 2. A stone is thrown vertically downwards and the formula d = 16t² + 4t gives the distance, d metres, that it falls in t seconds. How long does it take to fall 420 metres?
Solution:
From the given information,
\( 16t^2 + 4t = 420 \)
\( 4t^2 + t - 105 = 0 \)
\( 4t^2 - 20t + 21t - 105 = 0 \)
\( 4t(t - 5) + 21(t - 5) = 0 \)
\( (4t + 21)(t - 5) = 0 \)
\( t = -\frac{21}{4}, 5 \)

But, time cannot be negative.
Thus, the required time taken is 5 seconds.
In simple words: We substitute the distance 420 metres into the formula and solve to find it takes 5 seconds for the stone to fall.

📝 Teacher's Note: Remind students that time is always positive in physics problems. The negative solution has no physical meaning here.

🎯 Exam Tip: Write "time cannot be negative" clearly. Always substitute the given values into the formula as the first step.

Question 3. The product of the digits of a two digit number is 24. If its unit's digit exceeds twice its ten's digit by 2; find the number.
Solution:
Let the ten's and unit's digit of the required number be x and y respectively.
From the given information,
\( x \times y = 24 \)
\( y = \frac{24}{x} \) ...(1)
Also, \( y = 2x + 2 \)
\( \frac{24}{x} = 2x + 2 \) [Using (1)]
\( 24 = 2x^2 + 2x \)
\( 2x^2 + 2x - 24 = 0 \)
\( x^2 + x - 12 = 0 \)
\( (x + 4)(x - 3) = 0 \)
\( x = -4, 3 \)

The digit of a number cannot be negative, so, x = 3.
\( \therefore y = \frac{24}{3} = 8 \)

Thus, the required number is 38.
In simple words: The ten's digit is 3 and unit's digit is 8. Check: 3 × 8 = 24 ✓ and 8 = 2(3) + 2 = 8 ✓

📝 Teacher's Note: Help students understand "unit's digit exceeds twice ten's digit by 2" means unit's digit = 2 × ten's digit + 2. Use examples like 38: 8 exceeds 2×3 by 2.

🎯 Exam Tip: Always verify your answer by checking both conditions. Write "digit cannot be negative" when rejecting solutions.

Question 4. The product of the digits of a two digit number is 24. If its unit's digit exceeds twice its ten's digit by 2; find the number.
Solution:
The ages of two sisters are 11 years and 14 years.
Let in x number of years the product of their ages be 304.
\( \therefore (11 + x)(14 + x) = 304 \)
\( 154 + 11x + 14x + x^2 = 304 \)
\( x^2 + 25x - 150 = 0 \)
\( (x + 30)(x - 5) = 0 \)
\( x = -30, 5 \)

But, the number of years cannot be negative. So, x = 5.
Hence, the required number of years is 5 years.
In simple words: After 5 years, the sisters will be 16 and 19 years old. Their ages multiply to give 16 × 19 = 304.

📝 Teacher's Note: This question seems to have wrong problem statement - it starts with digits but talks about sisters' ages. Focus on the solution method shown.

🎯 Exam Tip: Read the problem carefully. Write "number of years cannot be negative" when rejecting solutions. Verify: (11+5) × (14+5) = 16 × 19 = 304.

Question 5. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his son's age. Find their present ages.
Solution:
Let the present age of the son be x years.
\( \therefore \) Present age of man = \( x^2 \) years

One year ago,
Son's age = (x – 1) years
Man's age = \( (x^2 - 1) \) years

It is given that one year ago; a man was 8 times as old as his son.
\( \therefore (x^2 - 1) = 8(x - 1) \)
\( x^2 - 8x - 1 + 8 = 0 \)
\( x^2 - 8x + 7 = 0 \)
\( (x - 7)(x - 1) = 0 \)
\( x = 7, 1 \)

If x = 1, then \( x^2 = 1 \), which is not possible as father's age cannot be equal to son's age.
So, x = 7.

Present age of son = x years = 7 years
Present age of man = \( x^2 \) years = 49 years
In simple words: The son is 7 years old now and the father is 49 years old. One year ago, they were 6 and 48, and 48 = 8 × 6.

📝 Teacher's Note: Help students see why x = 1 is rejected - it would make father and son the same age, which is impossible.

🎯 Exam Tip: Always check if your answer makes logical sense. Father's age must be greater than son's age. Verify both conditions given in the problem.

Question 6. The age of the father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.
Solution:
Let the present age of the son be x years.
Present age of father = \( 2x^2 \) years

Eight years hence,
Son's age = (x + 8) years
Father's age = \( (2x^2 + 8) \) years

It is given that eight years hence, the age of the father will be 4 years more than three times the age of the son.
\( 2x^2 + 8 = 3(x + 8) + 4 \)
\( 2x^2 + 8 = 3x + 24 + 4 \)
\( 2x^2 - 3x - 20 = 0 \)
\( 2x^2 - 8x + 5x - 20 = 0 \)
\( 2x(x - 4) + 5(x - 4) = 0 \)
\( (x - 4)(2x + 5) = 0 \)
\( x = 4, -\frac{5}{2} \)

But, the age cannot be negative, so, x = 4.
Present age of son = 4 years
Present age of father = \( 2(4)^2 \) years = 32 years
In simple words: The son is 4 years old and father is 32 years old now. After 8 years: son = 12, father = 40, and 40 = 3(12) + 4.

📝 Teacher's Note: Show students how "4 years more than three times" translates to 3(son's age) + 4 in the equation.

🎯 Exam Tip: Write the equation clearly: father's future age = 3 × son's future age + 4. Always verify your final answer by substituting back.

Question 7. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream and return downstream to the original point in 4 hours 30 minutes. Find the speed of the stream.
Solution:
Let the speed of the stream be x km/hr.
\( \therefore \) Speed of the boat downstream = (15 + x) km/hr
Speed of the boat upstream = (15 – x) km/hr

Time taken to go 30 km downstream = \( \frac{30}{15 + x} \) hr
Time taken to come back = \( \frac{30}{15 - x} \) hr

From the given information,
\( \frac{30}{15 + x} + \frac{30}{15 - x} = 4\frac{30}{60} \)
\( \frac{30}{15 + x} + \frac{30}{15 - x} = \frac{9}{2} \)

\( \frac{450 - 30x + 450 + 30x}{(15 + x)(15 - x)} = \frac{9}{2} \)
\( \frac{900}{225 - x^2} = \frac{9}{2} \)
\( \frac{100}{225 - x^2} = \frac{1}{2} \)
\( 225 - x^2 = 200 \)
\( x^2 = 25 \)
\( x = ±5 \)

But, x cannot be negative, so, x = 5.
Thus, the speed of the stream is 5 km/hr.
In simple words: The stream flows at 5 km/hr. This makes the boat go 20 km/hr downstream and 10 km/hr upstream.

📝 Teacher's Note: Help students understand that downstream speed = boat speed + stream speed, upstream speed = boat speed - stream speed.

🎯 Exam Tip: Convert 4 hours 30 minutes to 4.5 hours or 9/2 hours. Write the time formula: Time = Distance/Speed for both directions.

Question 8. Mr. Mehra sends his servant to the market to buy oranges worth Rs 15. The servant having eaten three oranges on the way. Mr. Mehra pays Rs 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.
Solution:
Number of oranges = y
Cost of one orange = Rs \( \frac{15}{y} \)

The servant ate 3 oranges, so Mr. Mehra received (y - 3) oranges.
So, x = y - 3 \( \Rightarrow \) y = x + 3 ...(1)

Cost of one orange paid by Mr. Mehra = Rs \( \frac{15}{y} + 0.25 \)
= Rs \( \frac{15}{x + 3} + \frac{1}{4} \) [Using (1)]

Now, Mr. Mehra pays a total of Rs 15.
\( \therefore \left(\frac{15}{x + 3} + \frac{1}{4}\right) \times x = 15 \)
In simple words: Mr. Mehra gets x oranges but pays for (x+3) oranges at a higher rate. The total payment is still Rs 15.

📝 Teacher's Note: Help students understand that the servant bought (x+3) oranges but gave only x to Mr. Mehra. The cost per orange increases because Mr. Mehra pays for fewer oranges.

🎯 Exam Tip: Set up the equation carefully: (cost per orange) × (number received) = total payment. Convert 25 paise to Rs 0.25.

Question 9. Rs 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.
Answer:
Given:
Total amount = Rs 250
If 25 more children, each gets 50 paise (Rs 0.5) less

Step 1: Let the number of children be \( x \).
Money received by each child = Rs \( \frac{250}{x} \)

Step 2: If there were 25 children more, then
Number of children = \( x + 25 \)
Money received by each child = Rs \( \frac{250}{x + 25} \)

Step 3: From the given information,
\[ \frac{250}{x} - \frac{250}{x + 25} = \frac{50}{100} = \frac{1}{2} \]

Step 4: Solving the equation
\[ \frac{250x + 6250 - 250x}{x(x + 25)} = \frac{1}{2} \]
\[ \frac{6250}{x^2 + 25x} = \frac{1}{2} \]
\[ x^2 + 25x - 12500 = 0 \]

Step 5: Factorizing
\[ (x + 125)(x - 100) = 0 \]
\[ x = -125, 100 \]

Since the number of children cannot be negative, \( x = 100 \).

The number of children = 100
In simple words: We set up an equation based on the fact that when 25 more children share the same money, each child gets 50 paise less. This gives us the answer.

📝 Teacher's Note: Show students how to convert 50 paise to Rs 0.5 or 1/2. Many students forget this conversion and get wrong answers.

🎯 Exam Tip: Always check if your answer makes sense. Here, with 100 children, each gets Rs 2.50. With 125 children, each gets Rs 2. The difference is 50 paise - correct!

Question 10. An employer finds that if he increased the weekly wages of each worker by Rs 5 and employs five workers less, he increases his weekly wage bill from Rs 3,150 to Rs 3,250. Taking the original weekly wage of each worker as Rs x; obtain an equation in x and then solve it to find the weekly wages of each worker.
Answer:
Given:
Original weekly wage of each worker = Rs \( x \)
Original weekly wage bill = Rs 3150
New weekly wage bill = Rs 3250

Step 1: Find number of workers
Number of workers = \( \frac{3150}{x} \)

Step 2: New conditions
New weekly wage of each worker = Rs \( (x + 5) \)
Number of workers = \( \frac{3250}{x + 5} \)

Step 3: From the given condition,
\[ \frac{3150}{x} - 5 = \frac{3250}{x + 5} \]

Step 4: Simplifying
\[ \frac{3150 - 5x}{x} = \frac{3250}{x + 5} \]
\[ 3150x - 5x^2 + 15750 - 25x = 3250x \]
\[ -5x^2 + 15750 - 125x = 0 \]
\[ x^2 + 25x - 3150 = 0 \]
\[ x^2 + 70x - 45x - 3150 = 0 \]
\[ x(x + 70) - 45(x + 70) = 0 \]
\[ (x + 70)(x - 45) = 0 \]
\[ x = -70, 45 \]

Since wage cannot be negative, \( x = 45 \).

The original weekly wage of each worker is Rs 45.
In simple words: We found that if each worker gets Rs 5 more and there are 5 fewer workers, the total bill goes up by Rs 100. This helped us find the original wage.

📝 Teacher's Note: Help students understand that fewer workers with higher pay can still cost more money overall. Use a simple example with small numbers first.

🎯 Exam Tip: Write "wage cannot be negative" to explain why you reject the negative solution. This shows you understand the real-world meaning.

Question 11. A trader bought a number of articles for Rs 1,200. Ten were damaged and he sold each of the remaining articles at Rs 2 more than what he paid for it, thus getting a profit of Rs 60 on whole transaction. Taking the number of articles he bought as x, form an equation in x and solve it.
Answer:
Given:
Number of articles bought = \( x \)
Total cost price = Rs 1200
Ten articles were damaged
Profit on whole transaction = Rs 60

Step 1: Find cost price of one article
Cost price of one article = Rs \( \frac{1200}{x} \)

Step 2: Find selling price and profit
Number of articles sold = \( x - 10 \)
Selling price of each article = Rs \( \left(\frac{1200}{x} + 2\right) \)
Total selling price = \( (x - 10) \left(\frac{1200}{x} + 2\right) \)

Step 3: Set up profit equation
Profit = Selling price - Cost price = Rs 60
\[ (x - 10) \left(\frac{1200}{x} + 2\right) - 1200 = 60 \]

Step 4: Simplifying
\[ 1200 + 2x - \frac{12000}{x} - 20 - 1200 = 60 \]
\[ 2x - \frac{12000}{x} - 80 = 0 \]
\[ 2x^2 - 80x - 12000 = 0 \]
\[ x^2 - 40x - 6000 = 0 \]
\[ x^2 - 100x + 60x - 6000 = 0 \]
\[ x(x - 100) + 60(x - 100) = 0 \]
\[ (x - 100)(x + 60) = 0 \]
\[ x = 100, -60 \]

Since number of articles cannot be negative, \( x = 100 \).

The number of articles bought = 100
In simple words: The trader bought 100 articles. Even though 10 got damaged, he made profit by selling the remaining 90 at Rs 2 more than cost price each.

📝 Teacher's Note: Explain that profit = selling price - cost price. Many students forget to subtract the cost price when calculating profit.

🎯 Exam Tip: Always write down what each variable represents. Here, clearly state that x is the number of articles bought, not sold.

Question 12. The total cost price of a certain number of identical articles is Rs 4800. By selling the articles at Rs 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.
Answer:
Given:
Total cost price = Rs 4800
Selling price of each article = Rs 100
Profit = Cost price of 15 articles

Step 1: Let the number of articles bought be \( x \)
Cost price of one article = Rs \( \frac{4800}{x} \)

Step 2: Find profit
Total selling price = Rs \( 100x \)
Profit = \( 100x - 4800 \)
Cost price of 15 articles = \( 15 \times \frac{4800}{x} = \frac{72000}{x} \)

Step 3: Set up equation
\[ 100x - 4800 = \frac{72000}{x} \]
\[ 100x^2 - 4800x = 72000 \]
\[ x^2 - 48x - 720 = 0 \]
\[ x^2 - 60x + 12x - 720 = 0 \]
\[ x(x - 60) + 12(x - 60) = 0 \]
\[ (x - 60)(x + 12) = 0 \]
\[ x = 60, -12 \]

Since number of articles cannot be negative, \( x = 60 \).

The number of articles bought = 60
In simple words: The trader bought 60 articles for Rs 4800 (Rs 80 each). By selling at Rs 100 each, he made Rs 20 profit per article. Total profit = Rs 1200, which equals cost of 15 articles.

📝 Teacher's Note: Show students how to check: 60 articles cost Rs 4800, so each costs Rs 80. Selling at Rs 100 gives Rs 20 profit each. 60 × Rs 20 = Rs 1200 profit.

🎯 Exam Tip: Always verify your answer by substituting back. Here, check that profit equals cost price of 15 articles: 15 × 80 = Rs 1200. ✓

Exercise 6E

Question 1. The distance by road between two towns A and B is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car. Calculate:
(i) the time taken by the car to reach town B from A, in terms of x;
(ii) the time taken by the train to reach town B from A, in terms of x.
(iii) If the train takes 2 hours less than the car, to reach town B, obtain an equation in x and solve it.
(iv) Hence, find the speed of the train.
Answer:
Given:
Speed of car = \( x \) km/hr
Speed of train = \( (x + 16) \) km/hr
Distance by road = 216 km
Distance by rail = 208 km

(i) Time taken by car = \( \frac{\text{Distance}}{\text{Speed}} = \frac{216}{x} \) hrs

(ii) Time taken by train = \( \frac{208}{x + 16} \) hrs

(iii) From the given information,
\[ \frac{216}{x} - \frac{208}{x + 16} = 2 \]
\[ \frac{216x + 3456 - 208x}{x(x + 16)} = 2 \]
\[ \frac{8x + 3456}{x(x + 16)} = 2 \]
\[ 4x + 1728 = x^2 + 16x \]
\[ x^2 + 12x - 1728 = 0 \]
\[ x^2 + 48x - 36x - 1728 = 0 \]
\[ x(x + 48) - 36(x + 48) = 0 \]
\[ (x + 48)(x - 36) = 0 \]
\[ x = -48, 36 \]

Since speed cannot be negative, \( x = 36 \) km/hr.

(iv) Speed of train = \( (36 + 16) = 52 \) km/hr

In simple words: The car goes at 36 km/hr and takes 6 hours. The train goes at 52 km/hr and takes 4 hours. The train is 2 hours faster.

📝 Teacher's Note: Remind students that Time = Distance ÷ Speed. Draw a simple table showing distance, speed, and time for both car and train.

🎯 Exam Tip: Write all four parts clearly with (i), (ii), (iii), (iv) labels. Show your working for the equation in part (iii) step by step.

Question 2. A trader buys x articles for a total cost of Rs 600.
(i) Write down the cost of one article in terms of x.
If the cost per article were Rs 5 more, the number of articles that can be bought for Rs 600 would be four less.
(ii) Write down the equation in x for the above situation and solve it for x.
Answer:
Given:
Number of articles = \( x \)
Total cost = Rs 600

(i) Cost of one article = Rs \( \frac{600}{x} \)

(ii) From the given information:
If cost per article increases by Rs 5, then
New cost per article = Rs \( \left(\frac{600}{x} + 5\right) \)
New number of articles = \( x - 4 \)

Setting up the equation:
\[ \frac{600}{x - 4} - \frac{600}{x} = 5 \]
\[ \frac{600x - 600x + 2400}{x(x - 4)} = 5 \]
\[ \frac{2400}{x(x - 4)} = 5 \]
\[ 2400 = 5x(x - 4) \]
\[ 480 = x(x - 4) \]
\[ x^2 - 4x - 480 = 0 \]
\[ x^2 - 24x + 20x - 480 = 0 \]
\[ x(x - 24) + 20(x - 24) = 0 \]
\[ (x - 24)(x + 20) = 0 \]
\[ x = 24, -20 \]

Since number of articles cannot be negative, \( x = 24 \).

The trader bought 24 articles.
In simple words: The trader bought 24 articles at Rs 25 each. If each cost Rs 30 instead, he could only buy 20 articles with the same Rs 600.

📝 Teacher's Note: Help students understand that when price goes up, you can buy fewer items with the same money. Use simple examples like buying chocolates.

🎯 Exam Tip: In part (i), write the answer clearly as "Rs 600/x". In part (ii), show all steps of solving the quadratic equation for full marks.

Question 3. A hotel bill for a number of people for overnight stay is Rs 4800. If there were 4 people more, the bill each person had to pay, would have reduced by Rs 200. Find the number of people staying overnight.
Answer:
Given:
Total hotel bill = Rs 4800
If 4 more people join, each person pays Rs 200 less

Step 1: Let the number of people staying overnight be x.

Step 2: Find hotel bill for each person.
Hotel bill for each person = Rs \( \frac{4800}{x} \)

Step 3: Set up equation from given information.
\( \frac{4800}{x} - \frac{4800}{x + 4} = 200 \)

Step 4: Solve the equation.
\( \frac{4800x + 4800 \times 4 - 4800x}{x(x + 4)} = 200 \)
\( \frac{96}{x^2 + 4x} = 1 \)
\( x^2 + 4x - 96 = 0 \)
\( x^2 + 12x - 8x - 96 = 0 \)
\( x(x + 12) - 8(x + 12) = 0 \)
\( (x - 8)(x + 12) = 0 \)
\( x = 8, -12 \)

Step 5: Since the number of people cannot be negative, x = 8.

The number of people staying overnight is 8.
In simple words: We made an equation showing that if 4 more people join the group, each person pays Rs 200 less. When we solve this, we get 8 people were originally staying.

📝 Teacher's Note: Help students see that when more people share the same bill, each person pays less. Use a pizza sharing example - if 4 friends share a pizza, each pays more than if 8 friends share the same pizza.

🎯 Exam Tip: Always write "let x be the number of people" first. Set up the equation carefully - original cost per person minus new cost per person equals the reduction. Check that your answer makes sense.

Question 4. An Aero plane travelled a distance of 400 km at an average speed of x km/hr. On the return journey, the speed was increased by 40 km/hr. Write down an expression for the time taken for: (i) the onward journey; (ii) the return journey. If the return journey took 30 minutes less than the onward journey, write down an equation in x and find its value.
Answer:
Given:
Distance = 400 km
Average speed of the aero plane = x km/hr
Speed while returning = (x + 40) km/hr

(i) We know: Time = \( \frac{\text{Distance}}{\text{Speed}} \)
Time taken for onward journey = \( \frac{400}{x} \) hrs

(ii) Time taken for return journey = \( \frac{400}{(x + 40)} \) hrs

From the given information, we have:
\( \frac{400}{x} - \frac{400}{x + 40} = \frac{30}{60} \)
\( \frac{400x + 16000 - 400x}{x(x + 40)} = \frac{1}{2} \)
\( \frac{16000}{x(x + 40)} = \frac{1}{2} \)
\( x^2 + 40x - 32000 = 0 \)
\( x^2 + 200x - 160x - 32000 = 0 \)
\( x(x + 200) - 160(x + 200) = 0 \)
\( (x + 200)(x - 160) = 0 \)
\( x = -200, 160 \)

Since the speed cannot be negative, x = 160.

The usual speed of the aeroplane is 160 km/hr.
In simple words: We used the time formula (distance ÷ speed) for both trips. The return trip was 30 minutes faster, so we made an equation and solved to find the original speed.

📝 Teacher's Note: Convert 30 minutes to hours first (30/60 = 1/2). Students often forget this conversion. Use a car trip example - if you drive faster on the way back, you save time.

🎯 Exam Tip: Write time = distance/speed clearly for both journeys. Convert minutes to hours (30 min = 0.5 hours). Always check that speed is positive in your final answer.

Question 5. Rs 6500 was divided equally among a certain number of persons. Had there been 15 persons more, each would have got Rs 30 less. Find the original number of persons.
Answer:
Given:
Total money which was divided = Rs 6500

Step 1: Let the original number of persons be x.

Step 2: Each person's share = Rs \( \frac{6500}{x} \)

Step 3: From the given information:
\( \frac{6500}{x} - \frac{6500}{x + 15} = 30 \)

Step 4: Solve the equation.
\( \frac{6500x + 6500 \times 15 - 6500x}{x(x + 15)} = 30 \)
\( \frac{3250}{x(x + 15)} = 1 \)
\( x^2 + 15x - 3250 = 0 \)
\( x^2 + 65x - 50x - 3250 = 0 \)
\( x(x + 65) - 50(x + 65) = 0 \)
\( (x + 65)(x - 50) = 0 \)
\( x = -65, 50 \)

Since the number of persons cannot be negative, the original number of persons is 50.
In simple words: We found that when 15 more people join the sharing, each person gets Rs 30 less. By solving this equation, we get the original number was 50 people.

📝 Teacher's Note: Use a chocolate sharing example. If you share 100 chocolates among 10 friends, each gets 10. If 5 more friends join, each gets less. Students understand this easily.

🎯 Exam Tip: Set up the equation as "original share minus new share equals the reduction". Always check that the number of people is positive. Show all working steps clearly.

Question 6. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time, it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Answer:
Given:
Distance = 1500 km
Plane left 30 minutes late
Speed increased by 250 km/hr

Step 1: Let the usual speed of plane be x km/hr.

Step 2: From the given information:
\( \frac{1500}{x} - \frac{1500}{x + 250} = \frac{30}{60} \)

Step 3: Solve the equation.
\( \frac{1500x + 1500 \times 250 - 1500x}{x(x + 250)} = \frac{1}{2} \)
\( \frac{1500 \times 250}{x^2 + 250x} = \frac{1}{2} \)
\( x^2 + 250x - 750000 = 0 \)
\( x^2 + 1000x - 750x - 750000 = 0 \)
\( x(x + 1000) - 750(x + 1000) = 0 \)
\( (x + 1000)(x - 750) = 0 \)
\( x = -1000, 750 \)

Since speed cannot be negative, x = 750.

The usual speed of plane is 750 km/hr.
In simple words: The plane was late by 30 minutes, so it flew faster to reach on time. We made an equation using this fact and found its normal speed is 750 km/hr.

📝 Teacher's Note: Explain that when you are late, you need to go faster to reach on time. Like running to catch a bus. The time saved by going faster equals the time you were late.

🎯 Exam Tip: Convert 30 minutes to 0.5 hours first. Write the equation as "normal time minus fast time equals time saved". Check that speed is positive.

Question 7. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If after 2 hours, they are 50 km apart, find the speed of each train.
Answer:
Given:
Two trains leave at same time
First train goes west, second goes north
First train is 5 km/hr faster
After 2 hours, they are 50 km apart

Step 1: Let the speed of the second train be x km/hr.
Then the speed of the first train is (x + 5) km/hr.

Step 2: After 2 hours:
Distance travelled by first train = 2(x + 5) km
Distance travelled by second train = 2x km

Step 3: By Pythagoras Theorem (they go at right angles):
\( 50^2 = [2(x + 5)]^2 + (2x)^2 \)
\( 2500 = 4(x + 5)^2 + 4x^2 \)
\( 2500 = 4(x^2 + 25 + 10x) + 4x^2 \)
\( 8x^2 + 40x - 2400 = 0 \)
\( x^2 + 5x - 300 = 0 \)
\( x^2 + 20x - 15x - 300 = 0 \)
\( (x + 20)(x - 15) = 0 \)
\( x = -20 \text{ or } x = 15 \)

Since speed cannot be negative, x = 15.

Speed of second train is 15 km/hr and speed of first train is 20 km/hr.
 

[Diagram: This diagram shows a right-angled triangle with two trains starting from point O, one going west (A) and one going north (B), with the distance AB = 50km after 2 hours.]

In simple words: The trains go in directions that make a right angle. We used Pythagoras theorem to find how far they travelled. Then we solved to get their speeds.

📝 Teacher's Note: Draw a right-angled triangle on the board. Show students that west and north make 90 degrees. Like walking straight and then turning left - you make an L shape.

🎯 Exam Tip: Remember that north and west make a right angle. Use Pythagoras theorem: \( c^2 = a^2 + b^2 \). Always check that both speeds are positive.

Question 8. The sum S of first n even natural numbers is given by the relation S = n(n + 1). Find n, if the sum is 420.
Answer:
Given:
S = n(n + 1)
S = 420

Step 1: Substitute the values.
n(n + 1) = 420
\( n^2 + n - 420 = 0 \)

Step 2: Factorize the equation.
\( n^2 + 21n - 20n - 420 = 0 \)
\( n(n + 21) - 20(n + 21) = 0 \)
\( (n + 21)(n - 20) = 0 \)
\( n = -21, 20 \)

Since n cannot be negative (number of terms), n = 20.

Therefore, n = 20.
In simple words: We substituted S = 420 in the given formula and solved the quadratic equation. The first 20 even numbers add up to 420.

📝 Teacher's Note: Remind students that n means "how many numbers". You cannot have negative numbers to count. The first few even numbers are 2, 4, 6, 8, 10...

🎯 Exam Tip: Always check which value of n makes sense. Since n is the number of terms, it must be positive. Write "n cannot be negative" to show the examiner you understand.

Question 9. The sum of the ages of a father and his son is 45 years. Five years ago, the product of their ages (in years) was 124. Determine their present ages.
Answer:
Let the present ages of father and his son be x years and (45 – x) years respectively.
Five years ago,
Father's age = (x – 5) years
Son's age = (45 – x – 5) years = (40 – x) years
From the given information, we have:
(x – 5)(40 – x) = 124
40x – \( x^2 \) – 200 + 5x = 124
\( x^2 \) – 45x + 324 = 0
\( x^2 \) – 36x – 9x + 324 = 0
x(x – 36) – 9(x – 36) = 0
(x – 36)(x – 9) = 0
x = 36, 9
If x = 9,
Father's age = 9 years, Son's age = (45 – x) = 36 years
This is not possible.
Hence, x = 36
Father's age = 36 years
Son's age = (45 – 36) years = 9 years
In simple words: We make an equation using the fact that 5 years ago their ages multiplied to give 124. Then we solve to find their present ages.

📝 Teacher's Note: Show students that a father cannot be younger than his son. This helps them pick the right answer from the two solutions.

🎯 Exam Tip: Always check if your answer makes sense in real life. Write "Father's age" and "Son's age" clearly in your final answer.

Question 10. In an auditorium, seats were arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. Find:
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Answer:
Let the number of rows in the original arrangement be x.
Then, the number of seats in each row in original arrangement = x
Total number of seats = x × x = \( x^2 \)
From the given information,
2x(x – 10) = \( x^2 \) + 300
2\( x^2 \) – 20x = \( x^2 \) + 300
\( x^2 \) – 20x – 300 = 0
(x – 30)(x + 10) = 0
x = 30, -10
Since, the number of rows or seats cannot be negative. So, x = 30.
(i) The number of rows in the original arrangement = x = 30
(ii) The number of seats after re-arrangement = \( x^2 \) + 300 = 900 + 300 = 1200
In simple words: We start with a square arrangement (same rows and columns). Then we double the rows and reduce seats per row by 10. This gives us 300 more seats total.

📝 Teacher's Note: Draw a simple grid to show students what "rows and columns" means. Show how doubling rows but reducing seats per row can still increase the total.

🎯 Exam Tip: Answer both parts clearly. Write "original arrangement" for part (i) and "after re-arrangement" for part (ii). Don't forget the negative solution is impossible.

Question 11. Mohan takes 16 days less than Manoj to do a piece of work. If both working together can do it in 15 days, in how many days will Mohan alone complete the work?
Answer:
Let the number of days in which Mohan completes the work be x.
Number of days in which Manoj completes the work = x + 16
In one day, Mohan completes \( \frac{1}{x} \) part of work.
In one day, Manoj completes \( \frac{1}{x + 16} \) part of work.
It is given that they both can do the work in 15 days.
\( \frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15} \)
\( \frac{x + 16 + x}{x(x + 16)} = \frac{1}{15} \)
\( \frac{2x + 16}{x^2 + 16x} = \frac{1}{15} \)
30x + 240 = \( x^2 \) + 16x
\( x^2 \) – 14x – 240 = 0
\( x^2 \) – 24x + 10x – 240 = 0
x(x – 24) + 10(x – 24) = 0
(x – 24)(x + 10) = 0
x = 24, -10
Since, the number of days cannot be negative. So, x = 24.
Thus, Mohan alone can complete the work in 24 days.
In simple words: We find how much work each person does in one day. Then we add their daily work to equal the combined daily work when working together.

📝 Teacher's Note: Explain that if someone does work in x days, they do 1/x of the work each day. This fraction concept is key to work problems.

🎯 Exam Tip: Always write "Let x = days for Mohan" clearly at the start. Show the equation \( \frac{1}{x} + \frac{1}{x+16} = \frac{1}{15} \) to get full marks.

Question 12. Two years ago, a man's age was three times the square of his son's age. In three years time, his age will be four times his son's age. Find their present ages.
Answer:
Let the age of son 2 years ago be x years.
Then, father's age 2 years ago = 3\( x^2 \) years
Present age of son = (x + 2) years
Present age of father = (3\( x^2 \) + 2) years
3 years hence:
Son's age = (x + 2 + 3) years = (x + 5) years
Father's age = (3\( x^2 \) + 2 + 3) years = (3\( x^2 \) + 5) years
From the given information,
3\( x^2 \) + 5 = 4(x + 5)
3\( x^2 \) – 4x – 15 = 0
3\( x^2 \) – 9x + 5x – 15 = 0
3x(x – 3) + 5(x – 3) = 0
(x – 3)(3x + 5) = 0
x = 3, \( x = -\frac{5}{3} \)
Since, age cannot be negative. So, x = 3.
Present age of son = (x + 2) years = 5 years
Present age of father = (3\( x^2 \) + 2) years = 29 years
In simple words: We use the relationship between their ages 2 years ago and 3 years from now to make an equation. Then we solve to find their present ages.

📝 Teacher's Note: Make a timeline showing "2 years ago", "present", and "3 years from now" to help students track the different ages clearly.

🎯 Exam Tip: Write the ages clearly for each time period. Always check that your final ages make sense - father should be much older than son.

Question 13. In a certain positive fraction, the denominator is greater than the numerator by 3. If 1 is subtracted from the numerator and the denominator both, the fraction reduces by. Find the fraction.
Answer:
Let the fraction be \( \frac{x}{x + 3} \).
When 1 is subtracted from both numerator and denominator, then
the fraction becomes \( \frac{x - 1}{x + 2} \).
From the given information, we have:
\( \frac{x}{x + 3} - \frac{1}{14} = \frac{x - 1}{x + 2} \)
\( \frac{14x - x - 3}{14(x + 3)} = \frac{x - 1}{x + 2} \)
\( \frac{13x - 3}{14(x + 3)} = \frac{x - 1}{x + 2} \)
(13x - 3)(x + 2) = 14(x - 1)(x + 3)
13\( x^2 \) + 26x - 3x - 6 = 14(\( x^2 \) - x + 3x - 3)
13\( x^2 \) + 23x - 6 = 14\( x^2 \) + 28x - 42
\( x^2 \) + 5x - 36 = 0
\( x^2 \) + 9x - 4x - 36 = 0
x(x + 9) - 4(x + 9) = 0
(x + 9)(x - 4) = 0
x = -9, 4
Since, x cannot be negative. So, x = 4.
Hence, the fraction is \( \frac{x}{x + 3} = \frac{4}{7} \).
In simple words: We set up the original fraction and the new fraction after subtracting 1. The difference between them gives us an equation to solve.

📝 Teacher's Note: The problem statement seems incomplete - it should specify by how much the fraction reduces. Assume it reduces by 1/14 based on the working shown.

🎯 Exam Tip: Always check that your fraction is positive and makes sense. Write the final answer as a simple fraction like 4/7.

Question 14. In a two digit number, the ten's digit is bigger. The product of the digits is 27 and the difference between two digits is 6. Find the number.
Answer:
Given, the difference between two digits is 6 and the ten's digit is bigger than the unit's digit.
So, let the unit's digit be x and ten's digit be (x + 6).
From the given condition, we have:
x(x + 6) = 27
\( x^2 \) + 6x - 27 = 0
\( x^2 \) + 9x - 3x - 27 = 0
x(x + 9) - 3(x + 9) = 0
(x + 9)(x - 3) = 0
x = -9, 3
Since, the digits of a number cannot be negative. So, x = 3.
Unit's digit = 3
Ten's digit = 9
Thus, the number is 93.
In simple words: We use the fact that the digits multiply to 27 and their difference is 6. This gives us an equation to find both digits.

📝 Teacher's Note: Remind students that digits can only be 0-9. Also explain that the ten's digit is the left digit in a two-digit number.

🎯 Exam Tip: Write clearly "unit's digit = " and "ten's digit = " in your answer. State the final two-digit number at the end.

Question 15. Some school children went on an excursion by a bus to a picnic spot at a distance of 300 km. While returning, it was raining and the bus had to reduce its speed by 5 km/hr and it took two hours longer for returning. Find the time taken to return.
Answer:
Distance = 300 km
Let the original speed of the bus be x km/hr.
While returning, speed of the bus = (x - 5) km/hr
From the given information, we have:
\( \frac{300}{x - 5} - \frac{300}{x} = 2 \)
\( \frac{300x - 300x + 1500}{x(x - 5)} = 2 \)
\( \frac{750}{x(x - 5)} = 1 \)
\( x^2 \) - 5x - 750 = 0
\( x^2 \) - 30x + 25x - 750 = 0
x(x - 30) + 25(x - 30) = 0
(x - 30)(x + 25) = 0
x = 30, -25
Since, speed cannot be negative. So, x = 30.
Speed of the bus while returning = 25 km/hr
Time taken by the bus to return = \( \frac{300}{25} \) hrs = 12 hrs
In simple words: We use the formula time = distance ÷ speed. The difference in times gives us an equation to solve for the original speed.

📝 Teacher's Note: Draw a simple diagram showing the journey there and back. Explain that slower speed means more time for the same distance.

🎯 Exam Tip: Write the time formula clearly. Show the equation with fractions. Always check that speed is positive and reasonable.

Question 16. Rs.480 is divided equally among 'x' children. If the number of children were 20 more, then each would have got Rs.12 less. Find 'x'.
Answer:
Amount each child gets when there are x children = \( \frac{480}{x} \) Rs.
Amount each child gets when there are (x + 20) children = \( \frac{480}{x + 20} \) Rs.
From the given information:
\( \frac{480}{x} - \frac{480}{x + 20} = 12 \)
\( \frac{480(x + 20) - 480x}{x(x + 20)} = 12 \)
\( \frac{480x + 9600 - 480x}{x(x + 20)} = 12 \)
\( \frac{9600}{x(x + 20)} = 12 \)
9600 = 12x(x + 20)
800 = x(x + 20)
\( x^2 \) + 20x - 800 = 0
\( x^2 \) + 40x - 20x - 800 = 0
x(x + 40) - 20(x + 40) = 0
(x + 40)(x - 20) = 0
x = -40, 20
Since, the number of children cannot be negative. So, x = 20.
In simple words: We compare how much each child gets in both situations. The difference of Rs.12 helps us make an equation to find the original number of children.

📝 Teacher's Note: Show students that more children means less money for each child. Use simple numbers like "If 4 children share Rs.20, each gets Rs.5" to explain the concept.

🎯 Exam Tip: Write clearly that each child gets 480/x rupees. Set up the equation showing the Rs.12 difference. Check your answer makes sense.

Question 17. A bus covers a distance of 240 km at a uniform speed. Due to heavy rain its speed gets reduced by 10 km/h and as such it takes two hrs longer to covers the total distance. Assuming the uniform speed to be 'x' km/h, form an equation and solve it to evaluate 'x'.
Answer:
Given:
Distance = 240 km
Original speed = x km/h
Reduced speed = (x - 10) km/h
Extra time taken = 2 hours

Time taken by bus to cover total distance with speed x km/h = \( \frac{240}{x} \) hrs

Time taken by bus to cover total distance with speed (x - 10) km/h = \( \frac{240}{x-10} \) hrs

According to the given condition, we have:
\( \frac{240}{x-10} - \frac{240}{x} = 2 \)

\( \Rightarrow 240\left(\frac{1}{x-10} - \frac{1}{x}\right) = 2 \)

\( \Rightarrow \frac{1}{x-10} - \frac{1}{x} = \frac{1}{120} \)

\( \Rightarrow \frac{x-x+10}{x(x-10)} = \frac{1}{120} \)

\( \Rightarrow \frac{10}{x^2-10x} = \frac{1}{120} \)

\( \Rightarrow x^2 - 10x = 1200 \)

\( \Rightarrow x^2 - 10x - 1200 = 0 \)

\( \Rightarrow (x - 40)(x + 30) = 0 \)

\( \Rightarrow x - 40 = 0 \) or \( x + 30 = 0 \)

\( \Rightarrow x = 40 \) or \( x = -30 \)

Since speed cannot be negative, x = 40 km/h.

The original speed of the bus is 40 km/h.
In simple words: We set up an equation using the fact that slower speed takes more time. We used the formula: time = distance ÷ speed. Then we solved the quadratic equation.

📝 Teacher's Note: Show students how time increases when speed decreases. Use simple examples like walking vs running to school. The key is setting up the time equation correctly.

🎯 Exam Tip: Always write "Given" first, then form the equation step by step. Remember that speed cannot be negative, so reject negative answers. Show all algebraic steps clearly.

Question 18. The sum of the ages of Vivek and his younger brother Amit is 47 years. The product of their ages in years is 550. Find their ages.
Answer:
Given:
Sum of ages = 47 years
Product of ages = 550
Vivek is the elder brother

Let the age of Vivek = x years
\( \Rightarrow \) the age of Amit = 47 - x years

The product of their ages in years is 550
\( \Rightarrow x(47 - x) = 550 \)

\( \Rightarrow 47x - x^2 = 550 \)

\( \Rightarrow x^2 - 47x + 550 = 0 \)

\( \Rightarrow x^2 - 25x - 22x + 550 = 0 \)

\( \Rightarrow x(x - 25) - 22(x - 25) = 0 \)

\( \Rightarrow (x - 25)(x - 22) = 0 \)

\( \Rightarrow x = 25 \) or \( x = 22 \)

Given that Vivek is the elder brother:
\( \therefore x = 25 \) years = age of Vivek
Age of Amit = 47 - 25 = 22 years

Vivek's age is 25 years and Amit's age is 22 years.
In simple words: We used algebra to find two numbers that add to 47 and multiply to 550. Since Vivek is older, he gets the bigger age (25 years).

📝 Teacher's Note: Remind students that when we get two solutions, we must check which one makes sense. Here, we know Vivek is elder, so he gets the larger age.

🎯 Exam Tip: Always state clearly who is elder or younger. Write "Let x = " clearly and show the factoring steps. Check your answer by substituting back into the original equations.