OP Malhotra Class 10 Maths Solutions Chapter 5 Quadratic Equations Exercise 5 (E)

On Numbers

Question 1. The sum of the squares of two consecutive positive even integers is 100. Find the integers.
Answer: Let the first positive even integer be \( 2x \). The next consecutive positive even integer will be \( 2x + 2 \). The problem states that the sum of their squares is 100. So, we write the equation: \( (2x)^2 + (2x + 2)^2 = 100 \) Now, we expand and simplify the equation: \( 4x^2 + (4x^2 + 8x + 4) = 100 \) \( 8x^2 + 8x + 4 = 100 \) Move 100 to the left side: \( 8x^2 + 8x + 4 - 100 = 0 \) \( 8x^2 + 8x - 96 = 0 \) Divide the entire equation by 8 to simplify:
\( \implies x^2 + x - 12 = 0 \) Factor the quadratic equation:
\( \implies x^2 + 4x - 3x - 12 = 0 \)
\( \implies x(x + 4) - 3(x + 4) = 0 \)
\( \implies (x + 4)(x - 3) = 0 \) This gives two possible values for \( x \): Either \( x + 4 = 0 \), which means \( x = -4 \). This is not possible because the problem asks for positive even integers, and \( x \) must lead to a positive result. Or \( x - 3 = 0 \), which means \( x = 3 \). This is a valid value for \( x \). Using \( x = 3 \): The first even number is \( 2x = 2 \times 3 = 6 \). The second even number is \( 2x + 2 = 6 + 2 = 8 \). Thus, the two consecutive positive even integers are 6 and 8. You can check this: \( 6^2 + 8^2 = 36 + 64 = 100 \).In simple words: We find two even numbers that come one after the other. When we square each of them and add the squares, the total is 100. These numbers are 6 and 8.

🎯 Exam Tip: Always check if the solution to a quadratic equation makes sense in the context of the word problem, especially when conditions like "positive" or "even" are given.

Question 2. Find two rational numbers which differ by 3 and the sum of whose squares is 117.
Answer: Let the first rational number be \( x \). Since the two numbers differ by 3, the second rational number can be \( x - 3 \) (or \( x + 3 \), but \( x-3 \) is used here for simplicity in initial setup). According to the condition, the sum of their squares is 117: \( x^2 + (x - 3)^2 = 117 \) Expand the equation: \( x^2 + (x^2 - 6x + 9) = 117 \) Combine like terms:
\( \implies 2x^2 - 6x + 9 = 117 \) Move 117 to the left side:
\( \implies 2x^2 - 6x + 9 - 117 = 0 \)
\( \implies 2x^2 - 6x - 108 = 0 \) Divide the entire equation by 2 to simplify:
\( \implies x^2 - 3x - 54 = 0 \) Now, factor the quadratic equation. We need two numbers that multiply to -54 and add to -3. These numbers are -9 and 6.
\( \implies x^2 - 9x + 6x - 54 = 0 \) Factor by grouping:
\( \implies x(x - 9) + 6(x - 9) = 0 \)
\( \implies (x - 9)(x + 6) = 0 \) This gives two possible values for \( x \): Either \( x - 9 = 0 \), which means \( x = 9 \). Or \( x + 6 = 0 \), which means \( x = -6 \). Now we find the second number for each case: Case 1: If \( x = 9 \). The first number is 9. The second number is \( x - 3 = 9 - 3 = 6 \). The numbers are 9 and 6. They differ by 3. Check: \( 9^2 + 6^2 = 81 + 36 = 117 \). This is correct. Case 2: If \( x = -6 \). The first number is -6. The second number is \( x - 3 = -6 - 3 = -9 \). The numbers are -6 and -9. They differ by 3. Check: \( (-6)^2 + (-9)^2 = 36 + 81 = 117 \). This is also correct. Both pairs of numbers (9, 6) and (-6, -9) satisfy the conditions. Rational numbers are numbers that can be written as a simple fraction.In simple words: We are looking for two numbers that are 3 apart, and when you square each of them and add the squares, you get 117. The pairs of numbers are 9 and 6, or -6 and -9.

🎯 Exam Tip: Remember that rational numbers can be positive or negative, so always consider both possible solutions from a quadratic equation unless the problem specifies otherwise.

Question 3. What number increased by its reciprocal equals \( \frac{65}{8} \)?
Answer: Let the unknown number be \( x \). The reciprocal of the number is \( \frac{1}{x} \). According to the condition, the number increased by its reciprocal equals \( \frac{65}{8} \): \( x + \frac{1}{x} = \frac{65}{8} \) To solve this, first eliminate the denominators by multiplying all terms by \( 8x \):
\( \implies 8x(x) + 8x(\frac{1}{x}) = 8x(\frac{65}{8}) \)
\( \implies 8x^2 + 8 = 65x \) Rearrange the equation into standard quadratic form \( ax^2 + bx + c = 0 \):
\( \implies 8x^2 - 65x + 8 = 0 \) Now, we factor this quadratic equation. We need two numbers that multiply to \( 8 \times 8 = 64 \) and add to -65. These numbers are -64 and -1.
\( \implies 8x^2 - 64x - x + 8 = 0 \) Factor by grouping:
\( \implies 8x(x - 8) - 1(x - 8) = 0 \)
\( \implies (x - 8)(8x - 1) = 0 \) This gives two possible values for \( x \): Either \( x - 8 = 0 \), which means \( x = 8 \). Or \( 8x - 1 = 0 \), which means \( 8x = 1 \), so \( x = \frac{1}{8} \). Let's check both solutions: If \( x = 8 \), its reciprocal is \( \frac{1}{8} \). Then \( 8 + \frac{1}{8} = \frac{64}{8} + \frac{1}{8} = \frac{65}{8} \). This is correct. If \( x = \frac{1}{8} \), its reciprocal is 8. Then \( \frac{1}{8} + 8 = \frac{1}{8} + \frac{64}{8} = \frac{65}{8} \). This is also correct. The number can be 8 or \( \frac{1}{8} \).In simple words: We need to find a number where if you add it to its flip-side (its reciprocal), the total is \( \frac{65}{8} \). The number can be 8, or it can be \( \frac{1}{8} \).

🎯 Exam Tip: When dealing with reciprocals, remember that if \( x \) is a solution, \( \frac{1}{x} \) might also be a solution due to the symmetric nature of the equation.

Question 4. The sum of the numerator and denominator of a certain fraction is 8. If 2 is added to the The denominator, the value of the fraction is increased by \( \frac{4}{35} \). Find the fraction.
Answer: Let the numerator of the fraction be \( x \). Since the sum of the numerator and denominator is 8, the denominator will be \( 8 - x \). So, the original fraction is \( \frac{x}{8-x} \). The question implies that 2 is added to *both* the numerator and the denominator, resulting in a new fraction \( \frac{x+2}{(8-x)+2} = \frac{x+2}{10-x} \). According to the condition, the new fraction is equal to the original fraction plus \( \frac{4}{35} \): \( \frac{x+2}{10-x} = \frac{x}{8-x} + \frac{4}{35} \) To solve this, we can first move the original fraction to the left side: \( \frac{x+2}{10-x} - \frac{x}{8-x} = \frac{4}{35} \) Find a common denominator on the left side, which is \( (10-x)(8-x) \):
\( \implies \frac{(x+2)(8-x) - x(10-x)}{(10-x)(8-x)} = \frac{4}{35} \) Expand the terms in the numerator:
\( \implies \frac{(8x - x^2 + 16 - 2x) - (10x - x^2)}{80 - 10x - 8x + x^2} = \frac{4}{35} \) Simplify the numerator and denominator:
\( \implies \frac{8x - x^2 + 16 - 2x - 10x + x^2}{x^2 - 18x + 80} = \frac{4}{35} \)
\( \implies \frac{-4x + 16}{x^2 - 18x + 80} = \frac{4}{35} \) Divide the numerator on the left by 4: \( \frac{4(-x + 4)}{x^2 - 18x + 80} = \frac{4}{35} \) Now, we can cross-multiply, cancelling the 4 from both numerators:
\( \implies \frac{-x + 4}{x^2 - 18x + 80} = \frac{1}{35} \)
\( \implies 35(4 - x) = x^2 - 18x + 80 \)
\( \implies 140 - 35x = x^2 - 18x + 80 \) Rearrange into standard quadratic form:
\( \implies x^2 - 18x + 35x + 80 - 140 = 0 \)
\( \implies x^2 + 17x - 60 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -60 and add to 17. These numbers are 20 and -3.
\( \implies x^2 + 20x - 3x - 60 = 0 \) Factor by grouping:
\( \implies x(x + 20) - 3(x + 20) = 0 \)
\( \implies (x + 20)(x - 3) = 0 \) This gives two possible values for \( x \): Either \( x + 20 = 0 \), which means \( x = -20 \). This is not possible for a "numerator" in this context unless specified for negative fractions. Or \( x - 3 = 0 \), which means \( x = 3 \). This is a valid value for the numerator. Using \( x = 3 \): The numerator is 3. The denominator is \( 8 - x = 8 - 3 = 5 \). So, the fraction is \( \frac{3}{5} \).In simple words: We need to find a fraction where its top and bottom numbers add up to 8. If we add 2 to both the top and bottom of this fraction, the new fraction is \( \frac{4}{35} \) larger than the original. The fraction is \( \frac{3}{5} \).

🎯 Exam Tip: Always double-check word problems for any implicit conditions, like numerators usually being positive for "a fraction" unless negative numbers are explicitly mentioned in context.

Question 5. There are three consecutive integers such that the square of the first increased by the product of the other two given 154. Find the integers.
Answer: Let the three consecutive integers be \( x \), \( x + 1 \), and \( x + 2 \). Consecutive integers follow each other in order, like 5, 6, 7. According to the problem, the square of the first integer increased by the product of the other two is 154: \( x^2 + (x + 1)(x + 2) = 154 \) Expand the product term: \( x^2 + (x^2 + 2x + x + 2) = 154 \) Simplify the equation: \( x^2 + x^2 + 3x + 2 = 154 \)
\( \implies 2x^2 + 3x + 2 = 154 \) Rearrange into standard quadratic form:
\( \implies 2x^2 + 3x + 2 - 154 = 0 \)
\( \implies 2x^2 + 3x - 152 = 0 \) Now, we solve this quadratic equation using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = 2 \), \( b = 3 \), and \( c = -152 \). First, calculate the discriminant \( D = b^2 - 4ac \): \( D = (3)^2 - 4 \times 2 \times (-152) \) \( D = 9 + 1216 \) \( D = 1225 \) Now, substitute the values into the quadratic formula: \( x = \frac{-3 \pm \sqrt{1225}}{2 \times 2} \) \( x = \frac{-3 \pm 35}{4} \) This gives two possible values for \( x \): \( x_1 = \frac{-3 + 35}{4} = \frac{32}{4} = 8 \) \( x_2 = \frac{-3 - 35}{4} = \frac{-38}{4} = -9.5 \) Since we are looking for integers, \( x = -9.5 \) is not a valid solution. So, we take \( x = 8 \). The first integer is \( x = 8 \). The second integer is \( x + 1 = 8 + 1 = 9 \). The third integer is \( x + 2 = 8 + 2 = 10 \). The three consecutive integers are 8, 9, and 10. Check: \( 8^2 + (9 \times 10) = 64 + 90 = 154 \). This is correct.In simple words: We have three numbers in a row. If you square the first number and add it to the product of the other two numbers, you get 154. The numbers are 8, 9, and 10.

🎯 Exam Tip: Always define your variables clearly (e.g., "let the consecutive integers be x, x+1, x+2") and ensure your final answer consists of integers if the problem asks for them.

Question 6. The sum of two numbers is 8 and 15 times the sum of their reciprocals is also 8. Find the numbers.
Answer: Let the first number be \( x \). Since the sum of the two numbers is 8, the second number will be \( 8 - x \). The reciprocals of these numbers are \( \frac{1}{x} \) and \( \frac{1}{8-x} \). According to the second condition, 15 times the sum of their reciprocals is 8: \( 15 \left( \frac{1}{x} + \frac{1}{8-x} \right) = 8 \) First, simplify the sum of reciprocals inside the parenthesis: \( 15 \left( \frac{8-x+x}{x(8-x)} \right) = 8 \) \( 15 \left( \frac{8}{x(8-x)} \right) = 8 \) Multiply 15 by 8 in the numerator: \( \frac{120}{x(8-x)} = 8 \) Now, cross-multiply: \( 120 = 8x(8-x) \) Divide both sides by 8 to simplify:
\( \implies 15 = x(8-x) \) Expand the right side: \( 15 = 8x - x^2 \) Rearrange into standard quadratic form:
\( \implies x^2 - 8x + 15 = 0 \) Factor the quadratic equation. We need two numbers that multiply to 15 and add to -8. These numbers are -5 and -3.
\( \implies x^2 - 5x - 3x + 15 = 0 \) Factor by grouping:
\( \implies x(x - 5) - 3(x - 5) = 0 \)
\( \implies (x - 5)(x - 3) = 0 \) This gives two possible values for \( x \): Either \( x - 5 = 0 \), which means \( x = 5 \). Or \( x - 3 = 0 \), which means \( x = 3 \). If the first number is \( x = 5 \), then the second number is \( 8 - x = 8 - 5 = 3 \). If the first number is \( x = 3 \), then the second number is \( 8 - x = 8 - 3 = 5 \). In both cases, the two numbers are 3 and 5.In simple words: We have two numbers. They add up to 8. If you take their reciprocals, add them, and then multiply the sum by 15, you get 8 again. The two numbers are 3 and 5.

🎯 Exam Tip: When setting up equations with two unknowns that have a given sum, expressing one in terms of the other (e.g., \( y = \text{Sum} - x \)) often simplifies the problem to a single-variable quadratic equation.

Question 7. A two digit number is such that the product of its digits is 12. When 36 is added to this number, the digits interchange their places. Find the number.
Answer: Let the units digit of the two-digit number be \( x \). The problem states that the product of its digits is 12. So, if the units digit is \( x \), the tens digit must be \( \frac{12}{x} \). The original number can be written as \( 10 \times (\text{tens digit}) + (\text{units digit}) \): Original Number \( = 10 \left( \frac{12}{x} \right) + x = \frac{120}{x} + x \) When the digits interchange their places, the new units digit is \( \frac{12}{x} \) and the new tens digit is \( x \). The number with interchanged digits is \( 10 \times (\text{new tens digit}) + (\text{new units digit}) \): New Number \( = 10x + \frac{12}{x} \) According to the condition, when 36 is added to the original number, the digits interchange: Original Number \( + 36 = \) New Number \( \frac{120}{x} + x + 36 = 10x + \frac{12}{x} \) Rearrange the terms to form a quadratic equation. It's helpful to first get rid of fractions by multiplying everything by \( x \): \( 120 + x^2 + 36x = 10x^2 + 12 \) Move all terms to one side to set the equation to zero:
\( \implies 10x^2 - x^2 - 36x + 12 - 120 = 0 \)
\( \implies 9x^2 - 36x - 108 = 0 \) Divide the entire equation by 9 to simplify:
\( \implies x^2 - 4x - 12 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -12 and add to -4. These numbers are -6 and 2.
\( \implies x^2 - 6x + 2x - 12 = 0 \) Factor by grouping:
\( \implies x(x - 6) + 2(x - 6) = 0 \)
\( \implies (x - 6)(x + 2) = 0 \) This gives two possible values for \( x \): Either \( x - 6 = 0 \), which means \( x = 6 \). Or \( x + 2 = 0 \), which means \( x = -2 \). Since a digit must be positive, \( x = -2 \) is not possible. So, the units digit is \( x = 6 \). The tens digit is \( \frac{12}{x} = \frac{12}{6} = 2 \). The original number is 26. Check: Product of digits \( 2 \times 6 = 12 \). Correct. Original number \( 26 + 36 = 62 \). The digits have interchanged. Correct.In simple words: We're looking for a two-digit number. If you multiply its two digits, you get 12. If you add 36 to this number, its digits swap places. The number is 26.

🎯 Exam Tip: When working with two-digit numbers, represent them as \( 10 \times (\text{tens digit}) + (\text{units digit}) \) for clear algebraic formulation. Remember that digits must be non-negative integers (0-9).

Question 8. Five times a certain whole number is equal to three less than twice the square of the number. Find the number.
Answer: Let the whole number be \( x \). The problem states that five times the number is equal to three less than twice its square. So, \( 5x = 2x^2 - 3 \). Rearrange the equation into standard quadratic form \( ax^2 + bx + c = 0 \):
\( \implies 2x^2 - 5x - 3 = 0 \) Factor the quadratic equation. We need two numbers that multiply to \( 2 \times (-3) = -6 \) and add to -5. These numbers are -6 and 1.
\( \implies 2x^2 - 6x + x - 3 = 0 \) Factor by grouping:
\( \implies 2x(x - 3) + 1(x - 3) = 0 \)
\( \implies (x - 3)(2x + 1) = 0 \) This gives two possible values for \( x \): Either \( x - 3 = 0 \), which means \( x = 3 \). Or \( 2x + 1 = 0 \), which means \( 2x = -1 \), so \( x = -\frac{1}{2} \). Since the problem asks for a "whole number," \( x = -\frac{1}{2} \) is not a valid solution (whole numbers are non-negative integers: 0, 1, 2, 3...). Thus, the number is 3. Check: Five times the number is \( 5 \times 3 = 15 \). Twice the square of the number minus three is \( 2 \times (3^2) - 3 = 2 \times 9 - 3 = 18 - 3 = 15 \). The conditions match.In simple words: We need to find a whole number. When you multiply this number by five, you get the same result as when you take twice its square and then subtract three. The number is 3.

🎯 Exam Tip: Always pay attention to keywords like "whole number," "natural number," or "integer," as they restrict the possible solutions and help in eliminating invalid results from quadratic equations.

Question 9. Three consecutive 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: Let the middle number be \( x \). Since they are consecutive numbers, the first number will be \( x - 1 \) and the third number will be \( x + 1 \). The problem states that "the square of the middle number exceeds the difference of the squares of the other two by 60." This means: (Square of middle number) = (Difference of squares of other two) + 60 \( x^2 = [(x + 1)^2 - (x - 1)^2] + 60 \) Expand the squared terms: \( x^2 = [(x^2 + 2x + 1) - (x^2 - 2x + 1)] + 60 \) Simplify the terms inside the square brackets: \( x^2 = [x^2 + 2x + 1 - x^2 + 2x - 1] + 60 \) \( x^2 = [4x] + 60 \) So, the quadratic equation is:
\( \implies x^2 = 4x + 60 \) Rearrange into standard quadratic form:
\( \implies x^2 - 4x - 60 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -60 and add to -4. These numbers are -10 and 6.
\( \implies x^2 - 10x + 6x - 60 = 0 \) Factor by grouping:
\( \implies x(x - 10) + 6(x - 10) = 0 \)
\( \implies (x - 10)(x + 6) = 0 \) This gives two possible values for \( x \): Either \( x - 10 = 0 \), which means \( x = 10 \). Or \( x + 6 = 0 \), which means \( x = -6 \). Since the problem asks for "numbers" and typically for such problems, we consider positive values first. Let's assume positive numbers are implied. A negative middle number would also yield valid consecutive integers. However, usually these problems imply natural numbers unless stated. If \( x = -6 \) was taken: numbers would be -7, -6, -5. Let's use \( x = 10 \). The middle number is 10. The first number is \( x - 1 = 10 - 1 = 9 \). The third number is \( x + 1 = 10 + 1 = 11 \). The three consecutive numbers are 9, 10, and 11. Check: Square of middle number \( = 10^2 = 100 \). Difference of squares of other two \( = 11^2 - 9^2 = 121 - 81 = 40 \). Does \( 100 = 40 + 60 \)? Yes, it does.In simple words: We are looking for three numbers that follow each other. If you square the middle number, it is 60 more than the difference between the squares of the other two numbers. The numbers are 9, 10, and 11.

🎯 Exam Tip: Using \( x-1, x, x+1 \) for consecutive numbers simplifies the algebraic expressions, especially when dealing with squares and differences, due to cancellation of terms.

Question 10. The sum of the ages (in years) of a son and his father is 35 and the product of their ages is 150. Find their ages.
Answer: Let the age of the son be \( x \) years. Since the sum of their ages is 35 years, the father's age will be \( 35 - x \) years. The problem states that the product of their ages is 150: \( x(35 - x) = 150 \) Expand the equation: \( 35x - x^2 = 150 \) Rearrange into standard quadratic form:
\( \implies x^2 - 35x + 150 = 0 \) Factor the quadratic equation. We need two numbers that multiply to 150 and add to -35. These numbers are -30 and -5.
\( \implies x^2 - 30x - 5x + 150 = 0 \) Factor by grouping:
\( \implies x(x - 30) - 5(x - 30) = 0 \)
\( \implies (x - 30)(x - 5) = 0 \) This gives two possible values for \( x \): Either \( x - 30 = 0 \), which means \( x = 30 \). Or \( x - 5 = 0 \), which means \( x = 5 \). Now, let's consider these two possibilities for the son's age: If the son's age is 30 years, then the father's age would be \( 35 - 30 = 5 \) years. This is not logical, as a son cannot be older than his father. If the son's age is 5 years, then the father's age would be \( 35 - 5 = 30 \) years. This is logical. So, the age of the son is 5 years, and the age of the father is 30 years. Check: Sum of ages \( = 5 + 30 = 35 \). Correct. Product of ages \( = 5 \times 30 = 150 \). Correct.In simple words: A son and his father have a combined age of 35 years. If you multiply their ages, you get 150. The son is 5 years old, and the father is 30 years old.

🎯 Exam Tip: Always evaluate the logical consistency of your solutions, especially in word problems involving real-world quantities like age, where some mathematical solutions might be physically impossible.

Question 11. Anjali was born in 1985 A.D. In the year x² A.D., she was (x – 5) years old. Find the value of x.
Answer: Anjali was born in 1985 A.D. Her age in any given year is (Current Year - Birth Year). According to the problem, in the year \( x^2 \) A.D., Anjali's age was \( (x - 5) \) years. So, we can write the equation: \( (x - 5) = x^2 - 1985 \) Rearrange the equation into standard quadratic form:
\( \implies x^2 - x - 5 + 1985 = 0 \)
\( \implies x^2 - x - 1980 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -1980 and add to -1. These numbers are -45 and 44.
\( \implies x^2 - 45x + 44x - 1980 = 0 \) Factor by grouping:
\( \implies x(x - 45) + 44(x - 45) = 0 \)
\( \implies (x - 45)(x + 44) = 0 \) This gives two possible values for \( x \): Either \( x - 45 = 0 \), which means \( x = 45 \). Or \( x + 44 = 0 \), which means \( x = -44 \). Since \( x^2 \) represents a year, it must be a positive value. If \( x = -44 \), then \( x^2 = (-44)^2 = 1936 \). In this year (1936), Anjali's age would be \( x-5 = -44-5 = -49 \) years, which is not possible. If \( x = 45 \), then \( x^2 = 45^2 = 2025 \). In this year (2025), Anjali's age would be \( x-5 = 45-5 = 40 \) years. Let's check: If Anjali was born in 1985, in 2025 her age would be \( 2025 - 1985 = 40 \) years. This matches. So, the value of \( x \) is 45.In simple words: Anjali was born in the year 1985. We're given that in a specific year, \( x^2 \), her age was \( x - 5 \). We need to find the value of \( x \). The correct value for \( x \) is 45.

🎯 Exam Tip: When dealing with dates and ages, ensure that both the resulting year and the calculated age are positive and logical in the context of the problem.

Question 12. The product of their ages is less than the mother's age by 18 years. Find their ages.
Answer: Let the age of the daughter be \( x \) years. The solution indicates that the difference in their ages is 21 years. So, the mother's age is \( x + 21 \) years. According to the condition, the product of their ages is less than the mother's age by 18 years. This means: (Product of ages) = (Mother's age) - 18 \( x(x + 21) = (x + 21) - 18 \) Expand and simplify: \( x^2 + 21x = x + 3 \) Rearrange into standard quadratic form:
\( \implies x^2 + 21x - x - 3 = 0 \)
\( \implies x^2 + 20x - 3 = 0 \) Oops, let me recheck the source. The source has: \( \frac{x(x+21)}{12} = x + 21 – 18 \) which simplifies to \( \frac{x(x+21)}{12} = x + 3 \) This indicates a different initial condition for the problem, probably linked to another part of the question not provided. The phrasing "The product of their ages is less than the mother's age by 18 years" in the question provided leads to my derived quadratic \( x^2 + 20x - 3 = 0 \). However, the source solution clearly follows: \( \frac{x(x+21)}{12} = x + 3 \) This is the setup I must follow as per Iron Rule 6. This means the question statement provided (which appears standalone) is incomplete or refers to a different problem. I will proceed with the setup implied by the solution provided in the OCR. From \( \frac{x(x+21)}{12} = x + 3 \): Multiply both sides by 12: \( x(x + 21) = 12(x + 3) \) Expand both sides: \( x^2 + 21x = 12x + 36 \) Rearrange into standard quadratic form:
\( \implies x^2 + 21x - 12x - 36 = 0 \)
\( \implies x^2 + 9x - 36 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -36 and add to 9. These numbers are 12 and -3.
\( \implies x^2 + 12x - 3x - 36 = 0 \) Factor by grouping:
\( \implies x(x + 12) - 3(x + 12) = 0 \)
\( \implies (x + 12)(x - 3) = 0 \) This gives two possible values for \( x \): Either \( x + 12 = 0 \), which means \( x = -12 \). An age cannot be negative. Or \( x - 3 = 0 \), which means \( x = 3 \). This is a valid age. So, the daughter's age is \( x = 3 \) years. The mother's age is \( x + 21 = 3 + 21 = 24 \) years.In simple words: We are looking for the ages of a mother and daughter. We assume the difference between their ages is 21 years, and using a specific condition (likely related to their ages and a fraction of their product), we find their ages. The daughter is 3 years old, and the mother is 24 years old.

🎯 Exam Tip: Always double-check the exact phrasing of a word problem to correctly set up the initial equation. Be cautious of solutions that seem to refer to an unstated problem or conditions. When an age is calculated, it must always be a positive value.

Question 13. Reena is x years old while her father Mr. Sunil is x² years old. 5 years hence, Mr. Sunil will be three times as old as Reena. Find their present ages.
Answer: Let Reena's current age be \( x \) years. Her father, Mr. Sunil's current age is \( x^2 \) years. "5 years hence" means 5 years from now. Reena's age 5 years from now will be \( x + 5 \) years. Mr. Sunil's age 5 years from now will be \( x^2 + 5 \) years. According to the condition, 5 years from now, Mr. Sunil will be three times as old as Reena: \( x^2 + 5 = 3(x + 5) \) Expand the right side: \( x^2 + 5 = 3x + 15 \) Rearrange into standard quadratic form:
\( \implies x^2 - 3x + 5 - 15 = 0 \)
\( \implies x^2 - 3x - 10 = 0 \) Factor the quadratic equation. We need two numbers that multiply to -10 and add to -3. These numbers are -5 and 2.
\( \implies x^2 - 5x + 2x - 10 = 0 \) Factor by grouping:
\( \implies x(x - 5) + 2(x - 5) = 0 \)
\( \implies (x - 5)(x + 2) = 0 \) This gives two possible values for \( x \): Either \( x - 5 = 0 \), which means \( x = 5 \). Or \( x + 2 = 0 \), which means \( x = -2 \). Since age cannot be negative, \( x = -2 \) is not a possible solution. So, Reena's present age is \( x = 5 \) years. Mr. Sunil's present age is \( x^2 = (5)^2 = 25 \) years. Check: In 5 years, Reena will be \( 5 + 5 = 10 \) years old. In 5 years, Mr. Sunil will be \( 25 + 5 = 30 \) years old. Is \( 30 = 3 \times 10 \)? Yes, it is. The conditions match.In simple words: Reena is \( x \) years old, and her father is \( x^2 \) years old. In 5 years, her father will be three times Reena's age. Reena is currently 5 years old, and her father is 25 years old.

🎯 Exam Tip: Pay close attention to "present age," "x years hence," or "x years ago" phrases, as they are crucial for correctly setting up age-related word problems.

MENSURATION

Question 14. Form a quadratic equation from the following information, taking x as width where x ∈ Ν. (i) The area of a rectangle whose length is five more than twice its width is 75. (ii) Solve the equation and find its length.
Answer: Let the width of the rectangle be \( x \). The length of the rectangle is "five more than twice its width," so Length \( = 2x + 5 \). The area of the rectangle is given as 75. The formula for the area of a rectangle is Length \( \times \) Width. So, \( (2x + 5) \times x = 75 \). (i) Form the quadratic equation: Expand the equation: \( 2x^2 + 5x = 75 \) Rearrange into standard quadratic form:
\( \implies 2x^2 + 5x - 75 = 0 \) (ii) Solve the equation and find its length: Factor the quadratic equation. We need two numbers that multiply to \( 2 \times (-75) = -150 \) and add to 5. These numbers are 15 and -10.
\( \implies 2x^2 + 15x - 10x - 75 = 0 \) Factor by grouping:
\( \implies x(2x + 15) - 5(2x + 15) = 0 \)
\( \implies (2x + 15)(x - 5) = 0 \) This gives two possible values for \( x \): Either \( 2x + 15 = 0 \), which means \( 2x = -15 \), so \( x = -\frac{15}{2} \). A width cannot be negative. Or \( x - 5 = 0 \), which means \( x = 5 \). This is a valid width. So, the width of the rectangle is \( x = 5 \). Now, find the length using Length \( = 2x + 5 \): Length \( = 2(5) + 5 = 10 + 5 = 15 \). The width is 5 units, and the length is 15 units. Check: Area \( = 15 \times 5 = 75 \). This matches the given area.In simple words: We have a rectangle where the length is found by doubling the width and adding 5. The total area is 75. We need to find the length and width. The width is 5 units and the length is 15 units.

🎯 Exam Tip: Always make sure your final answers for dimensions like length or width are positive, as negative values are not physically possible. Clearly state both parts of the answer if the question asks for multiple values.

Question 15. The two sides of a right-angled triangle containing the right angle are 5x cm and (3x – 1) cm. If the area of the triangle be 60 cm², calculate the lengths of the sides of the triangle.
Answer: For a right-angled triangle, the two sides containing the right angle are its base and height. Let these sides be \( \text{base} = 5x \) cm and \( \text{height} = (3x - 1) \) cm. The area of a triangle is given by the formula \( \frac{1}{2} \times \text{base} \times \text{height} \). The area is given as 60 cm². So, we can write the equation: \( \frac{1}{2} \times 5x \times (3x - 1) = 60 \) Multiply both sides by 2 to clear the fraction: \( 5x(3x - 1) = 120 \) Expand the left side: \( 15x^2 - 5x = 120 \) Rearrange into standard quadratic form:
\( \implies 15x^2 - 5x - 120 = 0 \) Divide the entire equation by 5 to simplify:
\( \implies 3x^2 - x - 24 = 0 \) Factor the quadratic equation. We need two numbers that multiply to \( 3 \times (-24) = -72 \) and add to -1. These numbers are -9 and 8.
\( \implies 3x^2 - 9x + 8x - 24 = 0 \) Factor by grouping:
\( \implies 3x(x - 3) + 8(x - 3) = 0 \)
\( \implies (x - 3)(3x + 8) = 0 \) This gives two possible values for \( x \): Either \( x - 3 = 0 \), which means \( x = 3 \). Or \( 3x + 8 = 0 \), which means \( 3x = -8 \), so \( x = -\frac{8}{3} \). A length cannot be negative. So, we take \( x = 3 \). Now, calculate the lengths of the sides: First side \( = 5x = 5 \times 3 = 15 \) cm. Second side \( = 3x - 1 = 3 \times 3 - 1 = 9 - 1 = 8 \) cm. These are the two sides containing the right angle. To calculate the lengths of *the* sides of the triangle, we also need the hypotenuse. Using the Pythagorean theorem, \( \text{hypotenuse}^2 = \text{side1}^2 + \text{side2}^2 \): Hypotenuse \( = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \) cm. So, the lengths of the sides of the triangle are 8 cm, 15 cm, and 17 cm.In simple words: We have a right-angled triangle with two sides next to the right angle being \( 5x \) and \( (3x - 1) \) cm long. The total area is 60 cm². We need to find the length of all three sides. The sides are 8 cm, 15 cm, and 17 cm.

🎯 Exam Tip: Remember to calculate all three sides of a right-angled triangle (including the hypotenuse using the Pythagorean theorem) if the question asks for "the lengths of the sides of the triangle," not just the given two expressions.

Question 16. The hypotenuse of a right angle triangle is 13 cm and the difference between the other two sides is 7 cm. (i) Taking x as the length of the shorter of the two sides, write an equation in 'x' that (ii) Solve the equation obtained in (i) above, and hence find the two unknown sides of the triangle. (ICSE)
Answer: (i) Write an equation in 'x': Let the length of the shorter of the two other sides be \( x \) cm. Since the difference between the two sides is 7 cm, the longer side will be \( x + 7 \) cm. The hypotenuse of the right-angled triangle is 13 cm. According to the Pythagorean Theorem (\( a^2 + b^2 = c^2 \), where \( c \) is the hypotenuse): \( x^2 + (x + 7)^2 = 13^2 \) Expand the equation: \( x^2 + (x^2 + 14x + 49) = 169 \) Combine like terms:
\( \implies 2x^2 + 14x + 49 = 169 \) Rearrange into standard quadratic form:
\( \implies 2x^2 + 14x + 49 - 169 = 0 \)
\( \implies 2x^2 + 14x - 120 = 0 \) Divide the entire equation by 2 to simplify:
\( \implies x^2 + 7x - 60 = 0 \) (ii) Solve the equation and find the two unknown sides: Factor the quadratic equation \( x^2 + 7x - 60 = 0 \). We need two numbers that multiply to -60 and add to 7. These numbers are 12 and -5.
\( \implies x^2 + 12x - 5x - 60 = 0 \) Factor by grouping:
\( \implies x(x + 12) - 5(x + 12) = 0 \)
\( \implies (x + 12)(x - 5) = 0 \) This gives two possible values for \( x \): Either \( x + 12 = 0 \), which means \( x = -12 \). A length cannot be negative. Or \( x - 5 = 0 \), which means \( x = 5 \). This is a valid length for the shorter side. So, the shorter side is \( x = 5 \) cm. The longer side is \( x + 7 = 5 + 7 = 12 \) cm. Thus, the two unknown sides of the triangle are 5 cm and 12 cm. Check: \( 5^2 + 12^2 = 25 + 144 = 169 \). And \( 13^2 = 169 \). The Pythagorean theorem holds true.In simple words: For a right triangle, the longest side (hypotenuse) is 13 cm. The other two sides have a difference of 7 cm. If we let the shorter side be \( x \), we can set up an equation. The two unknown sides are 5 cm and 12 cm.

🎯 Exam Tip: Clearly state the quadratic equation you formed in part (i) before solving it in part (ii). Always verify that side lengths are positive, and remember to state the lengths of *both* unknown sides.

Question 17. The length of a verandah is 3 m more than its breadth. The numerical value of its area is equal to the numerical value of its perimeter. (i) Taking as the breadth of the verandah, write an equation in 'x' that represents the above statement. (ii) Solve the equation in (i) above and hence find the dimension of the verandah.
Answer: (i) Write an equation in 'x': Let the breadth of the verandah be \( x \) m. The length is 3 m more than its breadth, so Length \( = (x + 3) \) m. Area of the verandah \( = \text{Length} \times \text{Breadth} = (x + 3)x = x^2 + 3x \). Perimeter of the verandah \( = 2(\text{Length} + \text{Breadth}) = 2((x + 3) + x) = 2(2x + 3) = 4x + 6 \). According to the problem, the numerical value of its area is equal to the numerical value of its perimeter: Area \( = \) Perimeter \( x^2 + 3x = 4x + 6 \) Rearrange into standard quadratic form:
\( \implies x^2 + 3x - 4x - 6 = 0 \)
\( \implies x^2 - x - 6 = 0 \) (ii) Solve the equation and find the dimensions: Factor the quadratic equation \( x^2 - x - 6 = 0 \). We need two numbers that multiply to -6 and add to -1. These numbers are -3 and 2.
\( \implies x^2 - 3x + 2x - 6 = 0 \) Factor by grouping:
\( \implies x(x - 3) + 2(x - 3) = 0 \)
\( \implies (x - 3)(x + 2) = 0 \) This gives two possible values for \( x \): Either \( x - 3 = 0 \), which means \( x = 3 \). Or \( x + 2 = 0 \), which means \( x = -2 \). Since breadth cannot be negative, \( x = -2 \) is not a possible solution. So, the breadth of the verandah is \( x = 3 \) m. The length of the verandah is \( x + 3 = 3 + 3 = 6 \) m. The dimensions of the verandah are 3 m by 6 m. Check: Area \( = 6 \times 3 = 18 \) sq m. Perimeter \( = 2(6 + 3) = 2(9) = 18 \) m. The numerical values of area and perimeter are equal.In simple words: A verandah is a rectangular space where its length is 3 meters more than its width. The area of this verandah is numerically the same as its perimeter. We need to find its dimensions. The width is 3 meters and the length is 6 meters.

🎯 Exam Tip: When setting up equations involving area and perimeter, ensure you correctly use the formulas for both and that your units for dimensions (length, breadth) are consistent and positive.

Question 18. Two squares have sides x cm and (x + 4) cm. The sum of their areas is 656 sq. cm. Express this as an algebraic equation in x and solve the equation to find the sides of the squares.
Answer: Let the side of the first square be \( x \) cm. Let the side of the second square be \( (x + 4) \) cm. The area of the first square is \( x^2 \) sq. cm. The area of the second square is \( (x + 4)^2 \) sq. cm. The problem states that the sum of their areas is 656 sq. cm. (i) Express this as an algebraic equation in \( x \): \( x^2 + (x + 4)^2 = 656 \) Expand the equation: \( x^2 + (x^2 + 8x + 16) = 656 \) Combine like terms:
\( \implies 2x^2 + 8x + 16 = 656 \) Rearrange into standard quadratic form:
\( \implies 2x^2 + 8x + 16 - 656 = 0 \)
\( \implies 2x^2 + 8x - 640 = 0 \) Divide the entire equation by 2 to simplify:
\( \implies x^2 + 4x - 320 = 0 \) (ii) Solve the equation to find the sides of the squares: Factor the quadratic equation \( x^2 + 4x - 320 = 0 \). We need two numbers that multiply to -320 and add to 4. These numbers are 20 and -16.
\( \implies x^2 + 20x - 16x - 320 = 0 \) Factor by grouping:
\( \implies x(x + 20) - 16(x + 20) = 0 \)
\( \implies (x + 20)(x - 16) = 0 \) This gives two possible values for \( x \): Either \( x + 20 = 0 \), which means \( x = -20 \). A side length cannot be negative. Or \( x - 16 = 0 \), which means \( x = 16 \). This is a valid side length. So, the side of the first square is \( x = 16 \) cm. The side of the second square is \( x + 4 = 16 + 4 = 20 \) cm. The sides of the squares are 16 cm and 20 cm. Check: Area of first square \( = 16^2 = 256 \) sq cm. Area of second square \( = 20^2 = 400 \) sq cm. Sum of areas \( = 256 + 400 = 656 \) sq cm. This matches the given sum.In simple words: We have two squares. One has a side length of \( x \) cm, and the other has a side length of \( (x + 4) \) cm. When you add their areas together, you get 656 square cm. The side lengths of the squares are 16 cm and 20 cm.

🎯 Exam Tip: For geometry problems, ensure all calculated dimensions (like side lengths) are positive. Also, make sure to answer all parts of the question, which in this case includes both forming the equation and finding the side lengths.

Question 19. The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x m. Use the perimeter 180 m to write the value of the breadth in terms of x. Use the values of length, breadth, and the area to write an equation in x. Solve the equation to calculate the length and breadth of the plot.
Answer: The perimeter of the plot is 180 m, and the area is 1800 m². Let the length of the plot be \( x \) m. We know that for a rectangle, Perimeter \( = 2 \times (\text{Length} + \text{Breadth}) \). So, \( 180 = 2 \times (x + \text{Breadth}) \). Dividing by 2, we get \( 90 = x + \text{Breadth} \). This means the breadth \( = (90 - x) \) m. Now, the Area \( = \text{Length} \times \text{Breadth} \). So, \( 1800 = x (90 - x) \). Multiplying \( x \) inside the bracket gives \( 1800 = 90x - x^2 \). Rearranging the terms to form a quadratic equation: \( x^2 - 90x + 1800 = 0 \) To solve this, we can factorize: \( x^2 - 60x - 30x + 1800 = 0 \) \( x(x - 60) - 30(x - 60) = 0 \) \( (x - 60)(x - 30) = 0 \) This gives two possible values for \( x \): Either \( x - 60 = 0 \implies x = 60 \) Or \( x - 30 = 0 \implies x = 30 \) If the length \( x = 60 \) m, then the breadth \( = 90 - 60 = 30 \) m. If the length \( x = 30 \) m, then the breadth \( = 90 - 30 = 60 \) m. Both pairs (60m length, 30m breadth) and (30m length, 60m breadth) satisfy the conditions. The problem asks for length and breadth, which are 60 m and 30 m respectively.In simple words: We used the perimeter to find how length and breadth are related, then used the area to make a quadratic equation. Solving it gave us the length as 60 m and the breadth as 30 m.

🎯 Exam Tip: Remember that length is usually considered greater than breadth. Always check if both solutions are positive and make sense in the real-world context of the problem.

Question 20. A rectangle has an area of 24 cm². If its length is x cm, write down its breadth in terms of x. Given that its perimeter is 20 cm, form an equation in x and solve it.
Answer: The area of the rectangle is 24 cm², and its perimeter is 20 cm. Let the length of the rectangle be \( x \) cm. We know that Perimeter \( = 2 \times (\text{Length} + \text{Breadth}) \). So, \( 20 = 2 \times (x + \text{Breadth}) \). Dividing by 2, we get \( 10 = x + \text{Breadth} \). This means the breadth \( = (10 - x) \) cm. Now, the Area \( = \text{Length} \times \text{Breadth} \). So, \( 24 = x (10 - x) \). Multiplying \( x \) inside the bracket gives \( 24 = 10x - x^2 \). Rearranging the terms to form a quadratic equation: \( x^2 - 10x + 24 = 0 \) To solve this, we can factorize: \( x^2 - 6x - 4x + 24 = 0 \) \( x(x - 6) - 4(x - 6) = 0 \) \( (x - 6)(x - 4) = 0 \) This gives two possible values for \( x \): Either \( x - 6 = 0 \implies x = 6 \) Or \( x - 4 = 0 \implies x = 4 \) If length \( x = 6 \) cm, then breadth \( = 10 - 6 = 4 \) cm. If length \( x = 4 \) cm, then breadth \( = 10 - 4 = 6 \) cm. Since length is typically greater than or equal to breadth, the length is 6 cm and the breadth is 4 cm.In simple words: We used the perimeter formula to write breadth in terms of length, then used the area formula to create a quadratic equation. Solving it gave us the length as 6 cm and breadth as 4 cm.

🎯 Exam Tip: Always make sure your final length and breadth values are consistent with the problem's conditions (e.g., length being greater than or equal to breadth). A diagram can sometimes help visualize the problem.

Question 21. A rectangular garden 10 m by 16 m is to be surrounded by a concrete wall of uniform width. If the area of the wall is 120 square metres, assuming the width of the wall to be x, form an equation in x and solve it to find the value of x.
Answer: The rectangular garden has a length of 16 m and a breadth of 10 m. The concrete wall surrounding it has a uniform width of \( x \) m. This means the outer dimensions of the garden including the wall will be: Outer length \( = (16 + 2x) \) m (adding \( x \) to each side) Outer breadth \( = (10 + 2x) \) m (adding \( x \) to each side) The area of the wall is the difference between the outer area and the inner garden area. Area of wall \( = (\text{Outer Area}) - (\text{Inner Area}) \) We are given that the Area of the wall \( = 120 \) m². So, \( 120 = (16 + 2x)(10 + 2x) - (16 \times 10) \). \( 120 = (160 + 32x + 20x + 4x^2) - 160 \). \( 120 = 4x^2 + 52x \). Rearranging the terms to form a quadratic equation: \( 4x^2 + 52x - 120 = 0 \). Divide the entire equation by 4 to simplify: \( x^2 + 13x - 30 = 0 \). To solve this, we can factorize: \( x^2 + 15x - 2x - 30 = 0 \) \( x(x + 15) - 2(x + 15) = 0 \) \( (x + 15)(x - 2) = 0 \) This gives two possible values for \( x \): Either \( x + 15 = 0 \implies x = -15 \) Or \( x - 2 = 0 \implies x = 2 \) Since the width of the wall cannot be negative, we reject \( x = -15 \). Therefore, the width of the wall \( x = 2 \) m.In simple words: We found the outside size of the garden with the wall, then subtracted the garden's own area from this bigger area to get the wall's area. This made an equation that we solved to find the wall's width is 2 meters.

🎯 Exam Tip: When dealing with a surrounding path or wall, remember to add 2x to both the length and breadth for the outer dimensions, as the width 'x' is added on both sides of each dimension.

Question 22. A man covers a distance of 200 km travelling with a uniform speed of 'x' km per hour. The distance could have been covered in 2 hours less, had the speed been (x + 5) km/hr. Calculate the value of x.
Answer: The total distance to cover is 200 km. Let the uniform speed be \( x \) km/hr. Time taken in the first case \( = \frac{\text{Distance}}{\text{Speed}} = \frac{200}{x} \) hours. If the speed were \( (x + 5) \) km/hr, the time taken would be \( = \frac{200}{x+5} \) hours. We are told that the second time is 2 hours less than the first time. So, \( \frac{200}{x} - \frac{200}{x+5} = 2 \). To solve this equation, find a common denominator: \( \frac{200(x+5) - 200x}{x(x+5)} = 2 \). Simplify the numerator: \( \frac{200x + 1000 - 200x}{x^2 + 5x} = 2 \). \( \frac{1000}{x^2 + 5x} = 2 \). Now, cross-multiply: \( 1000 = 2(x^2 + 5x) \). Divide both sides by 2: \( 500 = x^2 + 5x \). Rearrange into a quadratic equation: \( x^2 + 5x - 500 = 0 \). To solve this, we can factorize: \( x^2 + 25x - 20x - 500 = 0 \) \( x(x + 25) - 20(x + 25) = 0 \) \( (x + 25)(x - 20) = 0 \) This gives two possible values for \( x \): Either \( x + 25 = 0 \implies x = -25 \) Or \( x - 20 = 0 \implies x = 20 \) Since speed cannot be negative, we reject \( x = -25 \). Therefore, the uniform speed \( x = 20 \) km/hr.In simple words: We set up an equation by comparing the time taken at two different speeds. The speed was \( x \), and then \( x+5 \). The difference in time was 2 hours. Solving the equation gave us the original speed as 20 km/hr.

🎯 Exam Tip: In distance-speed-time problems, always ensure your units are consistent (km/hr, hours, km). Remember that speed and time must be positive values in real-world scenarios.

Question 23. An express train makes a run of 240 km at a certain speed. Another train, whose speed is 12 km/hr less than the first train takes an hour longer to make the same trip. Find the speed of the express train in km/hr.
Answer: The distance covered by both trains is 240 km. Let the speed of the express train be \( x \) km/hr. Time taken by the express train \( = \frac{\text{Distance}}{\text{Speed}} = \frac{240}{x} \) hours. The speed of the other train is 12 km/hr less than the express train, so its speed is \( (x - 12) \) km/hr. Time taken by the other train \( = \frac{240}{x-12} \) hours. The other train takes 1 hour longer than the express train. So, \( \frac{240}{x-12} - \frac{240}{x} = 1 \). To solve this equation, find a common denominator: \( \frac{240x - 240(x-12)}{x(x-12)} = 1 \). Simplify the numerator: \( \frac{240x - 240x + 2880}{x^2 - 12x} = 1 \). \( \frac{2880}{x^2 - 12x} = 1 \). Now, cross-multiply: \( 2880 = x^2 - 12x \). Rearrange into a quadratic equation: \( x^2 - 12x - 2880 = 0 \). To solve this, we can factorize: \( x^2 - 60x + 48x - 2880 = 0 \) \( x(x - 60) + 48(x - 60) = 0 \) \( (x - 60)(x + 48) = 0 \) This gives two possible values for \( x \): Either \( x - 60 = 0 \implies x = 60 \) Or \( x + 48 = 0 \implies x = -48 \) Since speed cannot be negative, we reject \( x = -48 \). Therefore, the speed of the express train \( = 60 \) km/hr.In simple words: We wrote down the time each train took using their speeds and the distance. Since one train took 1 hour longer, we made an equation and solved it to find the express train's speed, which was 60 km/hr.

🎯 Exam Tip: When setting up equations for word problems, clearly define your variables. Always check the feasibility of solutions in the context of the problem, especially for quantities like speed, time, or age which must be positive.

Question 24. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km/hr more, it would have taken 30 minutes less for the journey. Find the original speed of the train.
Answer: The total distance covered is 90 km. Let the original speed of the train be \( x \) km/hr. Time taken at original speed \( = \frac{90}{x} \) hours. If the speed were 15 km/hr more, the new speed would be \( (x + 15) \) km/hr. Time taken at increased speed \( = \frac{90}{x+15} \) hours. The problem states that the journey would take 30 minutes less with the increased speed. First, convert 30 minutes to hours: \( 30 \text{ minutes} = \frac{30}{60} = \frac{1}{2} \) hour. So, the equation is: \( \frac{90}{x} - \frac{90}{x+15} = \frac{1}{2} \). Find a common denominator: \( \frac{90(x+15) - 90x}{x(x+15)} = \frac{1}{2} \). Simplify the numerator: \( \frac{90x + 1350 - 90x}{x^2 + 15x} = \frac{1}{2} \). \( \frac{1350}{x^2 + 15x} = \frac{1}{2} \). Now, cross-multiply: \( 1350 \times 2 = x^2 + 15x \). \( 2700 = x^2 + 15x \). Rearrange into a quadratic equation: \( x^2 + 15x - 2700 = 0 \). To solve this, we can factorize: \( x^2 + 60x - 45x - 2700 = 0 \) \( x(x + 60) - 45(x + 60) = 0 \) \( (x + 60)(x - 45) = 0 \) This gives two possible values for \( x \): Either \( x + 60 = 0 \implies x = -60 \) Or \( x - 45 = 0 \implies x = 45 \) Since speed cannot be negative, we reject \( x = -60 \). Therefore, the original speed of the train \( = 45 \) km/hr.In simple words: We compared the time taken at the train's usual speed with the time taken if it went faster. The faster journey was 30 minutes shorter. We made an equation and solved it to find the original speed was 45 km/hr.

🎯 Exam Tip: Always remember to convert time units to be consistent (e.g., minutes to hours) before setting up your equation in distance-speed-time problems.

Question 25. A plane left 30 minutes later than the scheduled time and in order to reach its destination, 1500 km away it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Answer: The total distance to the destination is 1500 km. Let the usual speed of the plane be \( x \) km/hr. Scheduled time taken \( = \frac{1500}{x} \) hours. The plane left 30 minutes late, so it had to increase its speed by 250 km/hr. The new speed is \( (x + 250) \) km/hr. Time taken with increased speed \( = \frac{1500}{x+250} \) hours. Since the plane arrived on time despite leaving 30 minutes late, it means the actual travel time with increased speed is 30 minutes less than the scheduled time. Convert 30 minutes to hours: \( 30 \text{ minutes} = \frac{30}{60} = \frac{1}{2} \) hour. So, the equation is: \( \frac{1500}{x} - \frac{1500}{x+250} = \frac{1}{2} \). Find a common denominator: \( \frac{1500(x+250) - 1500x}{x(x+250)} = \frac{1}{2} \). Simplify the numerator: \( \frac{1500x + 375000 - 1500x}{x^2 + 250x} = \frac{1}{2} \). \( \frac{375000}{x^2 + 250x} = \frac{1}{2} \). Now, cross-multiply: \( 375000 \times 2 = x^2 + 250x \). \( 750000 = x^2 + 250x \). Rearrange into a quadratic equation: \( x^2 + 250x - 750000 = 0 \). To solve this, we can factorize: \( x^2 + 1000x - 750x - 750000 = 0 \) \( x(x + 1000) - 750(x + 1000) = 0 \) \( (x + 1000)(x - 750) = 0 \) This gives two possible values for \( x \): Either \( x + 1000 = 0 \implies x = -1000 \) Or \( x - 750 = 0 \implies x = 750 \) Since speed cannot be negative, we reject \( x = -1000 \). Therefore, the usual speed of the plane \( = 750 \) km/hr.In simple words: The plane was late, so it flew faster to make up for lost time. The difference between the usual travel time and the faster travel time was 30 minutes. We used this to set up and solve an equation, finding the usual speed was 750 km/hr.

🎯 Exam Tip: When a delay is involved, the time difference between the original planned journey and the actual journey is equal to the delay itself, if the destination is reached on schedule.

Question 26. A boat takes 1 hour longer to go 36 km up a river than to return. If the river flows at 3 km/hr, find the rate at which the boat travels in still water.
Answer: The distance for both upstream and downstream journeys is 36 km. The speed of the river (current) is 3 km/hr. Let the speed of the boat in still water be \( x \) km/hr. When going upstream (against the current), the effective speed of the boat \( = (x - 3) \) km/hr. Time taken to go upstream \( = \frac{36}{x-3} \) hours. When going downstream (with the current), the effective speed of the boat \( = (x + 3) \) km/hr. Time taken to return downstream \( = \frac{36}{x+3} \) hours. The boat takes 1 hour longer to go upstream than to return downstream. So, \( \frac{36}{x-3} - \frac{36}{x+3} = 1 \). Find a common denominator: \( \frac{36(x+3) - 36(x-3)}{(x-3)(x+3)} = 1 \). Simplify the numerator: \( \frac{36x + 108 - 36x + 108}{x^2 - 9} = 1 \). \( \frac{216}{x^2 - 9} = 1 \). Now, cross-multiply: \( 216 = x^2 - 9 \). Rearrange into a quadratic equation: \( x^2 - 9 - 216 = 0 \). \( x^2 - 225 = 0 \). This can be written as \( x^2 - 15^2 = 0 \), which is a difference of squares. \( (x - 15)(x + 15) = 0 \). This gives two possible values for \( x \): Either \( x - 15 = 0 \implies x = 15 \) Or \( x + 15 = 0 \implies x = -15 \) Since speed cannot be negative, we reject \( x = -15 \). Therefore, the speed of the boat in still water \( = 15 \) km/hr.In simple words: We calculated how long the boat would take to go upstream (slower) and downstream (faster). The upstream journey took one hour more. We set up an equation with these times and found the boat's speed in calm water to be 15 km/hr.

🎯 Exam Tip: For boat and stream problems, remember that upstream speed is \( (\text{boat speed} - \text{stream speed}) \) and downstream speed is \( (\text{boat speed} + \text{stream speed}) \).

Profit and Loss

Question 27. A man purchased some horses for Rs. 3000. Three of them died, and he sold the rest at Rs. 65 more than what he paid for each horse and thus gains 6% on his outlay. How many horses did he buy?
Answer: The total cost price of horses is Rs. 3000. Let the number of horses purchased be \( x \). The cost price of each horse \( = \frac{3000}{x} \) Rs. Three horses died, so the remaining horses are \( (x - 3) \). The man sold the remaining horses at Rs. 65 more than what he paid for each. So, the selling price of each remaining horse \( = \left( \frac{3000}{x} + 65 \right) \) Rs. The total selling price \( = (x - 3) \times \left( \frac{3000}{x} + 65 \right) \) Rs. He gained 6% on his outlay (total cost price). Total gain \( = 6\% \text{ of } 3000 = \frac{6}{100} \times 3000 = 180 \) Rs. Total Selling Price \( = \text{Total Cost Price} + \text{Gain} = 3000 + 180 = 3180 \) Rs. Now, we can form the equation: \( (x - 3) \times \left( \frac{3000}{x} + 65 \right) = 3180 \). Expand the left side: \( (x - 3) \left( \frac{3000 + 65x}{x} \right) = 3180 \). \( \frac{(x - 3)(3000 + 65x)}{x} = 3180 \). \( (x - 3)(3000 + 65x) = 3180x \). \( 3000x + 65x^2 - 9000 - 195x = 3180x \). \( 65x^2 + (3000 - 195)x - 9000 = 3180x \). \( 65x^2 + 2805x - 9000 = 3180x \). Move all terms to one side to form a quadratic equation: \( 65x^2 + 2805x - 3180x - 9000 = 0 \). \( 65x^2 - 375x - 9000 = 0 \). Divide the entire equation by 5 to simplify: \( 13x^2 - 75x - 1800 = 0 \). To solve this, we can factorize using the product-sum method (factors of \( 13 \times -1800 = -23400 \) that sum to -75): \( 13x^2 - 195x + 120x - 1800 = 0 \) \( 13x(x - 15) + 120(x - 15) = 0 \) \( (x - 15)(13x + 120) = 0 \) This gives two possible values for \( x \): Either \( x - 15 = 0 \implies x = 15 \) Or \( 13x + 120 = 0 \implies 13x = -120 \implies x = -\frac{120}{13} \) Since the number of horses cannot be negative, we reject \( x = -\frac{120}{13} \). Therefore, the number of horses purchased \( = 15 \).In simple words: We set up equations for the cost of each horse, how many were left, and how much each remaining horse was sold for. We also figured out the total profit. Combining these helped us find the original number of horses bought, which was 15.

🎯 Exam Tip: For profit and loss problems, carefully distinguish between cost price, selling price, total items, and items remaining. Always account for any losses or damages before calculating total selling price and profit/loss percentage.

Question 28. A trader bought a number of articles for Rs. 1200. Ten were damaged and he sold each of the rest at Rs. 2 more than what he paid for it, thus getting a profit of Rs. 60 on the whole transaction. Taking the number of articles he bought as x, form an equation in x and solve.
Answer: The total cost price for the articles is Rs. 1200. Let the number of articles bought by the trader be \( x \). The cost price of each article \( = \frac{1200}{x} \) Rs. Ten articles were damaged, so the remaining articles are \( (x - 10) \). He sold each of the remaining articles at Rs. 2 more than what he paid for it. So, the selling price of each remaining article \( = \left( \frac{1200}{x} + 2 \right) \) Rs. The total gain on the whole transaction is Rs. 60. Total Selling Price \( = \text{Total Cost Price} + \text{Gain} = 1200 + 60 = 1260 \) Rs. The total selling price is also the number of remaining articles multiplied by the selling price per article: Total Selling Price \( = (x - 10) \times \left( \frac{1200}{x} + 2 \right) \). So, we can form the equation: \( (x - 10) \times \left( \frac{1200}{x} + 2 \right) = 1260 \). Expand the left side: \( (x - 10) \left( \frac{1200 + 2x}{x} \right) = 1260 \). \( \frac{(x - 10)(1200 + 2x)}{x} = 1260 \). \( (x - 10)(1200 + 2x) = 1260x \). \( 1200x + 2x^2 - 12000 - 20x = 1260x \). \( 2x^2 + (1200 - 20)x - 12000 = 1260x \). \( 2x^2 + 1180x - 12000 = 1260x \). Move all terms to one side to form a quadratic equation: \( 2x^2 + 1180x - 1260x - 12000 = 0 \). \( 2x^2 - 80x - 12000 = 0 \). Divide the entire equation by 2 to simplify: \( x^2 - 40x - 6000 = 0 \). To solve this, we can factorize: \( x^2 - 100x + 60x - 6000 = 0 \) \( x(x - 100) + 60(x - 100) = 0 \) \( (x - 100)(x + 60) = 0 \) This gives two possible values for \( x \): Either \( x - 100 = 0 \implies x = 100 \) Or \( x + 60 = 0 \implies x = -60 \) Since the number of articles cannot be negative, we reject \( x = -60 \). Therefore, the number of articles bought \( = 100 \).In simple words: We found the cost of one article, then figured out how many were left after some were damaged. We used the profit amount to get the total selling price and then set up an equation. Solving it showed that the trader bought 100 articles.

🎯 Exam Tip: Be careful with the "per item" calculations. The cost price per item is based on the total items bought, but the selling price per item for profit calculation applies only to the undamaged items sold.

Self Evaluation and Revision (LATEST ICSE QUESTIONS)

Question 1. Solve for x and give your answer correct to 2 decimal places : \( x^2 - 10x + 6 = 0 \).
Answer: We need to solve the quadratic equation \( x^2 - 10x + 6 = 0 \) using the quadratic formula, and give the answer correct to 2 decimal places. The general quadratic equation is \( ax^2 + bx + c = 0 \). Comparing our equation, we have \( a = 1 \), \( b = -10 \), and \( c = 6 \). First, calculate the discriminant \( D = b^2 - 4ac \). \( D = (-10)^2 - 4 \times 1 \times 6 \) \( D = 100 - 24 \) \( D = 76 \) Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). \( x = \frac{-(-10) \pm \sqrt{76}}{2 \times 1} \) \( x = \frac{10 \pm \sqrt{76}}{2} \) To simplify \( \sqrt{76} \), we can write \( \sqrt{76} = \sqrt{4 \times 19} = 2\sqrt{19} \). So, \( x = \frac{10 \pm 2\sqrt{19}}{2} \). Divide both terms in the numerator by 2: \( x = 5 \pm \sqrt{19} \). Now, we need the value of \( \sqrt{19} \). \( \sqrt{19} \approx 4.3589 \). So, \( x_1 = 5 + 4.3589 \approx 9.3589 \). Rounded to 2 decimal places, \( x_1 = 9.36 \). And \( x_2 = 5 - 4.3589 \approx 0.6411 \). Rounded to 2 decimal places, \( x_2 = 0.64 \). Thus, the solutions for \( x \) are 9.36 and 0.64.In simple words: We used a special formula to solve the equation. We found two answers for \( x \), which are about 9.36 and 0.64 when rounded to two decimal places.

🎯 Exam Tip: When using the quadratic formula, always calculate the discriminant first to ensure there are real roots. Pay close attention to rounding instructions and carry enough decimal places in intermediate steps for accuracy.

Question 2. Solve using the quadratic formula \( x^2 - 4 + 1 = 0 \).
Answer: The given equation is \( x^2 - 4 + 1 = 0 \). This simplifies to \( x^2 - 3 = 0 \). Wait, the OCR seems to have mistyped the quadratic equation from the source. It is most likely \( x^2 - 4x + 1 = 0 \) as is standard for these types of problems. Let's assume the question meant \( x^2 - 4x + 1 = 0 \) to follow the typical quadratic formula structure. The general quadratic equation is \( ax^2 + bx + c = 0 \). Comparing our equation \( x^2 - 4x + 1 = 0 \), we have \( a = 1 \), \( b = -4 \), and \( c = 1 \). First, calculate the discriminant \( D = b^2 - 4ac \). \( D = (-4)^2 - 4 \times 1 \times 1 \) \( D = 16 - 4 \) \( D = 12 \) Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). \( x = \frac{-(-4) \pm \sqrt{12}}{2 \times 1} \) \( x = \frac{4 \pm \sqrt{12}}{2} \) To simplify \( \sqrt{12} \), we can write \( \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \). So, \( x = \frac{4 \pm 2\sqrt{3}}{2} \). Divide both terms in the numerator by 2: \( x = 2 \pm \sqrt{3} \). Now, we need the value of \( \sqrt{3} \). \( \sqrt{3} \approx 1.732 \). So, \( x_1 = 2 + 1.732 \approx 3.732 \). Rounded to 3 decimal places, \( x_1 = 3.732 \). And \( x_2 = 2 - 1.732 \approx 0.268 \). Rounded to 3 decimal places, \( x_2 = 0.268 \). Thus, the solutions for \( x \) are 3.732 and 0.268.In simple words: We used the quadratic formula to solve the equation. We found two answers for \( x \), which are about 3.732 and 0.268.

🎯 Exam Tip: Always double-check the quadratic equation given in the question. If there's an obvious typo, like a missing 'x' term, clarify or solve for the most probable intended equation, clearly stating your assumption if needed. The quadratic formula is reliable for all quadratic equations.

Question 3. Solve the equation \( 3x^2 - x - 7 = 0 \) and give your answer correct to two decimal places.
Answer: We need to solve the quadratic equation \( 3x^2 - x - 7 = 0 \) using the quadratic formula, and give the answer correct to two decimal places. The general quadratic equation is \( ax^2 + bx + c = 0 \). Comparing our equation, we have \( a = 3 \), \( b = -1 \), and \( c = -7 \). First, calculate the discriminant \( D = b^2 - 4ac \). \( D = (-1)^2 - 4 \times 3 \times (-7) \) \( D = 1 + 84 \) \( D = 85 \) Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). \( x = \frac{-(-1) \pm \sqrt{85}}{2 \times 3} \) \( x = \frac{1 \pm \sqrt{85}}{6} \) Now, we need the value of \( \sqrt{85} \). \( \sqrt{85} \approx 9.2195 \). So, \( x_1 = \frac{1 + 9.2195}{6} = \frac{10.2195}{6} \approx 1.7032 \). Rounded to two decimal places, \( x_1 = 1.70 \). And \( x_2 = \frac{1 - 9.2195}{6} = \frac{-8.2195}{6} \approx -1.3699 \). Rounded to two decimal places, \( x_2 = -1.37 \). Thus, the solutions for \( x \) are 1.70 and -1.37.In simple words: We solved the given quadratic equation using the formula. The two answers for \( x \) are approximately 1.70 and -1.37, rounded to two decimal places.

🎯 Exam Tip: Be careful with negative signs, especially when substituting values into the quadratic formula. Double-check your calculations for the discriminant and the final square root approximation.

Question 4. Solve the following equation and give your answer up to two decimal places : \( x^2 - 5x - 10 = 0 \).
Answer: We need to solve the quadratic equation \( x^2 - 5x - 10 = 0 \) using the quadratic formula, and give the answer correct up to two decimal places. The general quadratic equation is \( ax^2 + bx + c = 0 \). Comparing our equation, we have \( a = 1 \), \( b = -5 \), and \( c = -10 \). First, calculate the discriminant \( D = b^2 - 4ac \). \( D = (-5)^2 - 4 \times 1 \times (-10) \) \( D = 25 + 40 \) \( D = 65 \) Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). \( x = \frac{-(-5) \pm \sqrt{65}}{2 \times 1} \) \( x = \frac{5 \pm \sqrt{65}}{2} \) Now, we need the value of \( \sqrt{65} \). \( \sqrt{65} \approx 8.0622 \). So, \( x_1 = \frac{5 + 8.0622}{2} = \frac{13.0622}{2} \approx 6.5311 \). Rounded to two decimal places, \( x_1 = 6.53 \). And \( x_2 = \frac{5 - 8.0622}{2} = \frac{-3.0622}{2} \approx -1.5311 \). Rounded to two decimal places, \( x_2 = -1.53 \). Thus, the solutions for \( x \) are 6.53 and -1.53.In simple words: We used the quadratic formula to find the two possible values of \( x \) for the given equation. The answers, rounded to two decimal places, are 6.53 and -1.53.

🎯 Exam Tip: Always state the quadratic formula clearly before substituting values. This helps prevent errors and shows your understanding of the method.

Question 5. Solve the equation \( 2x - \frac{1}{x} = 7 \). Write your answer correct to two decimal places.
Answer: We need to solve the equation \( 2x - \frac{1}{x} = 7 \) and give the answer correct to two decimal places. First, clear the fraction by multiplying the entire equation by \( x \) (assuming \( x \ne 0 \)): \( 2x \times x - \frac{1}{x} \times x = 7 \times x \). \( 2x^2 - 1 = 7x \). Rearrange into a standard quadratic equation form \( ax^2 + bx + c = 0 \): \( 2x^2 - 7x - 1 = 0 \). Here, \( a = 2 \), \( b = -7 \), and \( c = -1 \). First, calculate the discriminant \( D = b^2 - 4ac \). \( D = (-7)^2 - 4 \times 2 \times (-1) \) \( D = 49 + 8 \) \( D = 57 \) Now, use the quadratic formula \( x = \frac{-b \pm \sqrt{D}}{2a} \). \( x = \frac{-(-7) \pm \sqrt{57}}{2 \times 2} \) \( x = \frac{7 \pm \sqrt{57}}{4} \) Now, we need the value of \( \sqrt{57} \). \( \sqrt{57} \approx 7.5498 \). So, \( x_1 = \frac{7 + 7.5498}{4} = \frac{14.5498}{4} \approx 3.6374 \). Rounded to two decimal places, \( x_1 = 3.64 \). And \( x_2 = \frac{7 - 7.5498}{4} = \frac{-0.5498}{4} \approx -0.1374 \). Rounded to two decimal places, \( x_2 = -0.14 \). Thus, the solutions for \( x \) are 3.64 and -0.14.In simple words: We changed the given equation into a standard quadratic form, then used the quadratic formula to solve for \( x \). The answers are about 3.64 and -0.14, when rounded.

🎯 Exam Tip: When an equation contains fractions with variables, always multiply by the least common multiple of the denominators to eliminate the fractions and simplify it into a standard quadratic form before solving.

Question 6. The bill for a number of people for overnight stay is Rs. 4800. If there were 4 more, the bill each person had to pay would have reduced by Rs. 200. Find the number of people staying overnight.
Answer: The total bill for the overnight stay is Rs. 4800. Let the original number of people be \( x \). The share of each person originally \( = \frac{4800}{x} \) Rs. If there were 4 more persons, the new number of people would be \( (x + 4) \). The new share of each person \( = \frac{4800}{x+4} \) Rs. The problem states that the bill each person had to pay would have reduced by Rs. 200. So, the original share minus the new share equals Rs. 200: \( \frac{4800}{x} - \frac{4800}{x+4} = 200 \). Factor out 4800: \( 4800 \left( \frac{1}{x} - \frac{1}{x+4} \right) = 200 \). Divide both sides by 4800: \( \frac{1}{x} - \frac{1}{x+4} = \frac{200}{4800} \). Simplify the fraction on the right: \( \frac{200}{4800} = \frac{2}{48} = \frac{1}{24} \). So, \( \frac{1}{x} - \frac{1}{x+4} = \frac{1}{24} \). Find a common denominator on the left side: \( \frac{(x+4) - x}{x(x+4)} = \frac{1}{24} \). Simplify the numerator: \( \frac{4}{x^2 + 4x} = \frac{1}{24} \). Now, cross-multiply: \( 4 \times 24 = x^2 + 4x \). \( 96 = x^2 + 4x \). Rearrange into a quadratic equation: \( x^2 + 4x - 96 = 0 \). To solve this, we can factorize: \( x^2 + 12x - 8x - 96 = 0 \) \( x(x + 12) - 8(x + 12) = 0 \) \( (x + 12)(x - 8) = 0 \) This gives two possible values for \( x \): Either \( x + 12 = 0 \implies x = -12 \) Or \( x - 8 = 0 \implies x = 8 \) Since the number of people cannot be negative, we reject \( x = -12 \). Therefore, the number of people staying overnight \( = 8 \).In simple words: We wrote expressions for the cost per person with the original number of people and with 4 extra people. The difference in these costs was Rs. 200. Solving the resulting equation showed that 8 people originally stayed overnight.

🎯 Exam Tip: In problems involving shared costs, the "per person" cost is usually inversely proportional to the number of people. Setting up the difference in per-person costs is a common strategy.

Question 7. An aeroplane 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: The total distance travelled is 400 km. Let the average speed of the aeroplane on the onward journey be \( x \) km/hr. (i) Time taken for the onward journey \( = \frac{\text{Distance}}{\text{Speed}} = \frac{400}{x} \) hours. (ii) On the return journey, the speed was increased by 40 km/hr. So, the speed on the return journey \( = (x + 40) \) km/hr. Time taken for the return journey \( = \frac{400}{x+40} \) hours. We are given that the return journey took 30 minutes less than the onward journey. First, convert 30 minutes to hours: \( 30 \text{ minutes} = \frac{30}{60} = \frac{1}{2} \) hour. So, the equation is: Time for onward journey - Time for return journey = 30 minutes (in hours) \( \frac{400}{x} - \frac{400}{x+40} = \frac{1}{2} \). To solve this, find a common denominator: \( \frac{400(x+40) - 400x}{x(x+40)} = \frac{1}{2} \). Simplify the numerator: \( \frac{400x + 16000 - 400x}{x^2 + 40x} = \frac{1}{2} \). \( \frac{16000}{x^2 + 40x} = \frac{1}{2} \). Now, cross-multiply: \( 16000 \times 2 = x^2 + 40x \). \( 32000 = x^2 + 40x \). Rearrange into a quadratic equation: \( x^2 + 40x - 32000 = 0 \). To solve this, we can factorize: \( x^2 + 200x - 160x - 32000 = 0 \) \( x(x + 200) - 160(x + 200) = 0 \) \( (x + 200)(x - 160) = 0 \) This gives two possible values for \( x \): Either \( x + 200 = 0 \implies x = -200 \) Or \( x - 160 = 0 \implies x = 160 \) Since speed cannot be negative, we reject \( x = -200 \). Therefore, the value of \( x \) (original speed) \( = 160 \) km/hr.In simple words: We wrote the time taken for each part of the journey. Since the return trip was 30 minutes faster, we made an equation to find the plane's original speed, which was 160 km/hr.

🎯 Exam Tip: Break down word problems into smaller parts. First, write expressions for individual quantities like "time for onward journey", then use the given conditions (like "30 minutes less") to form the main equation.

Question 8. In an auditorium, seats were arranged in rows and columns. The number of rows were equal to the number of seats in each row. When the number of rows were doubled and the number of seats in each row were 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 rearrangement.
Answer: Let the original number of rows be \( x \). Since the number of rows was equal to the number of seats in each row, the original number of columns is also \( x \). Original total number of seats \( = \text{Number of rows} \times \text{Number of columns} = x \times x = x^2 \). After rearrangement: The number of rows was doubled, so new number of rows \( = 2x \). The number of seats in each row (columns) was reduced by 10, so new number of columns \( = (x - 10) \). New total number of seats \( = 2x \times (x - 10) \). The total number of seats increased by 300. So, New total number of seats = Original total number of seats + 300. \( 2x(x - 10) = x^2 + 300 \). Expand the left side: \( 2x^2 - 20x = x^2 + 300 \). Rearrange into a quadratic equation: \( 2x^2 - x^2 - 20x - 300 = 0 \). \( x^2 - 20x - 300 = 0 \). To solve this, we can factorize: \( x^2 - 30x + 10x - 300 = 0 \) \( x(x - 30) + 10(x - 30) = 0 \) \( (x - 30)(x + 10) = 0 \) This gives two possible values for \( x \): Either \( x - 30 = 0 \implies x = 30 \) Or \( x + 10 = 0 \implies x = -10 \) Since the number of rows cannot be negative, we reject \( x = -10 \). (i) Therefore, the number of rows in the original arrangement \( = 30 \). (ii) The number of seats in the auditorium after rearrangement \( = 2x(x - 10) \). Substitute \( x = 30 \): Number of seats after rearrangement \( = 2 \times 30 \times (30 - 10) \) \( = 60 \times 20 \) \( = 1200 \).In simple words: We used \( x \) for the original number of rows and columns. We wrote equations for the total seats before and after changes. The difference in seats led to an equation which we solved to find the original number of rows was 30, and the new total number of seats was 1200.

🎯 Exam Tip: Clearly define your variables and write expressions for both the initial and changed conditions. Remember that physical quantities like the number of rows or seats cannot be negative.

Question 9. P and Q are centres of circles of radius 9 cm and 2 cm respectively. PQ = 17 cm. R is the centre of a circle of radius x cm, which touches the above circles externally. Given that \( \angle PRQ = 90^\circ \), write an equation in x and solve it.
Answer: Let the radii of the circles with centers P, Q, and R be \( r_P, r_Q, r_R \) respectively. Given: Radius of circle with center P, \( r_P = 9 \) cm. Radius of circle with center Q, \( r_Q = 2 \) cm. The distance between P and Q, \( PQ = 17 \) cm. The circle with center R has radius \( x \) cm, so \( r_R = x \). Since the circle with center R touches the circles with centers P and Q externally: The distance between P and R, \( PR = r_P + r_R = (9 + x) \) cm. The distance between Q and R, \( QR = r_Q + r_R = (2 + x) \) cm. We are given that \( \angle PRQ = 90^\circ \). This means that triangle PRQ is a right-angled triangle with the right angle at R. According to the Pythagorean theorem, for a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, PQ is the hypotenuse. So, \( PQ^2 = PR^2 + QR^2 \). Substitute the given values and expressions: \( (17)^2 = (9 + x)^2 + (2 + x)^2 \). \( 289 = (81 + 18x + x^2) + (4 + 4x + x^2) \). Combine like terms: \( 289 = 2x^2 + 22x + 85 \). Rearrange into a quadratic equation by moving all terms to one side: \( 2x^2 + 22x + 85 - 289 = 0 \). \( 2x^2 + 22x - 204 = 0 \). Divide the entire equation by 2 to simplify: \( x^2 + 11x - 102 = 0 \). To solve this, we can factorize: \( x^2 + 17x - 6x - 102 = 0 \) \( x(x + 17) - 6(x + 17) = 0 \) \( (x + 17)(x - 6) = 0 \) This gives two possible values for \( x \): Either \( x + 17 = 0 \implies x = -17 \) Or \( x - 6 = 0 \implies x = 6 \) Since the radius of a circle cannot be negative, we reject \( x = -17 \). Therefore, the radius of the circle with centre R, \( x = 6 \) cm.In simple words: We used the given information about the circles touching each other to find the lengths of the sides of the triangle PRQ. Since it was a right-angled triangle, we used the Pythagorean theorem to set up an equation. Solving this equation showed that the radius \( x \) is 6 cm.

🎯 Exam Tip: For problems involving circles touching externally, the distance between their centers is the sum of their radii. When a right angle is mentioned, immediately think of the Pythagorean theorem.

Question 10. By increasing the speed of car by 10 km/ hr, the time of journey for a distance of 72 km is reduced by 36 min. Find the original speed of the car.
Answer: Let the car's original speed be \( x \) km/hr. When the speed is increased by 10 km/hr, the new speed becomes \( (x + 10) \) km/hr. The distance to cover is 72 km. The time taken is calculated by dividing distance by speed. So, the original time is \( \frac{72}{x} \) hours, and the new time is \( \frac{72}{x+10} \) hours. The problem states that the new time is 36 minutes less than the original time. We convert 36 minutes to hours by dividing by 60, which gives \( \frac{36}{60} = \frac{3}{5} \) hours.
Therefore, we can write the equation: \( \frac{72}{x} - \frac{72}{x+10} = \frac{3}{5} \)
\( \implies \frac{72(x+10) - 72x}{x(x+10)} = \frac{3}{5} \)
\( \implies \frac{72x + 720 - 72x}{x^2 + 10x} = \frac{3}{5} \)
\( \implies \frac{720}{x^2 + 10x} = \frac{3}{5} \)
Now, we cross-multiply:
\( \implies 3(x^2 + 10x) = 720 \times 5 \)
\( \implies 3x^2 + 30x = 3600 \)
Divide the entire equation by 3:
\( \implies x^2 + 10x = 1200 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 + 10x - 1200 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -1200 and add up to 10. These numbers are 40 and -30.
\( \implies x^2 + 40x - 30x - 1200 = 0 \)
\( \implies x(x + 40) - 30(x + 40) = 0 \)
\( \implies (x + 40)(x - 30) = 0 \)
This gives two possible values for \( x \):
Either \( x + 40 = 0 \implies x = -40 \)
Or \( x - 30 = 0 \implies x = 30 \)
Since speed cannot be a negative value, we reject \( x = -40 \).
So, the original speed of the car is 30 km/hr. Knowing the speed helps predict journey times accurately.
In simple words: The car went faster and saved 36 minutes. By setting up an equation comparing the old and new journey times, we found the car's first speed was 30 km/hr.

🎯 Exam Tip: Remember to convert all units to be consistent (e.g., minutes to hours) before setting up the equation, and reject any negative solutions for physical quantities like speed.

Question 11. A shopkeeper buys a certain number of books for Rs. 720. If the cost per book was Rs. 5 less, the number of books that could be bought for Rs. 720 would be 2 more. Taking the original cost of each book to be x, write an equation in x and solve it.
Answer: Let the original cost of each book be \( x \) rupees. The shopkeeper buys books for a total of Rs. 720. So, the original number of books bought is \( \frac{720}{x} \).
If the cost per book was Rs. 5 less, the new cost per book would be \( (x - 5) \) rupees. With this lower price, the shopkeeper could buy 2 more books for Rs. 720. So, the new number of books is \( \frac{720}{x-5} \).
The condition given is that the new number of books is 2 more than the original number of books.
So, we write the equation: \( \frac{720}{x-5} = \frac{720}{x} + 2 \)
Rearrange the terms to one side:
\( \implies \frac{720}{x-5} - \frac{720}{x} = 2 \)
Find a common denominator:
\( \implies \frac{720x - 720(x-5)}{x(x-5)} = 2 \)
\( \implies \frac{720x - 720x + 3600}{x^2 - 5x} = 2 \)
\( \implies \frac{3600}{x^2 - 5x} = 2 \)
Cross-multiply:
\( \implies 2(x^2 - 5x) = 3600 \)
Divide by 2:
\( \implies x^2 - 5x = 1800 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 - 5x - 1800 = 0 \)
To solve this, we look for two numbers that multiply to -1800 and add up to -5. These numbers are -45 and 40.
\( \implies x^2 - 45x + 40x - 1800 = 0 \)
\( \implies x(x - 45) + 40(x - 45) = 0 \)
\( \implies (x - 45)(x + 40) = 0 \)
This gives two possible values for \( x \):
Either \( x - 45 = 0 \implies x = 45 \)
Or \( x + 40 = 0 \implies x = -40 \)
Since the cost of a book cannot be negative, we reject \( x = -40 \).
So, the original cost of each book was Rs. 45. This quadratic equation demonstrates how changes in price affect quantity in commerce.
In simple words: A shopkeeper spent Rs. 720 on books. If each book cost Rs. 5 less, he could buy 2 more. By solving the equation, we found the original cost of each book was Rs. 45.

🎯 Exam Tip: Always define your variable clearly at the beginning. When dealing with costs or quantities, reject any negative solutions, as these are not practical in real-world scenarios.

Question 12. Solve the following quadratic equation for x and give your answer correct to 2 decimal places : \( x^2 – 3x − 9 = 0 \)
Answer: We need to solve the quadratic equation \( x^2 - 3x - 9 = 0 \).
This equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -3 \), and \( c = -9 \).
We can use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-3)^2 - 4(1)(-9) \)
\( D = 9 + 36 \)
\( D = 45 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-3) \pm \sqrt{45}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{45}}{2} \)
We can simplify \( \sqrt{45} \) as \( \sqrt{9 \times 5} = 3\sqrt{5} \).
\( x = \frac{3 \pm 3\sqrt{5}}{2} \)
Now, we approximate \( \sqrt{5} \approx 2.236 \):
\( x = \frac{3 \pm 3(2.236)}{2} \)
\( x = \frac{3 \pm 6.708}{2} \)
This gives two solutions for \( x \):
\( x_1 = \frac{3 + 6.708}{2} = \frac{9.708}{2} = 4.854 \)
\( x_2 = \frac{3 - 6.708}{2} = \frac{-3.708}{2} = -1.854 \)
Rounding to two decimal places:
\( x_1 \approx 4.85 \)
\( x_2 \approx -1.85 \)
These are the two roots that satisfy the equation, showing how numbers can balance each other out in a polynomial function.
In simple words: We solved the equation \( x^2 - 3x - 9 = 0 \) using the quadratic formula. The two answers for \( x \) are about 4.85 and -1.85.

🎯 Exam Tip: When asked to give answers correct to a certain number of decimal places, make sure to perform the rounding at the very last step. Use parentheses carefully with negative numbers in the quadratic formula.

Question 13. Five years ago, a woman's age was the square of her son's age. Ten years hence her age will be twice that of her son's age. Find :
(i) The age of her son five years ago.
(ii) The present age of the woman.
Answer: Let \( x \) be the son's age five years ago.
Then, according to the problem, the woman's age five years ago was \( x^2 \).

(i) **The age of her son five years ago.**
To find the present ages:
Son's present age = \( x + 5 \)
Woman's present age = \( x^2 + 5 \)

Now, consider "ten years hence" (10 years from the present):
Son's age ten years hence = \( (x + 5) + 10 = x + 15 \)
Woman's age ten years hence = \( (x^2 + 5) + 10 = x^2 + 15 \)

The problem states that ten years hence, the woman's age will be twice that of her son's age.
So, we form the equation: \( x^2 + 15 = 2(x + 15) \)
Expand the right side:
\( x^2 + 15 = 2x + 30 \)
Rearrange into a standard quadratic equation:
\( x^2 - 2x + 15 - 30 = 0 \)
\( x^2 - 2x - 15 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -15 and add up to -2. These numbers are -5 and 3.
\( \implies x^2 - 5x + 3x - 15 = 0 \)
\( \implies x(x - 5) + 3(x - 5) = 0 \)
\( \implies (x - 5)(x + 3) = 0 \)
This gives two possible values for \( x \):
Either \( x - 5 = 0 \implies x = 5 \)
Or \( x + 3 = 0 \implies x = -3 \)
Since age cannot be a negative value, we reject \( x = -3 \).
Therefore, the age of her son five years ago was 5 years. This is a common way to use algebra to model age relationships.

(ii) **The present age of the woman.**
Using \( x = 5 \):
Son's present age = \( x + 5 = 5 + 5 = 10 \) years
Woman's present age = \( x^2 + 5 = 5^2 + 5 = 25 + 5 = 30 \) years
In simple words: Five years ago, the son's age was 5 and his mother's age was \( 5^2 = 25 \). So, the son's current age is 10, and his mother's current age is 30.

🎯 Exam Tip: Always define your variables clearly and be careful with "ago," "hence," and "present" calculations. Age, like speed, cannot be negative, so reject negative solutions.

Question 14. Solve the following quadratic equation for x and give your answer correct to two decimal places: \( 5x (x + 2) = 3 \)
Answer: We need to solve the equation \( 5x (x + 2) = 3 \).
First, expand the equation:
\( \implies 5x^2 + 10x = 3 \)
Rearrange into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( \implies 5x^2 + 10x - 3 = 0 \)
Here, \( a = 5 \), \( b = 10 \), and \( c = -3 \).
We will use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (10)^2 - 4(5)(-3) \)
\( D = 100 + 60 \)
\( D = 160 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-10 \pm \sqrt{160}}{2(5)} \)
\( x = \frac{-10 \pm \sqrt{160}}{10} \)
We can approximate \( \sqrt{160} \approx 12.649 \) (keeping more decimal places for accuracy before final rounding).
\( x = \frac{-10 \pm 12.649}{10} \)
This gives two solutions for \( x \):
\( x_1 = \frac{-10 + 12.649}{10} = \frac{2.649}{10} = 0.2649 \)
\( x_2 = \frac{-10 - 12.649}{10} = \frac{-22.649}{10} = -2.2649 \)
Rounding to two decimal places:
\( x_1 \approx 0.26 \)
\( x_2 \approx -2.26 \)
These solutions demonstrate how algebraic equations can have positive and negative roots.
In simple words: We solved the equation \( 5x(x+2)=3 \) by turning it into a quadratic equation. The two answers for \( x \) are about 0.26 and -2.26.

🎯 Exam Tip: Always expand and simplify the equation into the standard \( ax^2 + bx + c = 0 \) form before applying the quadratic formula. Ensure proper rounding to the specified decimal places at the very end.

Question 15. Some students planned a picnic. The budget for the food was Rs. 480. As eight of them failed to join the party the cost of the food for each member increased by Rs. 10. Find how many students went for the picnic?
Answer: Let the number of students who originally planned to go for the picnic be \( y \).
The total budget for food was Rs. 480.
So, the original share of food cost for each student was \( \frac{480}{y} \) rupees.

Eight students failed to join the party. So, the number of students who actually went for the picnic is \( y - 8 \).
The cost of food for each member who went increased by Rs. 10.
So, the new cost per student is \( \frac{480}{y-8} \) rupees.

The problem states that the new cost per person is Rs. 10 more than the original cost per person.
So, we form the equation: \( \frac{480}{y-8} = \frac{480}{y} + 10 \)
Rearrange the terms:
\( \implies \frac{480}{y-8} - \frac{480}{y} = 10 \)
Find a common denominator:
\( \implies \frac{480y - 480(y-8)}{y(y-8)} = 10 \)
\( \implies \frac{480y - 480y + 3840}{y^2 - 8y} = 10 \)
\( \implies \frac{3840}{y^2 - 8y} = 10 \)
Cross-multiply:
\( \implies 10(y^2 - 8y) = 3840 \)
Divide by 10:
\( \implies y^2 - 8y = 384 \)
Rearrange into a standard quadratic equation:
\( \implies y^2 - 8y - 384 = 0 \)
To solve this, we look for two numbers that multiply to -384 and add up to -8. These numbers are -24 and 16.
\( \implies y^2 - 24y + 16y - 384 = 0 \)
\( \implies y(y - 24) + 16(y - 24) = 0 \)
\( \implies (y - 24)(y + 16) = 0 \)
This gives two possible values for \( y \):
Either \( y - 24 = 0 \implies y = 24 \)
Or \( y + 16 = 0 \implies y = -16 \)
Since the number of students cannot be negative, we reject \( y = -16 \).
So, the original number of students who planned the picnic was 24.
The question asks for the number of students who *went* for the picnic.
Number of students who went = \( y - 8 = 24 - 8 = 16 \).
Understanding how costs per person change with group size is a practical application of quadratic equations.
In simple words: The picnic food cost Rs. 480. When 8 students dropped out, the cost for each remaining student went up by Rs. 10. We found that 16 students actually went to the picnic.

🎯 Exam Tip: Clearly distinguish between the initial number of students and the actual number who attended. Always answer the specific question asked, which might be a derived value, not just the variable you solved for.

Question 16. Solve the following quadratic equation and give the answer correct to two significant figures \( 4x^2 – 7x + 2 = 0 \).
Answer: We need to solve the quadratic equation \( 4x^2 - 7x + 2 = 0 \).
This equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 4 \), \( b = -7 \), and \( c = 2 \).
We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-7)^2 - 4(4)(2) \)
\( D = 49 - 32 \)
\( D = 17 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-7) \pm \sqrt{17}}{2(4)} \)
\( x = \frac{7 \pm \sqrt{17}}{8} \)
We approximate \( \sqrt{17} \approx 4.123 \) (keeping more decimal places for accuracy before final rounding).
\( x = \frac{7 \pm 4.123}{8} \)
This gives two solutions for \( x \):
\( x_1 = \frac{7 + 4.123}{8} = \frac{11.123}{8} = 1.390375 \)
\( x_2 = \frac{7 - 4.123}{8} = \frac{2.877}{8} = 0.359625 \)
Rounding to two significant figures:
\( x_1 \approx 1.4 \)
\( x_2 \approx 0.36 \)
These two values represent the points where the quadratic curve intersects the x-axis.
In simple words: We solved the equation \( 4x^2 - 7x + 2 = 0 \) using a formula. The two answers for \( x \) are about 1.4 and 0.36.

🎯 Exam Tip: Pay close attention to the rounding instruction (e.g., "two significant figures"). Make sure to use enough decimal places during intermediate calculations to maintain accuracy for the final rounded answer.

Question 17. The speed of an express train is x km/h and the speed of an ordinary train is 12 km/h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km, find the speed of the express train.
Answer: Let the speed of the express train be \( x \) km/h.
The speed of the ordinary train is 12 km/h less than the express train, so its speed is \( (x - 12) \) km/h.
The distance covered by both trains is 240 km.
Time taken by the express train = \( \frac{\text{Distance}}{\text{Speed}} = \frac{240}{x} \) hours.
Time taken by the ordinary train = \( \frac{\text{Distance}}{\text{Speed}} = \frac{240}{x-12} \) hours.
The problem states that the ordinary train takes one hour longer than the express train.
So, we form the equation: \( \frac{240}{x-12} = \frac{240}{x} + 1 \)
Rearrange the terms:
\( \implies \frac{240}{x-12} - \frac{240}{x} = 1 \)
Find a common denominator:
\( \implies \frac{240x - 240(x-12)}{x(x-12)} = 1 \)
\( \implies \frac{240x - 240x + 240 \times 12}{x^2 - 12x} = 1 \)
\( \implies \frac{2880}{x^2 - 12x} = 1 \)
Cross-multiply:
\( \implies x^2 - 12x = 2880 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 - 12x - 2880 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -2880 and add up to -12. These numbers are -60 and 48.
\( \implies x^2 - 60x + 48x - 2880 = 0 \)
\( \implies x(x - 60) + 48(x - 60) = 0 \)
\( \implies (x - 60)(x + 48) = 0 \)
This gives two possible values for \( x \):
Either \( x - 60 = 0 \implies x = 60 \)
Or \( x + 48 = 0 \implies x = -48 \)
Since speed cannot be a negative value, we reject \( x = -48 \).
So, the speed of the express train is 60 km/h. This problem illustrates how quadratic equations are used in physics to analyze motion.
In simple words: An express train is 12 km/h faster than an ordinary train. The ordinary train takes 1 hour more to travel 240 km. We found the express train's speed is 60 km/h.

🎯 Exam Tip: In speed-distance-time problems, convert all units to be consistent (e.g., all to hours, km). Remember that a train's speed must be a positive value.

Question 18. Without solving the following quadratic equation, find the value of 'p' for which the roots are equal. \( px^2 – 4x + 3 = 0 \).
Answer: We are given the quadratic equation \( px^2 - 4x + 3 = 0 \).
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be equal, its discriminant (D) must be zero. The discriminant is given by the formula \( D = b^2 - 4ac \).
In our given equation, we have:
\( a = p \)
\( b = -4 \)
\( c = 3 \)
Now, we set the discriminant to zero:
\( D = b^2 - 4ac = 0 \)
Substitute the values of \( a, b, \) and \( c \):
\( (-4)^2 - 4(p)(3) = 0 \)
\( 16 - 12p = 0 \)
Now, solve for \( p \):
\( 16 = 12p \)
\( p = \frac{16}{12} \)
Simplify the fraction:
\( p = \frac{4}{3} \)
So, the value of \( p \) for which the roots are equal is \( \frac{4}{3} \). This concept is fundamental to understanding the nature of quadratic roots.
In simple words: For the equation \( px^2 - 4x + 3 = 0 \) to have equal answers, a special number called the discriminant must be zero. By setting this to zero, we found that \( p \) must be \( \frac{4}{3} \).

🎯 Exam Tip: Remember that for equal roots, the discriminant \( (D = b^2 - 4ac) \) must always be zero. This is a key condition for quadratic equations.

Question 19. Solve the following equation: \( x - \frac { 18 }{ x } = 6 \). Give your answer correct to two significant figures.
Answer: We need to solve the equation \( x - \frac{18}{x} = 6 \).
First, multiply the entire equation by \( x \) to eliminate the fraction:
\( \implies x \cdot x - \frac{18}{x} \cdot x = 6 \cdot x \)
\( \implies x^2 - 18 = 6x \)
Rearrange into the standard quadratic form \( ax^2 + bx + c = 0 \):
\( \implies x^2 - 6x - 18 = 0 \)
Here, \( a = 1 \), \( b = -6 \), and \( c = -18 \).
We will use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-6)^2 - 4(1)(-18) \)
\( D = 36 + 72 \)
\( D = 108 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-6) \pm \sqrt{108}}{2(1)} \)
\( x = \frac{6 \pm \sqrt{108}}{2} \)
We can simplify \( \sqrt{108} \) as \( \sqrt{36 \times 3} = 6\sqrt{3} \).
\( x = \frac{6 \pm 6\sqrt{3}}{2} \)
Divide both terms in the numerator by 2:
\( x = 3 \pm 3\sqrt{3} \)
Now, we approximate \( \sqrt{3} \approx 1.732 \) (keeping more decimal places for accuracy).
\( x = 3 \pm 3(1.732) \)
\( x = 3 \pm 5.196 \)
This gives two solutions for \( x \):
\( x_1 = 3 + 5.196 = 8.196 \)
\( x_2 = 3 - 5.196 = -2.196 \)
Rounding to two significant figures:
\( x_1 \approx 8.2 \)
\( x_2 \approx -2.2 \)
These solutions show how a rational equation can lead to a quadratic equation and its roots.
In simple words: We solved the equation \( x - \frac{18}{x} = 6 \) by first removing the fraction, turning it into a quadratic equation. The two answers for \( x \) are about 8.2 and -2.2.

🎯 Exam Tip: When an equation involves fractions with the variable in the denominator, multiply the entire equation by the common denominator to clear the fractions. Remember to check for extraneous solutions if the original equation had variables in the denominator, though here both solutions are valid.

Question 20. 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: Let \( x \) be the original number of children.
The total amount of money to be divided is Rs. 480.
So, the amount each child gets originally is \( \frac{480}{x} \) rupees.

If the number of children were 20 more, the new number of children would be \( x + 20 \).
In this case, each child would have got Rs. 12 less.
The new amount each child gets is \( \frac{480}{x+20} \) rupees.

The problem states that the new amount per child is Rs. 12 less than the original amount per child.
So, we form the equation: \( \frac{480}{x+20} = \frac{480}{x} - 12 \)
Rearrange the terms:
\( \implies 12 = \frac{480}{x} - \frac{480}{x+20} \)
Divide the entire equation by 12:
\( \implies 1 = \frac{40}{x} - \frac{40}{x+20} \)
Find a common denominator:
\( \implies 1 = \frac{40(x+20) - 40x}{x(x+20)} \)
\( \implies 1 = \frac{40x + 800 - 40x}{x^2 + 20x} \)
\( \implies 1 = \frac{800}{x^2 + 20x} \)
Cross-multiply:
\( \implies x^2 + 20x = 800 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 + 20x - 800 = 0 \)
To solve this, we look for two numbers that multiply to -800 and add up to 20. These numbers are 40 and -20.
\( \implies x^2 + 40x - 20x - 800 = 0 \)
\( \implies x(x + 40) - 20(x + 40) = 0 \)
\( \implies (x + 40)(x - 20) = 0 \)
This gives two possible values for \( x \):
Either \( x + 40 = 0 \implies x = -40 \)
Or \( x - 20 = 0 \implies x = 20 \)
Since the number of children cannot be negative, we reject \( x = -40 \).
So, \( x = 20 \). The original number of children was 20. This problem shows how quadratic equations are used to model changes in shared costs.
In simple words: When Rs. 480 was shared among some children, each got a certain amount. If there were 20 more children, each would get Rs. 12 less. We found that originally there were 20 children.

🎯 Exam Tip: Always make sure your equations correctly represent the "less" or "more" conditions. Remember that quantities like the number of people cannot be negative, so discard any negative solutions.

Question 21. Without solving the following quadratic equation, find the value of 'm' for which the given equation has real and equal roots. \( x^2 + 2 (m – 1) x + (m + 5) = 0 \).
Answer: We are given the quadratic equation \( x^2 + 2(m - 1)x + (m + 5) = 0 \).
For a quadratic equation \( ax^2 + bx + c = 0 \) to have real and equal roots, its discriminant (D) must be zero. The discriminant is given by the formula \( D = b^2 - 4ac \).
In our given equation, we have:
\( a = 1 \)
\( b = 2(m - 1) \)
\( c = m + 5 \)
Now, we set the discriminant to zero:
\( D = b^2 - 4ac = 0 \)
Substitute the values of \( a, b, \) and \( c \):
\( (2(m - 1))^2 - 4(1)(m + 5) = 0 \)
\( 4(m - 1)^2 - 4(m + 5) = 0 \)
Divide the entire equation by 4:
\( (m - 1)^2 - (m + 5) = 0 \)
Expand \( (m - 1)^2 \):
\( m^2 - 2m + 1 - (m + 5) = 0 \)
\( m^2 - 2m + 1 - m - 5 = 0 \)
Combine like terms:
\( m^2 - 3m - 4 = 0 \)
To solve this quadratic equation for \( m \), we can factor it. We need two numbers that multiply to -4 and add up to -3. These numbers are -4 and 1.
\( \implies m^2 - 4m + m - 4 = 0 \)
\( \implies m(m - 4) + 1(m - 4) = 0 \)
\( \implies (m - 4)(m + 1) = 0 \)
This gives two possible values for \( m \):
Either \( m - 4 = 0 \implies m = 4 \)
Or \( m + 1 = 0 \implies m = -1 \)
So, the values of \( m \) for which the equation has real and equal roots are 4 and -1. This condition is crucial for problems involving perfect squares in quadratic forms.
In simple words: For the equation \( x^2 + 2(m-1)x + (m+5) = 0 \) to have equal answers, its special discriminant value must be zero. Solving this showed that \( m \) can be either 4 or -1.

🎯 Exam Tip: When the roots are real and equal, the discriminant is always zero. Be careful with algebraic expansion, especially \( (a-b)^2 \), and simplify the resulting quadratic equation for 'm' correctly.

Question 22. 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: Let the original speed of the car be \( x \) km/h.
The distance covered is 400 km.
Original time taken = \( \frac{\text{Distance}}{\text{Speed}} = \frac{400}{x} \) hours.

If the speed had been 12 km/h more, the new speed would be \( (x + 12) \) km/h.
New time taken = \( \frac{400}{x+12} \) hours.

The problem states that the new time taken is 1 hour 40 minutes less than the original time.
Convert 1 hour 40 minutes to hours: \( 1 \text{ hour } + 40 \text{ minutes} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3} \) hours.

So, we form the equation: \( \frac{400}{x} - \frac{400}{x+12} = \frac{5}{3} \)
Find a common denominator on the left side:
\( \implies \frac{400(x+12) - 400x}{x(x+12)} = \frac{5}{3} \)
\( \implies \frac{400x + 4800 - 400x}{x^2 + 12x} = \frac{5}{3} \)
\( \implies \frac{4800}{x^2 + 12x} = \frac{5}{3} \)
Cross-multiply:
\( \implies 5(x^2 + 12x) = 4800 \times 3 \)
\( \implies 5x^2 + 60x = 14400 \)
Divide the entire equation by 5:
\( \implies x^2 + 12x = 2880 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 + 12x - 2880 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -2880 and add up to 12. These numbers are 60 and -48.
\( \implies x^2 + 60x - 48x - 2880 = 0 \)
\( \implies x(x + 60) - 48(x + 60) = 0 \)
\( \implies (x + 60)(x - 48) = 0 \)
This gives two possible values for \( x \):
Either \( x + 60 = 0 \implies x = -60 \)
Or \( x - 48 = 0 \implies x = 48 \)
Since speed cannot be negative, we reject \( x = -60 \).
So, the original speed of the car is 48 km/h. This problem highlights how improving speed reduces travel time for a fixed distance.
In simple words: A car traveled 400 km. If it went 12 km/h faster, it would save 1 hour and 40 minutes. We found the car's first speed was 48 km/h.

🎯 Exam Tip: Always convert mixed time units (hours and minutes) into a single unit (hours or minutes) before using them in equations. Speed cannot be negative, so discard such solutions.

Question 23.
(i) Solve the following equation and calculate the answer correct to two decimal places: \( x^2 - 5x - 10 = 0 \).
(ii) Without solving the following quadratic equation, find the value of 'p' for which the given equation has real and equal roots \( x^2 + (p − 3)x + p = 0 \).
Answer:
(i) We need to solve the quadratic equation \( x^2 - 5x - 10 = 0 \).
This equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -5 \), and \( c = -10 \).
We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-5)^2 - 4(1)(-10) \)
\( D = 25 + 40 \)
\( D = 65 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-5) \pm \sqrt{65}}{2(1)} \)
\( x = \frac{5 \pm \sqrt{65}}{2} \)
We approximate \( \sqrt{65} \approx 8.062 \) (keeping more decimal places for accuracy).
\( x = \frac{5 \pm 8.062}{2} \)
This gives two solutions for \( x \):
\( x_1 = \frac{5 + 8.062}{2} = \frac{13.062}{2} = 6.531 \)
\( x_2 = \frac{5 - 8.062}{2} = \frac{-3.062}{2} = -1.531 \)
Rounding to two decimal places:
\( x_1 \approx 6.53 \)
\( x_2 \approx -1.53 \)
These roots are important for understanding the behavior of this particular parabola.

(ii) We are given the quadratic equation \( x^2 + (p - 3)x + p = 0 \).
For the roots of a quadratic equation \( ax^2 + bx + c = 0 \) to be real and equal, its discriminant (D) must be zero.
In this equation, we have:
\( a = 1 \)
\( b = p - 3 \)
\( c = p \)
Now, we set the discriminant to zero:
\( D = b^2 - 4ac = 0 \)
Substitute the values of \( a, b, \) and \( c \):
\( (p - 3)^2 - 4(1)(p) = 0 \)
Expand \( (p - 3)^2 \):
\( p^2 - 6p + 9 - 4p = 0 \)
Combine like terms:
\( p^2 - 10p + 9 = 0 \)
To solve this quadratic equation for \( p \), we can factor it. We need two numbers that multiply to 9 and add up to -10. These numbers are -9 and -1.
\( \implies p^2 - 9p - p + 9 = 0 \)
\( \implies p(p - 9) - 1(p - 9) = 0 \)
\( \implies (p - 9)(p - 1) = 0 \)
This gives two possible values for \( p \):
Either \( p - 9 = 0 \implies p = 9 \)
Or \( p - 1 = 0 \implies p = 1 \)
So, the values of \( p \) for which the equation has real and equal roots are 1 and 9. This concept ensures that the quadratic equation has exactly one distinct solution.
In simple words: For part (i), we solved the equation \( x^2 - 5x - 10 = 0 \) to get answers of about 6.53 and -1.53. For part (ii), we found that \( p \) must be 1 or 9 for the equation \( x^2 + (p-3)x + p = 0 \) to have equal solutions.

🎯 Exam Tip: For part (i), show all steps of the quadratic formula clearly and round only at the final answer. For part (ii), remember the condition \( D = 0 \) for real and equal roots and factor the resulting quadratic in 'p' correctly.

Question 24. A shopkeeper purchases a certain number of books for Rs. 960. If the cost per book was Rs. 8 less, the number of books that could be purchased for Rs. 960 would be 4 more. Write an equation, taking the original cost of each book to be Rs. x, and solve it to find the original cost of the books.
Answer: Let the original cost of each book be \( x \) rupees.
The total amount spent by the shopkeeper is Rs. 960.
So, the original number of books purchased is \( \frac{960}{x} \).

If the cost per book was Rs. 8 less, the new cost per book would be \( (x - 8) \) rupees.
With this new price, the shopkeeper could buy 4 more books for Rs. 960.
So, the new number of books purchased is \( \frac{960}{x-8} \).

According to the problem, the new number of books is 4 more than the original number of books.
So, we form the equation: \( \frac{960}{x-8} = \frac{960}{x} + 4 \)
Rearrange the terms:
\( \implies \frac{960}{x-8} - \frac{960}{x} = 4 \)
Find a common denominator:
\( \implies \frac{960x - 960(x-8)}{x(x-8)} = 4 \)
\( \implies \frac{960x - 960x + 7680}{x^2 - 8x} = 4 \)
\( \implies \frac{7680}{x^2 - 8x} = 4 \)
Cross-multiply:
\( \implies 4(x^2 - 8x) = 7680 \)
Divide the entire equation by 4:
\( \implies x^2 - 8x = 1920 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 - 8x - 1920 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -1920 and add up to -8. These numbers are -48 and 40.
\( \implies x^2 - 48x + 40x - 1920 = 0 \)
\( \implies x(x - 48) + 40(x - 48) = 0 \)
\( \implies (x - 48)(x + 40) = 0 \)
This gives two possible values for \( x \):
Either \( x - 48 = 0 \implies x = 48 \)
Or \( x + 40 = 0 \implies x = -40 \)
Since the cost of a book cannot be negative, we reject \( x = -40 \).
So, the original cost of each book was Rs. 48. This shows the inverse relationship between price and quantity purchased when total budget is fixed.
In simple words: A shopkeeper bought books for Rs. 960. If each book cost Rs. 8 less, he could get 4 more books. We found that each book originally cost Rs. 48.

🎯 Exam Tip: Carefully set up the fractions for quantities, ensuring the correct variable is in the denominator. Remember that costs must be positive values in real-world problems.

Question 25. Solve for x using the quadratic formula. Write your answer correct to two significant figures. \( (x – 1)^2 – 3x + 4 = 0 \).
Answer: We need to solve the equation \( (x - 1)^2 - 3x + 4 = 0 \).
First, expand \( (x - 1)^2 \):
\( \implies x^2 - 2x + 1 - 3x + 4 = 0 \)
Combine like terms to get the standard quadratic form \( ax^2 + bx + c = 0 \):
\( \implies x^2 - 5x + 5 = 0 \)
Here, \( a = 1 \), \( b = -5 \), and \( c = 5 \).
We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-5)^2 - 4(1)(5) \)
\( D = 25 - 20 \)
\( D = 5 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-5) \pm \sqrt{5}}{2(1)} \)
\( x = \frac{5 \pm \sqrt{5}}{2} \)
We approximate \( \sqrt{5} \approx 2.236 \) (keeping more decimal places for accuracy).
\( x = \frac{5 \pm 2.236}{2} \)
This gives two solutions for \( x \):
\( x_1 = \frac{5 + 2.236}{2} = \frac{7.236}{2} = 3.618 \)
\( x_2 = \frac{5 - 2.236}{2} = \frac{2.764}{2} = 1.382 \)
Rounding to two significant figures:
\( x_1 \approx 3.6 \)
\( x_2 \approx 1.4 \)
These solutions define the points where the parabola intersects the x-axis, highlighting the method for solving non-standard quadratic forms.
In simple words: We solved the equation \( (x-1)^2 - 3x + 4 = 0 \) by first expanding it into a simple quadratic form, then using the quadratic formula. The two answers for \( x \) are about 3.6 and 1.4.

🎯 Exam Tip: Always expand and simplify expressions like \( (x-1)^2 \) before applying the quadratic formula. Be careful with rounding to the specified number of significant figures at the end of your calculation.

Question 26. A two-digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
Answer: Let the two-digit number be represented as \( 10x + y \), where \( x \) is the tens digit and \( y \) is the units digit.

The first condition is that the product of its digits is 6:
\( x \cdot y = 6 \implies y = \frac{6}{x} \) (Equation 1)

The second condition is that if 9 is added to the number, the digits interchange their places.
The number with interchanged digits is \( 10y + x \).
So, we form the equation: \( (10x + y) + 9 = 10y + x \)
Rearrange the terms:
\( 10x - x + y - 10y + 9 = 0 \)
\( 9x - 9y + 9 = 0 \)
Divide the entire equation by 9:
\( x - y + 1 = 0 \) (Equation 2)

Now, substitute Equation 1 into Equation 2:
\( x - \frac{6}{x} + 1 = 0 \)
Multiply the entire equation by \( x \) to eliminate the fraction:
\( \implies x^2 - 6 + x = 0 \)
Rearrange into standard quadratic form:
\( \implies x^2 + x - 6 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to -6 and add up to 1. These numbers are 3 and -2.
\( \implies x^2 + 3x - 2x - 6 = 0 \)
\( \implies x(x + 3) - 2(x + 3) = 0 \)
\( \implies (x + 3)(x - 2) = 0 \)
This gives two possible values for \( x \):
Either \( x + 3 = 0 \implies x = -3 \)
Or \( x - 2 = 0 \implies x = 2 \)
Since \( x \) is a digit of a positive number, it must be a positive integer. We reject \( x = -3 \).
So, \( x = 2 \).
Now, find \( y \) using Equation 1:
\( y = \frac{6}{x} = \frac{6}{2} = 3 \)
The original two-digit number is \( 10x + y = 10(2) + 3 = 20 + 3 = 23 \).
This problem demonstrates how systems of equations are used to solve digit-based puzzles.
In simple words: We found a two-digit number where the digits multiplied to 6. When 9 was added to the number, its digits swapped places. The number is 23.

🎯 Exam Tip: When dealing with two-digit numbers, represent them as \( 10x + y \). Make sure to check the conditions for digits (must be integers from 0-9, and the tens digit cannot be 0). Reject negative or fractional solutions for digits.

Question 27. Find the value of 'k' for which \( x = 3 \) is a solution of the quadratic equation, \( (k + 2)x^2 – kx + 6 = 0 \). Thus find the other root of the equation.
Answer: We are given the quadratic equation \( (k + 2)x^2 - kx + 6 = 0 \).
We are told that \( x = 3 \) is a solution to this equation. This means if we substitute \( x = 3 \) into the equation, it will be satisfied.
Substitute \( x = 3 \):
\( (k + 2)(3)^2 - k(3) + 6 = 0 \)
\( (k + 2)(9) - 3k + 6 = 0 \)
Distribute the 9:
\( 9k + 18 - 3k + 6 = 0 \)
Combine like terms:
\( 6k + 24 = 0 \)
Solve for \( k \):
\( 6k = -24 \)
\( k = \frac{-24}{6} \)
\( k = -4 \)
So, the value of \( k \) is -4.

Now we need to find the other root. Substitute \( k = -4 \) back into the original quadratic equation:
\( (-4 + 2)x^2 - (-4)x + 6 = 0 \)
\( -2x^2 + 4x + 6 = 0 \)
Divide the entire equation by -2 to simplify:
\( x^2 - 2x - 3 = 0 \)
To find the roots of this simplified quadratic equation, we can factor it. We need two numbers that multiply to -3 and add up to -2. These numbers are -3 and 1.
\( \implies x^2 - 3x + x - 3 = 0 \)
\( \implies x(x - 3) + 1(x - 3) = 0 \)
\( \implies (x - 3)(x + 1) = 0 \)
This gives two possible values for \( x \):
Either \( x - 3 = 0 \implies x = 3 \) (This is the given root)
Or \( x + 1 = 0 \implies x = -1 \)
So, the other root of the equation is -1. This process shows how knowing one root can help find unknown coefficients and other roots.
In simple words: Given that \( x=3 \) is a solution to the equation \( (k+2)x^2 - kx + 6 = 0 \), we found that \( k \) must be -4. Then, by putting \( k=-4 \) back into the equation, we found the other solution for \( x \) is -1.

🎯 Exam Tip: If a value is a solution to an equation, substituting it into the equation should make the equation true. Use this property to find unknown coefficients. Once the coefficient is found, substitute it back to solve for the remaining roots.

Question 28. Sum of two natural numbers is 8 and the difference of their reciprocal is \( \frac { 2 }{ 15 } \). Find the numbers.
Answer: Let the two natural numbers be \( x \) and \( y \).

The first condition is that their sum is 8:
\( x + y = 8 \implies y = 8 - x \) (Equation 1)

The second condition is that the difference of their reciprocals is \( \frac{2}{15} \).
Since natural numbers are positive, their reciprocals will also be positive. We assume \( x > y \) to ensure the difference is positive, so \( \frac{1}{y} - \frac{1}{x} = \frac{2}{15} \). Or, if \( y > x \), then \( \frac{1}{x} - \frac{1}{y} = \frac{2}{15} \). Let's follow the solution provided, which seems to imply \( x \) is larger or \( y \) is larger.
Let's assume \( \frac{1}{x} - \frac{1}{y} = \frac{2}{15} \) based on the source solution's steps.

Substitute Equation 1 into the reciprocal difference equation:
\( \frac{1}{x} - \frac{1}{8-x} = \frac{2}{15} \)
Find a common denominator on the left side:
\( \implies \frac{(8-x) - x}{x(8-x)} = \frac{2}{15} \)
\( \implies \frac{8 - 2x}{8x - x^2} = \frac{2}{15} \)
Divide both sides by 2 (numerator on left, numerator on right):
\( \implies \frac{4 - x}{8x - x^2} = \frac{1}{15} \)
Cross-multiply:
\( \implies 15(4 - x) = 1(8x - x^2) \)
\( \implies 60 - 15x = 8x - x^2 \)
Rearrange into a standard quadratic equation:
\( \implies x^2 - 8x - 15x + 60 = 0 \)
\( \implies x^2 - 23x + 60 = 0 \)
To solve this, we can factor the quadratic equation. We need two numbers that multiply to 60 and add up to -23. These numbers are -20 and -3.
\( \implies x^2 - 20x - 3x + 60 = 0 \)
\( \implies x(x - 20) - 3(x - 20) = 0 \)
\( \implies (x - 20)(x - 3) = 0 \)
This gives two possible values for \( x \):
Either \( x - 20 = 0 \implies x = 20 \)
Or \( x - 3 = 0 \implies x = 3 \)

Now, we check these values with the condition \( x + y = 8 \).
If \( x = 20 \), then \( y = 8 - 20 = -12 \). This is not a natural number, so we reject this solution.
If \( x = 3 \), then \( y = 8 - 3 = 5 \). Both 3 and 5 are natural numbers.

The numbers are 3 and 5. This showcases how setting up simultaneous equations and solving them is effective.
In simple words: We searched for two natural numbers that add up to 8, and where the difference of their flip-over values (reciprocals) is \( \frac{2}{15} \). The two numbers are 3 and 5.

🎯 Exam Tip: Clearly define your variables. When dealing with "difference of reciprocals," be careful about the order of subtraction (it might be \( \frac{1}{x} - \frac{1}{y} \) or \( \frac{1}{y} - \frac{1}{x} \)). Always verify that your solutions meet all conditions (e.g., "natural numbers" means positive integers). Reject solutions that don't make sense in context.

Question 29. Solve the quadratic equation \( x^2 – 3(x + 3) = 0 \); Give your answer correct two significant figures.
Answer: We need to solve the quadratic equation \( x^2 - 3(x + 3) = 0 \).
First, expand the equation:
\( \implies x^2 - 3x - 9 = 0 \)
This equation is in the standard form \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -3 \), and \( c = -9 \).
We use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
First, calculate the discriminant \( D = b^2 - 4ac \):
\( D = (-3)^2 - 4(1)(-9) \)
\( D = 9 + 36 \)
\( D = 45 \)
Now, substitute the values into the quadratic formula:
\( x = \frac{-(-3) \pm \sqrt{45}}{2(1)} \)
\( x = \frac{3 \pm \sqrt{45}}{2} \)
We can simplify \( \sqrt{45} \) as \( \sqrt{9 \times 5} = 3\sqrt{5} \).
\( x = \frac{3 \pm 3\sqrt{5}}{2} \)
Now, we approximate \( \sqrt{5} \approx 2.236 \) (keeping more decimal places for accuracy).
\( x = \frac{3 \pm 3(2.236)}{2} \)
\( x = \frac{3 \pm 6.708}{2} \)
This gives two solutions for \( x \):
\( x_1 = \frac{3 + 6.708}{2} = \frac{9.708}{2} = 4.854 \)
\( x_2 = \frac{3 - 6.708}{2} = \frac{-3.708}{2} = -1.854 \)
Rounding to two significant figures:
\( x_1 \approx 4.9 \)
\( x_2 \approx -1.9 \)
These two solutions represent the points where the quadratic function crosses the x-axis.
In simple words: We solved the equation \( x^2 - 3(x+3) = 0 \) by expanding it first and then using the quadratic formula. The two answers for \( x \) are about 4.9 and -1.9.

🎯 Exam Tip: Always expand and simplify equations into the standard \( ax^2 + bx + c = 0 \) form before applying the quadratic formula. Pay close attention to rounding instructions like "two significant figures."

Question 30. 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: Let the original speed of the bus be \( x \) km/h. The total distance for the journey is 240 km. The time taken at the original speed is \( \frac{240}{x} \) hours. Due to heavy rain, the speed reduces by 10 km/h, making the new speed \( (x - 10) \) km/h. The time taken at this reduced speed is \( \frac{240}{x - 10} \) hours. The problem states that the bus takes 2 hours longer at the reduced speed. Understanding how speed, distance, and time relate is key to solving these types of real-world problems. So, we can write the equation as: \( \frac{240}{x - 10} - \frac{240}{x} = 2 \) To solve this, we find a common denominator for the fractions on the left side: \( \frac{240x - 240(x - 10)}{x(x - 10)} = 2 \) \( \frac{240x - 240x + 2400}{x^2 - 10x} = 2 \) \( \frac{2400}{x^2 - 10x} = 2 \) Now, we cross-multiply: \( 2400 = 2(x^2 - 10x) \) Divide both sides by 2: \( 1200 = x^2 - 10x \) Rearrange this into a standard quadratic equation: \( x^2 - 10x - 1200 = 0 \) We can solve this by factoring. We need two numbers that multiply to -1200 and add up to -10. These numbers are -40 and 30. \( x^2 - 40x + 30x - 1200 = 0 \) \( x(x - 40) + 30(x - 40) = 0 \) \( (x - 40)(x + 30) = 0 \) This gives us two possible values for \( x \): \( x - 40 = 0 \implies x = 40 \) \( x + 30 = 0 \implies x = -30 \) Since the speed of a bus cannot be a negative value, we must reject \( x = -30 \). Therefore, the uniform speed of the bus is 40 km/h.
In simple words: First, we write down equations for the time taken by the bus at its normal speed and then at its slower speed because of rain. The problem tells us the slower journey took 2 hours more. We set up an equation from this and solve it to find the bus's usual speed, making sure our answer is a positive number.

🎯 Exam Tip: Always make sure to define your variables clearly in word problems. When solving for physical quantities like speed, time, or distance, remember to reject any negative solutions as they are not practical in real-world scenarios.