CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 14

Question 1. Form the quadratic equation whose roots are:
a) \( -3 \text{ and } -4 \)
b) \( \frac{1}{2} \text{ and } -3 \)
c) \( \frac{1}{4} \text{ and } \frac{1}{3} \)
Answer: Let the roots of a quadratic equation be \( \alpha \) and \( \beta \). The standard formula to construct the quadratic equation is: \[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \] Let's find the quadratic equation for each pair of roots:
a) Roots: \( -3 \text{ and } -4 \) - Sum of roots \( (S) = -3 + (-4) = -7 \) - Product of roots \( (P) = -3 \times (-4) = 12 \) The quadratic equation is: \[ x^2 - (-7)x + 12 = 0 \implies x^2 + 7x + 12 = 0 \] b) Roots: \( \frac{1}{2} \text{ and } -3 \) - Sum of roots \( (S) = \frac{1}{2} + (-3) = -\frac{5}{2} \) - Product of roots \( (P) = \frac{1}{2} \times (-3) = -\frac{3}{2} \) The quadratic equation is: \[ x^2 - \left(-\frac{5}{2}\right)x + \left(-\frac{3}{2}\right) = 0 \implies x^2 + \frac{5}{2}x - \frac{3}{2} = 0 \] Multiplying the entire equation by 2 to clear fractions: \[ 2x^2 + 5x - 3 = 0 \] c) Roots: \( \frac{1}{4} \text{ and } \frac{1}{3} \) - Sum of roots \( (S) = \frac{1}{4} + \frac{1}{3} = \frac{7}{12} \) - Product of roots \( (P) = \frac{1}{4} \times \frac{1}{3} = \frac{1}{12} \) The quadratic equation is: \[ x^2 - \frac{7}{12}x + \frac{1}{12} = 0 \] Multiplying the entire equation by 12 to clear fractions: \[ 12x^2 - 7x + 1 = 0 \]
In simple words: Add the two numbers to find the sum, and multiply them to find the product. Put these into \( x^2 - (\text{Sum})x + \text{Product} = 0 \), then multiply the whole equation to clear any fractions.

Exam Tip: If the roots are fractional, always multiply the final equation by the least common multiple of the denominators to write it in standard integer form.

Question 2. 300 apples are distributed equally among a certain number of students. Had there been 10 more students, each would have received one apple less. Find the number of students.
Answer: Let the original number of students be \( x \). The number of apples each student gets initially is \( \frac{300}{x} \). If there are 10 more students, the total number of students becomes \( x + 10 \), and the share of each student is \( \frac{300}{x + 10} \). According to the given condition, each student receives 1 apple less in the second case: \[ \frac{300}{x} - \frac{300}{x + 10} = 1 \] \[ 300 \left( \frac{1}{x} - \frac{1}{x + 10} \right) = 1 \] \[ 300 \left( \frac{x + 10 - x}{x(x + 10)} \right) = 1 \] \[ \frac{3000}{x^2 + 10x} = 1 \] \[ x^2 + 10x - 3000 = 0 \] We solve this quadratic equation by splitting the middle term: \[ x^2 + 60x - 50x - 3000 = 0 \] \[ x(x + 60) - 50(x + 60) = 0 \] \[ (x - 50)(x + 60) = 0 \] Since the number of students must be a positive integer, we choose \( x = 50 \). Thus, the number of students is 50.
In simple words: Setting up the fraction of apples per student leads to the quadratic equation \( x^2 + 10x - 3000 = 0 \). Solving this gives 50 students.

Exam Tip: Clearly write down the expression for the difference in apple shares, ensuring that the larger fraction is written first.

Question 3. Solve for x: \( p^2x^2 + (p^2 - q^2)x - q^2 = 0 \).
Answer: Let's solve the given quadratic equation by expanding the middle term: \[ p^2x^2 + p^2x - q^2x - q^2 = 0 \] Now, group the terms to factorize: \[ p^2x(x + 1) - q^2(x + 1) = 0 \] \[ (p^2x - q^2)(x + 1) = 0 \] Set each factor to zero to find the roots: - \( x + 1 = 0 \implies x = -1 \) - \( p^2x - q^2 = 0 \implies x = \frac{q^2}{p^2} \) (provided \( p \ne 0 \)) Thus, the solutions are \( x = -1 \) and \( x = \frac{q^2}{p^2} \).
In simple words: We can solve this algebraic equation by grouping the terms, which easily factors into \( (p^2x - q^2)(x + 1) = 0 \).

Exam Tip: Factorization by grouping is much simpler and faster than using the quadratic formula when dealing with literal coefficients like \( p \) and \( q \).

Question 4. Find the solutions by the method of completion of squares. \( 3x^2 + 23x + 20 = 0 \).
Answer: To solve \( 3x^2 + 23x + 20 = 0 \) by completing the squares, we divide the entire equation by the coefficient of \( x^2 \), which is 3: \[ x^2 + \frac{23}{3}x + \frac{20}{3} = 0 \] Now, we add and subtract the square of half the coefficient of \( x \) - that is \( \left( \frac{1}{2} \times \frac{23}{3} \right)^2 = \left( \frac{23}{6} \right)^2 = \frac{529}{36} \): \[ x^2 + \frac{23}{3}x + \frac{529}{36} - \frac{529}{36} + \frac{20}{3} = 0 \] \[ \left( x + \frac{23}{6} \right)^2 = \frac{529}{36} - \frac{20}{3} \] \[ \left( x + \frac{23}{6} \right)^2 = \frac{529 - 240}{36} = \frac{289}{36} \] Taking the square root on both sides: \[ x + \frac{23}{6} = \pm \frac{17}{6} \] - Case 1: \( x = \frac{17}{6} - \frac{23}{6} = -\frac{6}{6} = -1 \) - Case 2: \( x = -\frac{17}{6} - \frac{23}{6} = -\frac{40}{6} = -\frac{20}{3} \) Thus, the solutions are \( x = -1 \) and \( x = -\frac{20}{3} \).
In simple words: Divide the whole equation by 3, make a perfect square on the left side, and find the two roots: \( -1 \) and \( -20/3 \).

Exam Tip: Make sure the coefficient of \( x^2 \) is exactly 1 before adding and subtracting the square of half the \( x \)-coefficient.

Question 5. Find the roots by factorization: \( 3\sqrt{3}x^2 + 19x + 10\sqrt{3} = 0 \).
Answer: We split the middle term \( 19x \) into two numbers whose product is \( (3\sqrt{3}) \times (10\sqrt{3}) = 90 \) and whose sum is 19. These numbers are 9 and 10: \[ 3\sqrt{3}x^2 + 9x + 10x + 10\sqrt{3} = 0 \] Rewrite \( 9 \) as \( 3\sqrt{3} \times \sqrt{3} \) to factor by grouping: \[ 3\sqrt{3}x(x + \sqrt{3}) + 10(x + \sqrt{3}) = 0 \] \[ (3\sqrt{3}x + 10)(x + \sqrt{3}) = 0 \] Set each factor to zero: - \( x + \sqrt{3} = 0 \implies x = -\sqrt{3} \) - \( 3\sqrt{3}x + 10 = 0 \implies x = -\frac{10}{3\sqrt{3}} = -\frac{10\sqrt{3}}{9} \) The roots are \( x = -\sqrt{3} \) and \( x = -\frac{10\sqrt{3}}{9} \).
In simple words: Split \( 19x \) into \( 9x + 10x \) and group the terms. This factors the equation into \( (3\sqrt{3}x + 10)(x + \sqrt{3}) = 0 \).

Exam Tip: Pay special attention to factoring out irrational numbers like \( \sqrt{3} \) from regular terms like 9 to simplify grouping.

Question 6. Find the solutions by quadratic equation formula: \( 3x^2 - 13x - 100 = 0 \).
Answer: Comparing \( 3x^2 - 13x - 100 = 0 \) with standard form \( ax^2 + bx + c = 0 \), we have \( a = 3 \), \( b = -13 \), and \( c = -100 \). First, find the discriminant \( D \): \[ D = b^2 - 4ac = (-13)^2 - 4(3)(-100) \] \[ D = 169 + 1200 = 1369 \] Now, apply the quadratic formula: \[ x = \frac{-b \pm \sqrt{D}}{2a} \] Since \( \sqrt{1369} = 37 \): \[ x = \frac{13 \pm 37}{6} \] - Case 1: \( x = \frac{13 + 37}{6} = \frac{50}{6} = \frac{25}{3} \) - Case 2: \( x = \frac{13 - 37}{6} = \frac{-24}{6} = -4 \) The solutions are \( x = \frac{25}{3} \) and \( x = -4 \).
In simple words: Calculate the discriminant (1369) and use the quadratic formula to find the roots: \( 25/3 \) and \( -4 \).

Exam Tip: Clearly show the step-by-step calculation of the discriminant \( b^2 - 4ac \) to secure partial marking even if a calculation error occurs later.

Question 7. Solve for x: \( 12abx^2 - (9a^2 - 8b^2)x - 6a = 0 \).
Answer: Let's solve the mathematically consistent standard version of this equation, which has the constant term as \( 6ab \): \[ 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \] Expand the terms in the middle: \[ 12abx^2 - 9a^2x + 8b^2x - 6ab = 0 \] Now factor the terms by grouping: \[ 3ax(4bx - 3a) + 2b(4bx - 3a) = 0 \] \[ (3ax + 2b)(4bx - 3a) = 0 \] Set each factor to zero to find the roots: - \( 3ax + 2b = 0 \implies x = -\frac{2b}{3a} \) - \( 4bx - 3a = 0 \implies x = \frac{3a}{4b} \) (assuming \( a, b \ne 0 \)) The solutions are \( x = -\frac{2b}{3a} \) and \( x = \frac{3a}{4b} \).
In simple words: Distribute the middle term to group and factor the expression into \( (3ax + 2b)(4bx - 3a) = 0 \).

Exam Tip: Be cautious of the minus sign in front of the parenthesis when expanding \( -(9a^2 - 8b^2)x \).

Question 8. Solve for x: \( (a + b)^2x^2 + 8(a^2 - b^2)x + 16(a - b)^2 = 0 \).
Answer: To simplify the equation, let's substitute \( A = a + b \) and \( B = a - b \): \[ A^2x^2 + 8ABx + 16B^2 = 0 \] This is a perfect square trinomial of the form \( (Ax + 4B)^2 = 0 \). Let's verify by expanding: \[ (Ax + 4B)^2 = (Ax)^2 + 2(Ax)(4B) + (4B)^2 = A^2x^2 + 8ABx + 16B^2 \] Since \( (Ax + 4B)^2 = 0 \): \[ Ax + 4B = 0 \] \[ x = -\frac{4B}{A} \] Substitute back the original values of \( A \) and \( B \): \[ x = -\frac{4(a - b)}{a + b} = \frac{4(b - a)}{a + b} \] The equal roots are \( x = \frac{4(b - a)}{a + b} \).
In simple words: Treat \( a+b \) and \( a-b \) as single variables. The equation is a perfect square \( (Ax + 4B)^2 = 0 \), which simplifies directly to \( x = \frac{4(b - a)}{a + b} \).

Exam Tip: Recognizing algebraic identities in quadratic equations avoids complex expansions and tedious calculations.

Question 9. The speed of a boat in still water is 11 km/h. It can go 12 km upstream and come back in 2 hours 45 minutes. Find the speed of the stream.
Answer: Let the speed of the stream be \( y \) km/h. Then: - Speed upstream = \( 11 - y \) km/h - Speed downstream = \( 11 + y \) km/h Using \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Time upstream = \( \frac{12}{11 - y} \) hours - Time downstream = \( \frac{12}{11 + y} \) hours The total time is 2 hours 45 minutes, which is \( 2 \frac{45}{60} = \frac{11}{4} \) hours. \[ \frac{12}{11 - y} + \frac{12}{11 + y} = \frac{11}{4} \] \[ 12 \left( \frac{11 + y + 11 - y}{(11 - y)(11 + y)} \right) = \frac{11}{4} \] \[ 12 \left( \frac{22}{121 - y^2} \right) = \frac{11}{4} \] Dividing both sides by 11: \[ 12 \left( \frac{2}{121 - y^2} \right) = \frac{1}{4} \] \[ \frac{24}{121 - y^2} = \frac{1}{4} \] \[ 96 = 121 - y^2 \] \[ y^2 = 25 \implies y = 5 \] (since speed must be positive) The speed of the stream is 5 km/h.
In simple words: Set up the sum of upstream and downstream times to equal \( 11/4 \) hours. Solving the equation gives the stream's speed as 5 km/h.

Exam Tip: Make sure to convert mixed time units (hours and minutes) completely to fractional hours before setting up the equation.

Question 10. The product of the digits of a 2 digit number is 15. If 18 is added to the number, the digits interchange their places. Find the number.
Answer: Let the two-digit number be represented by \( 10u + v \), where \( u \) is the tens digit and \( v \) is the units digit. According to the given conditions: 1. The product of the digits is 15: \[ u \times v = 15 \implies v = \frac{15}{u} \] (Equation 1) 2. Adding 18 interchanges the digits: \[ (10u + v) + 18 = 10v + u \] \[ 9u - 9v + 18 = 0 \] \[ u - v + 2 = 0 \implies v = u + 2 \] (Equation 2) Substitute Equation 1 into Equation 2: \[ u + 2 = \frac{15}{u} \] \[ u^2 + 2u - 15 = 0 \] We solve the quadratic equation by factoring: \[ (u + 5)(u - 3) = 0 \] Since digits of a number must be positive integers, we choose \( u = 3 \). Then \( v = 3 + 2 = 5 \). The number is 35.
In simple words: Using the digit equations, we find the tens digit is 3 and the units digit is 5, giving the number 35.

Exam Tip: Always use the standard expansion \( 10u + v \) to represent a two-digit number in place value problems.

Question 11. Divide 29 into 2 parts so that the sum of their squares is 425.
Answer: Let the first part be \( x \). The second part is therefore \( 29 - x \). The sum of their squares is 425: \[ x^2 + (29 - x)^2 = 425 \] \[ x^2 + (841 - 58x + x^2) = 425 \] \[ 2x^2 - 58x + 416 = 0 \] Divide the entire equation by 2: \[ x^2 - 29x + 208 = 0 \] We solve this quadratic equation by splitting the middle term: \[ (x - 13)(x - 16) = 0 \] This gives \( x = 13 \) or \( x = 16 \). Therefore, the two parts are 13 and 16.
In simple words: Let the two parts be \( x \) and \( 29 - x \). Setting the sum of their squares to 425 gives the parts 13 and 16.

Exam Tip: Dividing the equation by 2 simplifies the constant term to 208, making it much easier to find the factors.

Question 12. The denominator of a fraction is one more than twice the numerator. If the sum of the fraction and its reciprocal is \( \frac{16}{21} \), find the fraction.
Answer: Let the numerator be \( n \). The denominator of the fraction is given as \( 2n + 1 \). The fraction is \( \frac{n}{2n + 1} \). The sum of the fraction and its reciprocal is given as \( 2 \frac{16}{21} = \frac{58}{21} \): \[ \frac{n}{2n + 1} + \frac{2n + 1}{n} = \frac{58}{21} \] Let's substitute \( y = \frac{n}{2n + 1} \): \[ y + \frac{1}{y} = \frac{58}{21} \] \[ \frac{y^2 + 1}{y} = \frac{58}{21} \] \[ 21y^2 - 58y + 21 = 0 \] We solve this quadratic equation by factoring: \[ 21y^2 - 49y - 9y + 21 = 0 \] \[ 7y(3y - 7) - 3(3y - 7) = 0 \] \[ (7y - 3)(3y - 7) = 0 \] This gives \( y = \frac{3}{7} \) or \( y = \frac{7}{3} \). Since \( y = \frac{n}{2n + 1} \) must be less than 1 (as denominator is larger than numerator), we choose \( y = \frac{3}{7} \). Matching this to \( \frac{n}{2n + 1} \), we have \( n = 3 \) and \( 2n + 1 = 7 \). Thus, the fraction is \( \frac{3}{7} \).
In simple words: We can solve the equation by representing the fraction as \( y \). Solving the quadratic equation \( 21y^2 - 58y + 21 = 0 \) gives the fraction as \( 3/7 \).

Exam Tip: Substituting the entire algebraic fraction as a single variable \( y \) keeps the equation simple and easy to factor.

Question 13. The hypotenuse of a right triangle is 6m more than twice the shortest side. If the third side is 2 metre less than the hypotenuse, find the sides of the triangle. 
Answer: Let the length of the shortest side be \( x \) meters. Based on the conditions: - Hypotenuse = \( 2x + 6 \) meters - Third side = \( (2x + 6) - 2 = 2x + 4 \) meters Using Pythagoras theorem: \[ (\text{Shortest side})^2 + (\text{Third side})^2 = (\text{Hypotenuse})^2 \] \[ x^2 + (2x + 4)^2 = (2x + 6)^2 \] \[ x^2 + (4x^2 + 16x + 16) = 4x^2 + 24x + 36 \] \[ x^2 - 8x - 20 = 0 \] Factorize the quadratic equation: \[ (x - 10)(x + 2) = 0 \] Since side lengths must be positive, we choose \( x = 10 \). The dimensions of the triangle's sides are: - Shortest side = \( 10 \text{ m} \) - Third side = \( 2(10) + 4 = 24 \text{ m} \) - Hypotenuse = \( 2(10) + 6 = 26 \text{ m} \).
In simple words: We use Pythagoras theorem on sides \( x \), \( 2x+4 \), and \( 2x+6 \) to get a simple quadratic equation. Solving it gives the side lengths as 10 m, 24 m, and 26 m.

Exam Tip: When factoring \( x^2 - 8x - 20 = 0 \), always check the signs to ensure that \( (x-10)(x+2) \) is correct.

Question 14. The hypotenuse of a right triangle is \( 3\sqrt{5} \) cm. If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be 15 cm. Find the length of each side. [Ans.: 3cm, 6cm]
Answer: Let the smaller side be \( a \) cm and the larger side be \( b \) cm. From the first right-angled triangle condition: \[ a^2 + b^2 = (3\sqrt{5})^2 = 45 \] (Equation 1) From the second condition, when the smaller side is tripled (\( 3a \)) and the larger side is doubled (\( 2b \)), the new hypotenuse is 15 cm: \[ (3a)^2 + (2b)^2 = 15^2 \] \[ 9a^2 + 4b^2 = 225 \] (Equation 2) From Equation 1, we express \( b^2 = 45 - a^2 \). Substituting this into Equation 2: \[ 9a^2 + 4(45 - a^2) = 225 \] \[ 9a^2 + 180 - 4a^2 = 225 \] \[ 5a^2 = 45 \implies a^2 = 9 \implies a = 3 \] (since \( a > 0 \)) Now, find \( b \): \[ b^2 = 45 - 9 = 36 \implies b = 6 \] (since \( b > 0 \)) Therefore, the lengths of the sides are 3 cm and 6 cm.
In simple words: We write two equations using the Pythagorean theorem. Solving the system of equations gives the smaller side as 3 cm and the larger side as 6 cm.

Exam Tip: Substituting variables like \( a^2 \) or \( b^2 \) avoids dealing with square roots in simultaneous equations.

Question 15. The lengths of the sides forming right angle of a right angled ∆ is \( 5x \) cm and \( (3x - 1) \) cm. Area of the triangle is 60 cm². Find the hypotenuse.
Answer: Let's represent the perpendicular sides of the right-angled triangle. - Base = \( 5x \) - Height = \( 3x - 1 \) Since the area of the right triangle is 60 \( \text{cm}^2 \): \[ \frac{1}{2} \times 5x \times (3x - 1) = 60 \] \[ 5x(3x - 1) = 120 \] \[ 15x^2 - 5x - 120 = 0 \] Dividing the equation by 5: \[ 3x^2 - x - 24 = 0 \] Splitting the middle term: \[ 3x^2 - 9x + 8x - 24 = 0 \] \[ 3x(x - 3) + 8(x - 3) = 0 \] \[ (3x + 8)(x - 3) = 0 \] Since \( x \) must be a positive number for the side lengths to be positive, we choose \( x = 3 \). The side lengths are: - First side = \( 5(3) = 15 \text{ cm} \) - Second side = \( 3(3) - 1 = 8 \text{ cm} \) Using Pythagoras theorem to find the hypotenuse: \[ \text{Hypotenuse} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17 \text{ cm} \].
In simple words: We solve the area equation to find \( x = 3 \). The side lengths are 15 cm and 8 cm, which gives the hypotenuse as 17 cm.

Exam Tip: Always write the formula for the area of a right-angled triangle, \( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \), at the start of your work.

Question 16. A takes 6 days less than the time taken by B to complete a work. If they together can complete the work in 4 days, find the time taken by each to finish the work. 
Answer: Let the time taken by B to complete the work be \( x \) days. Then, the time taken by A is \( x - 6 \) days. In one day, B does \( \frac{1}{x} \) of the work, and A does \( \frac{1}{x - 6} \) of the work. Since they complete the work together in 4 days, their combined work rate is: \[ \frac{1}{x} + \frac{1}{x - 6} = \frac{1}{4} \] \[ \frac{x - 6 + x}{x(x - 6)} = \frac{1}{4} \] \[ \frac{2x - 6}{x^2 - 6x} = \frac{1}{4} \] \[ 8x - 24 = x^2 - 6x \] \[ x^2 - 14x + 24 = 0 \] We solve the quadratic equation by factoring: \[ (x - 12)(x - 2) = 0 \] Since A's time \( x - 6 \) must be positive, \( x \) must be greater than 6. Thus, we choose \( x = 12 \) days. The individual completion times are: - B's time = 12 days - A's time = \( 12 - 6 = 6 \) days.
In simple words: We write the sum of their individual daily work rates to equal \( 1/4 \). Solving the resulting quadratic equation gives B's time as 12 days and A's time as 6 days.

Exam Tip: Always verify that the final solutions are valid in context (here, \( x=2 \) is rejected because A's time would be negative).

Question 17. A person on a tour had Rs. 12000 for his daily expenses. In order to extend his journey for 2 more days he had to cut down his daily expenses by Rs. 300. Find the duration of the tour he planned first.
Answer: Let the original duration of the tour be \( d \) days. Original daily expense = Rs. \( \frac{12000}{d} \) New duration of the tour = \( d + 2 \) days New daily expense = Rs. \( \frac{12000}{d + 2} \) According to the problem, the difference in daily expenses is Rs. 300: \[ \frac{12000}{d} - \frac{12000}{d + 2} = 300 \] Dividing both sides of the equation by 300: \[ \frac{40}{d} - \frac{40}{d + 2} = 1 \] \[ 40 \left( \frac{d + 2 - d}{d(d + 2)} \right) = 1 \] \[ \frac{80}{d^2 + 2d} = 1 \] \[ d^2 + 2d - 80 = 0 \] Factoring the quadratic equation: \[ (d + 10)(d - 8) = 0 \] Since the duration must be positive, we choose \( d = 8 \). The duration of the tour first planned was 8 days.
In simple words: Set up the daily budget equation based on the total sum of Rs. 12000. Solving \( d^2 + 2d - 80 = 0 \) gives the initial duration of the tour as 8 days.

Exam Tip: Be careful with divisions - dividing large constants like 12000 by 300 makes the arithmetic steps much simpler.

Question 18. A man sold an article for Rs. 96 gaining as many as percent as the cost price (in Rupees) is. Find the cost price of the article. 
Answer: Let the cost price (C.P.) of the article be Rs. \( x \). The profit percentage is given to be equal to the numerical value of the cost price, which is \( x\% \). Therefore, the profit amount is: \[ \text{Profit} = x\% \text{ of C.P.} = \frac{x}{100} \times x = \frac{x^2}{100} \] The selling price (S.P.) is Rs. 96: \[ \text{S.P.} = \text{C.P.} + \text{Profit} \] \[ 96 = x + \frac{x^2}{100} \] Multiply by 100 to clear the fraction: \[ x^2 + 100x - 9600 = 0 \] Factoring the quadratic equation: \[ (x + 160)(x - 60) = 0 \] Since the cost price must be positive, we reject \( x = -160 \). Thus, the cost price of the article is Rs. 60.
In simple words: We set up the relation \( x + \frac{x^2}{100} = 96 \) where \( x \) is the cost price. Solving this gives the cost price as Rs. 60.

Exam Tip: Be careful to convert the cost price percentage gain into a fraction (over 100) before writing down the final quadratic expression.

Question 19. A peacock is sitting on the top of a pillar which is 9 m high. From a point 27 m away from the bottom of the pillar, a snake is coming to its hole at the base of the pillar. Seeing the snake, the peacock pounces on it. If their speeds are equal, at what distance from hole, the snake caught?
Answer: Let \( AB \) be the pillar of height 9 m, with \( A \) being the top and \( B \) being the base (where the hole is). Let the snake start at point \( C \), which is 27 m away from \( B \). Suppose the snake is caught at point \( D \), at a distance of \( x \) meters from the hole \( B \). The distance traveled by the snake is: \[ CD = 27 - x \] Since the speeds of the peacock and the snake are equal, they cover equal distances in the same time: \[ AD = CD = 27 - x \] In right-angled triangle \( ABD \), using Pythagoras theorem: \[ AD^2 = AB^2 + BD^2 \] \[ (27 - x)^2 = 9^2 + x^2 \] \[ 729 - 54x + x^2 = 81 + x^2 \] \[ 54x = 729 - 81 \] \[ 54x = 648 \] \[ x = 12 \text{ m} \] The snake is caught at a distance of 12 meters from the hole.
In simple words: Since they travel at equal speeds, the peacock's flight distance and the snake's slithering distance are equal. Setting their distances equal using Pythagoras theorem gives \( x = 12 \) meters.

Exam Tip: Representing the scenario with a neat right triangle sketch makes setting up the Pythagoras equation straightforward.

Question 20. One fourth of a herd of camels was seen in the forest. Twice the square root of the herd gone to mountains and the remaining 15 camels were seen on the bank of the river. Find the total number of camels.
Answer: Let the total number of camels be \( x \). Based on the conditions given: - Camels in the forest = \( \frac{x}{4} \) - Camels in the mountains = \( 2\sqrt{x} \) - Camels on the river bank = 15 Sum of these parts equals the total number of camels: \[ \frac{x}{4} + 2\sqrt{x} + 15 = x \] Substitute \( y = \sqrt{x} \implies x = y^2 \): \[ \frac{y^2}{4} + 2y + 15 = y^2 \] Multiplying the entire equation by 4: \[ y^2 + 8y + 60 = 4y^2 \] \[ 3y^2 - 8y - 60 = 0 \] Splitting the middle term to factorize: \[ 3y^2 - 18y + 10y - 60 = 0 \] \[ 3y(y - 6) + 10(y - 6) = 0 \] \[ (3y + 10)(y - 6) = 0 \] Since \( y = \sqrt{x} \) must be positive, we choose \( y = 6 \). Therefore: \[ x = y^2 = 6^2 = 36 \] The total number of camels is 36.
In simple words: Let the total number of camels be \( x \). We write the equation based on the fractions and square roots of \( x \), which simplifies to \( y = 6 \) (where \( y = \sqrt{x} \)), giving 36 camels in total.

Exam Tip: Substituting \( y = \sqrt{x} \) is a very helpful technique to convert a radical equation into a simple quadratic equation.

Question 21. Some boys planned a picnic. The budget for food was Rs. 3000. Five boys could not go for picnic and thus the cost of food for each member increased by Rs. 30. How many boys attended the picnic? 
Answer: Let the original number of boys who planned the picnic be \( x \). The initial share for food of each boy is Rs. \( \frac{3000}{x} \). Since 5 boys could not attend, the new number of boys is \( x - 5 \), and their share is Rs. \( \frac{3000}{x - 5} \). The difference between the individual shares is Rs. 30: \[ \frac{3000}{x - 5} - \frac{3000}{x} = 30 \] Dividing both sides of the equation by 30: \[ \frac{100}{x - 5} - \frac{100}{x} = 1 \] \[ 100 \left( \frac{x - (x - 5)}{x(x - 5)} \right) = 1 \] \[ \frac{500}{x^2 - 5x} = 1 \] \[ x^2 - 5x - 500 = 0 \] Factoring the quadratic equation: \[ (x - 25)(x + 20) = 0 \] Since the number of boys must be positive, we choose \( x = 25 \). The number of boys who attended the picnic is: \[ x - 5 = 25 - 5 = 20 \].
In simple words: Let \( x \) be the original number of boys. Solving \( x^2 - 5x - 500 = 0 \) gives \( x = 25 \). Thus, 20 boys actually attended the picnic.

Exam Tip: Read the final question carefully to make sure you answer with the number of boys who *attended* the picnic (\( x-5 \)), rather than the original number (\( x \)).

Question 22. Two taps running together can fill a cistern in \( 3\frac{1}{13} \) minutes. If one of the taps takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.  
Answer: Let the time taken by the faster tap alone to fill the cistern be \( x \) minutes. Then, the slower tap takes \( x + 3 \) minutes. Together, the taps fill the cistern in \( 3 \frac{1}{13} = \frac{40}{13} \) minutes, which means they fill \( \frac{13}{40} \) of the cistern in one minute: \[ \frac{1}{x} + \frac{1}{x + 3} = \frac{13}{40} \] \[ \frac{x + 3 + x}{x(x + 3)} = \frac{13}{40} \] \[ \frac{2x + 3}{x^2 + 3x} = \frac{13}{40} \] Cross-multiplying: \[ 40(2x + 3) = 13(x^2 + 3x) \] \[ 80x + 120 = 13x^2 + 39x \] \[ 13x^2 - 41x - 120 = 0 \] We solve the quadratic equation by splitting the middle term: \[ 13x^2 - 65x + 24x - 120 = 0 \] \[ 13x(x - 5) + 24(x - 5) = 0 \] \[ (13x + 24)(x - 5) = 0 \] Since time must be positive, we choose \( x = 5 \). The times taken by each pipe are: - Faster tap = 5 minutes - Slower tap = \( 5 + 3 = 8 \) minutes.
In simple words: We set up the work-rate equation \( \frac{1}{x} + \frac{1}{x+3} = \frac{13}{40} \). Solving this gives the individual times as 5 minutes and 8 minutes.

Exam Tip: Be accurate when computing the factors of \( 13 \times (-120) = -1560 \) to correctly split the middle term.

Question 23. The product of 3 consecutive even numbers is equal to 20 times their sum. Find the numbers. 
Answer: Let the three consecutive even numbers be represented by \( x - 2 \), \( x \), and \( x + 2 \). The sum of the numbers is: \[ (x - 2) + x + (x + 2) = 3x \] The product of the numbers is: \[ (x - 2) \times x \times (x + 2) = x(x^2 - 4) \] According to the problem: \[ x(x^2 - 4) = 20(3x) \] Since \( x \) is a non-zero even integer, we can divide both sides by \( x \): \[ x^2 - 4 = 60 \] \[ x^2 = 64 \implies x = 8 \] (considering positive integers) Thus, the three consecutive even numbers are: - \( x - 2 = 6 \) - \( x = 8 \) - \( x + 2 = 10 \)
In simple words: Let the consecutive even numbers be \( x-2 \), \( x \), and \( x+2 \). Setting their product equal to 20 times their sum simplifies to \( x = 8 \), which gives the numbers 6, 8, and 10.

Exam Tip: Representing consecutive even numbers symmetrically as \( x-2, x, x+2 \) simplifies the product-to-sum algebra significantly by letting you divide out \( x \) early on.

Question 24. The sum of the squares of two positive integers is 208. If the square of the larger number is 18 times the smaller number, find the numbers. 
Answer: Let the smaller positive integer be \( x \) and the larger positive integer be \( y \). We are given two conditions: 1. The sum of the squares is 208: \[ x^2 + y^2 = 208 \] (Equation 1) 2. The square of the larger number is 18 times the smaller number: \[ y^2 = 18x \] (Equation 2) Substitute Equation 2 into Equation 1: \[ x^2 + 18x = 208 \] \[ x^2 + 18x - 208 = 0 \] Factorize the quadratic equation: \[ (x + 26)(x - 8) = 0 \] Since \( x \) must be a positive integer, we reject the negative solution \( x = -26 \). Thus: \[ x = 8 \] Now find \( y^2 \) using Equation 2: \[ y^2 = 18(8) = 144 \implies y = 12 \] (since \( y > 0 \)) The numbers are 8 and 12.
In simple words: Substitute \( y^2 = 18x \) into the sum of squares equation to get \( x^2 + 18x - 208 = 0 \). Solving this gives the numbers as 8 and 12.

Exam Tip: Be careful to select only positive integers when given that specific constraint in the question.

Question 25. The length of a rectangle is 2 cm more than twice its breadth and its area is 60 cm². Find the diagonal of the rectangle.
Answer: Let the breadth of the rectangle be \( b \) cm. Then, the length is \( l = 2b + 2 \) cm. Since the area is 60 \( \text{cm}^2 \): \[ b(2b + 2) = 60 \] \[ 2b^2 + 2b - 60 = 0 \] Dividing by 2 on both sides: \[ b^2 + b - 30 = 0 \] We solve the quadratic equation by factoring: \[ (b + 6)(b - 5) = 0 \] Since breadth must be positive, we choose \( b = 5 \) cm. Now find the length: \[ l = 2(5) + 2 = 12 \text{ cm} \] We use Pythagoras theorem to find the diagonal \( d \): \[ d = \sqrt{l^2 + b^2} = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13 \text{ cm} \].
In simple words: Form the area equation \( b(2b+2) = 60 \), which gives breadth as 5 cm and length as 12 cm. Using Pythagoras theorem, the diagonal is 13 cm.

Exam Tip: Always state the formula for the diagonal of a rectangle \( \sqrt{l^2 + w^2} \) clearly before substituting the dimensions.