CBSE Class 10 Mathematics Quadratic Equations Assignment Set 13

Question 1. The sum of two numbers is 18 and their product is 45. Find the numbers.
Answer: Let one of the numbers be \( x \). Since their sum is 18, the other number is \( 18 - x \). We are given that their product is 45. This can be expressed as: \[ x(18 - x) = 45 \]
\( \implies 18x - x^2 = 45 \)
\( \implies x^2 - 18x + 45 = 0 \) We factor this quadratic equation by finding two numbers that multiply to 45 and add up to -18, which are -15 and -3:
\( \implies x^2 - 15x - 3x + 45 = 0 \)
\( \implies x(x - 15) - 3(x - 15) = 0 \)
\( \implies (x - 15)(x - 3) = 0 \) Setting each factor to zero, we get:
\( \implies x = 15 \) or \( x = 3 \) If \( x = 15 \), the other number is \( 18 - 15 = 3 \). If \( x = 3 \), the other number is \( 18 - 3 = 15 \). Thus, the two numbers are 3 and 15.
In simple words: To find two numbers that add up to 18 and multiply to 45, we create a quadratic equation. Solving it shows that the numbers are 3 and 15.
Exam Tip: Clearly define the two parts as \( x \) and \( (\text{Sum} - x) \) at the beginning of your solution to avoid working with multiple variables.

Question 2. Divide 29 into two parts so that the sum of the squares of the parts is 425
Answer: Let the first part be \( x \). The second part will then be \( 29 - x \). According to the problem, the sum of their squares is 425: \[ x^2 + (29 - x)^2 = 425 \]
\( \implies x^2 + (841 - 58x + x^2) = 425 \)
\( \implies 2x^2 - 58x + 841 - 425 = 0 \)
\( \implies 2x^2 - 58x + 416 = 0 \) Dividing the entire equation by 2:
\( \implies x^2 - 29x + 208 = 0 \) We need to find two numbers that multiply to 208 and add up to -29. These numbers are -13 and -16:
\( \implies x^2 - 13x - 16x + 208 = 0 \)
\( \implies x(x - 13) - 16(x - 13) = 0 \)
\( \implies (x - 13)(x - 16) = 0 \) This gives:
\( \implies x = 13 \) or \( x = 16 \) Thus, the two parts are 13 and 16.
In simple words: If you split 29 into two pieces whose squares add up to 425, those two pieces must be 13 and 16.

Exam Tip: Simplify the quadratic equation by dividing all terms by a common factor (like 2 in this case) before attempting to factorize it.

Question 3. Find two consecutive natural numbers whose product is 20.
Answer: Let the two consecutive natural numbers be \( x \) and \( x + 1 \). Given that their product is 20: \[ x(x + 1) = 20 \]
\( \implies x^2 + x - 20 = 0 \) Factoring the quadratic equation:
\( \implies x^2 + 5x - 4x - 20 = 0 \)
\( \implies x(x + 5) - 4(x + 5) = 0 \)
\( \implies (x + 5)(x - 4) = 0 \) This yields:
\( \implies x = -5 \) or \( x = 4 \) Since \( x \) must be a natural number, we reject the negative value \( x = -5 \). Therefore, \( x = 4 \), and the next consecutive number is \( 4 + 1 = 5 \). The required consecutive natural numbers are 4 and 5.
In simple words: We are looking for two regular counting numbers next to each other that multiply to 20. Solving the quadratic equation gives us 4 and 5.

Exam Tip: Natural numbers are strictly positive integers. Always write a brief reason for rejecting any negative value of \( x \).

Question 4. Find two consecutive numbers whose squares have the sum 85
Answer: Let the two consecutive numbers be \( x \) and \( x + 1 \). The sum of their squares is 85: \[ x^2 + (x + 1)^2 = 85 \]
\( \implies x^2 + x^2 + 2x + 1 = 85 \)
\( \implies 2x^2 + 2x + 1 - 85 = 0 \)
\( \implies 2x^2 + 2x - 84 = 0 \) Dividing the entire equation by 2:
\( \implies x^2 + x - 42 = 0 \) Factoring the equation:
\( \implies x^2 + 7x - 6x - 42 = 0 \)
\( \implies x(x + 7) - 6(x + 7) = 0 \)
\( \implies (x + 7)(x - 6) = 0 \) This gives:
\( \implies x = -7 \) or \( x = 6 \) Case 1: If \( x = 6 \), the consecutive numbers are 6 and 7. Case 2: If \( x = -7 \), the consecutive numbers are -7 and -6. Thus, the consecutive numbers are \( \pm 6 \) and \( \pm 7 \).
In simple words: Since the question did not specify "natural numbers," both positive pairs (6 and 7) and negative pairs (-7 and -6) are valid answers.

Exam Tip: Unless "natural numbers" or "positive integers" is specified, keep both negative and positive roots in your final answer.

Question 5. The sum of the squares of three consecutive natural numbers is 149. Find the numbers.
Answer: Let the three consecutive natural numbers be \( x - 1 \), \( x \), and \( x + 1 \). According to the given condition, the sum of their squares is 149: \[ (x - 1)^2 + x^2 + (x + 1)^2 = 149 \]
\( \implies (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 149 \) Combining like terms:
\( \implies 3x^2 + 2 = 149 \)
\( \implies 3x^2 = 147 \)
\( \implies x^2 = 49 \)
\( \implies x = \pm 7 \) Since the numbers must be natural numbers, we take the positive value \( x = 7 \). Thus, the three consecutive numbers are: \( x - 1 = 7 - 1 = 6 \) \( x = 7 \) \( x + 1 = 7 + 1 = 8 \) The required numbers are 6, 7, and 8.
In simple words: If you square three consecutive numbers and add them up to get 149, those numbers are 6, 7, and 8.

Exam Tip: Assuming the middle number as \( x \) simplifies the equation because the linear \( 2x \) terms cancel out during expansion.

Question 6. Find the whole number which when increased by 20 is equal to 69 times the reciprocal of the number.
Answer: Let the required whole number be \( x \). When increased by 20, the number becomes \( x + 20 \). This is equal to 69 times its reciprocal, which is \( \frac{69}{x} \). We can set up the equation: \[ x + 20 = \frac{69}{x} \] Multiplying both sides by \( x \):
\( \implies x^2 + 20x = 69 \)
\( \implies x^2 + 20x - 69 = 0 \) Factoring by finding two numbers that multiply to -69 and add up to 20 (which are 23 and -3):
\( \implies x^2 + 23x - 3x - 69 = 0 \)
\( \implies x(x + 23) - 3(x + 23) = 0 \)
\( \implies (x + 23)(x - 3) = 0 \) This gives:
\( \implies x = -23 \) or \( x = 3 \) Since \( x \) is a whole number, it cannot be negative. Therefore, we reject \( x = -23 \). The required whole number is 3.
In simple words: We want to find a positive number that, when we add 20 to it, matches 69 divided by that same number. That number is 3.

Exam Tip: Whole numbers start from 0 and include all positive integers. Negative integers are not whole numbers, so always discard negative roots in such cases.

Question 7. The sum of a number and its reciprocal is \( 2\frac{16}{21} \). Find the number.
Answer: Let the required number be \( x \). Its reciprocal is \( \frac{1}{x} \). The mixed fraction \( 2\frac{16}{21} \) can be converted to an improper fraction: \[ 2\frac{16}{21} = \frac{2 \times 21 + 16}{21} = \frac{58}{21} \] According to the problem: \[ x + \frac{1}{x} = \frac{58}{21} \]
\( \implies \frac{x^2 + 1}{x} = \frac{58}{21} \) Cross-multiplying:
\( \implies 21(x^2 + 1) = 58x \)
\( \implies 21x^2 + 21 = 58x \)
\( \implies 21x^2 - 58x + 21 = 0 \) To factorize, we need to split the middle term. Here, \( a \times c = 21 \times 21 = 441 \). We find two factors of 441 that sum to -58, which are -49 and -9:
\( \implies 21x^2 - 49x - 9x + 21 = 0 \)
\( \implies 7x(3x - 7) - 3(3x - 7) = 0 \)
\( \implies (3x - 7)(7x - 3) = 0 \) Setting each factor to zero:
\( \implies 3x - 7 = 0 \) or \( 7x - 3 = 0 \)
\( \implies x = \frac{7}{3} \) or \( x = \frac{3}{7} \) The required number is \( \frac{7}{3} \) or \( \frac{3}{7} \).
In simple words: Adding a number and its upside-down version gives us \( \frac{58}{21} \). The number we are looking for is either \( \frac{7}{3} \) or \( \frac{3}{7} \).

Exam Tip: In reciprocal problems, the two roots will always be reciprocals of each other. Mentioning both values covers all cases and secures full marks.

Question 8. The speed of a boat in still water is 15 km/hr. It can go 30 km upstream, and return downstream to the original point in \( 4\frac{1}{2} \) hrs. Find the speed of the stream.
Answer: Let the speed of the stream be \( x \) km/hr. The speed of the boat upstream is \( (15 - x) \) km/hr, and downstream is \( (15 + x) \) km/hr. Distance traveled in each direction is 30 km. The total time taken is \( 4\frac{1}{2} = \frac{9}{2} \) hours. Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): \[ \frac{30}{15 - x} + \frac{30}{15 + x} = \frac{9}{2} \] Dividing both sides by 3 to simplify:
\( \implies \frac{10}{15 - x} + \frac{10}{15 + x} = \frac{3}{2} \) Taking the common denominator:
\( \implies 10 \left[ \frac{(15 + x) + (15 - x)}{(15 - x)(15 + x)} \right] = \frac{3}{2} \)
\( \implies 10 \left[ \frac{30}{225 - x^2} \right] = \frac{3}{2} \)
\( \implies \frac{300}{225 - x^2} = \frac{3}{2} \) Cross-multiplying:
\( \implies 300 \times 2 = 3(225 - x^2) \)
\( \implies 600 = 675 - 3x^2 \)
\( \implies 3x^2 = 75 \)
\( \implies x^2 = 25 \)
\( \implies x = \pm 5 \) Since the speed of the stream cannot be negative, we reject \( x = -5 \). The speed of the stream is 5 km/hr.
In simple words: The current of the water slows down the boat when going upstream and speeds it up going downstream. Solving the time equation shows the river flows at 5 km/hr.

Exam Tip: Remember that upstream speed is always \( (\text{boat speed} - \text{stream speed}) \) and downstream speed is \( (\text{boat speed} + \text{stream speed}) \).

Question 9. Ram can row a boat 8 km downstream and return in 1 hour 40 minutes. If the speed of the stream is 2 km/hr. Find the speed of boat in still water.
Answer: Let the speed of the boat in still water be \( y \) km/hr. The speed of the stream is given as 2 km/hr. So, the downstream speed is \( (y + 2) \) km/hr and the upstream speed is \( (y - 2) \) km/hr. The total distance traveled in each direction is 8 km. Total time is 1 hour 40 minutes, which can be converted to hours: \[ 1\text{ hour } 40\text{ minutes} = 1 + \frac{40}{60} = 1 + \frac{2}{3} = \frac{5}{3}\text{ hours} \] Setting up the equation for total time: \[ \frac{8}{y + 2} + \frac{8}{y - 2} = \frac{5}{3} \]
\( \implies 8 \left[ \frac{(y - 2) + (y + 2)}{(y + 2)(y - 2)} \right] = \frac{5}{3} \)
\( \implies 8 \left[ \frac{2y}{y^2 - 4} \right] = \frac{5}{3} \)
\( \implies \frac{16y}{y^2 - 4} = \frac{5}{3} \) Cross-multiplying:
\( \implies 48y = 5(y^2 - 4) \)
\( \implies 5y^2 - 48y - 20 = 0 \) Factoring the quadratic equation:
\( \implies 5y^2 - 50y + 2y - 20 = 0 \)
\( \implies 5y(y - 10) + 2(y - 10) = 0 \)
\( \implies (5y + 2)(y - 10) = 0 \) This gives:
\( \implies y = -\frac{2}{5} \) or \( y = 10 \) Since speed cannot be negative, we reject \( y = -\frac{2}{5} \). The speed of the boat in still water is 10 km/hr.
In simple words: By writing the total rowing time as a sum of upstream and downstream trips, we find that Ram rows at a speed of 10 km/hr when the water is still.

Exam Tip: Be careful to convert mixed time units like "1 hour 40 minutes" completely into hours before using them in the equation.

Question 10. A passenger train takes 3 hrs less for a journey of 360 km. If its speed is increased by 10 km/hr from the usual speed. Find its usual speed.
Answer: Let the usual speed of the passenger train be \( v \) km/hr. The total distance of the journey is 360 km. The time taken at usual speed is \( \frac{360}{v} \) hours. If the speed is increased by 10 km/hr, the new speed is \( (v + 10) \) km/hr. The time taken at this new speed is \( \frac{360}{v + 10} \) hours. The difference in time is given as 3 hours: \[ \frac{360}{v} - \frac{360}{v + 10} = 3 \] Dividing by 3 to simplify:
\( \implies \frac{120}{v} - \frac{120}{v + 10} = 1 \)
\( \implies 120 \left[ \frac{(v + 10) - v}{v(v + 10)} \right] = 1 \)
\( \implies 120 \left[ \frac{10}{v^2 + 10v} \right] = 1 \)
\( \implies \frac{1200}{v^2 + 10v} = 1 \)
\( \implies v^2 + 10v - 1200 = 0 \) Factoring the quadratic equation:
\( \implies v^2 + 40v - 30v - 1200 = 0 \)
\( \implies v(v + 40) - 30(v + 40) = 0 \)
\( \implies (v + 40)(v - 30) = 0 \) This yields:
\( \implies v = -40 \) or \( v = 30 \) Since speed cannot be negative, we reject \( v = -40 \). The usual speed of the train is 30 km/hr.
In simple words: Going 10 km/hr faster cuts 3 hours off a 360 km trip. Solving the math show us the train's regular speed is 30 km/hr.

Exam Tip: When setting up speed differences, make sure you subtract the faster time from the slower time to keep the time difference value positive.

Question 11. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed.
Answer: Let the usual speed of the plane be \( x \) km/hr. The distance to the destination is 1500 km. The time taken at usual speed is \( \frac{1500}{x} \) hours. The increased speed is \( (x + 250) \) km/hr, and the time taken at this speed is \( \frac{1500}{x + 250} \) hours. The delay is 30 minutes, which is \( \frac{30}{60} = \frac{1}{2} \) hour. We can set up the equation: \[ \frac{1500}{x} - \frac{1500}{x + 250} = \frac{1}{2} \]
\( \implies 1500 \left[ \frac{(x + 250) - x}{x(x + 250)} \right] = \frac{1}{2} \)
\( \implies 1500 \left[ \frac{250}{x^2 + 250x} \right] = \frac{1}{2} \)
\( \implies \frac{375000}{x^2 + 250x} = \frac{1}{2} \) Cross-multiplying:
\( \implies x^2 + 250x = 750000 \)
\( \implies x^2 + 250x - 750000 = 0 \) To factorize, we look for two numbers that multiply to -750000 and add to 250. These are 1000 and -750:
\( \implies x^2 + 1000x - 750x - 750000 = 0 \)
\( \implies x(x + 1000) - 750(x + 1000) = 0 \)
\( \implies (x + 1000)(x - 750) = 0 \) This gives:
\( \implies x = -1000 \) or \( x = 750 \) Rejecting the negative speed, we have \( x = 750 \) km/hr. The usual speed of the plane is 750 km/hr.
In simple words: To make up for a 30-minute delay over 1500 km, the plane must speed up. Working through the quadratic equation tells us its normal speed is 750 km/hr.

Exam Tip: Be prepared to factor large quadratic equations. Finding factors of a large constant is easier if you temporarily ignore the trailing zeros and work with smaller values like 75 and 25.

Question 12. The area of a triangle is 30 sq. cm. Find the base if the altitude exceeds the base by 7 cm.
Answer: Let the base of the triangle be \( b \) cm. The altitude (height) exceeds the base by 7 cm, so the altitude is \( (b + 7) \) cm. The area of a triangle is given by the formula: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{altitude} \] Given that the area is 30 sq. cm: \[ \frac{1}{2} \times b \times (b + 7) = 30 \] Multiplying both sides by 2:
\( \implies b(b + 7) = 60 \)
\( \implies b^2 + 7b - 60 = 0 \) Factoring the quadratic equation:
\( \implies b^2 + 12b - 5b - 60 = 0 \)
\( \implies b(b + 12) - 5(b + 12) = 0 \)
\( \implies (b + 12)(b - 5) = 0 \) Setting the factors to zero:
\( \implies b = -12 \) or \( b = 5 \) Since the base of a triangle must be positive, we discard \( b = -12 \). The base of the triangle is 5 cm.
In simple words: A triangle with an area of 30 sq. cm and a height that is 7 cm longer than its base has a base of 5 cm.

Exam Tip: Always state the formula used for geometric shapes clearly before substituting the variables.

Question 13. The hypotenuse of a right angled triangle is 6 m more than twice the shortest side. If the third side is 2 m less than the hypotenuse Find sides of the triangle
Answer: Let the length of the shortest side be \( x \) meters. The hypotenuse is 6 m more than twice this shortest side: \[ \text{Hypotenuse} = 2x + 6 \] The third side is 2 m less than the hypotenuse: \[ \text{Third side} = (2x + 6) - 2 = 2x + 4 \] By Pythagoras' Theorem, the sum of the squares of the two shorter sides equals the square of the hypotenuse: \[ (\text{Shortest side})^2 + (\text{Third side})^2 = (\text{Hypotenuse})^2 \] \[ x^2 + (2x + 4)^2 = (2x + 6)^2 \] Expanding both sides:
\( \implies x^2 + (4x^2 + 16x + 16) = 4x^2 + 24x + 36 \)
\( \implies 5x^2 + 16x + 16 = 4x^2 + 24x + 36 \) Bringing all terms to one side:
\( \implies x^2 - 8x - 20 = 0 \) Factoring the quadratic equation:
\( \implies x^2 - 10x + 2x - 20 = 0 \)
\( \implies x(x - 10) + 2(x - 10) = 0 \)
\( \implies (x - 10)(x + 2) = 0 \) Setting the factors to zero:
\( \implies x = 10 \) or \( x = -2 \) Rejecting the negative length \( x = -2 \), we get \( x = 10 \) m. Now, calculating the other sides: Shortest side = 10 m Third side = \( 2(10) + 4 = 24 \) m Hypotenuse = \( 2(10) + 6 = 26 \) m The sides of the triangle are 10 m, 24 m, and 26 m.
In simple words: Using Pythagoras' Theorem on the described relationships between the sides, we find that the three sides of this right triangle measure 10 m, 24 m, and 26 m.

Exam Tip: Be careful to expand terms like \( (2x+4)^2 \) fully, using the identity \( (a+b)^2 = a^2 + 2ab + b^2 \), as missing the middle term is a very common mistake.

Question 14. The hypotenuse of a right angled triangle is 1 m less than twice the shortest side. If the third side is 1 m more than the shortest side. Find the sides of the triangle.
Answer: Let the shortest side of the triangle be \( x \) meters. The hypotenuse is 1 m less than twice the shortest side: \[ \text{Hypotenuse} = 2x - 1 \] The third side is 1 m more than the shortest side: \[ \text{Third side} = x + 1 \] Applying Pythagoras' Theorem: \[ (\text{Shortest side})^2 + (\text{Third side})^2 = (\text{Hypotenuse})^2 \] \[ x^2 + (x + 1)^2 = (2x - 1)^2 \] Expanding the terms:
\( \implies x^2 + (x^2 + 2x + 1) = 4x^2 - 4x + 1 \)
\( \implies 2x^2 + 2x + 1 = 4x^2 - 4x + 1 \) Subtracting \( 2x^2 + 2x + 1 \) from both sides:
\( \implies 2x^2 - 6x = 0 \)
\( \implies 2x(x - 3) = 0 \) This gives:
\( \implies x = 0 \) or \( x = 3 \) Since a side length cannot be 0, we take \( x = 3 \). Now, calculating the three sides: Shortest side = 3 m Third side = \( 3 + 1 = 4 \) m Hypotenuse = \( 2(3) - 1 = 5 \) m The sides of the triangle are 3 m, 4 m, and 5 m.
In simple words: Setting up Pythagoras' equation gives us a shortest side of 3 m. The other sides are then 4 m and 5 m.

Exam Tip: When factorizing \( 2x^2 - 6x = 0 \), factoring out \( 2x \) is much quicker than using the quadratic formula.

Question 15. A year ago, the father was 8 times as old as his son. Now his age is the square of his son's age. Find their present ages.
Answer: Let the son's age one year ago be \( s \) years. The father's age one year ago was \( 8s \) years. Therefore, their present ages are: Son's present age = \( s + 1 \) years Father's present age = \( 8s + 1 \) years We are given that the father's present age is equal to the square of the son's present age: \[ 8s + 1 = (s + 1)^2 \] Expanding and simplifying:
\( \implies 8s + 1 = s^2 + 2s + 1 \)
\( \implies s^2 - 6s = 0 \)
\( \implies s(s - 6) = 0 \) This gives:
\( \implies s = 0 \) or \( s = 6 \) If \( s = 0 \), the son's age one year ago would be 0, which is not possible for this context. So, we take \( s = 6 \). Now we can find their present ages: Son's present age = \( 6 + 1 = 7 \) years Father's present age = \( 8(6) + 1 = 49 \) years The present ages of the son and the father are 7 years and 49 years, respectively.
In simple words: Expressing their current ages in terms of their ages last year shows us that the son is currently 7 years old and the father is 49.

Exam Tip: Clearly define variable expressions for both "one year ago" and "present age" in a table format to avoid mixing up the time frames.

Question 16. A man is 3 times as old as his son and 6 years ago, the product of their ages was 288. Find their present ages.
Answer: Let the present age of the son be \( x \) years. The man's present age is 3 times the son's age, so it is \( 3x \) years. Six years ago: Son's age was \( x - 6 \) years Man's age was \( 3x - 6 \) years The product of their ages six years ago was 288: \[ (x - 6)(3x - 6) = 288 \] Expanding the bracket:
\( \implies 3x^2 - 6x - 18x + 36 = 288 \)
\( \implies 3x^2 - 24x + 36 - 288 = 0 \)
\( \implies 3x^2 - 24x - 252 = 0 \) Dividing the equation by 3:
\( \implies x^2 - 8x - 84 = 0 \) Factoring the equation:
\( \implies x^2 - 14x + 6x - 84 = 0 \)
\( \implies x(x - 14) + 6(x - 14) = 0 \)
\( \implies (x - 14)(x + 6) = 0 \) This gives:
\( \implies x = 14 \) or \( x = -6 \) Since age cannot be negative, we reject \( x = -6 \). Therefore, \( x = 14 \). Son's present age = 14 years Man's present age = \( 3 \times 14 = 42 \) years The present ages of the son and the father are 14 years and 42 years, respectively.
In simple words: The son is currently 14 years old and his father is 3 times older at 42. Six years ago, they were 8 and 36, which multiplies to 288.

Exam Tip: Check your final values by plugging them back into the "6 years ago" condition to verify that your calculations are correct.

Question 17. A two digit number is such that the product of the digits is 18. When 63 subtracted from the number, the digits interchange their places. Find the number.
Answer: Let the tens digit of the number be \( t \), and the units digit be \( u \). The product of the digits is 18: \[ t \times u = 18 \implies u = \frac{18}{t} \] The original two-digit number can be represented as \( 10t + u \). When 63 is subtracted from this number, the digits reverse, giving the new number \( 10u + t \): \[ (10t + u) - 63 = 10u + t \]
\( \implies 9t - 9u = 63 \) Dividing by 9:
\( \implies t - u = 7 \) Substituting \( u = \frac{18}{t} \) into this equation:
\( \implies t - \frac{18}{t} = 7 \) Multiplying the entire equation by \( t \):
\( \implies t^2 - 18 = 7t \)
\( \implies t^2 - 7t - 18 = 0 \) Factoring the equation:
\( \implies t^2 - 9t + 2t - 18 = 0 \)
\( \implies t(t - 9) + 2(t - 9) = 0 \)
\( \implies (t - 9)(t + 2) = 0 \) Since \( t \) represents a digit of a number, it must be a positive integer. Therefore, we take \( t = 9 \). Now, solving for \( u \): \[ u = \frac{18}{9} = 2 \] The original number is \( 10(9) + 2 = 92 \).
In simple words: We are looking for a two-digit number whose digits multiply to 18. Subtracting 63 swaps the digits around. That number is 92.

Exam Tip: Remember that a two-digit number is written algebraically as \( 10t + u \), where \( t \) is the tens digit and \( u \) is the units digit.

Question 18. A two digit number is such that the product of the digits is 14. When 45 is added to the number, the digits interchange their places. Find the number.
Answer: Let the tens digit of the number be \( t \) and the units digit be \( u \). The product of the digits is 14: \[ t \times u = 14 \implies u = \frac{14}{t} \] The original number is \( 10t + u \). When 45 is added, the digits are reversed, making the new number \( 10u + t \): \[ (10t + u) + 45 = 10u + t \]
\( \implies 9u - 9t = 45 \) Dividing by 9:
\( \implies u - t = 5 \) Substituting \( u = \frac{14}{t} \):
\( \implies \frac{14}{t} - t = 5 \) Multiplying the entire equation by \( t \):
\( \implies 14 - t^2 = 5t \)
\( \implies t^2 + 5t - 14 = 0 \) Factoring the quadratic equation:
\( \implies t^2 + 7t - 2t - 14 = 0 \)
\( \implies t(t + 7) - 2(t + 7) = 0 \)
\( \implies (t + 7)(t - 2) = 0 \) Since \( t \) is a positive digit, we discard \( t = -7 \) and take \( t = 2 \). Solving for the units digit \( u \): \[ u = \frac{14}{2} = 7 \] The original number is \( 10(2) + 7 = 27 \).
In simple words: The digits of the number multiply to 14, and adding 45 swaps their places. The number is 27.

Exam Tip: Always make sure your final digits are positive single digits from 1 to 9 before constructing the final answer.

Question 19. One fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and 15 camels were on the bank of river. Find the number of camels.
Answer: Let the total number of camels in the herd be \( x \). The number of camels in the forest is \( \frac{x}{4} \). The number of camels that went to the mountains is \( 2\sqrt{x} \). The number of camels on the river bank is 15. The sum of all these groups must equal the total number of camels \( x \): \[ \frac{x}{4} + 2\sqrt{x} + 15 = x \] Multiplying the entire equation by 4 to clear the fraction:
\( \implies x + 8\sqrt{x} + 60 = 4x \)
\( \implies 3x - 8\sqrt{x} - 60 = 0 \) Let \( \sqrt{x} = y \) (where \( y > 0 \)), so \( x = y^2 \). Substituting this in the equation:
\( \implies 3y^2 - 8y - 60 = 0 \) Splitting the middle term using factors of -180 that sum to -8, which are -18 and 10:
\( \implies 3y^2 - 18y + 10y - 60 = 0 \)
\( \implies 3y(y - 6) + 10(y - 6) = 0 \)
\( \implies (3y + 10)(y - 6) = 0 \) This gives:
\( \implies y = -\frac{10}{3} \) or \( y = 6 \) Since \( y = \sqrt{x} \) must be positive, we reject \( y = -\frac{10}{3} \) and take \( y = 6 \). Since \( y = 6 \): \[ \sqrt{x} = 6 \implies x = 36 \] The total number of camels in the herd is 36.
In simple words: Substituting a new variable for the square root turns the equation into a simple quadratic. Solving it shows there are 36 camels in the herd.

Exam Tip: Substituting \( y = \sqrt{x} \) simplifies equations with fractional powers, making them standard quadratics that are easier to solve.

Question 20. Out of a group of swans, \( \frac{7}{2} \) times the square root of the total number are playing on the shore of a pond. The two remaining ones are swimming in water. Find the total number of swans.
Answer: Let the total number of swans be \( x \). The number of swans playing on the shore is \( \frac{7}{2}\sqrt{x} \). The number of swans swimming in the water is 2. The sum of both groups equals the total number of swans \( x \): \[ \frac{7}{2}\sqrt{x} + 2 = x \] Let \( \sqrt{x} = y \) (where \( y > 0 \)), which means \( x = y^2 \). Substituting this into the equation: \[ \frac{7}{2}y + 2 = y^2 \] Multiplying the entire equation by 2:
\( \implies 7y + 4 = 2y^2 \)
\( \implies 2y^2 - 7y - 4 = 0 \) Factoring the quadratic equation:
\( \implies 2y^2 - 8y + y - 4 = 0 \)
\( \implies 2y(y - 4) + 1(y - 4) = 0 \)
\( \implies (2y + 1)(y - 4) = 0 \) This yields:
\( \implies y = -\frac{1}{2} \) or \( y = 4 \) Since \( y = \sqrt{x} \) must be positive, we reject the negative root and take \( y = 4 \). Now, solving for \( x \): \[ x = y^2 = 4^2 = 16 \] The total number of swans is 16.
In simple words: By writing the total swan count as the sum of the shore-playing swans and the water-swimming ones, we calculate that there are 16 swans in total.

Exam Tip: For problems involving animal counts, the final answer must be a positive integer and a perfect square, as \( \sqrt{x} \) represents a whole number of animals.

Question 21. Some students planned a picnic. The budget for food was Rs 500. But 5 of them failed to go and thus the cost of each member increased by Rs 5. How many students attended the picnic?
Answer: Let the number of students who originally planned the picnic be \( x \). The total budget for food is Rs. 500. The initial cost per student is \( \frac{500}{x} \) rupees. Since 5 students did not go, the number of students who actually went is \( (x - 5) \). The new cost per student is \( \frac{500}{x - 5} \) rupees. The price per student increased by Rs. 5, so: \[ \frac{500}{x - 5} - \frac{500}{x} = 5 \] Dividing by 5 to simplify:
\( \implies \frac{100}{x - 5} - \frac{100}{x} = 1 \) Taking the common denominator:
\( \implies 100 \left[ \frac{x - (x - 5)}{x(x - 5)} \right] = 1 \)
\( \implies 100 \left[ \frac{5}{x^2 - 5x} \right] = 1 \)
\( \implies \frac{500}{x^2 - 5x} = 1 \)
\( \implies x^2 - 5x - 500 = 0 \) Factoring the equation:
\( \implies x^2 - 25x + 20x - 500 = 0 \)
\( \implies x(x - 25) + 20(x - 25) = 0 \)
\( \implies (x - 25)(x + 20) = 0 \) This gives:
\( \implies x = 25 \) or \( x = -20 \) Since the number of students must be positive, we reject \( x = -20 \) and take \( x = 25 \). The number of students who actually attended the picnic is \( x - 5 = 25 - 5 = 20 \).
In simple words: Initially, 25 students planned to go. Since 5 stayed home, only 20 actually went to the picnic, which increased the cost for everyone.

Exam Tip: Read the final question carefully. It asks for the number of students who *attended* the picnic (\( x - 5 \)), not the number of students who *planned* it (\( x \)).

Question 22. A piece of cloth costs Rs 200. If the piece were 5 metres longer and the cost of each meter of cloth was Rs 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is its original price per metre?
Answer: Let the original length of the cloth be \( L \) meters. The total cost of the cloth is Rs. 200. The original price per meter is Rs. \( \frac{200}{L} \). If the cloth was 5 meters longer, the new length would be \( (L + 5) \) meters. If the price per meter was Rs. 2 less, the new price per meter would be Rs. \( \left(\frac{200}{L} - 2\right) \). Since the total cost remains Rs. 200: \[ (L + 5) \left( \frac{200}{L} - 2 \right) = 200 \] Expanding the bracket:
\( \implies 200 - 2L + \frac{1000}{L} - 10 = 200 \)
\( \implies -2L + \frac{1000}{L} - 10 = 0 \) Multiplying the entire equation by \( -L \) and dividing by 2 to simplify:
\( \implies L^2 + 5L - 500 = 0 \) Factoring the equation:
\( \implies L^2 + 25L - 20L - 500 = 0 \)
\( \implies L(L + 25) - 20(L + 25) = 0 \)
\( \implies (L + 25)(L - 20) = 0 \) This yields:
\( \implies L = -25 \) or \( L = 20 \) Since length must be positive, we take \( L = 20 \) meters. Now, the original price per meter is: \[ \text{Price per meter} = \frac{200}{20} = \text{Rs. } 10 \] The length of the piece of cloth is 20 meters, and its original price per meter is Rs. 10.
In simple words: Setting up an equation for the cost of the cloth shows us that the original piece was 20 meters long and cost Rs. 10 per meter.

Exam Tip: Ensure you calculate both requested values (length and price per meter) to get full marks for multi-part questions.

Question 23. Rs 250 are divided equally among a certain number of children. If there are 25 more children, each would have received 50 paisa less. Find the number of children.
Answer: Let the original number of children be \( n \). The total amount to be distributed is Rs. 250. The share of each child initially is Rs. \( \frac{250}{n} \). If there are 25 more children, the new number of children is \( (n + 25) \). The new share of each child is Rs. \( \frac{250}{n + 25} \). The difference between the two shares is 50 paise, which is Rs. \( \frac{50}{100} = \text{Rs. } 0.5 = \text{Rs. } \frac{1}{2} \). We can set up the equation: \[ \frac{250}{n} - \frac{250}{n + 25} = \frac{1}{2} \]
\( \implies 250 \left[ \frac{(n + 25) - n}{n(n + 25)} \right] = \frac{1}{2} \)
\( \implies 250 \left[ \frac{25}{n^2 + 25n} \right] = \frac{1}{2} \)
\( \implies \frac{6250}{n^2 + 25n} = \frac{1}{2} \) Cross-multiplying:
\( \implies n^2 + 25n = 12500 \)
\( \implies n^2 + 25n - 12500 = 0 \) Factoring the quadratic equation:
\( \implies n^2 + 125n - 100n - 12500 = 0 \)
\( \implies n(n + 125) - 100(n + 125) = 0 \)
\( \implies (n + 125)(n - 100) = 0 \) This yields:
\( \implies n = -125 \) or \( n = 100 \) Since the number of children must be a positive integer, we reject the negative root. The number of children is 100.
In simple words: Splitting Rs. 250 among 100 children gives each of them Rs. 2.50. Adding 25 more children lowers everyone's share by 50 paise.

Exam Tip: Always convert all monetary values into a single unit (either all Rupees or all Paise) before writing down your equation.

Question 24. A person on a tour has Rs 360 for his daily expenses. If he exceeds his tour by 4 days, he must cut down his daily expenses by Rs 3 per day. Find the original duration of the tour.
Answer: Let the original duration of the tour be \( d \) days. The total budget is Rs. 360. The original daily expense is Rs. \( \frac{360}{d} \). If the tour is extended by 4 days, the new duration is \( (d + 4) \) days. The new daily expense is Rs. \( \frac{360}{d + 4} \). The difference between the two daily expenses is Rs. 3: \[ \frac{360}{d} - \frac{360}{d + 4} = 3 \] Dividing by 3 to simplify:
\( \implies \frac{120}{d} - \frac{120}{d + 4} = 1 \) Taking the common denominator:
\( \implies 120 \left[ \frac{(d + 4) - d}{d(d + 4)} \right] = 1 \)
\( \implies 120 \left[ \frac{4}{d^2 + 4d} \right] = 1 \)
\( \implies \frac{480}{d^2 + 4d} = 1 \) Cross-multiplying:
\( \implies d^2 + 4d = 480 \)
\( \implies d^2 + 4d - 480 = 0 \) Factoring the equation:
\( \implies d^2 + 24d - 20d - 480 = 0 \)
\( \implies d(d + 24) - 20(d + 24) = 0 \)
\( \implies (d + 24)(d - 20) = 0 \) This yields:
\( \implies d = -24 \) or \( d = 20 \) Since the duration of the tour must be positive, we reject \( d = -24 \). The original duration of the tour is 20 days.
In simple words: The traveler originally planned a 20-day trip. Extending it by 4 days means spending Rs. 3 less each day to stay on budget.

Exam Tip: Be sure to write units like "days" or "km/hr" in your final answer to avoid minor marks deduction.

Question 25. The side of a square exceeds the side of another square by 4 cm and the sum of the areas of two squares is 400 square cm. Find the dimensions of the square.
Answer: Let the side of the smaller square be \( x \) cm. Since the side of the larger square exceeds the side of the smaller square by 4 cm, its side is \( (x + 4) \) cm. The area of a square is \( (\text{side})^2 \). The sum of their areas is 400 square cm: \[ x^2 + (x + 4)^2 = 400 \] Expanding and simplifying:
\( \implies x^2 + (x^2 + 8x + 16) = 400 \)
\( \implies 2x^2 + 8x + 16 - 400 = 0 \)
\( \implies 2x^2 + 8x - 384 = 0 \) Dividing the entire equation by 2:
\( \implies x^2 + 4x - 192 = 0 \) Factoring the quadratic equation:
\( \implies x^2 + 16x - 12x - 192 = 0 \)
\( \implies x(x + 16) - 12(x + 16) = 0 \)
\( \implies (x + 16)(x - 12) = 0 \) This gives:
\( \implies x = -16 \) or \( x = 12 \) Since a side length must be positive, we reject \( x = -16 \). Therefore, \( x = 12 \) cm. Side of the smaller square = 12 cm Side of the larger square = \( 12 + 4 = 16 \) cm The dimensions of the squares are 12 cm and 16 cm.
In simple words: The sides of the two squares are 12 cm and 16 cm. Their squared values (144 and 256) add up perfectly to 400.

Exam Tip: Always clearly state which variable corresponds to which geometric part (e.g., smaller vs. larger square) to prevent calculation mix-ups.