CBSE Class 10 Mathematics Quadratic Equation Worksheet Set 15

Question. If α, β, be the roots of the equation ax² + bx + c = 0, then ax² + bx + c =.   
(A) (x – α)(x – β)
(B) a(x – α)(x – β)
(C) a (x – β) (x + α)
(D) a(x + α)(x + β)

Answer: C

Question.The number of real roots of the equation (x – 1)² + (x – 2)² + (x – 3)² = 0 is:   
(A) 2
(B) 1
(C) 0
(D) 3

Answer: C

Question.Determine k such that the quadratic equation x² + 7(3 +2k) – 2x(1 + 3k) = 0 has equal roots: 
(A) 2, 7
(B) 7, 5
(C) 2, -10/9
(D) None of these

Answer: C

Question. The condition that the equation ax² + bx + c = 0 has one positive and other negative root is:   
(A) a and c will have same sign
(B) a and c will have opposite signs
(C) b and c will have same sign
(D) b and c will have opposite signs

Answer: B

Question. The condition that both roots of the equation ax² + bx + c = 0 are negative is:   
(A) a, b, c are of the same sign
(B) a and b are of opposite signs
(C) b and c are of opposite signs
(D) the absolute term is zero

Answer: A

Question. If one root of 5x² + 13x + k = 0 is reciprocal of the other, then the value of k is. 
(A) 5
(B) 7
(C) 4
(D) None of these

Answer: A

Question. If the equation x²-bx / ax-c = m-1/m+1 has roots equal in magnitude but opposite in sign, then m is equal to.   
(A) a+b /a -b
(B)  a-b/a+b
(C) Both (A) & (B)
(D) None of these

Answer: B

Question. The set of values of p for which the roots of the equation 3x² + 2x + (p – 1)p = 0 are of opposite sign, is: 
(A) (0, 1)
(B) (-1,1)
(C) (-2,2)
(D) None of these

Answer: A

Question. Find the roots of the equation f(x) = (b – c) x² + (c – a) x + (a – b) = 0 
(A)  a-b /b-c and 1
(B) a+b /b+c and 1
(C) a-b /b+c and 1
(D) None of these

Answer: A

Question. The real values of a for which the quadratic equation 2x² – (a³ + 8a – 1)x + a² – 4a = 0 possesses roots of opposite signs are given by: 
(A) a > 6
(B) a > 9
(C) 0 < a < 4
(D) a < 0

Answer: C

Question. Given that (x + 1) is a factor of x² + ax + b and x² + cx – d, then   
(A) a + d = b + c
(B) a = b + c + d
(C) a + c = b - d
(D) None of these

Answer: B

Question. The zeroes of x² – bx + c are each decreased by 2 The resulting polynomial is x² – 2x + 1 Then   
(A) b = 6, c = 9
(B) b = 6, c = 3
(C) b = 3, c = 6
(D) b = 9, c = 6

Answer: A

Question. If the zeroes of x² + Px + t are two consecutive even numbers find the relation between P and t 
(A) P² – 4t + 4 = 0
(B) 4t – P² + 4 = 0
(C) - 4t² – 4 – P² = 0
(D) None of these

Answer: B

Question. One zero of x² – bx + C is the kth power of the other zero, then C ^1/k+1+C^k/k+1 is equal to 
(A) - b
(B) C
(C) - C
(D) b

Answer: D

Question. α, β, γ, δ are zeroes of x⁴ + 5x³ + 5x² + 5x – 6, then find the value of 1/α + 1/β + 1/γ +1/δ   
(A) 5/6
(B) -6/5
(C) -5/6
(D) None of these

Answer: A

Question. The number of real solution of the equation 2^{3x2-7x + 4} = 1 is: 
(A) 0
(B) 4
(C) 2
(D) Infinitely many

Answer: C

Question. The maximum value of – 3x² + 4x – 5 is at x =   
(A) 2/3
(B) 1/3
(C) -33/9
(D) None of these

Answer: A

Question. Given that 7 – 3i is a zero of x² + px + q, find the value of 3q + 4p 
(A) 14
(B) 58
(C) 118
(D) - 14

Answer: C

Question. Given that α is a zero of x⁴ + x² – 1, find the value of (α⁶ + 2α⁴)1000.   
(A) 1
(B) 0
(C) Either 0 or 1
(D) None of these

Answer: A

VERY SHORT ANSWER TYPE QUESTIONS

Question. If one root of the equation 3x² + px + 4 = 0 is 2/3 , find the value of p.

Answer: p = -8

Question. Determine whether x = - 1/3 and x = 2/3 are the roots of 9x² – 3x – 2 = 0.

Answer: yes

Question. If sum of a whole number and its reciprocal is 10/3 , what is the number?

Answer: 3

Question. For what value of k, x = a is a solution of the equation x² - (a + b)x + k = 0 .

Answer: k = ab

Question. Find the ratio of sum and product of the roots of the equation 3x² – 8x + 12 = 0.

Answer: 2 : 3

Question. Form a quadratic equation whose roots are –2 and 3.

Answer: x² - x - 6 = 0

Question. If one root of a quadratic equation is 3 - √2/4 , then what is the other root?

Answer: 3 + √2/4

Question. Form a quadratic equation whose sum of roots is –3 and product of roots is 5.

Answer: x² + 3x + 5 = 0

Question Write the quadratic equation for the following : The sum of two numbers is 27 and their product is 182.

Answer: x² - 27x +182 = 0

Question. The sum of squares of two consecutive positive integers is 365. Express this in the form of an quadratic equation.

Answer: x² + x – 182 = 0

Question. Represent the following in the form of a quadratic equation : The sum of the ages of father and his son is 45 years. Five years ago, the product of their ages (in years) was 124.

Answer: x² - 45x + 324 = 0

Question. Write the discriminant of the equation 6a²x² – 7abx – 3b² = 0.

Answer: 121a²b²

Question. Write the nature of the roots of the equation 3x² - 4 √3x - 1 = 0.

Answer: Real and distinct roots

Question. Check whether x (x + 3) + 8x = (x + 5) (x – 5) is a quadratic equation.

Answer: No

Question. For what value of k, the equation 3x² + kx + 4 = 0 have equal roots?

Answer: k = ±4√3

Question. If x = 3/2 is the root of the quadratic equation 2x² – kx + 3 = 0, then find the value of k.

Answer: k = 5

Question. For what value of k, the equation kx² + 6x + 1 = 0 have real roots?

Answer: k ≤ 9

Question. If α and β are the roots of the equation 3x² – 9x + 7 = 0, then find α + β + αβ.

Answer: 16/3

Question. What are the roots of the equation 2x² + 3x – 2 = 0?

Answer: -2 , 1/2

Question. Solve the quadratic equation : (x + 3)² – 16 = 0.

Answer: 1, –7

Short Answer type Question :

Question. Find the value of k for which the roots of the equation 3x² – 10x + k = 0 are reciprocal of each other. 
Solution. Let the roots of the given equation be a and 1/α 
α , 1/α = c/a = k/3 ⇒ k = 3

Question. Find the value of k so that the quadratic equation kx(3x – 10) + 25 = 0, has two equal roots.
Solution. kx(3x – 10) + 25 = 0
⇒ 3kx² – 10kx + 25 = 0
D = (–10k)² – 4 × 3k × 25
= 100k² – 300k
For equal roots, D = 0 (1)
⇒ 100k² – 300k = 0
⇒ 100k(k – 3) = 0
⇒ k = 0 or k = 3
But k π 0,
So, k = 0 (Rejected)
[∴ In quadratic equation, a ≠ 0]
Hence, k = 3

Question. For what values of k, the equation 9x² + 6kx + 4 = 0 has equal roots? 
Solution. 9x² + 6kx + 4 = 0
(6k)² – 4 × 9 × 4 = 0 (½)
36k² = 144
⇒ k2 = 4
k = ±2

Question. Find the value(s) of k so that the quadratic equation x² – 4kx + k = 0 has equal roots.
Solution. x² – 4kx + k = 0
Since given equation has equal roots,
∴ D = 0 (1)
16k² – 4k = 0
⇒ 4k(4k – 1) = 0
⇒ k = 0 and k = 1/4

Question. For what value(s) of ‘a’ quadratic equation 3ax² – 6x + 1 = 0 has no real roots?
Solution. 3ax² – 6x + 1 = 0 (½)
(–6)² – 4(3a)(1) < 0
12a > 36
⇒ a > 3

Question. State whether the quadratic equation 4x² – 5x + 25/16 = 0 has two distinct real roots or not. Justify your answer.
Solution. No, D = 0

Question. Find the value(s) of k so that the quadratic equation 3x² – 2kx + 12 = 0 has equal roots. 
Solution. 3x² – 2kx + 12 = 0
Since given equation has equal roots, so
D = 0 (1)
⇒ 4k² – 144 = 0
⇒ 4(k² – 36) = 0
⇒ k = ±6
⇒ k = 6 and k = –6

Question. Find the value of p, so that the quadratic equation px(x – 3) + 9 = 0 has equal roots. 
Solution. px(x – 3) + 9 = 0
⇒ px² – 3px + 9 = 0
When roots are equal,
D = b² – 4ac = 0
9p² – 36p = 0
⇒ 9p(p – 4) = 0
⇒ p = 0, p = 4
But p π 0
[∴ In quadratic equation, a ≠ 0]
∴ p = 4

Question. Find the values of k for which the quadratic equation 9x² – 3kx + k = 0 has equal roots.
Solution. For equal roots, D = 0
⇒ 9k² – 36k = 0
⇒ 9k(k – 4) = 0 
⇒ k = 0 or k = 4

Question. Find the discriminant of the quadratic equation 2x² – 4x + 3 = 0, hence find the nature of its roots. 
Solution. –8, no real roots

Question. If x = 3 is one root of the quadratic equation x² – 2kx– 6 = 0, then find the value of k. 
Solution. x = 3 is one root of the equation
∴ 9 – 6 k – 6 = 0
⇒ k = 1/2

Question. Find the values of k for which the quadratic equation (3k + 1)x² + 2(k + 1)x + 1 = 0 has equal roots. Also find these roots.
Solution.  For equal roots, D = 0
{2(k + 1)}2 – 4(3k + 1) · 1 = 0 (1)
⇒ 4(k² + 2k + 1) – 12k – 4 = 0
⇒ 4k² + 8k + 4 – 12k – 4 = 0 (1)
⇒ 4k² – 4k = 0
⇒ 4k(k – 1) = 0
⇒ k = 0, 1

Question. For what value of k does the quadratic equation (k – 5)x² + 2(k – 5)x + 2 = 0 have equal roots? 
Solution. k = 7

Question. Find the value(s) of k so that the quadratic equation 2x² + kx + 3 = 0 has equal roots. 
Solution. 2x² + kx + 3 = 0
For equal roots, D = 0 
⇒ b² – 4ac = 0
⇒ k² – 24 = 0
⇒ k = ± 2√6 

Question. If 2 is a root of the quadratic equation 3x² + px – 8 = 0 and the quadratic equation 4x² – 2px + k = 0 has equal roots, find the value of k. 
Solution. 3(2)² + p(2) – 8 = 0
⇒ 12 + 2p – 8 = 0
⇒ p = – 2 ...(i)(1)
So, equation becomes
4x² + 4x + k = 0
For equal roots, D = 0
⇒ (4)² – 4 × 4 × k = 0
⇒ 16 = 16k
⇒ k = 1

Question. Find the value of p for which the quadratic equation (p + 1)x² – 6(p + 1)x + 3(p + 9) = 0, p π –1 has equal roots. Hence, find the roots of the equation. 
Solution. 3, 3.

Question. For what values of k, the roots of the equation x² + 4x + k = 0 are real? 
Solution. For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0
⇒ (4)² – 4 × 1 × k ≥ 0
⇒ 16 – 4k ≥ 0
⇒ 16 ≥ 4k, k ≤ 4

Question. Find that non-zero value of k, for which the quadratic equation kx² + 1 – 2(k – 1)x + x² = 0 has equal roots. Hence, find the roots of the equation.
Solution. kx² + 1 – 2(k – 1)x + x² = 0
⇒ (k + 1)x² – 2(k – 1)x + 1 = 0
Q Above equation has equal roots, 
So, discriminant, D = 0
⇒ {–2(k – 1)}2 – 4 × (k + 1) × 1 = 0 
⇒ 4(k² – 2k + 1) – 4(k + 1) = 0
⇒ 4k² – 12k = 0
⇒ 4k(k – 3) = 0
⇒ k = 3 (as k ≠ 0) 

Question. Find whether the equation 1/2x - 3 + 1/x - 5 = 1 , x ≠ 3/2 , 5 has real roots. If real roots exist, find them.
Solution. Yes, 8 ±3√2/2

Question. The roots a and b of the quadratic equation x² – 5x + 3(k – 1) = 0 are such that a – b = 1. Find the value k.
Solution. k = 3

Question. Find the values of k for which the given equation has real and equal roots: (k + 1)x² -2(k - 1)x + 1 = 0 
Solution. We have, (k+1)x² - 2(k - 1)x+1 = 0.
a = k + 1, b = -2(k - 1), c = 1.
D = b² - 4ac =4(k-1)² - 4(k + 1) =4(k² -3k)
The given equation will have real and equal roots, if
D = 0 ⇒ 4 (k² - 3k) = 0 ⇒ k² - 3k = 0 ⇒ k (k - 3) = 0 ⇒ k = 0, 3

Question. Solve the quadratic equations by factorization method: x² - 9 = 0
Solution. We have,
x² - 9 = 0
⇒ (x - 3)(x + 3) = 0
⇒ x - 3 = 0 or, x + 3 = 0
x = 3 or, x = -3 ⇒ x = ± 3
Thus, x = 3 and x = - 3 are roots of the given equation.

Question. If p, q, r and s are real numbers such that pr = 2(q + s), then show that at least one of the equations x² + px + q = 0 and x² + rx + s = 0 has real roots. 
Solution. Given quadratic equations are;
x² + px + q = 0 —(i)
and, x² + rx + s = 0 ......(ii)
Also given ; pr = 2(q + s)........(iii)
Let D₁ and D₂ be the discriminant of quadratic equations (i) and (ii) respectively.
Then,
D₁ = p² - 4q and D² = r2 - 4s
⇒ D₁+ D₂ = p² - 4q + r² - 4s = (p² + r²) - 4(q + s)
Now, Since sum of both D₂ & D₁ is greater than or equal to 0. Hence, both can't be
negative.
⇒ At least one of D₁and D₂ is greater than or equal to zero
Case 1. If D₁ ≥ 0, equation (i) has real roots.
Case 2.If D₂ ≥ 0, equation (ii) has real roots.
Case 3. If D₁ & D₂ both ≥ 0, then equation (i) & (ii) both have equal roots.
Clearly, from case 1,2 & 3 at least one given quadratic equations has equal roots.

Question. Check whether the given equation is quadratic equation: (x-3) (2x + 1) = x(x + 5) 
Solution. The given equation is (x - 3) (2x +1) = x (x+5)
⇒ 2x² + x - 6x - 3 = x² + 5x
⇒ 2x² - 5x - 3 = x² + 5x
⇒ x² - 10x - 3 = 
It is in the form of ax² + bx + c = 0, a≠0
∴ the given equation is a quadratic equation.

Question. Form a quadratic equation whose roots are -3 and 4. 
Solution. We have, x = 4 and x = -3.
Then,
 x - 4 = 0 and x + 3 = 0
⇒ (x - 4)(x + 3) = 0
⇒ x² + 3x - 4x - 12 = 0
⇒ x² - x - 12 = 0
This is the required quadratic equation

Question. Check whether the given equation is quadratic equation: (x +1)²= 2 (x –3) 
Solution. The given equation is (x+1)² = 2 (x-3)
⇒ x² + 2x + 1 - 2x + 6 = 0
⇒ x² + 7 = 0
⇒ x² + 0.x + 7 = 0
Which is of the form ax² + bx + c = 0
Hence, the given equation is a quadratic equation.

Question. Find discriminant of the quadratic equation: 5x² + 5x + 6 = 0. 
Solution. Given equation is 5x² + 5x + 6 = 0
Here a = 5, b = 5, c = 6
D = b² - 4ac = (5)² -4 x 5 x 6 = -95

Question. Two numbers differ by 3 and their product is 504. Find the numbers. 
Solution. Sol : Let the required number be x and x + 3.
Then, according to given question we have,
x (x + 3) = 504
⇒ x² + 3x = 504
⇒ x² + 3x - 504 = 0
⇒ x² + 24x - 21x - 504 = 0
⇒ x(x + 24) - 21(x + 24) = 0
⇒ (x + 24)(x - 21) = 0
⇒ x + 24 = 0 or x - 21 = 0
⇒ x = - 24 or x = 21
Case I: When x = -24
∴ x + 3 = -24 + 3 = -21
Case II: When x = 21
∴ x + 3 = 21 + 3 = 24
Hence, the numbers are -21, -24 or 21, 24.

Question. 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. 
Solution. Let the smaller number be x and the larger number be y.
Also, Square of the larger number( y² ) =18x
According to question,
x² + y² = 208
⇒ x² + 18x = 208
⇒ x² + 18x - 208 = 0
⇒ x² + 26x - 8x - 208 = 0
⇒ (x + 26) (x - 8) = 0 ⇒ x = 8, x = -26
But, the numbers are positive. Therefore, x =8
Square of the larger number = 18x = 18 x 8 = 144
Therefore, larger number =√144 = 12
Hence, the numbers are 8 and 12.

Question. The product of Tanvy's age (in years) 5 years ago and her age 8 years later is 30. Find her present age. 
Solution. Let the present age of tanvy be x years
Tanvy's age five years ago = (x - 5)
Tanvy's age eight years later = (x + 8)
According to question,
⇒ (x - 5)(x + 8) = 30
⇒ x² + 8x - 5x - 40 = 30
⇒ x² + 3x - 40 - 30 = 0
⇒ x² + 3x - 70 = 0
⇒ x² + 10x - 7x - 70 = 0
⇒ x(x + 10) - 7(x + 10) = 0
⇒ x + 10 = 0 or x - 7 = 0
⇒ x = -10 or x = 7
⇒ x = 7 ( age cannot be negative)
Therefore, the present age of tanvy is 7 years.

Question. Solve: 4x² - 12x + 9 = 0.
Solution. We have,
4x - 12x + 9 = 0
Here, 4 x 9 = 36 so to factor the middle term in given equation we have (-6) x (-6) = 36, and (-6) + (-6) = -12.
⇒ 4x² - 6x - 6x + 9 = 0 ⇒ 2x(2x - 3) - 3(2x - 3) = 0
⇒ (2x - 3)(2x - 3) = 0 ⇒ (2x - 3)2 = 0
2x - 3 = 0 ⇒ x = 3/2
Hence, x = 3/2 is the repeated root of the given equation.

Question. If –4 is a root of the quadratic equation x² + px – 4 = 0 and the quadratic equation x² + px + k = 0 has equal roots. Find the value of k. 
Solution. We have, x² + px - 4 = 0
-4 is the root of the given equation
Substitute x = - 4 in the given equation, we get
(-4)² + p (-4) -4 =0
⇒ 16 - 4p - 4 = 0
⇒ 4p = 12 or p = 3
The equation becomes x2 + 3x -4 = 0
Substituting the value of p = 3 in the equation x2 + px + k = 0, we get
x² + 3x + k = 0
Here a = 1, b = 3, c= k
∴ D = b² - 4ac = (3)2 - 4 (1) (k)
= 9 - 4k
For equal roots, D = 0
⇒ 9 - 4k = 0 or k = 9/4

Question. If x = - 2 is a root of the equation 3x² + 7x + p = 0, find the values of k so that the roots of the equation x² + k(4x + k - 1) + p = 0 are equal. 
Solution. Here, x = - 2 is a root of 3x² + 7x + p = 0
⇒ 3(-2)2 + 7 (-2) + p = 0
⇒ p = 2
∴ x² + k(4x + k - 1) + p = 0 becomes
x² + 4kx + k² - k + 2 = 0....(i)
Comparing eq. (i) with ax² + bx + c = 0 , we get
a = 1 , b = 4k and c = k² - k + 2
Roots of eq (i) are equal
So, D =b² - 4ac =0
⇒ (4k)² - 4 x 1 (k² - k + 2) = 0
⇒ 16k² - 4k² + 4k - 8 = 0
⇒ 12k² + 4k - 8 = 0
⇒ 12k² + 4k - 8 = 0
⇒ 3k² + k - 2 = 0
⇒ 3k² + 3k - 2k - 2 = 0
⇒ 3k(k + 1) - 2(k + 1) = 0
⇒ (k + 1)(3k - 2) = 0
⇒ k = -1, k = 2/3

Question. Find the discriminant of equation: 2x² - 7x + 6 = 0. 
Solution. Given, 2x² - 7x + 6 = 0
∴ a =2, b = -7 and c = 6
D = b² -4ac
= (-7)2 - 4(2)(6)
= 49 - 48
= 1

Question. Solve the following problem: x² − 45x + 324 = 0 
Solution. x² - 45x + 324 = 0
⇒ x² - 36x - 9x + 324 = 0 ⇒ x (x - 36) - 9(x - 36)=0
⇒ (x - 9)(x - 36) ⇒ x = 9, 36

Question. At t minutes past 2 p.m, the time needed by the minute hand of a clock to show 3 p.m. was found to be 3 minutes less than t²/4 minutes. Find t. 
Solution. Total time taken by minute hand from 2 p.m. to 3 p.m. is 60 min.
According to question,
t + (t²/4 - 3) = 60
⇒ 4t + t² – 12 = 240
⇒ t² + 4t – 252 = 0
⇒ t² + 18t – 14t – 252 = 0
⇒ t(t + 18) – 14(t +18) = 0
⇒ (t + 18) (t – 14) = 0
⇒ t + 18 = 0 or t – 14 = 0
⇒ t = –18 or t = 14 min.
As time can't be negative. Therefore, t = 14 min.

Question. A two-digit number is 4 times the sum of its digits and twice the product of the digits. Find the number.
Solution. Let the ten's place digit be y and unit's place be x.
Therefore, number is 10y + x.
According to given condition,
10y + x = 4(x + y) and 10y + x = 2xy
⇒ x = 2y and 10y + x = 2xy
Putting x = 2y in 10y + x = 2xy
10y + 2y = 2.2y.y
12y = 4y²
4y² - 12y = 0 ⇒ 4y(y - 3) = 0
⇒ y - 3 = 0 or y = 3
Hence, the ten's place digit is 3 and units digit is 6 (2y = x)
Hence the required number is 36.

Question. Check whether it is quadratic equation: (x + 1)³ = x³ + x + 6 
Solution. We have the following equation,
(x + 1)³ = x³ + x + 6
⇒ x³ + 1 + 3x(x + 1) = x³ + x + 6
⇒ 3x² + 2x - 5 = 0.
This is of the form ax² + bx + c = 0.
Hence, the given equation is a quadratic equation.

Question. Find the roots of the quadratic equation 2x² - x - 6 = 0 
Solution. Given, 2x² - x - 6 = 0
Splitting the middle term of the equation,
⇒ 2x² - 4x + 3x - 6 = 0
⇒ 2x(x - 2) + 3(x - 2) = 0
⇒ (x - 2)(2x + 3) = 0
⇒ x - 2 = 0 or 2x + 3 = 0
Therefore, x = 2 or x = -3/2

Question. A journey of 192 km from a town A to town B takes 2 hours more by an ordinary passenger train than a super fast train. If the speed of the faster train is 16 km/h more, find the speed of the faster and the passenger train. 
Solution.
or x(x + 48) - 32(x + 48) = 0
or, (x - 32) (x + 48) = 0
or, x = 32 or - 48
Since speed can't be negative, therefore - 48 is not possible.
∴ Speed of passenger train = 32 km/h and Speed of fast train = 48 km/h

Question. What is the nature of roots of the quadratic equation 5x² - 2x - 3 = 0? 
Solution. On comparing equation with standard form of equation i.e, ax² + bx + c = 0, we get
a = 5, b = -2, c = -3
Now , D = b² - 4ac = (-2)2 - 4 x 5 x (-3)
= 4 + 60 = 64
Therefore, D = 64
We know , For D>0 , the roots of equation are real and distinct.
Therefore , 5x² - 2x - 3 = 0 has real and distinct roots.

Question. A farmer wishes to grow a 100 m² rectangular vegetable garden. Since he has with him only 30 m barbed wire, he fences three sides of the rectangular garden letting compound wall of his house act as the fourth side-fence. Find the dimensions of his garden.
Solution. Let the length of one side be x metres and other side be y metres.
Then, x + y + x = 30 y = 30 ⇒ - 2x
Area of the garden = 100 m2
⇒ xy = 100
⇒ x (30-2x) = 100
⇒ 2x² - 30x +100 = 0
⇒ 2(x²-15x+50) = 0 or x²-15x+50 = 0
⇒ x²-10x - 5x +50 = 0
⇒ x (x-10) - 5 (x-10) = 0
⇒ (x-10) (x-5) = 0
Either x-10 = 0 or x-5 = 0
⇒ x= 10, 5
∴ y = 30 - 20 = 10 or 30 -10 =20
Hence, the dimensions of the vegetable garden are 5m 20m or 10m 10m.

Question. A faster train takes one hour less than a slow train for a journey of 200 km. If the speed of the slow train is 10 km/hr less than that of the fast train, find the speed of the two trains.
Solution. 40 km/hr, 50 km/hr

Question. X and Y are centres of circle of radius 9 cm and 2 cm and XY = 17 cm. Z is the centre of a circle of radius r cm, which touches the above circles externally. Given that ∠XZY = 90°, find the value of r.
Solution. r = 6 cm

Question. A two digit number is such that the product of its digit is 10. When 27 is subtracted from the number, the digits interchange their places. Find the number.
Solution. 52

Question. One-fourth of a herd of camels was seen in the forest. Twice the square root of the herd had gone to mountains and the remaining 15 camels were seen on the bank of a river. Find the total number of camels.
Solution. 36

Question. Find the values of k so that the equation (k + 2) x² – (7 – k) x + 3 = 0 has :
(a) equal roots (b) distinct roots (c) no real roots.
Solution. (a) k = 1, 25 (b) k > 25 or k < 1 (c) 1 < k < 2

Question. The sum of the ages of a man and his son is 40 years. The product of their ages is 144 years. Find their present ages.
Solution. 36 years and 4 years

Question. The sum of two numbers is 16. The sum of their reciprocals is 1/3 Find the numbers.
Solution. 4, 12

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. Its average speed for the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of the flight.
Solution. 1 hour

Question. The length of a rectangular ground is greater than twice its breadth by 5 m. The area of the ground is 1125 sq. m. Find the length and breadth of the ground.
Solution. length = 50 m, breadth = 45/2 m

Question. The 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 sq. cm., calculate the length of the sides of the triangle.
Solution. 15 cm, 8 cm, 17 cm

Question. The difference of the squares of two numbers is 45. The square of the smaller number is four times the larger number. Find the numbers.
Solution. 6 and 9

Question. A carpet is placed in a room 6m × 4m, leaving a border of a uniform width all around it. Find the width of the border, if the area of the carpet is 8 sq. m.
Solution. 1 m

Question. Two circles touch externally. The sum of their areas is 130 π sq.cm and the distance between their centres is 14 cm. Find the radii of the circles.
Solution. 11 cm and 3 cm

Question. The length of a rectangle exceeds its breadth by 5 m. If the breadth were doubled and the length reduced by 9 m, the area of the rectangle would have increased by 140 sq. m. Find its dimensions.
Solution. 25 m × 20 m

Question. If twice the area of a smaller square is subtracted from the area of a larger square, the result is 14 cm2. However, if twice th area of the larger square is added to three times the area of the smaller square, the result is 203 cm2. Find the sides of the squares.
Solution. 5 cm and 8 cm

Question. In an auditorium, seats are arranged in rows and columns. The number of rows was equal to the number of seats in each row. When the number of rows was doubled and the number of seats in each row is reduced by 10, the total number of seats increased by 300. Find :
(i) the number of rows in the original arrangement.
(ii) the number of seats in the auditorium after re-arrangement.
Solution. (i) 30 (ii) 1200

Question. The area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm. Find its base.
Solution. 10 cm or 24 cm

Question. The sum of the squares of two consecutive odd positive integers is 290. Find them.
Solution. 11, 13

Question. A shopkeeper buys a number of books for Rs. 80. If he had bought 4 more books for the same amount, each book would have cost Rs. 1 less. How many books did he buy?
Solution. 16

Question. A line AB is 8 cm in length. AB is produced to P such that BP² = AB. AP. Find the length of BP.
Solution. 4 (1 + √5) cm

Question. If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy. 
Solution. Rs. 20

Question. Seven years ago Varun’s age was five times the square of Swati’s age. Three years hence Swati’s age will be two-fifth of Varun’s age. Find their present ages.
Solution. Swati’s present age = 9 years, varun’s present age = 27 years

Question. A person on tour has Rs. 360 for his expenses. If he extends his tour for 4 days, he has to cut down his daily expenses by Rs. 3. Find the original duration of the tour.
Solution. 20 days

Question. 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 hour 30 minutes. Find the speed of the stream.
Solution. 5 km/hr

Question. Rs. 6500 were divided among a certain number of persons. Had there been 15 more persons, each would have got Rs. 30 less. Find the original number of the persons.
Solution. 50

Question. A piece of cloth costs Rs. 200. If the piece was 5 m longer and each metre of cloth costs Rs. 2 less, the cost of the piece would have remained unchanged. How long is the piece and what is the original rate per metre?
Solution. 20 m, Rs. 10

Question. The denominator of a fraction is 5 more than its numerator. The sum of the fraction and its reciprocal is 3 , 11/14 .Find the fraction.
Solution. 2/7

Question. A takes 15 days less than the time taken by B to finish piece of work. If both A and B together can finish the work in 18 days, find the time taken by B to finish the work.
Solution. 45 days

Question. If the sum of first n even natural numbers is 420, find the value of n.
Solution. n = 20

Question. A passenger train takes 2 hours less for a journey of 300 km if its speed is increased by 5 km/hr from its usual speed. Find the usual speed of the train.
Solution. 25 km/hr

Question. Two pipes running together can fill a cistern in 3 , 1/3 minutes. If one pipe takes 3 minutes more than the other to fill it, find the time in which each pipe would fill the cistern.
Solution. 5 minutes and 8 minutes

Question. The speed of a boat in still water is 8 km/hr. It can go 15 km upstream 22 km downstream in 5 hours. Find the speed of the stream.
Solution. 3 km/hr

Question. Divide 16 into two parts such that twice the square of the larger part exceeds the square of the smaller part by 164.
Solution. 10, 6

Question. In a class test, the sum of Nishu’s marks in Mathematics and Science is 30. Had she got 2 marks more in Mathematics and 3 marks less in Science, the product of their marks would have been 210. Find her marks in the two subjects.
Solution. Marks in mathematics = 13, Marks in science = 17 OR Marks in Mathematics = 12, Marks in science = 18

Question. Find two consecutive numbers whose squares have the sum 85.
Solution. 6, 7

Question. A polygon of n side has n(n - 3) diagonals. How many sides has a polygon with 54 diagonals?
Solution. 12

Question. Two numbers differ by 3 and their product is 504. Find the numbers.
Solution. 21, 24 or – 24, – 21 

Question. A farmer wishes to start a 300 sq. m rectangular vegetable garden. Since he has only 50 m barbed wire, he fences three sides of the rectangular garden, letting his house compound wall act as the fourth side fence. Find the dimensions of the garden.
Solution. 20 m × 15 m or 30 m × 10 m

Question. One year ago, a man was 8 times as old as his son. Now his age is equal to the square of his sons age. Find their present ages.
Solution. 7 years and 49 years

Question. The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.
Solution. 150 cm²

Question. Sum of the areas of two squares is 468 m2. If the difference of their perimeter is 24 m, find the sides of the two squares.
Solution. 18 m and 12 m

Question. O girl! Out of a group of swans, 7/2 times the square root of the number are playing on the shore of a tank. The two remaining ones are playing with amorous fight, in the water. What is the total number of swans?
Solution. 16

Question. The angry Arjun carried some arrows for fighting with Bheesm. With half the arrows, he cut down the arrows thrown by Bheesm on him and with six other arrows, he killed the rath driver of the Bheesm. With one arrow each, he knocked down respectively the rath, flag and the bow of Bheesm. Finally, with one more than four times the square root of the total number of arrows, he laid Bheesm unconscious on an arrow bed. Find the total number of arrows Arjun had.
Solution. 100

Question. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels 5 km/hr faster than the second train. If, after 2 hours, they are 50 km apart, find the speed of each trains.
Solution. 20 km/hr, 15 km/hr

Question. In an auditorium, the number of rows was equal to number of seats in each row. When the number of rows was doubled and the number of seats in each row was reduced by 10, the total number of seats increased by 300. How many rows were there?
Solution. 30 rows

Question. Some students planned a picnic. The budget the food was Rs. 500. But, 5 of them failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic?
Solution. 20

Question. A train covers a distance of 90 km at a uniform speed. Had the speed been 15 km per hour more, it would have taken 30 minutes less for the journey. Find the original speed of the train. 
Solution. 45 km/hr

Question. Out of a number of Saras birds, one fourth the number are moving about in lotus plants; 1/9 th coupled (along) with 1/4 th as well as 7 times the square root of the number move on a hill; 56 birds remain in vakula trees. What is the total number of birds?
Solution. 576

Question. A two digit number is such that the product of the digits is 14. When 45 is added to the number, then the digits are reversed. Find the number.
Solution. 27

Question. The length of the hypotenuse of a right-angled triangle exceeds the base by 1 cm and also exceeds twice the length of the altitude by 3 cm. Find the length of each side of the triangle.
Solution. 5 cm, 12 cm, 13 cm

Question. Find the values of k for which the equation x² + 5kx + 16 = 0 has no real roots.
Solution. - 8 /5 <k< 8/5

Question. Solve the equation 2x² – 5x + 3 = 0 by the method of completing the square.
Solution. 1 , 3/2

Question. The sum of two numbers is 16 and the sum of their reciprocals is 1/3. Find the numbers.
Solution. 4, 12

Question. A two digit number is four times the sum of its digits and twice the product of its digits. Find the number.
Solution. 36

Question. Solve for x using quadratic formula :
a²b²x² + b²x - a²x - 1 = 0
Solution. - 1/ a² , 1/b²

Question. Divide 26 into two parts, whose sum of squares is 346. Frame an equation for the given statement.
Solution. x² + (26 – x)² = 346

Question. Solve for x : 1/ x + 4 - 1/x - 7 = 11/30 ; x ≠ - 4, 7
Solution. 1, 2

Question. Solve for x : 4x² – 4ax + a² – b² = 0.
Solution. a + b /2 , x = a - b /2 

Question. A plane left 40 minutes late due to bad weather and in order to reach its destination, 1600 km away in time, it had to increase its speed by 400 km/hr from its usual speed. Find the usual speed of the plane.
Solution. 800 km/hr

Question. Find the values of k so that the quadratic equation x² – 2x (1 + 3k) + 7 (3 + 2k) = 0 has equal roots.
Solution. 2 or - 10/9 

Question. Represent the following situation in the form a quadratic equation:
(a) Abdul and Sneha together have 30 oranges. Both of them ate 3 oranges each and the product of the number of oranges they have now is 120. We would like to find out how many oranges they had initially.
(b) The area of a rectangular plot is 428 m2. The length of the plot (in metres) is two more than twice its breadth. We need to find the length and breadth of the plot.
Solution. (a) Let the number of number of left oranges with Abdul and Sneha be x and y respectively.
Then, x + y = 30 ⇒ y = 30 – x
The number of oranges left with both Abdul and Sneha are x – 3 and y – 3 respectively.
The product of number of left oranges = 120
⇒ (x – 3) (y – 3) = 120
⇒ (x – 3) (30 – x – 3) = 120 (Q y = 30 – x)
⇒ (x – 3) (27 – x) = 120
⇒ 27x – x² – 81 + 3x = 120 ⇒ x² – 30x + 201 = 0
(b) Let the breadth of the plot be x. Then the length of the plot = 2x + 2
Since, area of the plot = 428 m² (Given)
∴ x(2x + 2) = 428
⇒ 2x² + 2x – 428 = 0
⇒ x² + x – 214 = 0

Question. Check whether the following are quadratic equations:
(a) x(x + 2) – 3 = (x + 4) x (b) (x + 2)³ = x³ – 4x² + 2
Solution. (a) Since x(x + 2) – 3 = x(x + 4)
⇒ x² + 2x – 3 = x² + 4x
⇒ 2x + 3 = 0
This is linear equation not a quadratic equation.
(b) (x + 2)³ = x³ – 4x² + 2
⇒ x³ + 6x² + 12x + 8 = x³ – 4x² + 2
⇒ 10x² + 12x + 6 = 0
⇒ 5x² + 6x + 3 = 0
This is a quadratic equation.

Question. Write the set of values of k for which the quadratic equation 2x² + kx + 8 has real roots. 
Solution. For real roots, D ≥ 0
⇒ b² – 4ac ≥ 0 ⇒ k² – 4(2) (8) ≥ 0
⇒ k² – 64 ≥ 0 ⇒ k² ≥ 64 ⇒ k < –8 and k > 8

Question. Find the value of p, for which one root of the quadratic equation px² – 14x + 8 = 0 is 6 times the other.
Solution. Let the roots of the given equation be a and 6a.
Thus the quadratic equation is (x – a)(x – 6a) = 0
⇒ x² – 7ax + 6a² = 0 ...(i)
Given equation can be written as
x² - 14/P x + 8/P = 0
Comparing the coefficients in (i) and (ii) 7α= 14/P and 6α² = 8/P
Solving to get p = 3.

Question. State whether the equation (x + 1) (x – 2) + x = 0 has two distinct real roots or not. Justify your answer.
Solution. We have (x + 1) (x – 2) + x = 0 ⇒ x² – x – 2 + x = 0 ⇒ x² – 2 = 0
∴ D = b² – 4ac = 0 – 4(1) (–2) = 8 > 0
∴ Given equation has two distinct real roots.

Question. If –5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal roots, then find the value of k. 
Solution. ∴ –5 is a root of the equation 2x2 + px – 15 = 0
∴ 2 (–5)² + p (–5) – 15 = 0
⇒ 50 – 5p – 15 = 0 or 5p = 35 or p = 7
Again p (x² + x) + k = 0 or 7x² + 7x + k = 0 has equal roots
∴ D = 0
i.e., b² – 4ac = 0 or 49 – 4 × 7k = 0 ⇒ k = 49/28 = 7/4

Question. If the equation (1 + m²) x² + 2mcx + c² – a² = 0 has equal roots, show that c² = a² (1 + m²). 
Solution. The given equation is (1 + m2) x2 + (2mc) x + (c² – a²) = 0
Here, A = 1 + m², B = 2mc and C = c2 – a2
Since the given equation has equal roots, therefore
D = 0 ⇒ B² – 4AC = 0
⇒ (2mc)² – 4 (1 + m2) (c² – a²) = 0
⇒ 4m²c² – 4(c² – a² + m²c² – m²a²) = 0
⇒ m²c² – c² + a² – m²c² + m²a² = 0 [Dividing throughout by 4]
⇒ –c² + a² (1 + m²) = 0 ⇒ c² = a2 (1 + m²) Hence Proved.

Question. Find the value of k for the quadratic equation kx (x – 2) + 6 = 0, so that it has two equal roots.
Solution. We have kx (x – 2) + 6 = 0 
⇒ kx² – 2kx + 6 = 0
Here, a = k, b = –2k, c = 6
For equal roots, D = 0
i.e., b² – 4ac = 0 ⇒ (–2k)2 – 4 × k × 6 = 0
⇒ 4k² – 24k = 0 ⇒ 4k (k – 6) = 0
Either 4k = 0 or k – 6 = 0 ⇒ k = 0 or k = 6
But k ≠ 0 (because if k = 0, then given equation will not be a quadratic equation).
So, k = 6.

Question. Form a quadratic equation whose roots are 2 and 3.
Solution. Sum of roots = 2 + 3 = 5
Product of roots = 2 × 3 = 6
So, quadratic equation can be formed as x² – (sum of roots)x + product of roots = 0
x² – 5x + 6 = 0

Question. What will be the nature of roots of quadratic equation 2x² + 4x – 7 = 0?
Solution. ∴ 2x² + 4x – 7 = 0
Here, a = 2, b = 4, c = –7
D = b² – 4ac = 16 – 4 × 2 × (–7) = 16 + 56 = 72 > 0
Hence, roots of quadratic equation are real and unequal.

Question. Find the value(s) of k for which the equation x² + 5kx + 16 = 0 has real and equal roots.
Solution. For roots to be real and equal, b² – 4ac = 0 
⇒ (5k)² – 4 × 1 × 16 = 0 ⇒ k = ± 8/5

Question. If the roots of the equation (a – b) x² + (b – c) x + (c – a) = 0 are equal, prove that 2a = b + c.
Solution. Since the equation (a – b) x² + (b – c) x + (c – a) = 0 has equal roots, therefore discriminant
D = (b – c)² – 4 (a – b) (c – a) = 0
⇒ b² + c² – 2bc – 4 (ac – a² – bc + ab) = 0
⇒ b² + c² – 2bc – 4ac + 4a² + 4bc – 4ab = 0
⇒ 4a² + b² + c² – 4ab + 2bc – 4ac = 0
⇒ (2a)² + (–b)² + (–c)² + 2 (2a) (–b) + 2 (–b) (–c) + 2 (–c) 2a = 0
⇒ (2a – b – c)2 = 0
⇒ 2a – b – c = 0
⇒ 2a = b + c. Hence Proved.

Question. If ax² + bx + c = 0 has equal roots, find the value of c.
Solution. For equal roots D = 0 
i.e., b² – 4ac = 0 ⇒ b² = 4ac ⇒ c = b²/4a

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: For any quadratic equation, if the roots are \( \alpha \) and \( \beta \), the equation is given by: \[ x^2 - (\alpha + \beta)x + \alpha\beta = 0 \] 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: Use the formula \( x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0 \) to build the equation for each pair of numbers, then multiply to clear any fractions.

Exam Tip: Always multiply the final equation by the least common multiple of any denominators to present the answer in standard integer coefficient 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 new count is \( x + 10 \), and the number of apples each receives 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 positive, we reject the negative solution \( x = -60 \). 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 apples per student, 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 splitting the middle term: \[ p^2x^2 + p^2x - q^2x - q^2 = 0 \] Factor out the common terms from each group: \[ p^2x(x + 1) - q^2(x + 1) = 0 \] \[ (p^2x - q^2)(x + 1) = 0 \] Now set each factor to zero: - \( x + 1 = 0 \implies x = -1 \) - \( p^2x - q^2 = 0 \implies x = \frac{q^2}{p^2} \) (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: Do not use the quadratic formula here - grouping terms is far quicker and minimizes the risk of mistakes with algebraic signs.

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 the method of completing the squares: First, divide the entire equation by the coefficient of \( x^2 \), which is 3: \[ x^2 + \frac{23}{3}x + \frac{20}{3} = 0 \] Now, add and subtract the square of half the coefficient of \( x \) - that is \( \left( \frac{1}{2} \times \frac{23}{3} \right)^2 = \left( \frac{23}{6} \right)^2 = \frac{529}{36} \): \[ x^2 + \frac{23}{3}x + \frac{529}{36} - \frac{529}{36} + \frac{20}{3} = 0 \] \[ \left( x + \frac{23}{6} \right)^2 = \frac{529}{36} - \frac{240}{36} \] \[ \left( x + \frac{23}{6} \right)^2 = \frac{289}{36} \] 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: \( x = -\frac{17}{6} - \frac{23}{6} = -\frac{40}{6} = -\frac{20}{3} \) The solutions are \( x = -1 \) and \( x = -\frac{20}{3} \).
In simple words: Divide the entire equation by 3, complete the perfect square on the left side, and take square roots to find the two answers: \( -1 \) and \( -20/3 \).

Exam Tip: Verify your completed square calculation by expanding \( \left(x + \frac{23}{6}\right)^2 \) to check if it matches the original terms.

Question 5. Find the roots by factorization: \( 3\sqrt{3}x^2 + 19x + 10\sqrt{3} = 0 \).
Answer: We need to 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: \[ 3\sqrt{3}x(x + \sqrt{3}) + 10(x + \sqrt{3}) = 0 \] \[ (3\sqrt{3}x + 10)(x + \sqrt{3}) = 0 \] Now 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 the middle term \( 19x \) into \( 9x + 10x \). Group the terms to find the factors, which gives the roots \( -\sqrt{3} \) and \( -\frac{10}{3\sqrt{3}} \).

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

Question 6. Find the solutions by quadratic equation formula: \( 3x^2 - 13x - 100 = 0 \).
Answer: For the quadratic equation \( 3x^2 - 13x - 100 = 0 \), comparing with standard form shows \( a = 3 \), \( b = -13 \), and \( c = -100 \). We calculate the discriminant \( D \): \[ D = b^2 - 4ac = (-13)^2 - 4(3)(-100) \] \[ D = 169 + 1200 = 1369 \] Now, we 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: Using the quadratic formula with the computed discriminant of 1369, we calculate the roots to be \( 25/3 \) and \( -4 \).

Exam Tip: Finding the square root of large numbers like 1369 can be simplified by identifying closest tens square numbers (such as \( 30^2 = 900 \) and \( 40^2 = 1600 \)).

Question 7. Solve for x: \( 12abx^2 - (9a^2 - 8b^2)x - 6ab = 0 \).
Answer: Let's expand the middle term of the equation: \[ 12abx^2 - 9a^2x + 8b^2x - 6ab = 0 \] Factor the terms by grouping: \[ 3ax(4bx - 3a) + 2b(4bx - 3a) = 0 \] \[ (3ax + 2b)(4bx - 3a) = 0 \] Now set each factor to zero: - \( 3ax + 2b = 0 \implies x = -\frac{2b}{3a} \) - \( 4bx - 3a = 0 \implies x = \frac{3a}{4b} \) (assuming \( a, b \ne 0 \)) The solutions are \( x = -\frac{2b}{3a} \) and \( x = \frac{3a}{4b} \).
In simple words: Expand the middle term to group the variables, which lets us factor the expression easily as \( (3ax + 2b)(4bx - 3a) = 0 \).

Exam Tip: Be cautious of minus signs when distributing variables across parenthetical expressions.

Question 8. Solve for x: \( (a + b)^2x^2 + 8(a^2 - b^2)x + 16(a - b)^2 = 0 \).
Answer: Let's substitute \( A = a + b \) and \( B = a - b \) to make the quadratic equation simpler: \[ A^2x^2 + 8ABx + 16B^2 = 0 \] This is a perfect square trinomial of the form \( (Ax + 4B)^2 = 0 \). Let's check by expanding: \[ (Ax + 4B)^2 = (Ax)^2 + 2(Ax)(4B) + (4B)^2 = A^2x^2 + 8ABx + 16B^2 \] Since \( (Ax + 4B)^2 = 0 \): \[ Ax + 4B = 0 \] \[ x = -\frac{4B}{A} \] Now, 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: We can recognize the equation as a perfect square of the form \( (Ax + 4B)^2 = 0 \), where \( A = a+b \) and \( B = a-b \), simplifying the solution to \( x = \frac{4(b - a)}{a + b} \).

Exam Tip: Always look for perfect square patterns when coefficients contain matched algebraic expressions.

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, we find: - Speed upstream = \( 11 - y \) km/h - Speed downstream = \( 11 + y \) km/h Using the formula \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \): - Time upstream = \( \frac{12}{11 - y} \) hours - Time downstream = \( \frac{12}{11 + y} \) hours The total time taken 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: Summing the times for both journeys to equal \( 11/4 \) hours yields a quadratic equation that simplifies to \( y^2 = 25 \), meaning the stream's speed is 5 km/h.

Exam Tip: Always convert the time unit completely into hours (e.g. 2 hours 45 minutes as \( 11/4 \) hours) before setting up the speed 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: By using the relations between the digits, we find the tens digit is 3 and the units digit is 5, making the number 35.

Exam Tip: Remember that digits can only take integer values from 0 to 9, so reject any fractional or negative values immediately.

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 \( 29 - x \). According to the problem, 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 of 29 are 13 and 16.
In simple words: Let the two parts be \( x \) and \( 29 - x \). Setting the sum of their squares to 425 yields the parts 13 and 16.

Exam Tip: Reducing coefficients by dividing by a common factor (like 2) makes factoring larger numbers much easier.

Question 12. The denominator of a fraction is one more than twice the denominator. If the sum of the fraction and its reciprocal is \( 2\frac{16}{21} \), 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 doubles, 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 a right angled ∆ is 5x cm and (3x – 4) cm. Area of the triangle is 60 cm². Find the hypotenuse. 
Answer: Let's represent the perpendicular sides of the right-angled triangle. According to the standard parameters (using \( 3x - 1 \) for the second side to match the given answer of 17 cm): - 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: 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: Be sure to write out the formula for the area of a triangle before starting your equation steps.

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 snake, the peacock pounces on it. If their speeds are equal, at what distance from hole, the snake caught? [Ans.: 12 m]
Answer: Let \( A \) be the top of the pillar of height 9 m, \( B \) be the base of the pillar (the hole), and \( C \) be the initial position of the snake which is 27 m away from \( B \). Let the snake be caught at a point \( D \) at a distance \( x \) meters from the base of the pillar \( B \). The distance traveled by the snake is: \[ CD = 27 - x \] Since the speeds of the peacock and the snake are equal, and they start simultaneously, the distance they travel must be equal. Therefore, the distance flown by the peacock \( AD \) is equal to \( CD \): \[ AD = 27 - x \] In right-angled triangle \( ABD \), by 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. Squaring both sides of the relation yields \( x = 12 \) meters.

Exam Tip: A simple geometric sketch of the right-angled triangle is highly recommended for these word problems to clarify variables.

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 problem description: - Camels in the forest = \( \frac{x}{4} \) - Camels in the mountains = \( 2\sqrt{x} \) - Camels on the river bank = 15 Sum of all camels equals the total: \[ \frac{x}{4} + 2\sqrt{x} + 15 = x \] Let's substitute \( y = \sqrt{x} \implies x = y^2 \): \[ \frac{y^2}{4} + 2y + 15 = y^2 \] Multiply the entire equation by 4: \[ y^2 + 8y + 60 = 4y^2 \] \[ 3y^2 - 8y - 60 = 0 \] We solve this quadratic equation by splitting the middle term: \[ 3y^2 - 18y + 10y - 60 = 0 \] \[ 3y(y - 6) + 10(y - 6) = 0 \] \[ (3y + 10)(y - 6) = 0 \] 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 lost food for each member increased by Rs. 30. How many boys attended the picnic? 
Answer: Let the original number of boys be \( x \). The share of each boy originally is Rs. \( \frac{3000}{x} \). Since 5 boys could not attend, the number of boys who went is \( x - 5 \), and their share is Rs. \( \frac{3000}{x - 5} \). The difference in individual shares is Rs. 30: \[ \frac{3000}{x - 5} - \frac{3000}{x} = 30 \] Dividing the entire equation by 30: \[ \frac{100}{x - 5} - \frac{100}{x} = 1 \] \[ 100 \left( \frac{x - (x - 5)}{x(x - 5)} \right) = 1 \] \[ \frac{500}{x^2 - 5x} = 1 \] \[ x^2 - 5x - 500 = 0 \] We solve the quadratic equation by factoring: \[ (x - 25)(x + 20) = 0 \] 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: Be sure to read the final question carefully - the question asks for the number of boys who *attended* the picnic (\( x-5 \)), not the original number planned (\( x \)).

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 \( 2k - 2 \), \( 2k \), and \( 2k + 2 \). The sum of the numbers is: \[ (2k - 2) + 2k + (2k + 2) = 6k \] The product of the numbers is: \[ (2k - 2) \times 2k \times (2k + 2) = 2k(4k^2 - 4) = 8k(k^2 - 1) \] According to the problem: \[ 8k(k^2 - 1) = 20(6k) \] Dividing both sides by \( 4k \) (assuming \( k \ne 0 \)): \[ 2(k^2 - 1) = 30 \] \[ k^2 - 1 = 15 \] \[ k^2 = 16 \implies k = 4 \] Substitute \( k = 4 \) back to find the numbers: - First number = \( 2(4) - 2 = 6 \) - Second number = \( 2(4) = 8 \) - Third number = \( 2(4) + 2 = 10 \) The consecutive even numbers are 6, 8, and 10.
In simple words: Let the consecutive even numbers be \( 2k-2 \), \( 2k \), and \( 2k+2 \). Setting their product equal to 20 times their sum simplifies to \( k = 4 \), which gives the numbers 6, 8, and 10.

Exam Tip: Representing consecutive even numbers symmetrically as \( 2k-2, 2k, 2k+2 \) allows you to eliminate common terms easily when dividing.

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 choose \( x = 8 \). 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 equation for the area \( b(2b+2) = 60 \), 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 \( \sqrt{l^2 + b^2} \) clearly before substituting the dimensions.