Class 10 Math Worksheets and Problems: Quadratic Equations ... - Free Printable

(1) Let Ria’s current age be x years.
3 years ago, her age was (x - 3).
5 years later, her age will be (x + 5).
Given: (x - 3)(x + 5) = 240
Expand: x² + 5x - 3x - 15 = 240 → x² + 2x - 15 = 240
→ x² + 2x - 255 = 0
Factor: (x + 17)(x - 15) = 0
Solutions: x = -17 (invalid), x = 15
Answer: 15 years

(2) Let number of cookies bought be x.
Cost per cookie = 40/x
If he bought (x + 2) cookies for Rs.40, cost per cookie = 40/(x + 2)
Given: 40/x - 40/(x + 2) = 1
Multiply both sides by x(x + 2):
40(x + 2) - 40x = x(x + 2)
→ 40x + 80 - 40x = x² + 2x
→ 80 = x² + 2x → x² + 2x - 80 = 0
Factor: (x + 10)(x - 8) = 0
Solutions: x = -10 (invalid), x = 8
Answer: 8 cookies

(3) Let length of copper wire be x meters.
Original cost: 3x = 240 → x = 80
New length: (x + 4) = 84 m
New price per meter: (3 - 1) = Rs.2
New cost: 84 × 2 = 168 ≠ 240 — contradiction.
Re-read: “total cost would remain unchanged” → 3x = 2(x + 4)
→ 3x = 2x + 8 → x = 8
Check: original cost = 3×8 = 24, new cost = 2×12 = 24 → matches.
Answer: 8 meters

(4) Consecutive odd numbers starting from 3: 3, 5, 7, ..., up to n terms.
This is an AP with first term a = 3, common difference d = 2.
Sum = n/2 [2a + (n-1)d] = 48
→ n/2 [6 + 2(n-1)] = 48
→ n/2 [2n + 4] = 48 → n(n + 2) = 48
→ n² + 2n - 48 = 0
Factor: (n + 8)(n - 6) = 0
Solutions: n = -8 (invalid), n = 6
Answer: 6

(5) Equation: 6y² - 13y - 5 = 0
Use quadratic formula: y = [13 ± √(169 + 120)] / 12 = [13 ± √289]/12 = [13 ± 17]/12
→ y = (13 + 17)/12 = 30/12 = 5/2
→ y = (13 - 17)/12 = -4/12 = -1/3
Answer: y = 5/2 or y = -1/3

(6) Let smaller number be x, larger be y.
Given: x² + y² = 832 and y² = 36x
Substitute: x² + 36x = 832 → x² + 36x - 832 = 0
Discriminant: 1296 + 3328 = 4624 = 68²
x = [-36 ± 68]/2 → x = 16 or x = -52 (discard negative)
Then y² = 36×16 = 576 → y = 24 (positive)
Answer: 16 and 24

(7) Let original price per sweet be x rupees.
Number of sweets for Rs.192 at original price: 192/x
At reduced price (x - 4), number of sweets: 192/(x - 4)
Given: 192/(x - 4) - 192/x = 8
Multiply by x(x - 4): 192x - 192(x - 4) = 8x(x - 4)
→ 192x - 192x + 768 = 8x² - 32x
→ 8x² - 32x - 768 = 0 → x² - 4x - 96 = 0
Factor: (x - 12)(x + 8) = 0
Solutions: x = 12 or x = -8 (invalid)
Answer: Rs.12

(8) Equation: x² + 5x - (a² + 3a - 4) = 0
Factor the constant term: -(a² + 3a - 4) = -(a + 4)(a - 1)
So equation becomes: x² + 5x - (a + 4)(a - 1) = 0
We need two numbers that multiply to -(a+4)(a-1) and add to 5.
Try: (x + (a + 4))(x - (a - 1)) = x² + (a + 4 - a + 1)x - (a+4)(a-1) = x² + 5x - (a+4)(a-1)
Perfect match.
So factors: (x + a + 4)(x - a + 1) = 0
Solutions: x = -a - 4 or x = a - 1
Answer: x = -a - 4, x = a - 1

(9) Equation: y² - 7√3 y + 36 = 0
Discriminant: (7√3)² - 4×1×36 = 147 - 144 = 3
Roots: [7√3 ± √3]/2 = (8√3)/2 = 4√3 and (6√3)/2 = 3√3
Answer: b. 3√3 and 4√3

(10) For no real roots, discriminant < 0.
a. -4x² + 7x - 4 = 0 → D = 49 - 64 = -15 < 0 → no real roots
b. -4x² + 7x - 2 = 0 → D = 49 - 32 = 17 > 0 → real roots
c. -2x² + 5x - 2 = 0 → D = 25 - 16 = 9 > 0 → real roots
d. 3x² + 6x + 2 = 0 → D = 36 - 24 = 12 > 0 → real roots
Answer: a. -4x² + 7x - 4 = 0

(11) Let number be 10a + b, where a and b are digits.
Given: a×b = 10
Also: (10a + b) - 27 = 10b + a
→ 10a + b - 27 = 10b + a
→ 9a - 9b = 27 → a - b = 3
Now solve: a×b = 10, a - b = 3
Possible digit pairs for ab=10: (2,5), (5,2), (1,10 invalid), (10,1 invalid)
Check: a=5, b=2 → a-b=3 → valid
Number: 10×5 + 2 = 52
Check: 52 - 27 = 25 → digits interchanged → correct
Answer: b. 52

(12) For two distinct real roots, discriminant > 0.
a. 3x² + 6x + 4 = 0 → D = 36 - 48 = -12 < 0 → no
b. -x² + 3x - 3 = 0 → D = 9 - 12 = -3 < 0 → no
c. x² + 3x + 1 = 0 → D = 9 - 4 = 5 > 0 → yes
d. 4x² + 8x + 4 = 0 → D = 64 - 64 = 0 → equal roots → no
Answer: c. x² + 3x + 1 = 0