Class 10 Quadratic Equations worksheet with questions and multiple-choice options for practice.

Let's solve each of the problems step by step.

---

(1) The product of Ria’s age (in years) 3 years ago with her age (in years) 5 years later is 240. Find her current age.

Let Ria's current age be $ x $.

- Age 3 years ago: $ x - 3 $
- Age 5 years later: $ x + 5 $

Given:
$$
(x - 3)(x + 5) = 240
$$

Expand:
$$
x^2 + 5x - 3x - 15 = 240 \Rightarrow x^2 + 2x - 15 = 240
\Rightarrow x^2 + 2x - 255 = 0
$$

Solve using quadratic formula:
$$
x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-255)}}{2(1)} = \frac{-2 \pm \sqrt{4 + 1020}}{2} = \frac{-2 \pm \sqrt{1024}}{2}
$$
$$
\sqrt{1024} = 32 \Rightarrow x = \frac{-2 \pm 32}{2}
$$

So:
- $ x = \frac{30}{2} = 15 $
- $ x = \frac{-34}{2} = -17 $ → Not possible (age can't be negative)

✔ Answer: 15 years

---

(2) Ashish buys number of cookies for Rs.40. If he had bought 2 more cookies for the same amount, each cookie would have cost Rs.1 less. How many cookies did he buy?

Let number of cookies bought = $ x $

Then price per cookie = $ \frac{40}{x} $

If he bought $ x+2 $ cookies, price per cookie = $ \frac{40}{x+2} $

Given:
$$
\frac{40}{x} - \frac{40}{x+2} = 1
$$

Multiply both sides by $ x(x+2) $:
$$
40(x+2) - 40x = x(x+2)
\Rightarrow 40x + 80 - 40x = x^2 + 2x
\Rightarrow 80 = x^2 + 2x
\Rightarrow x^2 + 2x - 80 = 0
$$

Factor:
$$
x^2 + 10x - 8x - 80 = 0 \Rightarrow x(x+10) -8(x+10) = 0 \Rightarrow (x-8)(x+10)=0
$$

So $ x = 8 $ or $ x = -10 $ → discard negative

✔ Answer: 8 cookies

---

(3) A piece of copper wire costs Rs.240. If it was 4 meters longer and price of each meter of copper wire costs Rs.3 less, the total cost would remain unchanged. Find the length of copper wire.

Let original length = $ x $ meters
Original price per meter = $ \frac{240}{x} $

New length = $ x + 4 $
New price per meter = $ \frac{240}{x} - 3 $

Total cost remains same:
$$
(x + 4)\left(\frac{240}{x} - 3\right) = 240
$$

Expand:
$$
(x + 4)\left(\frac{240 - 3x}{x}\right) = 240
\Rightarrow \frac{(x + 4)(240 - 3x)}{x} = 240
$$

Multiply both sides by $ x $:
$$
(x + 4)(240 - 3x) = 240x
$$

Expand left side:
$$
240x - 3x^2 + 960 - 12x = 240x
\Rightarrow -3x^2 + 228x + 960 = 240x
\Rightarrow -3x^2 -12x + 960 = 0
\Rightarrow x^2 + 4x - 320 = 0
$$

Solve:
$$
x = \frac{-4 \pm \sqrt{16 + 1280}}{2} = \frac{-4 \pm \sqrt{1296}}{2} = \frac{-4 \pm 36}{2}
$$

So:
- $ x = \frac{32}{2} = 16 $
- $ x = \frac{-40}{2} = -20 $ → invalid

✔ Answer: 16 meters

---

(4) The sum of the n consecutive natural odd numbers starting from 3 is 48. Find the value of n.

Odd numbers starting from 3: 3, 5, 7, 9, ...

This is an AP with:
- First term $ a = 3 $
- Common difference $ d = 2 $
- Sum $ S_n = 48 $

Sum of AP:
$$
S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}[2(3) + (n-1)(2)] = \frac{n}{2}[6 + 2n - 2] = \frac{n}{2}(2n + 4)
= n(n + 2)
$$

Set equal to 48:
$$
n(n + 2) = 48 \Rightarrow n^2 + 2n - 48 = 0
\Rightarrow n^2 + 8n - 6n - 48 = 0 \Rightarrow (n+8)(n-6)=0
$$

So $ n = 6 $ (since $ n > 0 $)

Check: 3 + 5 + 7 + 9 + 11 + 13 = 48 ✔

✔ Answer: n = 6

---

(5) Find solution of quadratic equation $ 6y^2 - 13y - 5 = 0 $

Use quadratic formula:
$$
y = \frac{13 \pm \sqrt{(-13)^2 - 4(6)(-5)}}{2(6)} = \frac{13 \pm \sqrt{169 + 120}}{12} = \frac{13 \pm \sqrt{289}}{12}
$$
$$
\sqrt{289} = 17 \Rightarrow y = \frac{13 \pm 17}{12}
$$

So:
- $ y = \frac{30}{12} = \frac{5}{2} $
- $ y = \frac{-4}{12} = -\frac{1}{3} $

✔ Answer: $ y = \frac{5}{2}, -\frac{1}{3} $

---

(6) The sum of square of two positive numbers is 832. If square of the larger number is 36 times the smaller number, find the numbers.

Let the smaller number be $ x $, larger be $ y $

Given:
1. $ x^2 + y^2 = 832 $
2. $ y^2 = 36x $

Substitute into first:
$$
x^2 + 36x = 832 \Rightarrow x^2 + 36x - 832 = 0
$$

Solve:
Discriminant $ D = 36^2 + 4(832) = 1296 + 3328 = 4624 $

$ \sqrt{4624} = 68 $

$$
x = \frac{-36 \pm 68}{2}
\Rightarrow x = \frac{32}{2} = 16 \quad (\text{since } x > 0)
$$

Then $ y^2 = 36 \times 16 = 576 \Rightarrow y = \sqrt{576} = 24 $

✔ Answer: Numbers are 16 and 24

---

(7) If the price of a sweet is reduced by Rs.4, Gita can buy 8 more sweets for Rs.192. Find the original price of sweet.

Let original price = $ x $ Rs.

Number of sweets she can buy at price $ x $: $ \frac{192}{x} $

At price $ x - 4 $: $ \frac{192}{x - 4} $

She buys 8 more:
$$
\frac{192}{x - 4} - \frac{192}{x} = 8
$$

Multiply both sides by $ x(x - 4) $:
$$
192x - 192(x - 4) = 8x(x - 4)
\Rightarrow 192x - 192x + 768 = 8x^2 - 32x
\Rightarrow 768 = 8x^2 - 32x
\Rightarrow 8x^2 - 32x - 768 = 0
\Rightarrow x^2 - 4x - 96 = 0
$$

Factor:
$$
x^2 - 12x + 8x - 96 = 0 \Rightarrow x(x - 12) + 8(x - 12) = 0 \Rightarrow (x + 8)(x - 12) = 0
$$

So $ x = 12 $ (price can't be negative)

✔ Answer: Rs.12

---

(8) Solve quadratic equation $ x^2 + 5x - (x^2 + 3x - 4) = 0 $ using factorization.

First simplify:
$$
x^2 + 5x - x^2 - 3x + 4 = 0 \Rightarrow 2x + 4 = 0
\Rightarrow 2x = -4 \Rightarrow x = -2
$$

Wait — this is linear, not quadratic! But let's check:

Original expression:
$$
x^2 + 5x - (x^2 + 3x - 4) = x^2 + 5x - x^2 - 3x + 4 = 2x + 4
$$

So equation becomes:
$$
2x + 4 = 0 \Rightarrow x = -2
$$

But it's not a quadratic anymore. Maybe typo?

Alternatively, if the equation was meant to be:
$$
x^2 + 5x - (x^2 + 3x - 4) = 0
$$
It simplifies to linear.

But perhaps the question has a typo? Or maybe we're supposed to recognize that it reduces.

Anyway, as written, solution is:
✔ Answer: $ x = -2 $

---

(9) Find the roots of $ y^2 - 7\sqrt{3}y + 36 = 0 $

Use quadratic formula:
$$
y = \frac{7\sqrt{3} \pm \sqrt{(7\sqrt{3})^2 - 4(1)(36)}}{2}
= \frac{7\sqrt{3} \pm \sqrt{147 - 144}}{2} = \frac{7\sqrt{3} \pm \sqrt{3}}{2}
$$

So:
- $ y = \frac{7\sqrt{3} + \sqrt{3}}{2} = \frac{8\sqrt{3}}{2} = 4\sqrt{3} $
- $ y = \frac{7\sqrt{3} - \sqrt{3}}{2} = \frac{6\sqrt{3}}{2} = 3\sqrt{3} $

✔ Answer: $ 3\sqrt{3} $ and $ 4\sqrt{3} $ → Option (b)

---

(10) Which of the following quadratic equations have no real roots?

We use discriminant $ D = b^2 - 4ac $. If $ D < 0 $, no real roots.

#### a. $ -4x^2 + 7x - 4 = 0 $

$ D = 7^2 - 4(-4)(-4) = 49 - 64 = -15 < 0 $ → No real roots ✔

#### b. $ -4x^2 + 7x - 2 = 0 $

$ D = 49 - 4(-4)(-2) = 49 - 32 = 17 > 0 $ → Real roots

#### c. $ -2x^2 + 5x - 2 = 0 $

$ D = 25 - 4(-2)(-2) = 25 - 16 = 9 > 0 $ → Real roots

#### d. $ 3x^2 + 6x + 2 = 0 $

$ D = 36 - 4(3)(2) = 36 - 24 = 12 > 0 $ → Real roots

Only a has no real roots.

✔ Answer: (a)

---

(11) A two-digit number is such that the product of its digits is 10. If 27 is subtracted from the number, the digits interchange their places. Find the number.

Let the number be $ 10a + b $, where $ a $ and $ b $ are digits.

Given:
1. $ a \times b = 10 $
2. $ 10a + b - 27 = 10b + a $

Simplify second equation:
$$
10a + b - 27 = 10b + a \Rightarrow 9a - 9b = 27 \Rightarrow a - b = 3
$$

Now solve:
- $ ab = 10 $
- $ a = b + 3 $

Substitute:
$$
(b + 3)b = 10 \Rightarrow b^2 + 3b - 10 = 0
\Rightarrow (b + 5)(b - 2) = 0 \Rightarrow b = 2 \text{ or } -5
$$

Take $ b = 2 $, then $ a = 5 $

Number = $ 10a + b = 52 $

Check:
- Product of digits: $ 5 \times 2 = 10 $ ✔
- $ 52 - 27 = 25 $ → digits reversed ✔

✔ Answer: 52 → Option (b)

---

(12) Which of the following quadratic equations has two distinct real roots?

We check discriminant $ D > 0 $

#### a. $ 3x^2 + 6x + 4 = 0 $

$ D = 36 - 4(3)(4) = 36 - 48 = -12 < 0 $ → No real roots

#### b. $ -x^2 + 3x - 3 = 0 $

$ D = 9 - 4(-1)(-3) = 9 - 12 = -3 < 0 $ → No real roots

#### c. $ x^2 + 3x + 1 = 0 $

$ D = 9 - 4 = 5 > 0 $ → Two distinct real roots ✔

#### d. $ 4x^2 + 8x + 4 = 0 $

$ D = 64 - 4(4)(4) = 64 - 64 = 0 $ → One real root (repeated)

So only c has two distinct real roots.

✔ Answer: (c)

---

✔ Final Answers Summary:

| Question | Answer |
|--------|--------|
| (1) | 15 years |
| (2) | 8 cookies |
| (3) | 16 meters |
| (4) | $ n = 6 $ |
| (5) | $ y = \frac{5}{2}, -\frac{1}{3} $ |
| (6) | 16 and 24 |
| (7) | Rs. 12 |
| (8) | $ x = -2 $ |
| (9) | b. $ 3\sqrt{3} $ and $ 4\sqrt{3} $ |
| (10) | a. $ -4x^2 + 7x - 4 = 0 $ |
| (11) | b. 52 |
| (12) | c. $ x^2 + 3x + 1 = 0 $ |

Let me know if you'd like these in a printable format or need explanations in simpler terms!