Quadratic Equations - 100 word, non-word problems with solutions - Free Printable

Here's the complete solution to all 12 problems from the worksheet, with detailed explanations.

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(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 years.

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

Given:
> (x – 3)(x + 5) = 240

Expand:
> x² + 5x – 3x – 15 = 240
> x² + 2x – 15 = 240
> x² + 2x – 255 = 0

Factorize:
We need two numbers that multiply to -255 and add to +2 → 17 and -15

> (x + 17)(x – 15) = 0
> ⇒ x = -17 or x = 15

Age cannot be negative → x = 15

✔ Answer: Ria’s current age is 15 years.

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

Let number of cookies = x

Then, cost per cookie = 40/x

If he bought x + 2 cookies, 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

Factorize:
> (x + 10)(x – 8) = 0
> ⇒ x = -10 (reject) or x = 8

✔ Answer: He bought 8 cookies.

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

Let original length = x meters
Original cost per meter = 240/x

New length = x + 4 meters
New cost per meter = 240/x – 3

Total cost remains same → New total cost = 240

So:
> (x + 4)(240/x – 3) = 240

Multiply out:
> (x + 4)(240 – 3x)/x = 240
> Multiply both sides by x:

> (x + 4)(240 – 3x) = 240x
> 240x – 3x² + 960 – 12x = 240x
> -3x² – 12x + 960 = 0
> Divide by -3:

> x² + 4x – 320 = 0

Factorize:
> (x + 20)(x – 16) = 0
> ⇒ x = -20 (reject) or x = 16

✔ Answer: Length of copper wire is 16 meters.

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(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 arithmetic sequence with:

- First term (a) = 3
- Common difference (d) = 2
- Sum of n terms = 48

Sum formula:
> Sₙ = n/2 [2a + (n–1)d]
> 48 = n/2 [2×3 + (n–1)×2]
> 48 = n/2 [6 + 2n – 2]
> 48 = n/2 [2n + 4]
> 48 = n(n + 2)

So:
> n² + 2n – 48 = 0

Factorize:
> (n + 8)(n – 6) = 0
> ⇒ n = -8 (reject) or n = 6

✔ Answer: n = 6

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(5) Find solution of quadratic equation 6y² – 13y – 5 = 0

Use factorization.

We need two numbers that multiply to 6×(-5) = -30 and add to -13 → -15 and +2

Split middle term:

> 6y² – 15y + 2y – 5 = 0
> 3y(2y – 5) + 1(2y – 5) = 0
> (3y + 1)(2y – 5) = 0

⇒ y = -1/3 or y = 5/2

✔ Answer: y = -1/3, 5/2

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(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 smaller number = x, larger = y

Given:
> x² + y² = 832 ...(1)
> y² = 36x ...(2)

Substitute (2) into (1):

> x² + 36x = 832
> x² + 36x – 832 = 0

Factorize:

Find factors of -832 that add to 36 → 52 and -16

> (x + 52)(x – 16) = 0
> ⇒ x = -52 (reject, since positive) or x = 16

Then y² = 36×16 = 576 → y = √576 = 24

✔ Answer: Numbers are 16 and 24

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(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 per sweet = x rupees

Number of sweets she could buy originally = 192/x

New price = x – 4

New number of sweets = 192/(x – 4)

Given:
> 192/(x – 4) – 192/x = 8

Multiply both sides by x(x – 4):

> 192x – 192(x – 4) = 8x(x – 4)
> 192x – 192x + 768 = 8x² – 32x
> 768 = 8x² – 32x
> Divide by 8:

> 96 = x² – 4x
> x² – 4x – 96 = 0

Factorize:
> (x – 12)(x + 8) = 0
> ⇒ x = 12 or x = -8 (reject)

✔ Answer: Original price = Rs.12

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(8) Solve quadratic equation x² + 5x – (a² + 3a – 4) = 0 using factorization.

First, factor the constant term: -(a² + 3a – 4)

Factor a² + 3a – 4 → (a + 4)(a – 1)

So equation becomes:
> x² + 5x – (a + 4)(a – 1) = 0

We want to factor as (x + p)(x + q) = 0, where p + q = 5 and p×q = -(a+4)(a–1)

Try splitting 5 into two parts whose product is -(a+4)(a–1)

Let’s try:
p = a + 4, q = - (a – 1) → p + q = a + 4 – a + 1 = 5 ✔
p×q = (a+4)(-a+1) = - (a+4)(a–1) ✔

So:
> (x + a + 4)(x – (a – 1)) = 0
> ⇒ x = -a – 4 or x = a – 1

✔ Answer: x = -a – 4, a – 1

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(9) Find the roots of the quadratic equation y² – 7√3 y + 36 = 0.

Use factorization.

Need two numbers that multiply to 36 and add to 7√3.

Try: 3√3 and 4√3

Check:
> 3√3 × 4√3 = 12 × 3 = 36 ✔
> 3√3 + 4√3 = 7√3 ✔

So:
> (y – 3√3)(y – 4√3) = 0
> ⇒ y = 3√3 or y = 4√3

✔ Answer: a. 3√3 and 4√3

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

Check discriminant D = b² – 4ac < 0

a. -4x² + 7x – 4 = 0 → D = 49 – 4(-4)(-4) = 49 – 64 = -15 < 0 → No real roots

b. -4x² + 7x – 2 = 0 → D = 49 – 4(-4)(-2) = 49 – 32 = 17 > 0 → Real roots

c. -2x² + 5x – 2 = 0 → D = 25 – 4(-2)(-2) = 25 – 16 = 9 > 0 → Real roots

d. 3x² + 6x + 2 = 0 → D = 36 – 4(3)(2) = 36 – 24 = 12 > 0 → Real roots

✔ Answer: a. -4x² + 7x – 4 = 0

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(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 tens digit = x, units digit = y

Number = 10x + y

Given:
> x × y = 10 ...(1)
> 10x + y – 27 = 10y + x ...(2)

Simplify (2):
> 10x + y – 27 = 10y + x
> 9x – 9y = 27
> x – y = 3 ...(3)

From (1): xy = 10
From (3): x = y + 3

Substitute:
> (y + 3)y = 10
> y² + 3y – 10 = 0
> (y + 5)(y – 2) = 0
> y = -5 (reject) or y = 2

Then x = 2 + 3 = 5

Number = 10x + y = 50 + 2 = 52

Check: 5×2=10 ✔; 52 – 27 = 25 → digits interchanged ✔

✔ Answer: b. 52

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

Check D = b² – 4ac > 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

✔ Answer: c. x² + 3x + 1 = 0

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## ✔ Final Answers Summary:

1. 15 years
2. 8 cookies
3. 16 meters
4. n = 6
5. y = -1/3, 5/2
6. 16 and 24
7. Rs.12
8. x = -a – 4, a – 1
9. a. 3√3 and 4√3
10. a. -4x² + 7x – 4 = 0
11. b. 52
12. c. x² + 3x + 1 = 0

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