CBSE Class 10 Mathematics Quadratic Equations Worksheet Set C - Free Printable

14. Given the quadratic equation: (a - b)x² + (b - c)x + (c - a) = 0.
Since the roots are equal, the discriminant must be zero.
Discriminant D = (b - c)² - 4(a - b)(c - a) = 0.
Expanding:
(b - c)² = b² - 2bc + c²
4(a - b)(c - a) = 4[ac - a² - bc + ab] = 4ac - 4a² - 4bc + 4ab
So,
b² - 2bc + c² - (4ac - 4a² - 4bc + 4ab) = 0
b² - 2bc + c² - 4ac + 4a² + 4bc - 4ab = 0
4a² + b² + c² - 4ab + 2bc - 4ac = 0
Rearranging:
(2a)² - 2*(2a)*b + b² + c² + 2bc - 4ac = 0
(2a - b)² + c² + 2bc - 4ac = 0
Note that c² + 2bc - 4ac = c(c + 2b - 4a). Alternatively, group differently:
4a² - 4ab - 4ac + b² + 2bc + c² = 0
4a² - 4a(b + c) + (b + c)² = 0
This is a perfect square: [2a - (b + c)]² = 0.
Therefore, 2a - (b + c) = 0, so 2a = b + c.

15. Let the two consecutive multiples of 7 be 7n and 7(n+1).
Sum of squares: [7n]² + [7(n+1)]² = 637.
49n² + 49(n² + 2n + 1) = 637
49n² + 49n² + 98n + 49 = 637
98n² + 98n + 49 = 637
Divide by 49: 2n² + 2n + 1 = 13
2n² + 2n - 12 = 0
n² + n - 6 = 0
(n + 3)(n - 2) = 0
n = -3 or n = 2.
For n = 2: multiples are 14 and 21.
For n = -3: multiples are -21 and -14.
The multiples are 14 and 21, or -21 and -14.

16. Let the speed of the ordinary train be v km/h.
Then the speed of the superfast train is (v + 16) km/h.
Time taken by ordinary train: 192/v hours.
Time taken by superfast train: 192/(v + 16) hours.
Difference in time: 192/v - 192/(v + 16) = 2.
192[(v + 16) - v] / [v(v + 16)] = 2
192 * 16 / [v(v + 16)] = 2
3072 = 2v(v + 16)
1536 = v² + 16v
v² + 16v - 1536 = 0
Solve using quadratic formula: v = [-16 ± √(256 + 6144)] / 2 = [-16 ± √6400]/2 = [-16 ± 80]/2.
v = (64)/2 = 32 or v = (-96)/2 = -48 (discard negative).
So, ordinary train speed = 32 km/h, superfast train speed = 32 + 16 = 48 km/h.

17. Let the side of the smaller square be x cm, and the larger square be y cm.
Area smaller = x², area larger = y².
Given: 2x² - y² = 14 ...(1)
And: 2y² - 3x² = 20 ...(2)
Multiply (1) by 2: 4x² - 2y² = 28 ...(3)
Add (2) and (3): (2y² - 3x²) + (4x² - 2y²) = 20 + 28 → x² = 48.
From (1): 2*48 - y² = 14 → 96 - y² = 14 → y² = 82.
Sides: x = √48 = 4√3 cm, y = √82 cm.

18. Let the total marks be M and number of questions be N.
Had Anu scored 10 more marks: (M + 10)/N = 30 → M + 10 = 30N ...(1)
If she scored 9 times her marks: 9M/N = 15 → 9M = 15N → 3M = 5N → M = (5/3)N ...(2)
Substitute (2) into (1): (5/3)N + 10 = 30N
10 = 30N - (5/3)N = (90N - 5N)/3 = 85N/3
30 = 85N → N = 30/85 = 6/17. Not integer. Recheck.
Equation (2): 9M = 15N → M = (15/9)N = (5/3)N. Correct.
Equation (1): M + 10 = 30N.
Substitute: (5/3)N + 10 = 30N
10 = 30N - (5/3)N = (90N - 5N)/3 = 85N/3
30 = 85N → N = 30/85 = 6/17. Still not integer.
Perhaps "9 times" refers to something else? Rethink.
Alternative interpretation: "if she had scored 9 times the marks she actually got, then her average would have been 15".
So, 9M / N = 15 → M = (15/9)N = (5/3)N. Same as before.
But N must be integer. Check calculation again.
From (1): M = 30N - 10.
Set equal: 30N - 10 = (5/3)N
Multiply both sides by 3: 90N - 30 = 5N
85N = 30 → N = 30/85 = 6/17. Still not integer.
There might be an error in the problem or interpretation. Assume the equations are correct and proceed.
From M = 30N - 10 and M = (5/3)N, we have 30N - 10 = (5/3)N → 90N - 30 = 5N → 85N = 30 → N = 6/17. Not possible for number of questions.
Perhaps "9 times" means something else? Or typo in problem. Skip for now.

19. (a) Let the original number of students be x, and the amount each was to get be y.
Total collection: xy = 1000.
After 20 more join, total students = x + 20, each gets y - 40.
So, (x + 20)(y - 40) = 1000.
From first equation, y = 1000/x.
Substitute: (x + 20)(1000/x - 40) = 1000
(x + 20)(1000 - 40x)/x = 1000
(x + 20)(1000 - 40x) = 1000x
Expand left: x*1000 - 40x² + 20*1000 - 800x = 1000x
1000x - 40x² + 20000 - 800x = 1000x
-40x² + 200x + 20000 = 1000x
-40x² - 800x + 20000 = 0
Divide by -40: x² + 20x - 500 = 0
Solve: x = [-20 ± √(400 + 2000)]/2 = [-20 ± √2400]/2 = [-20 ± 20√6]/2 = -10 ± 10√6.
Not integer. Recheck.
From (x + 20)(y - 40) = 1000 and xy = 1000.
So, xy - 40x + 20y - 800 = 1000
But xy = 1000, so 1000 - 40x + 20y - 800 = 1000
200 - 40x + 20y = 1000
-40x + 20y = 800
Divide by 20: -2x + y = 40 → y = 2x + 40.
Now, xy = 1000 → x(2x + 40) = 1000
2x² + 40x - 1000 = 0
x² + 20x - 500 = 0
Discriminant: 400 + 2000 = 2400 = 400*6, so x = [-20 ± 20√6]/2 = -10 ± 10√6. Still not integer.
Perhaps the total is not 1000? Or typo. Assume it's correct and proceed.
(b) The value depicted is cooperation or unity.

20. Let the rate of pipe A be 1/a tanks per minute, pipe B be 1/b tanks per minute.
Together: 1/a + 1/b = 1/11.5 = 2/23.
Pipe A takes 5 minutes more than pipe B: a = b + 5.
So, 1/(b+5) + 1/b = 2/23.
(b + b + 5)/(b(b+5)) = 2/23
(2b + 5)/(b² + 5b) = 2/23
Cross multiply: 23(2b + 5) = 2(b² + 5b)
46b + 115 = 2b² + 10b
0 = 2b² + 10b - 46b - 115
2b² - 36b - 115 = 0
Use quadratic formula: b = [36 ± √(1296 + 920)] / 4 = [36 ± √2216]/4.
√2216 = √(4*554) = 2√554. Not nice.
Calculate discriminant: 36² = 1296, 4*2*115=920, sum 2216.
√2216 ≈ 47.08, so b = [36 + 47.08]/4 ≈ 83.08/4 ≈ 20.77, or negative.
Then a = b + 5 ≈ 25.77.
But perhaps exact value. 2216 = 4*554, and 554=2*277, so √2216=2√554.
So b = [36 + 2√554]/4 = [18 + √554]/2.
Not nice. Maybe the 11.5 is 23/2, and we can solve.
From 1/a + 1/b = 2/23 and a = b + 5.
As above, 2b² - 36b - 115 = 0.
Discriminant 1296 + 920 = 2216. Factor 2216: 2216 ÷ 2 = 1108, ÷2=554, ÷2=277. 277 is prime.
So no rational solution. Perhaps the problem has a typo, or accept irrational.
Time for pipe B: b = [36 + √2216]/4 minutes, pipe A: a = b + 5 minutes.