GSEB std 10 science solution for Gujarati check Subject Chapters Wise::
જવાબ : Let us say, the speed of the train be x km/hr. Time taken to cover 360 km = 360/x hr. As per the question given, ⇒ (x + 5)(360-1/x) = 360 ⇒ 360 – x + 1800-5/x = 360 ⇒ x2 + 5x + 10x – 1800 = 0 ⇒ x(x + 45) -40(x + 45) = 0 ⇒ (x + 45)(x – 40) = 0 ⇒ x = 40, -45 As we know, the value of speed cannot be negative. Therefore, the speed of train is 40 km/h.
જવાબ : x2 – 3x – m(m + 3) = 0
D = b2 – 4ac
D = (- 3)2 – 4(1) [-m(m + 3)]
= 9 + 4m (m + 3)
= 4m2 + 12m + 9 = (2m + 3)2
જવાબ : ay2 + ay + 3 = 0
a(1)2 + a(1) + 3 = 0
2a = -3
a = −32
y2 + y + b = 0
12 + 1 + b = 0
b = -2
∴ ab =(−32)(−2) = 3
જવાબ : The given quadratic equation can be written as, 3x2 + 2kx – 3 = 0
જવાબ : The given quadratic equation can be written as px\frac{1}{2} – 25–√ px + 15 = 0
Here a = p, b = – 25–√ p, c = 15
For equal roots, D = 0
D = b2 – 4ac – 0 …[∵ Equal roots
0 = (-25–√p)2 – 4 × p × 15
0 = 4 × 5p2 – 60p
0 = 20p2 – 60p => 20p2 = 60p
p = 60p20p = 3 ∴ p = 3
જવાબ : We have, mx (x – 7) + 49 = 0
mx2 – 7mx + 49 = 0
Here, a = m, b = – 7m, c = 49
D = b2 – 4ac = 0 …[For equal roots
⇒ (-7m)2 – 4(m) (49) = 0
⇒ 49m2 – 4m (49) = 0
⇒ 49m (m – 4) = 0
⇒ 49m = 0 or m – 4 = 0
m = 0 (rejected) or m = 4
∴ m = 4
જવાબ : Here 4x2 + 3x + 5 = 0
But (2x+34)2 cannot be negative for any real value of x.
જવાબ : We have, px (x – 3) + 9 = 0
px2 – 3px + 9 = 0 Here a = p, b = -3p,
D = 0
b2 – 4ac = 0 ⇒ (-3p)2 – 4(p)(9) = 0
⇒ 9p2 – 36p = 0
⇒ 9p (p – 4) = 0
⇒ 9p = 0 or p – 4= 0
p = 0 (rejected) or p = 4
∴ p = 4 ……..(∵ Coeff. of x2 cannot be zero
36x2 – 12ax + (a2 – b2) = 0 (2011OD)
જવાબ : We have, 36x2 – 12ax + (a2 – b2) = 0
⇒ (36x2 – 12ax + a2) – b2 = 0
⇒ [(6x)2 – 2(6x)(a) + (a)2] – b2 = 0
⇒ (6x – a)2 – (b)2 = 0 …[∵ x2 – 2xy + y2 = (x – y)2
⇒ (6x – a + b) (6x – a – b) = 0 „[∵ x2 – y2 = (x + y)(x – y)
⇒ 6x – a + b = 0 or 6x – a – b = 0
⇒ 6x = a – b or 6x = a + b
⇒ x = a−b6 or a+b6
જવાબ : We have , px(x – 2) + 6 = 0
px2 – 2px + 6 = 0, p ≠ 0
Two equal roots …[Given
b2 – 4ac = 0 ….[a = p, b = -2p, c = 6
∴ (-2p)2 – 4(p)(6) = 0
4p2 – 24p = 0 ⇒ 4p(p – 6) = 0
4p = 0 or p – 6 = 0
p = 0 (rejected) or p = 6
Since p cannot be equal to 0.
…[Standard form of a quad. eq. ax2 + bx + c = 0, a ≠ 0
∴ P = 6
જવાબ : We have, x2 – 4kx + k = 0
Here a = 1, b = -4k:, c = k D = 0 …[Since, Equal roots
As b2 – 4ac = 0
⇒ (-4k)2 – 4(1) (k) = 0
⇒ 16k2 – 4k = 0 ⇒ 4k(4k – 1) = 0
⇒ 4k = 0 or 4k – 1 = 0
k = 0 (rejected) or 4k = 1
∴ k = ¼
જવાબ : 43–√x2 + 5x – 23–√ = 0
43–√x2 + 8x – 3x – 23–√ =0
4x(3–√x + 2) – 3–√(3–√x + 2) = 0
(3–√x + 2)(4x – 3–√) = 0
3–√x + 2 = 0 or 4x – 3–√ = 0
જવાબ : 2–√x2 + 7x + 52–√ = 0
⇒ 2–√x2 + 5x + 2x + 52–√ = 0
x(2–√x + 5) + 2–√(2–√x + 5) = 0
(2–√x + 5)(x + 2–√) = 0
જવાબ : We have, 2x2 + ax – a2 = 0
2x2 + 2ax – ax – a2 = 0
2x(x + a) – a(x + a) = 0
(x + a) (2x – a) = 0
x + a = 0 or 2x – a = 0
∴ x = -a or x = a2
Alternatively:
First calculate D = b2 – 4ac
Then apply x = −b±D√2a
We get x = -a, x = a/2
જવાબ : We have, x2 + k(2x + k – 1) + 2 = 0
x2 + 2kx + k2 – k + 2 = 0
Here a = 1, b = 2k, c = k2 – k + 2
D = 0 …[real and equal roots
∴ b2 – 4ac = 0
⇒ (2k)2 – 4 × 1(k2 – k + 2) = 0
⇒ 4k2 – 4 (k2 – k + 2) = 0
⇒ 4(k2 – k2 + k – 2) = 0 ⇒ 4(k – 2) = 0
⇒ k – 2 = 0 ⇒ k = 2
જવાબ : The given quadratic equation can be written as,
4x2 – 4a2x + (a44 – b4) = 0
(4x2 – 4a2x + a4) – b4 = 0
or (2x – a2)2 – (b2)2 = 0
⇒ (2x – a2 + b2) (2x – a2 – b2) = 0
⇒ (2x – a2 + b2) = 0 or (2x – a2 – b2) = 0
∴ x = a2−b22 or x = a2+b22
જવાબ : Given: 4x2 + px + 3 = 0
Here a = 4, b = p. (= 3 … [Equal roots
D = 0 (Equal roots)
As b2 – 4ac = 0
∴ (p)2 – 4(4)(3) = 0
= p2 – 48 = 0 ⇒ p2 = 48
∴ p = ±16×3−−−−−√=±43–√
જવાબ : Given quadratic equation can be written as
x2 – 2ax – 4b2 + a2 = 0.
(x2 – 2ax + a2) – 4b2 = 0 or (x – a)2 – (2b)2 = 0
As we know,
[a2 – b2 = (a + b)(a – b)]
∴ (x – a + 2b) (x – a – 2b) = 0
⇒ x – a + 2b = 0 or x – a – 2b = 0
⇒ x = a – 2b or x = a + 2b
⇒ x = a – 2b and x = a + 2b
જવાબ : The given quadratic equation can be written as
4x2 + 4bx + b2 – a22 = 0
⇒ (2x + b)2 – (a)2 = 0
⇒ (2x + b + a) (2x + b – a) = 0 …[x2 – y2 = (x + y)(x – y)
⇒ (2x + b + a) = 0 or (2x + b – a) = 0
⇒ 2x = -(a + b) or 2x = (a – b)
જવાબ : The given quadratic equation can be written as
(9x2 – 6b2x + b4) – a4 = 0
⇒ (3x – b2)2 – (a2)2 = 0
⇒ (3x – b2 + a2) (3x – b2 – a2) = 0 …[:: x2 – y2 = (x + y) (x – y)
⇒ 3x – b2 + a2 = 0 or 3x – b2 – a2 = 0
⇒ 3x = b2 – a2 or 3x = b2 + a2
જવાબ : We have, ax2 + 7x + b = 0
Here ‘a’ = a, ‘b’ = 7, ‘c’ = b
Now, α = 23 and β = -3 … [Given
જવાબ : Given equation is px2 – 14x + 8 = 0.
Here a = p b = -14 c = 8
Let roots be a and 6α.
જવાબ : We have, 2x2 + px – 15 =0
Since (-5) is a root of the given equation
∴ 2(-5)2 + p(-5) – 15 = 0
⇒ 2(25) – 5p – 15 = 0
⇒ 50 – 15 = 5p
⇒ 35 = 5p ⇒ p = 7 …(i)
Now, p(x2 + x) + k ⇒ px2 + px + k = 0
7x2 + 7x + k = 0 …[From (i)
Here, a = 7, b = 7, c = k
D = 0 …[Roots are equal
b2 – 4ac = 0
⇒ (7)2 – 4(7)k = 0 ⇒ 49 – 28k = 0
⇒ 49 = 28k ∴ k = 4928=74
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જવાબ : Let us say, the marks of Shefali in Maths be x.
Then, the marks in English will be 30 – x. As per the given question, (x + 2)(30 – x – 3) = 210 (x + 2)(27 – x) = 210 ⇒ -x2 + 25x + 54 = 210 ⇒ x2 – 25x + 156 = 0 ⇒ x2 – 12x – 13x + 156 = 0 ⇒ x(x – 12) -13(x – 12) = 0 ⇒ (x – 12)(x – 13) = 0 ⇒ x = 12, 13 Therefore, if the marks in Maths are 12, then marks in English will be 30 – 12 = 18 and the marks in Maths are 13, then marks in English will be 30 – 13 = 17.જવાબ : Let us say, present age of Rahman is x years.
Three years ago, Rehman’s age was (x – 3) years. Five years after, his age will be (x + 5) years. Given, the sum of the reciprocals of Rehman’s ages 3 years ago and after 5 years is equal to 1/3. ∴ 1/x-3 + 1/x-5 = 1/3 (x+5+x-3)/(x-3)(x+5) = 1/3 (2x+2)/(x-3)(x+5) = 1/3 ⇒ 3(2x + 2) = (x-3)(x+5) ⇒ 6x + 6 = x2 + 2x – 15 ⇒ x2 – 4x – 21 = 0 ⇒ x2 – 7x + 3x – 21 = 0 ⇒ x(x – 7) + 3(x – 7) = 0 ⇒ (x – 7)(x + 3) = 0 ⇒ x = 7, -3 As we know, age cannot be negative. Therefore, Rahman’s present age is 7 years.જવાબ : Let us say, the number of articles produced be x.
Therefore, cost of production of each article = Rs (2x + 3) Given, total cost of production is Rs.90 ∴ x(2x + 3) = 90 ⇒ 2x2 + 3x – 90 = 0 ⇒ 2x2 + 15x -12x – 90 = 0 ⇒ x(2x + 15) -6(2x + 15) = 0 ⇒ (2x + 15)(x – 6) = 0 Thus, either 2x + 15 = 0 or x – 6 = 0 ⇒ x = -15/2 or x = 6 As the number of articles produced can only be a positive integer, therefore, x can only be 6. Hence, number of articles produced = 6 Cost of each article = 2 × 6 + 3 = Rs 15.જવાબ : Let us say, the base of the right triangle be x cm.
Given, the altitude of right triangle = (x – 7) cm From Pythagoras theorem, we know, Base2 + Altitude2 = Hypotenuse2 ∴ x2 + (x – 7)2 = 132 ⇒ x2 + x2 + 49 – 14x = 169 ⇒ 2x2 – 14x – 120 = 0 ⇒ x2 – 7x – 60 = 0 ⇒ x2 – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x – 12)(x + 5) = 0 Thus, either x – 12 = 0 or x + 5 = 0, ⇒ x = 12 or x = – 5 Since sides cannot be negative, x can only be 12. Therefore, the base of the given triangle is 12 cm and the altitude of this triangle will be (12 – 7) cm = 5 cm.જવાબ : Let us say, the two consecutive positive integers be x and x + 1.
Therefore, as per the given questions, x2 + (x + 1)2 = 365 ⇒ x2 + x2 + 1 + 2x = 365 ⇒ 2x2 + 2x – 364 = 0 ⇒ x2 + x – 182 = 0 ⇒ x2 + 14x – 13x – 182 = 0 ⇒ x(x + 14) -13(x + 14) = 0 ⇒ (x + 14)(x – 13) = 0 Thus, either, x + 14 = 0 or x – 13 = 0, ⇒ x = – 14 or x = 13 since, the integers are positive, so x can be 13, only. ∴ x + 1 = 13 + 1 = 14 Therefore, two consecutive positive integers will be 13 and 14.જવાબ : Let us say, first number be x and the second number is 27 – x.
Therefore, the product of two numbers x(27 – x) = 182 ⇒ x2 – 27x – 182 = 0 ⇒ x2 – 13x – 14x + 182 = 0 ⇒ x(x – 13) -14(x – 13) = 0 ⇒ (x – 13)(x -14) = 0 Thus, either, x = -13 = 0 or x – 14 = 0 ⇒ x = 13 or x = 14 Therefore, if first number = 13, then second number = 27 – 13 = 14 And if first number = 14, then second number = 27 – 14 = 13 Hence, the numbers are 13 and 14.જવાબ : Let the speed of the stream = km/h
Speed of the boat in still water = 20 km/h
Now, speed of boat during downstream = (20 + x) km/h
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જવાબ : Let the length and breadth of the park be l and b.
Perimeter of the rectangular park = 2 (l + b) = 80 So, l + b = 40 Or, b = 40 – l Area of the rectangular park = l×b = l(40 – l) = 40l – l2 = 400 l2 – 40l + 400 = 0, which is a quadratic equation. Comparing the equation with ax2 + bx + c = 0, we get a = 1, b = -40, c = 400 Since, Discriminant = b2 – 4ac =(-40)2 – 4 × 400 = 1600 – 1600 = 0 Thus, b2 – 4ac = 0 Therefore, this equation has equal real roots. Hence, the situation is possible. Root of the equation, l = –b/2a l = (40)/2(1) = 40/2 = 20 Therefore, length of rectangular park, l = 20 m And breadth of the park, b = 40 – l = 40 – 20 = 20 m.
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જવાબ : Let’s say, the age of one friend be x years.
Then, the age of the other friend will be (20 – x) years. Four years ago, Age of First friend = (x – 4) years Age of Second friend = (20 – x – 4) = (16 – x) years As per the given question, we can write, (x – 4) (16 – x) = 48 16x – x2 – 64 + 4x = 48 – x2 + 20x – 112 = 0 x2 – 20x + 112 = 0 Comparing the equation with ax2 + bx + c = 0, we get a = 1, b = -20 and c = 112 Discriminant = b2 – 4ac = (-20)2 – 4 × 112 = 400 – 448 = -48 b2 – 4ac < 0 Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.જવાબ : Let the breadth of mango grove be l.
Length of mango grove will be 2l. Area of mango grove = (2l) (l)= 2l2 2l2 = 800 l2 = 800/2 = 400 l2 – 400 =0 Comparing the given equation with ax2 + bx + c = 0, we get a = 1, b = 0, c = 400 As we know, Discriminant = b2 – 4ac => (0)2 – 4 × (1) × ( – 400) = 1600 Here, b2 – 4ac > 0 Thus, the equation will have real roots. And hence, the desired rectangular mango grove can be designed. l = ±20 As we know, the value of length cannot be negative. Therefore, breadth of mango grove = 20 m Length of mango grove = 2 × 20 = 40 m(i) 2x2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0
જવાબ : (i) 2x2 + kx + 3 = 0
Comparing the given equation with ax2 + bx + c = 0, we get, a = 2, b = k and c = 3 As we know, Discriminant = b2 – 4ac = (k)2 – 4(2) (3) = k2 – 24 For equal roots, we know, Discriminant = 0 k2 – 24 = 0 k2 = 24 k = ±√24 = ±2√6 (ii) kx(x – 2) + 6 = 0 or kx2 – 2kx + 6 = 0 Comparing the given equation with ax2 + bx + c = 0, we get a = k, b = – 2k and c = 6 We know, Discriminant = b2 – 4ac = ( – 2k)2 – 4 (k) (6) = 4k2 – 24k For equal roots, we know, b2 – 4ac = 0 4k2 – 24k = 0 4k (k – 6) = 0 Either 4k = 0 or k = 6 = 0 k = 0 or k = 6 However, if k = 0, then the equation will not have the terms ‘x2‘ and ‘x‘. Therefore, if this equation has two equal roots, k should be 6 only.જવાબ : Let the sides of the two squares be x m and y m.
Therefore, their perimeter will be 4x and 4y respectively And area of the squares will be x2 and y2 respectively. Given, 4x – 4y = 24 x – y = 6 x = y + 6 Also, x2 + y2 = 468 ⇒ (6 + y2) + y2 = 468 ⇒ 36 + y2 + 12y + y2 = 468 ⇒ 2y2 + 12y + 432 = 0 ⇒ y2 + 6y – 216 = 0 ⇒ y2 + 18y – 12y – 216 = 0 ⇒ y(y +18) -12(y + 18) = 0 ⇒ (y + 18)(y – 12) = 0 ⇒ y = -18, 12 As we know, the side of a square cannot be negative. Hence, the sides of the squares are 12 m and (12 + 6) m = 18 m.જવાબ : Let us say, the average speed of passenger train = x km/h.
Average speed of express train = (x + 11) km/h Given, time taken by the express train to cover 132 km is 1 hour less than the passenger train to cover the same distance. Therefore, (132/x) – (132/(x+11)) = 1 132(x+11-x)/(x(x+11)) = 1 132 × 11 /(x(x+11)) = 1 ⇒ 132 × 11 = x(x + 11) ⇒ x2 + 11x – 1452 = 0 ⇒ x2 + 44x -33x -1452 = 0 ⇒ x(x + 44) -33(x + 44) = 0 ⇒ (x + 44)(x – 33) = 0 ⇒ x = – 44, 33 As we know, Speed cannot be negative. Therefore, the speed of the passenger train will be 33 km/h and thus, the speed of the express train will be 33 + 11 = 44 km/h.
hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
જવાબ : Let the time taken by the smaller pipe to fill the tank = x hr.
Time taken by the larger pipe = (x – 10) hr Part of tank filled by smaller pipe in 1 hour = 1/x Part of tank filled by larger pipe in 1 hour = 1/(x – 10) As given, the tank can be filled in = 75/8 hours by both the pipes together.
Therefore,
1/x + 1/x-10 = 8/75
x-10+x/x(x-10) = 8/75
⇒ 2x-10/x(x-10) = 8/75
⇒ 75(2x – 10) = 8x2 – 80x
⇒ 150x – 750 = 8x2 – 80x
⇒ 8x2 – 230x +750 = 0
⇒ 8x2 – 200x – 30x + 750 = 0
⇒ 8x(x – 25) -30(x – 25) = 0
⇒ (x – 25)(8x -30) = 0
⇒ x = 25, 30/8
Time taken by the smaller pipe cannot be 30/8 = 3.75 hours, as the time taken by the larger pipe will become negative, which is logically not possible.
Therefore, time taken individually by the smaller pipe and the larger pipe will be 25 and 25 – 10 =15 hours respectively.
જવાબ : Let us say, the larger and smaller number be x and y respectively.
As per the question given, x2 – y2 = 180 and y2 = 8x ⇒ x2 – 8x = 180 ⇒ x2 – 8x – 180 = 0 ⇒ x2 – 18x + 10x – 180 = 0 ⇒ x(x – 18) +10(x – 18) = 0 ⇒ (x – 18)(x + 10) = 0 ⇒ x = 18, -10 However, the larger number cannot considered as negative number, as 8 times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible. Therefore, the larger number will be 18 only. x = 18 ∴ y2 = 8x = 8 × 18 = 144 ⇒ y = ±√144 = ±12 ∴ Smaller number = ±12 Therefore, the numbers are 18 and 12 or 18 and -12.જવાબ : Let us say, the shorter side of the rectangle be x m.
Then, larger side of the rectangle = (x + 30) m
⇒ x2 + (x + 30)2 = (x + 60)2
⇒ x2 + x2 + 900 + 60x = x2 + 3600 + 120x
⇒ x2 – 60x – 2700 = 0
⇒ x2 – 90x + 30x – 2700 = 0
⇒ x(x – 90) + 30(x -90) = 0
⇒ (x – 90)(x + 30) = 0
⇒ x = 90, -30
However, side of the field cannot be negative. Therefore, the length of the shorter side will be 90 m.
and the length of the larger side will be (90 + 30) m = 120 m.
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