Vidya Sanskar, Munger

Class 10 study materials part 5

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Science
1) what is Matter? What is its characteristics?
2) What is the Mole concept?
3) Write their definition.
a. Atomic mass. b. Molecular mass c. Molar mass
4) Write and remember Atomic Masses.
5) How do calculating Molecular mass? With their example.
6) What is relation between molar and molar volumes?
7) Calculate number of moles in 392g of H2SO4.
8) Calculate number of moles in 44.8l of Co2 at STP.
9) Calculate number of atoms of the consitutent element in 53g of Co2Co3.
10) Calculate number of moles in 350cm3 of Co2 at 273k and 2atm pressure.
11) Write and remember periodic table.

Exercise 4.1 Page: 73
1. Check whether the following are quadratic equations:

(i) (x + 1)2 = 2(x – 3)

(ii) x2 – 2x = (–2) (3 – x)

(iii) (x – 2)(x + 1) = (x – 1)(x + 3)

(iv) (x – 3)(2x +1) = x(x + 5)

(v) (2x – 1)(x – 3) = (x + 5)(x – 1)

(vi) x2 + 3x + 1 = (x – 2)2

(vii) (x + 2)3 = 2x (x2 – 1)

(viii) x3 – 4×2 – x + 1 = (x – 2)3

Solutions:

(i) Given,

(x + 1)2 = 2(x – 3)

By using the formula for (a+b)2 = a2+2ab+b2

⇒ x2 + 2x + 1 = 2x – 6

⇒ x2 + 7 = 0

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

(ii)Given, x2 – 2x = (–2) (3 – x)

By using the formula for (a+b)2 = a2+2ab+b2

⇒ x2 – 2x = -6 + 2x

⇒ x2 – 4x + 6 = 0

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

(iii)Given, (x – 2)(x + 1) = (x – 1)(x + 3)

By using the formula for (a+b)2 = a2+2ab+b2

⇒ x2 – x – 2 = x2 + 2x – 3

⇒ 3x – 1 = 0

Since the above equation is not in the form of ax2 + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(iv)Given, (x – 3)(2x +1) = x(x + 5)

By using the formula for (a+b)2=a2+2ab+b2

⇒ 2×2 – 5x – 3 = x2 + 5x

⇒ x2 – 10x – 3 = 0

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

(v)Given, (2x – 1)(x – 3) = (x + 5)(x – 1)

By using the formula for (a+b)2=a2+2ab+b2

⇒ 2×2 – 7x + 3 = x2 + 4x – 5

⇒ x2 – 11x + 8 = 0

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

(vi)Given, x2 + 3x + 1 = (x – 2)2

By using the formula for (a+b)2=a2+2ab+b2

⇒ x2 + 3x + 1 = x2 + 4 – 4x

⇒ 7x – 3 = 0

Since the above equation is not in the form of ax2 + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(vii)Given, (x + 2)3 = 2x(x2 – 1)

By using the formula for (a+b)2 = a2+2ab+b2

⇒ x3 + 8 + x2 + 12x = 2×3 – 2x

⇒ x3 + 14x – 6×2 – 8 = 0

Since the above equation is not in the form of ax2 + bx + c = 0.

Therefore, the given equation is not a quadratic equation.

(viii)Given, x3 – 4×2 – x + 1 = (x – 2)3

By using the formula for (a+b)2 = a2+2ab+b2

⇒ x3 – 4×2 – x + 1 = x3 – 8 – 6×2 + 12x

⇒ 2×2 – 13x + 9 = 0

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

2. Represent the following situations in the form of quadratic equations:

The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
The product of two consecutive positive integers is 306. We need to find the integers.
Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken
Solutions:

(1)Let us consider,

Breadth of the rectangular plot = x m

Thus, the length of the plot = (2x + 1) m.

As we know,

Area of rectangle = length × breadth = 528 m2

Putting the value of length and breadth of the plot in the formula, we get,

(2x + 1) × x = 528

⇒ 2×2 + x =528

⇒ 2×2 + x – 528 = 0

Therefore, the length and breadth of plot, satisfies the quadratic equation, 2×2 + x – 528 = 0, which is the required representation of the problem mathematically.

(2)Let us consider,

The first integer number = x

Thus, the next consecutive positive integer will be = x + 1

Product of two consecutive integers = x × (x +1) = 306 ⇒ x2 + x = 306

⇒ x2 + x – 306 = 0

Therefore, the two integers x and x+1, satisfies the quadratic equation, x2 + x – 306 = 0, which is the required representation of the problem mathematically.

3) Let us consider,

Age of Rohan’s = x years

Therefore, as per the given question,

Rohan’s mother’s age = x + 26

After 3 years,

Age of Rohan’s = x + 3

Age of Rohan’s mother will be = x + 26 + 3 = x + 29

The product of their ages after 3 years will be equal to 360, such that

(x + 3)(x + 29) = 360

⇒ x2 + 29x + 3x + 87 = 360

⇒ x2 + 32x + 87 – 360 = 0

⇒ x2 + 32x – 273 = 0

Therefore, the age of Rohan and his mother, satisfies the quadratic equation, x2 + 32x – 273 = 0, which is the required representation of the problem mathematically.

4) Let us consider,

The speed of train = x km/h

And

Time taken to travel 480 km = 480/x km/hr

As per second condition, the speed of train = (x – 8) km/h

Also given, the train will take 3 hours to cover the same distance.

Therefore, time taken to travel 480 km = 480/(x+3) km/h

As we know,

Speed × Time = Distance

Therefore,

(x – 8)(480/(x + 3) = 480

⇒ 480 + 3x – 3840/x – 24 = 480

⇒ 3x – 3840/x = 24

⇒ 3×2 – 8x – 1280 = 0

Therefore, the speed of the train, satisfies the quadratic equation, 3×2 – 8x – 1280 = 0, which is the required representation of the problem mathematically.

Exercise 4.2 Page: 76
1. Find the roots of the following quadratic equations by factorisation:

(i) x2 – 3x – 10 = 0

(ii) 2×2 + x – 6 = 0

(iii) √2 x2 + 7x + 5√2 = 0

(iv) 2×2 – x +1/8 = 0

(v) 100×2 – 20x + 1 = 0

Solutions:

(i) Given, x2 – 3x – 10 =0

Taking LHS,

=>x2 – 5x + 2x – 10

=>x(x – 5) + 2(x – 5)

=>(x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 = 0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

=> x = 5 or x = -2

(ii)Given, 2×2 + x – 6 = 0

Taking LHS,

=> 2×2 + 4x – 3x – 6

=> 2x(x + 2) – 3(x + 2)

=> (x + 2)(2x – 3)

The roots of this equation, 2×2 + x – 6=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x + 2 = 0 or 2x – 3 = 0

=> x = -2 or x = 3/2

(iii)√2 x2 + 7x + 5√2=0

Taking LHS,

=> √2 x2 + 5x + 2x + 5√2

=> x (√2x + 5) + √2(√2x + 5)= (√2x + 5)(x + √2)

The roots of this equation, √2 x2 + 7x + 5√2=0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, √2x + 5 = 0 or x + √2 = 0

=> x = -5/√2 or x = -√2

(iv) 2×2 – x +1/8 = 0

Taking LHS,

=1/8 (16×2 – 8x + 1)

= 1/8 (16×2 – 4x -4x + 1)

= 1/8 (4x(4x – 1) -1(4x – 1))

= 1/8 (4x – 1)2

The roots of this equation, 2×2 – x + 1/8 = 0, are the values of x for which (4x – 1)2= 0

Therefore, (4x – 1) = 0 or (4x – 1) = 0

⇒ x = 1/4 or x = 1/4

(v)Given, 100×2 – 20x + 1=0

Taking LHS,

= 100×2 – 10x – 10x + 1

= 10x(10x – 1) -1(10x – 1)

= (10x – 1)2

The roots of this equation, 100×2 – 20x + 1=0, are the values of x for which (10x – 1)2= 0

∴ (10x – 1) = 0 or (10x – 1) = 0

⇒x = 1/10 or x = 1/10

2. Solve the problems given in Example 1.

Represent the following situations mathematically:

(i).John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124. We would like to find out how many marbles they had to start with.

(ii).A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ` 750. We would like to find out the number of toys produced on that day.

Solutions:

(i)

Let us say, the number of marbles John have = x.

Therefore, number of marble Jivanti have = 45 – x

After losing 5 marbles each,

Number of marbles John have = x – 5

Number of marble Jivanti have = 45 – x – 5 = 40 – x

Given that the product of their marbles is 124.

∴ (x – 5)(40 – x) = 124

⇒ x2 – 45x + 324 = 0

⇒ x2 – 36x – 9x + 324 = 0

⇒ x(x – 36) -9(x – 36) = 0

⇒ (x – 36)(x – 9) = 0

Thus, we can say,

x – 36 = 0 or x – 9 = 0

⇒ x = 36 or x = 9

Therefore,

If, John’s marbles = 36,

Then, Jivanti’s marbles = 45 – 36 = 9

And if John’s marbles = 9,

Then, Jivanti’s marbles = 45 – 9 = 36

(ii)

Let us say, number of toys produced in a day be x.

Therefore, cost of production of each toy = Rs(55 – x)

Given, total cost of production of the toys = Rs 750

∴ x(55 – x) = 750

⇒ x2 – 55x + 750 = 0

⇒ x2 – 25x – 30x + 750 = 0

⇒ x(x – 25) -30(x – 25) = 0

⇒ (x – 25)(x – 30) = 0

Thus, either x -25 = 0 or x – 30 = 0

⇒ x = 25 or x = 30

Hence, the number of toys produced in a day, will be either 25 or 30.

3. Find two numbers whose sum is 27 and product is 182.

Solutions: 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.

4. Find two consecutive positive integers, sum of whose squares is 365.

Solutions: 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

⇒ 2×2 + 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.

5. The altitude of a right triangle is 7 cm less than its base. If the hypotenuse is 13 cm, find the other two sides.

Solutions: 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

⇒ 2×2 – 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.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs.90, find the number of articles produced and the cost of each article.

Solutions: 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

⇒ 2×2 + 3x – 90 = 0

⇒ 2×2 + 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.

Exercise 4.3 Page: 87
1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

(i) 2×2 – 7x +3 = 0

(ii) 2×2 + x – 4 = 0
(iii) 4×2 + 4√3x + 3 = 0

(iv) 2×2 + x + 4 = 0

Solutions:

(i) 2×2 – 7x + 3 = 0

⇒ 2×2 – 7x = – 3

Dividing by 2 on both sides, we get

⇒ x2 -7x/2 = -3/2

⇒ x2 -2 × x ×7/4 = -3/2

On adding (7/4)2 to both sides of equation, we get

⇒ (x)2-2×x×7/4 +(7/4)2 = (7/4)2-3/2

⇒ (x-7/4)2 = (49/16) – (3/2)

⇒(x-7/4)2 = 25/16

⇒(x-7/4)2 = ±5/4

⇒ x = 7/4 ± 5/4

⇒ x = 7/4 + 5/4 or x = 7/4 – 5/4

⇒ x = 12/4 or x = 2/4

⇒ x = 3 or x = 1/2

(ii) 2×2 + x – 4 = 0

⇒ 2×2 + x = 4

Dividing both sides of the equation by 2, we get

⇒ x2 +x/2 = 2

Now on adding (1/4)2 to both sides of the equation, we get,

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = 2 + (1/4)2

⇒ (x + 1/4)2 = 33/16

⇒ x + 1/4 = ± √33/4

⇒ x = ± √33/4 – 1/4

⇒ x = ± √33-1/4

Therefore, either x = √33-1/4 or x = -√33-1/4

(iii) 4×2 + 4√3x + 3 = 0

Converting the equation into a2+2ab+b2 form, we get,

⇒ (2x)2 + 2 × 2x × √3 + (√3)2 = 0

⇒ (2x + √3)2 = 0

⇒ (2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2.

(iv) 2×2 + x + 4 = 0

⇒ 2×2 + x = -4

Dividing both sides of the equation by 2, we get

⇒ x2 + 1/2x = 2

⇒ x2 + 2 × x × 1/4 = -2

By adding (1/4)2 to both sides of the equation, we get

⇒ (x)2 + 2 × x × 1/4 + (1/4)2 = (1/4)2 – 2

⇒ (x + 1/4)2 = 1/16 – 2

⇒ (x + 1/4)2 = -31/16

As we know, the square of numbers cannot be negative.

Therefore, there is no real root for the given equation, 2×2 + x + 4 = 0.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

(i) 2×2 – 7x + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we get,

a = 2, b = -7 and c = 3

By using quadratic formula, we get,

Ncert solutions class 10 chapter 4-1

⇒ x = (7±√(49 – 24))/4

⇒ x = (7±√25)/4

⇒ x = (7±5)/4

⇒ x = (7+5)/4 or x = (7-5)/4

⇒ x = 12/4 or 2/4

∴ x = 3 or 1/2

(ii) 2×2 + x – 4 = 0

On comparing the given equation with ax2 + bx + c = 0, we get,

a = 2, b = 1 and c = -4

By using quadratic formula, we get,

Ncert solutions class 10 chapter 4-2

⇒x = -1±√1+32/4

⇒x = -1±√33/4

∴ x = -1+√33/4 or x = -1-√33/4

(iii) 4×2 + 4√3x + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we get

a = 4, b = 4√3 and c = 3

By using quadratic formula, we get,

Ncert solutions class 10 chapter 4-3

⇒ x = -4√3±√48-48/8

⇒ x = -4√3±0/8

∴ x = -√3/2 or x = -√3/2

(iv) 2×2 + x + 4 = 0

On comparing the given equation with ax2 + bx + c = 0, we get,

a = 2, b = 1 and c = 4

By using quadratic formula, we get

Ncert solutions class 10 chapter 4-4

⇒ x = -1±√1-32/4

⇒ x = -1±√-31/4

As we know, the square of a number can never be negative. Therefore, there is no real solution for the given equation.

3. Find the roots of the following equations:

(i) x-1/x = 3, x ≠ 0
(ii) 1/x+4 – 1/x-7 = 11/30, x = -4, 7

Solutions:

(i) x-1/x = 3

⇒ x2 – 3x -1 = 0

On comparing the given equation with ax2 + bx + c = 0, we get

a = 1, b = -3 and c = -1

By using quadratic formula, we get,

Ncert solutions class 10 chapter 4-5

⇒ x = 3±√9+4/2

⇒ x = 3±√13/2

∴ x = 3+√13/2 or x = 3-√13/2

(ii) 1/x+4 – 1/x-7 = 11/30

⇒ x-7-x-4/(x+4)(x-7) = 11/30

⇒ -11/(x+4)(x-7) = 11/30

⇒ (x+4)(x-7) = -30

⇒ x2 – 3x – 28 = 30

⇒ x2 – 3x + 2 = 0

We can solve this equation by factorization method now,

⇒ x2 – 2x – x + 2 = 0

⇒ x(x – 2) – 1(x – 2) = 0

⇒ (x – 2)(x – 1) = 0

⇒ x = 1 or 2

4. The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is 1/3. Find his present age.

Solutions: 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.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Solutions: 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.

6. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.

Solutions: Let us say, the shorter side of the rectangle be x m.

Then, larger side of the rectangle = (x + 30) m

Ncert solutions class 10 chapter 4-6

As given, the length of the diagonal is = x + 30 m

Therefore,

Ncert solutions class 10 chapter 4-7

⇒ 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.

7. The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Solution: 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.

8. A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.

Solution: 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.

9. Two water taps together can fill a tank in 9 3/8 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.

Solution: 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 9 3/8 = 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) = 8×2 – 80x

⇒ 150x – 750 = 8×2 – 80x

⇒ 8×2 – 230x +750 = 0

⇒ 8×2 – 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.

10. An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains.

Solution: 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.

11. Sum of the areas of two squares is 468 m2. If the difference of their perimeters is 24 m, find the sides of the two squares.

Solutions: 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.

Exercise 4.4 Page: 91
1. Find the nature of the roots of the following quadratic equations. If the real roots exist, find them;
(i) 2×2 – 3x + 5 = 0
(ii) 3×2 – 4√3x + 4 = 0
(iii) 2×2 – 6x + 3 = 0

Solutions:

(i) Given,

2×2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 2, b = -3 and c = 5

We know, Discriminant = b2 – 4ac

= ( – 3)2 – 4 (2) (5) = 9 – 40

= – 31

As you can see, b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2×2 – 3x + 5 = 0.

(ii) 3×2 – 4√3x + 4 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 3, b = -4√3 and c = 4

We know, Discriminant = b2 – 4ac

= (-4√3)2 – 4(3)(4)

= 48 – 48 = 0

As b2 – 4ac = 0,

Real roots exist for the given equation and they are equal to each other.

Hence the roots will be –b/2a and –b/2a.

–b/2a = -(-4√3)/2×3 = 4√3/6 = 2√3/3 = 2/√3

Therefore, the roots are 2/√3 and 2/√3.

(iii) 2×2 – 6x + 3 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 2, b = -6, c = 3

As we know, Discriminant = b2 – 4ac

= (-6)2 – 4 (2) (3)

= 36 – 24 = 12

As b2 – 4ac > 0,

Therefore, there are distinct real roots exist for this equation, 2×2 – 6x + 3 = 0.

Ncert solutions class 10 chapter 4-8

=( -(-6) ± √(-62-4(2)(3)) )/ 2(2)

= (6±2√3 )/4

= (3±√3)/2

Therefore the roots for the given equation are (3+√3)/2 and (3-√3)/2

2. Find the values of k for each of the following quadratic equations, so that they have two equal roots.
(i) 2×2 + kx + 3 = 0
(ii) kx (x – 2) + 6 = 0

Solutions:

(i) 2×2 + 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.

3. Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2? If so, find its length and breadth.

Solutions: 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

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.

Solution: 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.

5. Is it possible to design a rectangular park of perimeter 80 and area 400 m2? If so find its length and breadth.

Solution: 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.

Quadratic Equations
A 1 mark question was asked from Chapter 4 Quadratic Equations in the year 2018. However, in the year 2017, a total of 13 marks were asked from the topic Quadratic Equations. Therefore, students need to have a thorough understanding of the topic. The topics and sub-topics provided in this chapter include:

4.1 Introduction

If we equate the polynomial ax2+ bx + c, a ≠ 0 to zero, we get a quadratic equation. Quadratic equations come up when we deal with many real-life situations. In this chapter, students will study quadratic equations, and various ways of finding their roots. They will also see some applications of quadratic equations in daily life situations.

4.2 Quadratic Equations

A quadratic equation in the variable x is an equation of the form ax2+ bx + c = 0, where a, b, c are real numbers, a ≠ 0. In fact, any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is a quadratic equation. But when we write the terms of p(x) in descending order of their degrees, then we get the standard form of the equation. That is, ax2+ bx + c = 0, a ≠ 0 is called the standard form of a quadratic equation.

4.3 Solution of Quadratic Equations by Factorization

A real number α is called a root of the quadratic equation ax2 + bx + c = 0, a ≠ 0 if a α2+ bα + c = 0. We also say that x = α is a solution of the quadratic equation, or that α satisfies the quadratic equation. Note that the zeroes of the quadratic polynomial ax2+ bx + c and the roots of the quadratic equation ax2 + bx + c = 0 are the same.

4.4 Solution of a Quadratic Equation by Completing the Square

Finding the value that makes a quadratic equation a square trinomial is called completing the square. The square trinomial can then be solved easily by factorizing.

4.5 Nature of Roots

If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, there are no real roots for the given quadratic equation in this case. Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation. So, a quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2– 4ac = 0,
(iii) no real roots, if b2– 4ac < 0.

Questions

In a quadratic equation, x represents an unknown form and a, b, c are the known values. An equation to be quadratic “a” should not be equal to 0. The equation is of the form ax2 + bx + c = 0. The values of a, b, and c are always real numbers.

A quadratic equation can be calculated by completing the square.

A quadratic equation has:

Two different real roots.
No real roots.
Two equal roots.
Frequently Asked Questions on Chapter 4 – Quadratic Equations
Check whether the following are quadratic equations (x + 1)2 = 2(x – 3)?
(x + 1)2 = 2(x – 3)

By using the formula for (a+b)2=a2+2ab+b2

x2 + 2x + 1 = 2x – 6 x2 + 7

Since the above equation is in the form of ax2 + bx + c = 0.

Therefore, the given equation is quadratic equation.

Find the roots of the following quadratic equations by factorisation x2 – 3x – 10 = 0?
Given, x2 – 3x – 10 =0

Taking LHS,

x2 – 5x + 2x – 10

x(x – 5) + 2(x – 5)

(x – 5)(x + 2)

The roots of this equation, x2 – 3x – 10 =0 are the values of x for which (x – 5)(x + 2) = 0

Therefore, x – 5 = 0 or x + 2 = 0

x = 5 or x = -2

No, not all the chemical reaction is a redox reaction. Reactions like double decompositions, acid-base neutralisation reactions, precipitation reactions are non-redox reactions.

Find the roots of the following quadratic equations, if they exist, by the method of completing the square 4×2 + 4√3x + 3 = 0?
4×2 + 4√3x + 3 = 0

Converting the equation into a2+2ab+b2 form, we get,

(2x)2 + 2 × 2x × √3 + (√3)2 = 0

(2x + √3)2 = 0

(2x + √3) = 0 and (2x + √3) = 0

Therefore, either x = -√3/2 or x = -√3/2.

Find the nature of the roots of the following quadratic equations. If the real roots exist, find them 2×2 – 3x + 5 = 0?
Given,

2×2 – 3x + 5 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 2, b = -3 and c = 5

We know, Discriminant = b2 – 4ac

= ( – 3)2 – 4 (2) (5) = 9 – 40

= – 31

As you can see, b2 – 4ac < 0

Therefore, no real root is possible for the given equation, 2×2 – 3x + 5 = 0.

Find the values of k for each of the following quadratic equations, so that they have two equal roots 2×2 + kx + 3 = 0?
2×2 + 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

Exercise Page No 22

1.Multiple choice questions.

(i) Which of these statements is not a valid reason for the depletion of flora and fauna?

(a) Agricultural expansion.

(b) Large scale developmental projects.

(c) Grazing and fuelwood collection.

(d) Rapid industrialisation and urbanisation.

Answer:

Grazing and fuelwood collection

(ii) Which of the following conservation strategies do not directly involve community participation?

(a) Joint forest management

(b) Beej Bachao Andolan

(c) Chipko Movement

(d) Demarcation of Wildlife sanctuaries

Answer:

Demarcation of wildlife sanctuaries

2. Match the following animals with their category of existence.

Animals/Plants Category of existence
Black Buck Extinct
Asiatic Elephant Rare
Andaman wild pig Endangered
Himalayan Brown Bear Vulnerable
Pink Head Duck Endemic
Answer:

Animals/Plants Category of existence
Black Buck Endangered
Asiatic Elephant Vulnerable
Andaman wild pig Endemic
Himalayan Brown Bear Rare
Pink Head Duck Extinct
3. Match the following.

Reserved Forests Other forests and wastelands belonging to both government and private individuals and communities
Protected Forests Forests are regarded as most valuable as far as the conservation of forest and wildlife resources
Unclassed Forests Forest lands are protected from any further depletion
Answer:

Reserved Forests Forests are regarded as most valuable as far as the conservation of forest and wildlife resources
Protected Forests Forest lands are protected from any further depletion
Unclassed Forests Other forests and wastelands belonging to both Government and private individuals and communities
4. Answer the following questions in about 30 words.

(i) What is biodiversity? Why is biodiversity important for human lives?

Answer:

Biodiversity is made up of various types of life forms found on earth. It is a measure of variation at the ecosystem, species and genetic level. Biodiversity is abundant in Tropical areas. Tropical areas cover 10 per cent of the earth surface but it hosts 90% of the world species.

Contribution of biodiversity in human lives

Agriculture – Variety of Plant species meet our needs for food.
Their contribution to business and industry
Leisurely activities
Ecological services
(ii) How have human activities affected the depletion of flora and fauna? Explain

Answer:

Various river valley projects have affected the flora and fauna
Many illegal Mining projects have depleted the flora and fauna
Too many development projects for leisure activities in the forests have negatively affected.
Too many human activities in the forest area due to rising population and lack of space has created human-animal conflict.
5. Answer the following questions in about 120 words.

(i) Describe how communities have conserved and protected forests and wildlife in India?

Answer:

Chipko Movement:

The famous Chipko movement in the Himalayas has not only successfully resisted deforestation in several areas but has also shown that community afforestation with indigenous species can be enormously successful.

Certain societies revere a particular tree which they have preserved from time immemorial. The Mundas and the Santhal of Chota Nagpur region worship mahua (Bassia latifolia) and kadamba (Anthocaphalus cadamba) trees. The tribals of Odisha and Bihar worship the tamarind (Tamarindus indica) and mango (Mangifera indica) trees during weddings. To many of us, peepal and banyan trees are considered sacred.

In Sariska Tiger Reserve, Rajasthan, villagers have fought against mining by citing the Wildlife Protection Act. In many areas, villagers themselves are protecting habitats and are explicitly rejecting government involvement. The inhabitants of five villages in the Alwar district of Rajasthan have declared 1,200 hectares of forest as the Bhairodev Dakav ‘Sonchuri’, declaring their own set of rules and regulations which do not allow hunting and are protecting the wildlife against any outside encroachments.

(ii) Write a note on good practices towards conserving forest and wildlife.

Answer:

In India, joint forest management (JFM) programme furnishes a good example of involving local communities in the management and restoration of degraded forests. The programme has been in formal existence since 1988 when the state of Odisha passed the first resolution for joint forest management. JFM depends on the formation of local (village) institutions that undertake protection activities mostly on degraded forest land managed by the forest department. In return, the members of these communities are entitled to intermediary benefits like non-timber forest produces and share in the timber harvested by ‘successful protection’. The clear lesson from the dynamics of both environmental destruction and reconstruction in India is that local communities everywhere have to be involved in some kind of natural resource management. But there is still a long way to go before local communities are at the centre stage in decision-making. Accept only those economic or developmental activities, that are people-centric, environment-friendly and economically rewarding.