Wednesday, January 8, 2020

Grade 10 NCERT Math Solution: Chapter 3 - Pair Of Linear Equations In Two Variables

Exercise 3.1:

1. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be." (Isn't this interesting?) Represent this situation algebraically and graphically.

Solution:

Let present age of Aftab be x
And, present age of daughter is represented by y
Then Seven years ago,
Age of Aftab = x – 7
Age of daughter = y – 7
According to the question,
(x – 7)  = 7 (y – 7 )
x – 7 = 7 y – 49
x – 7y = – 49 + 7
x – 7y = – 42 …(i)
x = 7y – 42
Putting y = 5, 6 and 7, we get
x = 7 × 5 – 42 = 35 – 42 = – 7
x = 7 × 6 – 42 = 42 – 42 = 0
x = 7 × 7 – 42 = 49 – 42 = 7






Three years from now ,
Age of Aftab = x +3
Age of daughter = y +3
According to the question,
(x + 3) = 3 (y + 3)
x + 3 = 3y + 9
x – 3y = 9–3
x – 3y = 6 …(ii)
x = 3y + 6
Putting, y = –2,–1 and 0, we get
x = 3 × – 2 + 6 = –6 + 6 =0
x = 3 × – 1 + 6 = –3 + 6 = 3
x = 3 × 0 + 6 = 0 + 6 = 6






Algebraic representation
From equation (i) and (ii)
x – 7y = – 42 …(i)
x – 3y = 6 …(ii)
Graphical representation


















2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.

Solution:

Let cost of one bat = Rs x
Cost of one ball = Rs y
3 bats and 6 balls for Rs 3900 So that
3x + 6y = 3900 … (i)
Dividing equation by 3, we get
x + 2y = 1300
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = –1300, 0 and 1300 we get
x = 1300 – 2 (–1300) = 1300 + 2600 = 3900
x = 1300 –2(0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = – 1300






Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x + 2y = 1300 … (ii)
Subtracting 2y both side we get
x = 1300 – 2y
Putting y = – 1300, 0 and 1300 we get
x = 1300 – 2 (–1300) = 1300 + 2600 = 3900
x = 1300 – 2 (0) = 1300 – 0 = 1300
x = 1300 – 2(1300) = 1300 – 2600 = –1300






Algebraic representation
3x + 6y = 3900 … (i)
x + 2y = 1300 … (ii)
Graphical representation,





















3. The cost of 2 kg of apples and 1 kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.

Solution:

Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
2 x + y = 160 … (i)
2x = 160 - y
x = (160 – y)/2
Let y = 0 , 80 and 160,  we get
x = (160 – ( 0 )/2 = 80
x = (160 – 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0






Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2y = 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 ×  50 = 50
y = 150 – 2 × (100) = –50






Algebraic representation,
2x + y = 160 … (i)
4x + 2y = 300 … (ii)

Graphical representation,


Exercise 3.2:


1. Form the pair of linear equations in the following problems, and find their solutions graphically.

(i)  10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.

Solution:

Let number of boys = x
Number of girls = y
Given that total number of student is 10 so that
x + y = 10
Subtract y both side we get
x = 10 – y
Putting y = 0 , 5, 10 we get
x = 10 – 0 = 10
x = 10 – 5 = 5
x = 10 – 10 = 0






Given that If the number of girls is 4 more than the number of boys
So that
y = x + 4
Putting x = –4, 0, 4, and we get
y = – 4 + 4 = 0
y = 0 + 4 = 4
y = 4 + 4 = 8






Graphical representation

Both lines intersect at (3,7)

Hence the number of boys = 3 and number of girls = 7.


(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.

Solution:

Let cost of pencil = Rs x
Cost of pens = Rs y
5 pencils and 7 pens together cost Rs 50,
So we get
5x + 7y = 50
Subtracting 7y both sides we get
5x = 50 – 7y
Dividing by 5 we get



Putting value of y = 5 , 10 and 15 we get
x = 10 – 7 × 5/5 = 10 – 7 = 3
x = 10 – 7 × 10/5 = 10 – 14 = – 4
x = 10 – 7 × 15/5 = 10 – 21 = – 11





Given that 7 pencils and 5 pens together cost Rs 46
7x + 5y = 46
Subtracting 7x both side we get
5y = 46 – 7x
Dividing by 5 we get




y = 9.2 – 1.4x
Putting x = 0 , 2 and 4 we get
y = 9.2 – 1.4 × 0 = 9.2 – 0 = 9.2
y = 9.2 – 1.4 (2) = 9.2 – 2.8 = 6.4
y = 9.2 – 1.4 (4) = 9.2 – 5.6 = 3.6





Graphical representation:


















Both lines intersect at (3,5)

Hence the cost of one pencil = Rs 3  and cost of one pen = Rs 5.


2. On comparing the ratios a1/a2 , b1/b2 and c1/c2, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.

(i) 5x – 4y + 8 = 0
7x + 6y – 9 = 0

Solution:       

Comparing these equation with
a1x + b1y + c1 = 0
a2x + b2y + c= 0

We get
a1 = 5, b1 = –4, and c1 = 8
a2 =7, b2 = 6 and c2 = –9
a1/a2 = 5/7,
b1/b2 = –4/6 and
c1/c2 = 8/-9
Hence, a1/a2 ≠ b1/b2

Therefore, both are intersecting lines at one point.


(ii) 9x + 3y + 12 = 0
18x + 6y + 24 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + c= 0
We get
a1 = 9, b1 = 3, and c1 = 12
a2 = 18, b2 = 6 and c2 = 24
a1/a2 = 9/18 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2

Therefore, both lines are coincident.


(iii) 6x – 3y + 10 = 0
2x – y + 9 = 0
Comparing these equations with
a1x + b1y + c1 = 0
a2x + b2y + c= 0

We get
a1 = 6, b1 = –3, and c1 = 10
a2 = 2, b2 = –1 and c2 = 9
a1/a2 = 6/2 = 3/1
b1/b2 = –3/–1 = 3/1 and
c1/c2 = 12/24 = 1/2
Hence, a1/a2 = b1/b2 ≠ c1/c2

Therefore, both lines are parallel.


3. On comparing the ratios a1/a2 , b1/b2 and c1/c2 find out whether the following pair of linear equations are consistent, or inconsistent.
(i) 3x + 2y = 5 ; 2x – 3y = 7
(ii) 2x – 3y = 8 ; 4x – 6y = 9
(iii) 3/2x + 5/3y = 7 ; 9x – 10y = 14
(iv) 5x – 3y = 11 ; – 10x + 6y = –22
(v) 4/3x + 2y =8 ; 2x + 3y = 12

Solution:

(i)

3x + 2y = 5 ; 2x – 3y = 7
a1/a2 = 3/2
b1/b2 = –2/3 and
c1/c2 = 5/7
Hence, a1/a2 ≠ b1/b2
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.

(ii)

2x – 3y = 8 ; 4x – 6y = 9
a1/a2 = 2/4 = 1/2
b1/b2 = –3/–6 = 1/2 and
c1/c2 = 8/9
Hence, a1/a2 = b1/b2 ≠ c1/c2


Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.


(iii)

3/2x + 5/3y = 7 ; 9x – 10y = 14
a1/a2 = 3/2/9 = 1/6
b1/b2 = 5/3/10 = 1/6 and
c1/c2 = 7/14 = 1/2
Hence, a1/a2 ≠ b1/b2


Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.


(iv)

5x – 3y = 11 ; – 10x + 6y = –22
a1/a2 = 5/-10 = -1/2
b1/b2 = -3/6 = -1/2 and
c1/c2 = 11/-22 = -1/2
Hence, a1/a2 = b1/b2 = c1/c2


Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.


(v)

4/3x + 2y =8 ; 2x + 3y = 12
a1/a2 = 4/3/2 = 2/3
b1/b2 = 2/3 and
c1/c2 = 8/12 = 2/3
Hence, a1/a2 = b1/b2 = c1/c2

Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.


4. Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically:
(i) x + y = 5, 2x + 2y = 10
(ii) x – y = 8, 3x – 3y = 16
(iii) 2x + y – 6 = 0, 4x – 2y – 4 = 0
(iv) 2x – 2y – 2 = 0, 4x – 4y – 5 = 0

Solution:

x + y = 5; 2x + 2y = 10
a1/a2 = 1/2
b1/b2 = 1/2 and
c1/c2 = 5/10 = 1/2
Hence, a1/a2 = b1/b2 = c1/c2
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.

x + y = 5

x = 5  y






And, 2x + 2y = 10
x = 10 – 2y/2


Graphical representation:


(ii)

x – y = 8, 3x – 3y = 16
a1/a2 = 1/3
b1/b2 = –1/–3 = 1/3 and
c1/c2 = 8/16 = 1/2
Hence, a1/a2 = b1/b2 ≠ c1/c2

Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.


(iii)

2x + y – 6 = 0, 4x – 2y – 4 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = –1/2 and
c1/c2 = –6/–4 = 3/2
Hence, a1/a2 ≠ b1/b2

Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.
2x + y – 6 = 0
y = 6 – 2x








And, 4x – 2y – 4 = 0
y = 4x – 4/2






Graphical representation:

From the figure, it can be observed that these lines are intersecting each other at the only one point i.e., (2,2) which is the solution for the given pair of equations.


(iv)

2x – 2y – 2 = 0, 4x – 4y – 5 = 0
a1/a2 = 2/4 = 1/2
b1/b2 = –2/–4 = 1/2 and
c1/c2 = 2/5
Hence, a1/a2 = b1/b2 ≠ c1/c2

Therefore, these linear equations are parallel to each other and thus, have no possible solution. Hence, the pair of linear equations is inconsistent.


5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.

Solution:

Let length of rectangle = x m
Width of the rectangle = y m
According to the question,
y – x = 4 ... (i)
y + x = 36 ... (ii)
y – x = 4
y = x + 4







y + x = 36






Graphical representation:

From the figure, it can be observed that these lines are intersecting each other at only point i.e., (16, 20). Therefore, the length and width of the given garden is 20 m and 16 m respectively.


6. Given the linear equation 2x + 3y – 8 = 0, write another linear equations in two variables such that the geometrical representation of the pair so formed is:
(i) intersecting lines
(ii) parallel lines
(iii) coincident lines

Solution:

(i) Intersecting lines:
For this condition,
a1/a2 ≠ b1/b2
The second line such that it is intersecting the given line is
2x + 4y – 6 = 0 as
a1/a2 = 2/2 = 1
b1/b2 = 3/4 and
a1/a2 ≠ b1/b2


(ii) Parallel lines
For this condition,
a1/a2 = b1/b2 ≠ c1/c2
Hence, the second line can be
4x + 6y  8 = 0 as
a1/a2 = 2/4 = 1/2
b1/b2 = 3/6 = 1/2 and
c1/c2 = 8/8 = 1

and a1/a2 = b1/b2 ≠ c1/c2


(iii) Coincident lines
For coincident lines,
a1/a2 = b1/b2 = c1/c2
Hence, the second line can be
6x + 9y – 24 = 0 as
a1/a2 = 2/6 = 1/3
b1/b2 = 3/9 = 1/3 and
c1/c2 = –8/–24 = 1/3
and a1/a2 = b1/b2 = c1/c2


7. Draw the graphs of the equations x  y + 1 = 0 and 3x + 2y  12 = 0. Determine the coordinates of the vertices of the triangle formed by these lines and the x-axis, and shade the triangular region.

Answer

 y + 1 = 0

x = y  1





3x + 2y – 12 = 0
x = 12 – 2y/3





Graphical representation:





















Friday, December 27, 2019

Grade 10 NCERT Math Solution: Chapter 2 - Polynomials

Exercise 2.1

1. The graphs y = p(x) are given in the following figure, for some polynomials p(x). Find the number of zeros of p(x), in each case.
Solution:

(i)

The number of zeros is 0 as the graph does not intersect the x-axis at any point.

(ii)

The number of zeros is 1 as the graph intersects x-axis at only one point.

(iii)

The number of zeros is 3 as the graph intersects x-axis at 3 points.

(iv)

The number of zeros is 2 as the graph intersects x-axis at 2 points.

(v)

The number of zeros is 4 as the graph intersects x-axis at 4 points.

(vi)

The number of zeros is 3 as the graph intersects x-axis at 3 points.


Exercise 2.2:


1. Find the zeros of the following quadratic polynomials and verify the relationship between the zeros and coefficients.

(i) x² ˗ 2x ˗ 8

(ii) 4s² ˗ 4s + 1

(iii) 6x² ˗ 3 ˗ 7x

(iv) 4u² + 8u

(v) t² ˗ 15

(vi) 3x² ˗ x ˗ 4


Solution:

(i)

x² ˗ 2x ˗ 8 = x² ˗ 4x + 2x ˗ 8                          [ ˗ 4 + 2 = ˗ 2 and ˗ 4 × 2 = ˗ 8 ]
                = x(x ˗ 4) + 2(x ˗ 4)
                = (x ˗ 4)(x + 2)

To find the zero, equate the polynomial to 0.

(x ˗ 4)(x + 2) = 0

⇒ x = 4 and x = ˗ 2.

Hence the zeros of x² ˗ 2x ˗ 8 is 4 and ˗ 2.


Verifying the relationship between zeros and coefficients.



















(ii)

4s² ˗ 4s + 1 = ( 2s ˗ 1 )²
           
To find the zero, equate the polynomial to 0.

( 2s ˗ 1 )² = 0

⟹ ( 2s ˗ 1 )( 2s ˗ 1 ) = 0



Verifying the relationship between zeros and coefficients.





















(iii)

6x² ˗ 3 ˗ 7x = 6x² ˗ 7x ˗ 3
                   = 6x² + 2x ˗ 9x ˗ 3                         [ 2 ˗ 9 = ˗7 and 2 × ( ˗ 9 ) = ˗ 18 ]
                   = 2x(3x + 1) ˗ 3(3x + 1)
                   = (3x + 1)(2x ˗ 3)
                 
       
To find the zero, equate the polynomial to 0.











Verifying the relationship between zeros and coefficients.



















(iv)

4u² + 8u = 4u(u + 2 )

To find the zero, equate the polynomial to 0.

4u(u + 2 ) = 0

4u = 0 and  u + 2 = 0

u = 0  and u = ˗2

Verifying the relationship between zeros and coefficients.




















(v)

t² ˗ 15 = ( t + √15 ) ( t ˗ √15 )

To find the zero, equate the polynomial to 0.

( t + √15 ) ( t ˗ √15 ) = 0

t + √15 = 0  and  t ˗ √15 = 0

t = ˗√15   and  t =  √15

Verifying the relationship between zeros and coefficients.




















(vi)

3x² ˗ x ˗ 4 = 3x² + 3x ˗ 4x ˗ 4
                 = 3x( x + 1 ) ˗4( x + 1 )
                 = ( x + 1 ) ( 3x ˗ 4 )

To find the zero, equate the polynomial to 0.

( x + 1 ) ( 3x ˗ 4 ) = 0

x + 1 = 0  and  3x ˗ 4 = 0

x = ˗1  and  3x = 4



Verifying the relationship between zeros and coefficients.



















2. Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively.
















Solution:

(i)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.
∴ a = 4 , b = ˗1 and c = ˗4.

Hence the quadratic polynomial is 4x² ˗ x ˗ 4.


(ii)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.
















∴ a = 3, b = ˗3√2  and c = 1.

Hence the quadratic polynomial is 3x² ˗ 3√2x + 1.


(iii)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

















∴ a = 1 , b = 0  and c = √5.

Hence the quadratic polynomial is x²  + √5.


(iv)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

















∴ a = 1, b = ˗1  and c = 1

Hence the quadratic polynomial is x² ˗ x + 1.


(v)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.













∴ a = 4, b = 1  and c = 1.

Hence the quadratic polynomial is 4x² + x + 1.


(vi)

Let the polynomial be ax² + bx + c and its zeroes be ⍺ and β.

















∴ a = 1, b = ˗4  and c = 1.

Hence the quadratic polynomial is x² ˗ 4x + 1.


Exercise 2.3:


1. Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.

(i)

p(x) = x³ ˗ 3x² + 5x ˗ 3, g(x) = x² ˗ 2.

Solution:
















∴ Quotient = x ˗ 3  and Remainder = 7x ˗ 9.


(ii)

p(x) = x⁴ ˗ 3x² + 4x + 5, g(x) = x² + 1 ˗ x               

Solution:



















∴ Quotient = x² + x ˗ 3  and Remainder = 8


(iii)

p(x) = x⁴ ˗ 5x + 6, g(x) = 2 ˗ x²

Solution:

∴ Quotient = ˗x² ˗2  and Remainder = ˗5x + 10


2. Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial.

(i) t² ˗ 3, 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12

Solution:


















The remainder is 0 while dividing 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12 by t² ˗ 3.

∴ t² ˗ 3 is a factor of 2t⁴ + 3t³ ˗ 2t² ˗ 9t ˗ 12.


(ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2

Solution:

















The remainder is 0 while dividing 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1.

Hence x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.


(iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1

Solution:













The remainder is 0 while dividing x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1.

Hence x³ – 3x + 1 is a factor of x⁵ – 4x³ + x² + 3x + 1.


3. Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are √(5/3)
and –√(5/3).

Solution:

p(x) = 3x⁴ + 6x³ – 2x² – 10x – 5
Since the two zeroes are √(5/3) and –√(5/3).































To find the other zeros we equate (x + 1)² to 0.

(x + 1)² = 0

(x + 1)(x + 1) = 0

x = ˗1 and ˗1.

Hence the zeroes of the given polynomial are √(5/3), –√(5/3), ˗1 and ˗1.


4. On dividing x³ ˗ 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x ˗ 2 and
˗2x + 4, respectively. Find g(x).

Solution:

Dividend = x³ ˗ 3x² + x + 2

Divisor = g(x) = ?

Quotient = x ˗ 2

Remainder = ˗2x + 4

We know that

Dividend = Quotient × Divisor + Remainder

x³ ˗ 3x² + x + 2 = (x ˗ 2) × g(x) + (˗2x + 4 )

(x ˗ 2) × g(x) = ( x³ ˗ 3x² + x + 2 ) ˗ (˗2x + 4 )

                      =   x³ ˗ 3x² + x + 2 + 2x ˗ 4

                      =   x³ ˗ 3x² + 3x ˗ 2


















∴ g(x) = x² ˗ x + 1


5. Give examples of polynomial p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
(i) deg p(x) = deg q(x)
(ii) deg q(x) = deg r(x)
(iii) deg r(x) = 0

Solution:

(i)

Let us assume the division of 6x² + 2x + 2 by 2




Here, p(x) = 6x² + 2x + 2
g(x) = 2
q(x) = 3x² + x + 1
r(x) = 0
Degree of p(x) and q(x) is same i.e. 2.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
Or, 6x² + 2x + 2 = 2 × (3x² + x + 1) + 0
                           = 6x² + 2x + 2
Hence, division algorithm is satisfied.

(ii)

Let us assume the division of x³ + x by x²,









Here, p(x) = x³ + x
g(x) = x²
q(x) = x and r(x) = x
Clearly, the degree of q(x) and r(x) is the same i.e., 1.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + x = (x² ) × x + x
          = x³ + x
Thus, the division algorithm is satisfied.


(iii)

Let us assume the division of x³ + 1 by x².








Here, p(x) = x³ + 1
g(x) = x²
q(x) = x and r(x) = 1
Clearly, the degree of r(x) is 0.
Checking for division algorithm,
p(x) = g(x) × q(x) + r(x)
x³ + 1 = (x² ) × x + 1
          = x³ + 1
Thus, the division algorithm is satisfied.