CBSE Class X

Question 2:

Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes 1/2. if we only add 1 to the denominator. What is the fraction?

(ii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

(iv) Meena went to bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days,  while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Answer:

(i)Let the fraction be x/y.

According to the given information,

Subtracting equation (1) from equation (2), we obtain

x = 3 (3)

Substituting this value in equation (1), we obtain

3 − y = −2

−y = −5

y = 5

Hence, the fraction is 3/5.

(ii)Let present age of Nuri = x

and present age of Sonu = y

According to the given information,

(x – 5) = 3(y – 5)

x – 3y = −10                 (1)

(x + 10) = 2(y + 10)

x – 2y = 10                    (2)

Subtracting equation (1) from equation (2), we obtain

y = 20 (3)

Substituting it in equation (1), we obtain

x − 60 = −10

x = 50

Hence, age of Nuri = 50 years

And, age of Sonu = 20 years

(iii)Let the unit digit and tens digits of the number be x and y respectively. Then, number = 10y + x

Number after reversing the digits = 10x + y

According to the given information,

x + y = 9 (1)

9(10y + x) = 2(10x + y)

88y − 11x = 0

− x + 8y =0 (2)

Adding equation (1) and (2), we obtain

9y = 9

y = 1 (3)

Substituting the value in equation (1), we obtain

x = 8

Hence, the number is 10y + x = 10 × 1 + 8 = 18

(iv)Let the number of Rs 50 notes and Rs 100 notes be x and y respectively.

According to the given information,

x + y = 25    (1)

50x + 100y = 2000  (2)

Multiplying equation (1) by 50, we obtain

50x + 50y = 1250     (3)

Subtracting equation (3) from equation (2), we obtain

50 y = 750

y = 15

Substituting in equation (1), we have x = 10

Hence, Meena has 10 notes of Rs 50 and 15 notes of Rs 100.

(v)Let the fixed charge for first three days and each day charge thereafter be Rs x and Rs y respectively.

According to the given information,

x + 4y = 27     (1)

x + 2y = 21     (2)

Subtracting equation (2) from equation (1), we obtain

2y = 6

y = 3                (3)

Substituting in equation (1), we obtain

x + 12 = 27

x = 15

Hence, fixed charge = Rs 15

And Charge per day = Rs 3

Question 1:

Solve the following pair of linear equations by the elimination method and the substitution method:

(i) x + y = 5 and 2x – 3y = 4

(ii) 3x + 4y = 10 and 2x – 2y = 2

(iii) 3x + 4y = 10 and 2x – 2y = 2

Answer:

(i) By elimination method

x + y = 5       (1)

2x − 3y = 4  (2)

Multiplying equation (1) by 2, we obtain

2x + 2y = 10    (3)

Subtracting equation (2) from equation (3), we obtain

5y = 6

y = 6/5      (4)

Substituting the value in equation (1), we obtain

By substitution method

From equation (1), we obtain

x = 5 − y    (5)

Putting this value in equation (2), we obtain

2(5 − y) − 3y = 4

−5y = −6

y = 6/5

Substituting the value in equation (5), we obtain

(ii) By elimination method

3x + 4y = 10    (1)

2x − 2y = 2      (2)

Multiplying equation (2) by 2, we obtain

4x − 4y = 4     (3)

Adding equation (1) and (3), we obtain

7x = 14

x = 2               (4)

Substituting in equation (1), we obtain

6 + 4y = 10

4y = 4

y = 1

Hence, x = 2, y = 1

By substitution method

From equation (2), we obtain

x = 1 + y   (5)

Putting this value in equation (1), we obtain

3(1 + y) + 4y = 10

7y = 7

y = 1

Substituting the value in equation (5), we obtain

x = 1 + 1 = 2

∴ x = 2, y = 1

(iii) By elimination method

3x − 5y − 4 = 0     (1)

9x = 2y + 7

9x − 2y − 7 = 0      (2)

Multiplying equation (1) by 3, we obtain

9x − 15y − 12 = 0  (3)

Subtracting equation (3) from equation (2), we obtain

13y = −5

y = −5/13                 (4)

Substituting in equation (1), we obtain

By substitution method

From equation (1), we obtain

Putting this value in equation (2), we obtain

(iv)By elimination method

Subtracting equation (2) from equation (1), we obtain

5y = −15

y = −3                  (3)

Substituting this value in equation (1), we obtain

3x − 12 = −6

3x = 6

x = 2

Hence, x = 2, y = −3

By substitution method

From equation (2), we obtain

Question 3:

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km.

(v) A fraction becomes 9/11, if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes 5/6. Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Answer:

(i) Let the first number be x and the other number be y such that y > x.

According to the given information,

y = 3x           (1)

y − x = 26   (2)

On substituting the value of y from equation (1) into equation (2), we obtain

3x – x = 26

x = 13        (3)

Substituting this in equation (1), we obtain

y = 39

Hence, the numbers are 13 and 39.

(ii) Let the larger angle be x and smaller angle be y.

We know that the sum of the measures of angles of a supplementary pair is always 180º.

According to the given information,

x + y = 180°        (1)

x – y = 18°            (2)

From (1), we obtain

x = 180º − y (3)

Substituting this in equation (2), we obtain

180° − y – y = 18°

162° = 2y

81° = y                    (4)

Putting this in equation (3), we obtain

x = 180º − 81º

= 99º

Hence, the angles are 99º and 81º.

(iii) Let the cost of a bat and a ball be x and y respectively.

According to the given information,

7x + 6y = 3800   (1)

3x + 5y = 1750   (2)

From (1), we obtain

Hence, the cost of a bat is Rs 500 and that of a ball is Rs 50.

(iv)Let the fixed charge be Rs x and per km charge be Rs y.

According to the given information,

x + 10y = 105       (1)

x + 15y = 155       (2)

From (3), we obtain

x = 105 – 10y        (3)

Substituting this in equation (2), we obtain

105 – 10y + 15y = 155

5y = 50

y = 10                   (4)

Putting this in equation (3), we obtain

x = 105 – 10 × 10

x = 5

Hence, fixed charge = Rs 5

And per km charge = Rs 10

Charge for 25 km = x + 25y

= 5 + 250 = Rs 255

(v) Let the fraction be x/y.

According to the given information,

Substituting this in equation (2), we obtain

(vi) Let the age of Jacob be x and the age of his son be y.

According to the given information,

(x + 5)  = 3 (y + 5)

x – 3y = 10              (1)

(x – 5) = 7 (y – 5)

x – 7y = −30          (2)

From (1), we obtain

x = 3y + 10            (3)

Substituting this value in equation (2), we obtain

3y + 10 – 7y = −30

−4y = −40

y = 10                   (4)

Substituting this value in equation (3), we obtain

x = 3 × 10 + 10 = 40

Hence, the present age of Jacob is 40 years whereas the present age of his son is 10 years.

Question 2:

Solve 2x + 3y = 11 and 2x − 4y = − 24 and hence find the value of ‘m’ for which y = mx + 3.

Answer:

2x + 3y = 11        (1)

2x − 4y = −24    (2)

From equation (1), we obtain

Question 1:

Solve the following pair of linear equations by the substitution method.

Answer:
(i) x + y = 14 (1)
x − y = 4 (2)
From (1), we obtain
x = 14 − y (3)

Substituting this value in equation (2), we obtain

(14 – y) – y = 4

14 – 2y = 4

10 = 2y

y = 5     (4)

Substituting this in equation (3), we obtain

x = 9

∴ x = 9, y = 5

(ii) s – t = 3       (1)

From (1), we obtain

s = t + 3

Substituting this value in equation (2), we obtain

Substituting in equation (3), we obtain

s = 9

∴ s = 9, t = 6

(iii)3x − y = 3 (1)

9x − 3y = 9 (2)

From (1), we obtain

y = 3x − 3 (3)

Substituting this value in equation (2), we obtain

9x – 3(3x – 3) = 9

9x – 9x + 9 = 9

9 = 9

This is always true.

Hence, the given pair of equations has infinite possible solutions and the relation between these variables can be given by

y = 3x − 3

Therefore, one of its possible solutions is x = 1, y = 0.

(iv) 0.2x + 0.3y = 1.3       (1)

        0.4x + 0.5y = 2.3      (2)

From equation (1), we obtain

Substituting this value in equation (3), we obtain

x = 0

∴ x = 0, y = 0

Substituting this value in equation (3), we obtain

Question 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:

x − y + 1 = 0

x = y − 1

3x + 2y − 12 = 0

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are intersecting each other at point (2, 3) and x-axis at (−1, 0) and (4, 0). Therefore, the vertices of the triangle are (2, 3), (−1, 0), and (4, 0).

Question 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

Answer:

(i)Intersecting lines:

For this condition,

(ii) Parallel lines:

For this condition,

(iii)Coincident lines:

For coincident lines,

Question 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.

Answer:

Let the width of the garden be x and length be y.

According to the question,
y − x = 4 (1)
y + x = 36 (2)
y − x = 4
y = x + 4

y + x = 36

Hence, the graphic representation is as follows.

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.

Question 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

Answer:

(i)x + y = 5

2x + 2y = 10

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

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines are overlapping each other.

Therefore, infinite solutions are possible for the given pair of equations.

(ii)x − y = 8

3x − 3y = 16

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

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

Hence, the graphic representation is as follows.

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

(iv)2x − 2y − 2 = 0

       4x − 4y − 5 = 0

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

Question 3:

On comparing the ratios  find out whether the following pair of linear equations are consistent, or inconsistent.

Answer:

(i) 3x + 2y = 5
      2x − 3y = 7

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

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

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 − 3 y = 11

       − 10x + 6y = − 22

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.

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.

Question 2:

On comparing the ratios  find out whether the lines representing the following pairs of linear equations at a point, are parallel or coincident:

Answer:

(i) 5x − 4y + 8 = 0

      7x + 6y − 9 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

a₁ = 5, b₁ = -4, c₁ = 8

a₂ = 7, b₂ = 6, c₂ = -9

Hence, the lines representing the given pair of equations have a unique solution and the pair of lines intersects at exactly one point.

(ii) 9x + 3y + 12 = 0
      18x + 6y + 24 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

Hence, the lines representing the given pair of equations are coincident and there are infinite possible solutions for the given pair of equations.

(iii)6x − 3y + 10 = 0

2x − y + 9 = 0

Comparing these equations with a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0, we obtain

Hence, the lines representing the given pair of equations are parallel to each other and hence, these lines will never intersect each other at any point or there is no possible solution for the given pair of equations.

Question 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.

(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.

Answer:

(i) Let the number of girls be x and the number of boys be y.

According to the question, the algebraic representation is

x + y = 10

x − y = 4

For x + y = 10,

x = 10 − y

For x − y = 4,

x = 4 + y

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (7, 3).

Therefore, the number of girls and boys in the class are 7 and 3 respectively.

(ii) Let the cost of 1 pencil be Rs x and the cost of 1 pen be Rs y.

According to the question, the algebraic representation is

5x + 7y = 50

7x + 5y = 46

For 5x + 7y = 50,

7x + 5y = 46

Hence, the graphic representation is as follows.

From the figure, it can be observed that these lines intersect each other at point (3, 5).

Therefore, the cost of a pencil and a pen are Rs 3 and Rs 5 respectively.

Question 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.

Answer:

Let the cost of 1 kg of apples be Rs x.

And, cost of 1 kg of grapes = Rs y

According to the question, the algebraic representation is

2x + y = 160

4x + 2y = 300

For 2x + y = 160,

y = 160 – 2x

The solution table is

For 4x + 2y = 300,

The solution table is

The graphical representation is as follows.

Question 2:

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

Answer:

Let the cost of a bat be Rs x.

And, cost of a ball = Rs y

According to the question, the algebraic representation is

3x + 6y = 3900

X + 2y = 1300

For 3x + 6y = 3900

The solution table is

F or x + 2y = 1300

x = 1300 – 2y

The solution table is

The graphical representation is as follows.

Question 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.

Answer:
Let the present age of Aftab be x.

And, present age of his daughter = y

Seven years ago,

Age of Aftab = x − 7

Age of his daughter = y − 7

According to the question,

(x – 7) = 7 (y – 7)

x – 7 = 7y – 49

x – 7y = –42        (1)

Three years hence,

Age of Aftab = x + 3

Age of his daughter = y + 3

According to the question,

(x + 3) = 3 (y + 3)

x + 3 = 3y + 9

x – 3y = 6        (2)

Therefore, the algebraic representation is

x – 7y = –42

x – 3y = 6

For x – 7y = –42

x =  –42 + 7y

The solution table is

For x – 3y = 6,

x = 6 + 3y

The solution table is

The graphical representation is as follows.

Question 5:

If the polynomial x⁴ – 6x³ + 16x² + 10 is divided by another polynomial x² – 2x + k, the remainder comes out to be x + a, find k and a.

Answer:

By division algorithm,
Dividend = Divisor × Quotient + Remainder
Dividend − Remainder = Divisor × Quotient

x⁴ – 6x³ + 16x² – 25x + 10 – x – a = x⁴ – 6x³ + 16x² – 26x + 10 – a will be perfectly divisible by x² – 2x + k.

Let us divide x⁴ – 6x³ + 16x² – 26x + 10 – a by x² – 2x + k

 

It can be observed that (–10 + 2k) x + (10 – a – 8k + k²) will be 0.

Therefore, (–10 + 2k) = 0 and (10 – a – 8k + k²) = 0

For (–10 + 2k) = 0,

2k = 10

And thus, k = 5

For (10 – a – 8k + k²) = 0

10 – a – 8 × 5 + 25 = 0

10 − a − 40 + 25 = 0
− 5 − a = 0
Therefore, a = −5
Hence, k = 5 and a = −5

Question 4:

If two zeroes of the polynomial x⁴ – 6x³ – 26x² 138x – 35 are 2 ± √3, find other zeroes.

Answer:

Given that 2 + √3 and 2 – √3 are zeroes of the given polynomial.

Therefore, (x – 2 – √3) (x – 2 + √3) = x² + 4 – 4x – 3

= x² – 4x + 1 is a factor of the given polynomial

For finding the remaining zeroes of the given polynomial, we will find the quotient by dividing x⁴ – 6x³ – 26x² + 138x – 35 by x² – 4x + 1.

Clearly, x⁴ – 6x³ – 26x² + 138x – 35 = (x² – 4x + 1) (x² – 2x – 35)

It can be observed that (x² – 2x – 35) is also a factor of the given polynomial.

And (x² – 2x – 35) = (x – 7) (x + 5)

Therefore, the value of the polynomial is also zero when x – 7 = 0 or x + 5 = 0

or  x = 7 or –5

Hence, 7 and −5 are also zeroes of this polynomial.

Question 3:

If the zeroes of polynomial x³ − 3x² + x + 1 are a − b, a, a + b, find a and b.

Answer:

p(x) = x³ − 3x² + x + 1

Zeroes are a − b, a + a + b

Comparing the given polynomial with px³ + qx² + rx + t, we obtain p = 1, q = −3, r = 1, t = 1 

Sum of zeroes = a − b + a + a + b

The zeroes are 1 − b, 1, 1 + b

Multiplication of zeroes = 1(1 − b) (1 + b)

Hence, a = 1 and b = √2 or – √2

Question 2:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, − 7, − 14 respectively.

Answer:

Let the polynomial be ax³ + bx² + cx + d and the zeroes be α,β, and γ.

It is given that

α + β  +  γ = 2/1 = −b/a

α β  +  βγ + αγ  = −7 /1 = c/a

α βγ = −14/1 = −d/a

If a = 1, then b = −2, c = −7, d = 14

Hence, the polynomial is x³ − 2x² − 7x + 14.

Question 1:

Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:

Answer:

(i) p(x) = 2x³ + x² – 5x + 2.

Zeroes for this polynomials are 1/2, 1 –2.

Therefore, 1/2, 1, and −2 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we obtain a = 2, b = 1, c = −5, d = 2

Therefore, the relationship between the zeroes and the coefficients is verified.

(ii) p(x) = x³ – 4x² + 5x – 2

Zeroes for this polynomial are 2, 1, 1.

p(2) = 2³ – 4(2²) + 5(2) – 2

= 8 – 16 + 10 – 2 = 0

p(1) = 1³ – 4(1)² + 5(1) – 2

= 1 – 4 + 5 – 2 = 0

Therefore, 2, 1, 1 are the zeroes of the given polynomial.

Comparing the given polynomial with ax³ + bx² + cx + d, we obtain a = 1, b = −4, c = 5, d = −2.

Verification of the relationship between zeroes and coefficient of the given polynomial

Multiplication of zeroes taking two at a time = (2)(1) + (1)(1) + (2)(1) =2 + 1 + 2 = 5 = (5)/1 = c/a

Multiplication of zeroes = 2 × 1 × 1 = 2  

Hence, the relationship between the zeroes and the coefficients is verified.