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Category: Maths

  • Chapter 2 Polynomials RD Sharam Solution for Class 9th Maths

    Chapter 2: Exponents Of Real Numbers Exercise – 2.1

    Question: 1

    Simplify the following: 

    (i) 3(a4 b3)10 × 5 (a2 b2)3 

    (ii) (2x-2 y3)3 

    Solution:

    (i) 3(a4 b3)10 × 5 (a2 b2)3 

    = 3(a40 b30) × 5(a6 b6)

    = 15 (a46 b36) 

    (ii) (2x-2 y3)3 

    (23 × -2 × 3 y3 × 3) = 8x-6y9 

     

    Question: 2

    If a = 3 and b = – 2, find the values of: 

    (i) aa + bb 

    (ii) ab + ba 

    (iii) ab + ba 

    Solution:

    (i) We have,

    aa + bb 

    = 33 + (−2) −2

     = 33 + (−1/2)2 

    = 27 +1/4 

    = 109/4 

    (ii) ab + ba

     = 3−2 + (−2)3

     = (1/3)2 + (−2)3

     = 1/9 – 8

     = −(71/9) 

    (iii) We have, 

    ab + ba 

    = (3 + (−2))3(−2) 

    = (3 − 2))−6

     = 1−6 = 1 

     

    Question: 3

    Prove that:

    Solution:

    (i) To prove

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    Or, Therefore, LHS = RHS Hence proved

    (ii) To prove, 

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    Therefore, LHS = RHS Hence proved

    (iii) To prove, 

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    = xac−bc × xba−ca × xbc−ab

    = xac − bc + ba − ca + bc − ab

    = x0

    = 1

    Therefore, LHS = RHS

    Hence proved 

     

    Question: 4

    Prove that: 

    Solution:

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    Therefore, LHS = RHS Hence proved

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    Therefore, LHS = RHS Hence proved

     

    Question: 5

    Prove that:

    Solution:

    (i) To prove,

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    = abc

    Therefore, LHS = RHS Hence proved

    (ii) To prove,

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    Therefore, LHS = RHS

    Hence proved 

     

    Question: 6

    If abc = 1, show that

    Solution:

    To prove,

    Left hand side (LHS) = Right hand side (RHS) Considering LHS,

    We know abc = 1

    c = 1/ab

    By substituting the value c in equation (1), we get

    Therefore, LHS = RHS Hence proved 

     

    Question: 7

    Simplify: 

    Solution:

     

    Question: 8

    Solve the following equations for x: 

    (i) 72x+3 = 1 

    (ii) 2x+1 = 4x−3 

    (iii) 25x+3 = 8x+3 

    (iv) 42x = 1/32 

    (v) 4x−1 × (0.5)3−2x = (1/8)x 

    (vi) 23x−7 = 256 

    Solution:

    (i) We have,

    ⟹ 72x+3 = 1

    ⟹ 72x+3 = 70 

    ⟹ 2x + 3 = 0

    ⟹ 2x = -3

    ⟹ x = −3/2 

    (ii) We have,

    = 2x+1 = 4x−3

    = 2x+1 = 22x−6

    = x + 1 = 2x – 6

    = x = 7

    (iii) We have,

    = 25x+3 = 8x+3

    = 25x+3 = 23x+9

    = 5x + 3 = 3x + 9

    = 2x = 6

    = x = 3

    (iv) We have,

    = 42x = 1/32

    = 24x = 1/25

    = 24x = 2−5

    = 4x = – 5

    x = -5/4

    (v) We have,

    4x−1 × (0.5)3−2x = (1/8)x

    22x−2 × (1/2)3−2x = (1/2)3x

    22x−2 × 22x−3 = (1/2)3x

    22x−2+ 2x−3 = (1/2)3x

    24x−5 = 2−3x

    4x-5 = -3x

    7x = 5

    x = 5/7

    (vi) 23x−7 = 256

    23x−7 = 28

    3x – 7 = 8

    3x = 15

    x = 5

     

    Question: 9

    Solve the following equations for x: 

    (i) 22x − 2x+3 + 24 = 0 

    (ii) 32x+4 + 1 = 2 × 3x+2 

    Solution:

    (i) We have, ⟹ 22x − 2x+3 + 24 = 0 

    ⟹ 22x + 24 = 2x.23 

    ⟹ Let 2x = y 

    ⟹ y2 + 24 = y × 23 

    ⟹ y2 − 8y + 16 = 0 

    ⟹ y2 − 4y − 4y + 16 = 0 

    ⟹ y(y – 4) – 4(y – 4) = 0

    ⟹ y = 4

    ⟹ x2 = 22 

    ⟹ x = 2

    (ii) We have,

    32x+4 + 1 = 2 × 3x+2

    (3x+2)2 + 1 = 2 × 3x+2

    Let 3x+2 = y

    y2 + 1 = 2y

    y2 − 2y + 1 = 0

    y2 − y − y + 1 = 0

    y(y − 1) − 1(y − 1) = 0

    (y − 1)(y − 1) = 0

    y = 1

     

    Question: 10

    If 49392 = a4b2c3, find the values of a, b and c, where a, b and c, where a, b, and c are different positive primes. 

    Solution:

    Taking out the LCM, the factors are 24, 32 and 73 a4b2c3 = 24, 32 and 73

    a = 2, b = 3 and c = 7 [Since, a, b and c are primes]

     

    Question: 11

    If 1176 = 2a × 3b × 7c, Find a, b, and c. 

    Solution:

    Given that 2, 3 and 7 are factors of 1176.

    Taking out the LCM of 1176, we get 23 × 31 × 72 = 2a × 3b × 7c 

    By comparing, we get

    a = 3, b = 1 and c = 2.

     

    Question: 12

    Given 4725 = 3a × 5b × 7c, find 

    (i) The integral values of a, b and c 

    (ii) The value of 2−a × 3b × 7c 

    Solution:

    (i) Taking out the LCM of 4725, we get

    33 × 52 × 71 = 3a × 5b × 7c

    By comparing, we get

    a = 3, b = 2 and c = 1.

    (ii) The value of 2−a × 3b × 7c 

    Sol:

    2-a × 3b × 7c = 2−3 × 32 × 71

    2−3 × 32 × 71 = 1/8 × 9 × 7

    63/8

     

    Question: 13

    If a = xyp−1, b = xyq−1 and c = xyr−1, prove that aq−r br−p cp−q = 1 

    Solution:

    Given, a = xyp−1, b = xyq−1 and c = xyr−1 

    To prove, aq−rbr−pcp−q = 1

    Left hand side (LHS) = Right hand side (RHS)

    Considering LHS, = aq−r br−p cp−q …… (i)

    By substituting the value of a, b and c in equation (i), we get

    = (xyp−1)q−r(xyq−1)r−p(xyr−1)p−q 

    = xypq−pr−q+rxyqr−pq−r+pxyrp−rq−p+q 

    = xypq−pr−q+ r+qr−pq−r+p+rp−rq−p+q 

    = xy0 

    = 1

    Chapter 2: Exponents Of Real Numbers Exercise – 2.2

    Question: 1

    Assuming that x, y, z are positive real numbers, simplify each of the following

    Solution:

    = (243x10y5z10)1/5

    = (243)1/5x10/5y5/5z10/5

    = (35)1/5x2yz2

    = 3x2yz2

     

    Question: 2

    Simplify

    Solution:

    = (4-1)

    = 1/4

    = [(25)−3]1/5

    = (2−15)1/5

    = 2−3

    = 1/23 = 1/8

    = [(343)−2]1/3

    = (343)−2×1/3

    = (73)−2/3

    = (7−2)

    =(1/72)

    = (1/49)

    (iv) (0.001)1/3

    = (1/1000)1/3

    = (1/103)1/3

    = 721-20 × 525/2 – 21/2

    = 71 × 54/2

    = 71 × 52

    = 7 × 25

    = 175

     

    Question: 3

    Prove that

    (ii) 93/2 − 3 × 50 − (1/81)−1/2

    Solution:

    = ((3 × 5−3)1/2 ÷ (3−1)1/3(5)1/2) × (3 × 56)1/6

    = ((3)1/2(5−3)1/2 ÷ (3−1)1/3(5)1/2) × (3 × 56)1/6

    = ((3)1/2(5)−3/2 ÷ (3)−1/3(5)1/2) × ((3)1/6 × (5)6/6)

    = ((3)1/2 − (−1/3) × (5)−3/2−1/2) × ((3)1/6 × (5))

    = ((3)5/6 × (5)−2) × ((3)1/6 × (5))

    = ((3)5/6+1/6 × (5)−2+1)

    = ((3)6/6 × (5)−1)

    = ((3)1 × (5)−1)

    = ((3) × (5)−1)

    = ((3) × (1/5))

    = (3/5)

    (ii) 93/2 − 3 × 50 − (1/81)−1/2

    = (32)3/2 − 3 − (1/92)−1/2

    = 33 − 3 − (9)−2×−1/2

    = 27 − 3 − 9

    = 15

    = 24 − 3 × 23×2/3 + 4/3

    = 16 − 3 × 22 + 4/3

    = 16 − 3 × 4 + 4/3

    = 16 − 12 + 4/3

    = (12 + 4)/3

    = 16/3

    = 2 × 1 × 5

    = 10

    = 1/2 + 1/(0.1)1 − (3)2

    = 1/2 +1/(0.1) −9

    = 1/2 + 10 − 9

    = 1/2 + 1

    = 3/2

    = (5/4)2 + 5/4 + 5/4

    = 25/16 + 10/4

    = 25/16 + 40/16

    = (26 + 40)/16

    = 65/16

     

    Question: 4

    Show that

    (v) (xa−b)a + b (xb−c)b + c(xc − a)c + a = 1

    Solution:

    Left hand side (LHS) = Right hand side (RHS)

    Considering LHS,

    Therefore, LHS = RHS

    Hence proved

    Left hand side (LHS) = Right hand side (RHS)

    Considering LHS,

    Therefore, LHS = RHS

    Hence proved

    (v) (xa−b)a + b (xb−c)b + c(xc − a)c + a = 1

    (xa−b)a+b(xb−c)b+c(xc−a)c+a

     

    Question: 5

    If 2x = 3y = 12z, show that 1/z = 1/y + 2/x

    Solution:

    2x = 3y = (2 × 3 × 2)z

    2x = 3y = (22 × 3)z

    2x = 3y = (22z × 3z)

    2x = 3y = 12z = k

    2 = k1/x

    3 = k1/y

    12 = k1/z

    12 = 2 × 3 × 2

    12 = k1/z = k1/y × k1/x × k1/x

    k1/z = k2/x + 1/y

    1/z = 1/y + 2/x

     

    Question: 6

    If 2x = 3y = 6−z, show that 1/x + 1/y + 1/z = 0

    Solution:

     2x = 3y = 6−z

    2x = k

    2 = k1/x

    3y = k

    3 = k1/y

    6−z = k

    k = 1/6z

    6 = k−1/z

    2 × 3 = 6

    k1/x × k1/y = k−1/z

    1/x + 1/y = −1/z [by equating exponents]

    1/x + 1/y + 1/z = 0

     

    Question: 7

    If ax = by = cz and b2 = ac, then show that

    Solution:

    Let ax = by = cz = k

    a = k1/x, b = k1/y, c = k1/z

    Now,

    b2 = ac

    (k1/y)2 = k1/x × k1/z

    k2/y = k1/x + 1/z

    2/y = 1/x + 1/z

     

    Question: 8

    If 3x = 5y = (75)z, Show that

    Solution:

    3x = k

    3 = k1/x

    5y = k

    5 = k1/y

    75z = k

    75 = k1/z

    31 × 52 = 751

    k1/x × k2/y = k1/z

    1/x + 2/y = 1/z

     

    Question: 9

    If (27)x = 9/3x, find x

    Solution:

    We have,

    (27)x = 9/3x

    (33)x = 9/3x

    33x = 9/3x

    33x = 32/3x

    33x = 32−x

    3x = 2 − x [On equating exponents]

    3x + x = 2

    4x = 2

    x = 2/4

    x = 1/2

    Here the value of x is ½

     

    Question: 10

    Find the values of x in each of the following

    (ii) (23)4 = (22)x

    (iii) (3/5)x(5/3)2x = 125/27

    (iv) 5x−2 × 32x−3 = 135

    (v) 2x−7 × 5x−4 = 1250

    (vii) 52x+3 = 1

    Solution:

    We have

    = 4x = 4 [On equating exponent]

    x = 1

    Hence the value of x is 1

    (ii) (23)4 = (22)x

    We have

    (23)4 = (22)x

    = 23×4 = 22×x

    12 = 2x

    2x = 12 [On equating exponents]

    x = 6

    Hence the value of x is 6

    (iii) (3/5)x(5/3)2x = 125/27

    We have

    (3/5)x(5/3)2x = 125/27

    ⇒ 52x−x/32x−x  = 53/33

    ⇒ 5x/3x = 53/33

    ⇒ (5/3)x = (5/3)3

    x = 3 [on equating exponents]

    Hence the value of x is 3

    (iv) 5x−2 × 32x−3 = 135

    We have,

    5x−2 × 32x−3 = 135

    ⇒ 5x−2 × 32x−3 = 5 × 27

    ⇒ 5x−2 × 32x−3 = 51 × 33

    ⇒ x − 2 = 1, 2x − 3 = 3 [On equating exponents]

    ⇒ x = 2 + 1, 2x = 3 + 3

    ⇒ x = 3, 2x = 6

    ⇒ x = 3

    Hence the value of x is 3

    (v) 2x−7 × 5x−4 = 1250

    We have

    2x−7 × 5x−4 = 1250

    ⇒ 2x−7 × 5x−4 = 2 × 625

    ⇒ 2x−7 × 5x−4 = 2 × 54

    ⇒ x − 7 = 1

    ⇒ x = 8, x − 4 = 4

    ⇒ x = 8

    Hence the value of x is 8

    4x + 1 = -15

    4x = -15 – 1

    4x = -16

    x = (-16)/4

    x = – 4

    Hence the value of x is 4

    (vii) 52x+3 = 1

    52x+3 = 1 × 50

    2x + 3 = 0 [By equating exponents]

    2x = −3

    x = −3/2

    Hence the value of x is −3/2

    √x = 2 [By equating exponents]

    (√x)2 = (2)2

    x = 4

    Hence the value of x is 4

    x + 1 = – 6

    x = – 6 – 1

    x = -7

    Hence the value of x is 7

     

    Question: 11

    If x = 21/3 + 22/3, show that x3 − 6x = 6

    Solution:

    x3 − 6x = 6

    x = 21/3 + 22/3

    Putting cube on both the sides, we get

    x3  = (21/3 + 22/3)3

    As we know, (a + b)3 = a3 + b3 + 3ab(a + b)

    x3 = (21/3)3 + (22/3)3 + 3(21/3)(22/3)(21/3 + 22/3)

    x3 = (21/3)3 + (22/3)3 + 3(21/3+2/3)(x)

    x3 = (21/3)3 + (22/3)3 + 3(2)(x)

    x3 = 6 + 6x

    x3 – 6x = 6

    Hence proved

     

    Question: 12

    Determine (8x)x, if 9x+2 = 240 + 9x.

    Solution:

     9x+2 = 240 + 9x

    9x .92 = 240 + 9x

    Let 9x be y

    81y = 240 + y

    81y – y = 240

    80y = 240

    y = 3

    Since, y = 3

    Then,

    9x = 3

    32x = 3

    Therefore, x = ½

    (8x)x = (8 × 1/2)1/2

    = (4)1/2

    = 2

    Therefore (8x)x = 2

     

    Question: 13

    If 3x+1 = 9x-2, find the value of 21+x

    Solution:

    3x+1 = 9x-2

    3x+1 = 32x-4

    x + 1 = 2x – 4

    x = 5

    Therefore the value of 21+x = 21+5 = 26 = 64

     

    Question: 14

    If 34x = (81)-1 and (10)1/y = 0.0001, find the value of 2-x+4y.

    Solution:

    34x = (81)-1 and (10)1/y = 0.0001

    34x = (3)-4

    x = -1

    And, (10)1/y = 0.0001

    (10)1/y = (10)−4

    1/y = -4

    y = 1/−4

    To find the value of 2-x+4y, we need to substitute the value of x and y

    2-x+4y = 21+4(1/−4) = 21-1 = 20 = 1

     

    Question: 15

    If 53x = 125 and 10y = 0.001. Find x and y.

    Solution:

    53x = 125 and 10y = 0.001

    53x = 53

    x = 1

    Now,

    10y = 0.001

    10y = 10-3

    y = -3

    Therefore, the value of x = 1 and the value of y = – 3

     

    Question: 16

    Solve the following equations

    (i) 3x+1 = 27 × 34

    (iii) 3x−1 × 52y−3 = 225

    (iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y

    Solution:

    (i) 3x+1 = 27 × 34

    3x+1 = 33 × 34

    3x+1 = 33+4

    x + 1 = 3 + 4 [By equating exponents]

    x + 1 = 7

    x = 7 − 1

    x = 6

    4x = 3 (By equating exponents)

    (iii) 3x−1 × 52y−3 = 225

    3x−1 × 52y−3 = 32 × 52

    x − 1 = 2 [By equating exponents]

    x = 3

    3x−1 × 52y−3 = 32 × 52

    2y − 3 = 2 [By equating exponents]

    2y = 5

    y = 5/2

    (iv) 8x+1 = 16y+2 and (1/2)3+x = (1/4)3y

    (23)x+1 and (2−1)3+x = (2−2)3y

    3x + 3 = 4y + 8 and − 3 − x = −6y

    3x + 3 = 4y + 8 and 3 + x = 6y

    3x + 3 = 4y + 8 and y = (3+x)/6

    3x + 3 = 4y + 8… eq1

    Substitute eq2 in eq1

    3(3x + 3) = 6 + 2x + 24

    9x + 9 = 30 + 2x

    7x = 21

    x = 21/7

    x = 3

    Putting value of x in eq2

    y = 1

    (v) 4x−1 × (0.5)3−2x = (1/8)x

    22x−2 × (5/10)3−2x = (1/23)x

    22x−2 × (1/2)3−2x = 2−3x

    22x−2 × 2−3+2x = 2−3x

    2x − 2 − 3 + 2x = −3x [By equating exponents]

    4x + 3x = 5

    7x = 5x = 5/7

    (a/b)1/2 = (a/b)−(1−2x)1/2 = −1 + 2x [By equating exponents]

    1/2 + 1 = 2x

    2x = 3/2

    x = ¾

     

    Question: 17

    If a and b are distinct positive primes such that

    , find x and y

    Solution:

    (a6b−4)1/3 = axb2y

    a6/3b−4/3 = axb2y

    a2b−4/3 = axb2y

    x = 2, 2y = −4/3

     

    Question: 18

    If a and b are different positive primes such that

    (ii) (a + b)−1(a−1 + b−1) = axby, find x and y

    Solution:

    (a−1−2b2+4)7 ÷ (a3+2b−5−3) = axby

    (a−3b6)7 ÷ (a5b−8) = axby

    (a−21b42) ÷ (a5b−8) = axby

    (a−21−5b42+8) = axby

    (a−26b50) = axby

    x = −26, y = 50

    (ii) (a + b)−1(a−1 + b−1) = axby, find x and y

    (a + b)−1(a−1 + b−1)

    = 1/ab

    = (ab)−1 = a−1b−1

    By equating exponents

    x = −1, y = −1

    Therefore x + y + 2 = −1 − 1 + 2 = 0

     

    Question: 19

    If 2x × 3y × 5z = 2160, find x, y and z. Hence compute the value of 3x × 2−y × 5−z

    Solution:

    2x × 3y × 5z = 2160

    2x × 3y × 5z = 24 × 33 × 51

    x = 4, y = 3, z = 1

    3x × 2−y × 5−z = 34 × 2−3 × 5−1

    = 81/40

     

    Question: 20

    If 1176 = 2a × 3b × 7c, find the values of a, b and c.

    Solution:

    Hence compute the value of 2a × 3b × 7−c as a fraction

    1176 = 2a × 3b × 7c

    23 × 31 × 72 = 2a × 3b × 7c

    a = 3, b = 1, c = 2

    We have to find the value of 2a × 3b × 7−c

    2a × 3b × 7−c = 23 × 31 × 7−2

    = 24/49

     

    Question: 21

    Simplify

    Solution:

    (xa+b−c)a−b(xb+c−a)b−c(xc+a−b)c−a

     

    Question: 22

    Show that

    Solution:

    Hence, LHS = RHS

     

    Question: 23

    (i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−nbn−lcl−m = 1

    (ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm

    Solution:

    (i) If a = xm+nyl, b = xn+lym and c = xl+myn, prove that am−nbn−lcl−m = 1

    (xm+nyl)m−n(xn+lym)n−l(xl+myn)l−m

    = (x(m+n)(m−n)yl(m−n))(x(n+l)(n−l)ym(n−l))(x(l+m)(l−m)yn(l−m))

    = x0y0 = 1

    (ii) If x = am+n, y = an+l and z = al+m, prove that xmynzl = xnylzm

    LHS = xmynzl

    (am+n)m(an+l)n(al+m)l

    =a(m+n)na(n+l)la(l+m)m

    = xnylzm

  • Chapter 1 Number Systems RD Sharam Solution for Class 9th Maths

    Chapter 1: Number System Exercise – 1.1

    Question: 1

    Is 0 a rational number? Can you write it in the form P/Q, where P and Q are integers and Q ≠ 0?

    Solution:

    Yes, 0 is a rational number and it can be written in P ÷ Q form provided that Q?

    0 is an integer and it can be written various forms, for example

    0 ÷ 2, 0 ÷ 100, 0 ÷ 95 etc.

     

    Question: 2

    Find five rational numbers between 1 and 2

    Solution:

    Given that to find out 5 rational numbers between 1 and 2

    Rational number lying between 1 and 2

    = 3/2

    = 1 < 3/2 < 2

    Rational number lying between 1 and 3/2

    = 5/4

    = 1 < 5/4 < 3/2

    Rational number lying between 1 and 5/4

    Rational number lying between 3/2 and 2

    = 9/8

    = 1 < 9/8 < 5/4

    Rational number lying between 3/2 and 2

    = 7/4

    = 3/2 < 7/4 < 2

    Rational number lying between 7/4 and 2

    = 15/8

    = 7/4 < 15/8 < 2

    Therefore, 1 < 9/8 < 5/4 < 3/2 < 7/4 < 15/8 < 2

     

    Question: 3

    Find out 6 rational numbers between 3 and 4

    Solution:

    Given that to find out 6 rational numbers between 3 and 4

    We have,

    3 × 7/7 = 21/7 and

    4 × 6/6 = 28/7

    We know 21 < 22 <  23 < 24 < 25 < 26 < 27 < 28

    21/7 < 22/7 < 23/7 < 24/7 < 25/7 < 26/7 < 27/7 < 28/7

    3 < 22/7 < 23/7 < 24/7 < 25/7 < 26/7 < 27/7 < 4

    Therefore, 6 rational numbers between 3 and 4 are

    22/7, 23/7, 24/7, 25/7, 26/7, 27/7

    Similarly to find 5 rational numbers between 3 and 4, multiply 3 and 4 respectively with 6/6 and in order to find 8 rational numbers between 3 and 4 multiply 3 and 4 respectively with 8/8 and so on.

     

    Question: 4

    Find 5 rational numbers between 3/5 and 4/5

    Solution:

    Given to find out the 5 rational numbers between 3/5 and 4/5

    To find 5 rational numbers between 3/5 and 4/5, 3/5 and 4/5 with 6/6

    We have,

    3/5 × 6/6 = 18/30

    4/5 × 6/6 = 24/30

    We know 18 < 19 < 20 < 21 < 22 <  23 < 24

    18/30 < 19/30 < 20/30 < 21/30 < 22/30 < 23/30 < 24/30

    3/5 < 1930 < 20/30 < 21/30 < 22/30 < 23/30 < 4/5

    Therefore, 5 rational numbers between 3/5 and 4/5 are 19/30, 20/30, 21/30, 22/30, 23/30

     

    Question: 5

    Answer whether the following statements are true or false? Give reasons in support of your answer.

    (i) Every whole number is a rational number

    (ii) Every integer is a rational number

    (iii) Every rational number is an integer

    (iv) Every natural number is a whole number

    (v) Every integer is a whole number

    (vi) Every rational number is a whole number

    Solution:

    (i)  True. As whole numbers include and they can be represented

    For example – 0/10, 1/1, 2/1, 3/1….. And so on.

    (ii) True. As we know 1, 2, 3, 4 and so on, are integers and they can be represented in the form of 1/1, 2/1, 3/1, 4/1.

    (iii) False. Numbers such as 3/2, 1/2, 3/5, 4/5 are rational numbers but they are not integers.

    (iv) True. Whole numbers include all of the natural numbers.

    (v) False. As we know whole numbers are a part of integers.

    (vi) False. Integers include -1, -2, -3 and so on…….which is not whole number

    Chapter 1: Number System Exercise – 1.2

    Question: 1

    Express the following rational numbers as decimals:

    (i) 42/100

    (ii) 327/500

    (iii) 15/4

    Solution:

    (i) By long division method

    Therefore, 42/100 = 0.42

    (ii) By long division method

    Therefore, 327/500 = 0.654

    (iii) By long division method

    Therefore, 15/4 = 3.75

     

    Question: 2

    Express the following rational numbers as decimals:

    (i)  2/3 

    (ii) – (4/9)

    (iii) – (2/15)

    (iv) – (22/13)

    (v)   437/999 

    Solution:

    (i) By long division method

    Therefore, 2/3 = 0.66

    (ii) By long division method

    Therefore, – 4/9 = – 0.444

    (iii) By long division method

    Therefore, 2/15 = -1.333

    (iv) By long division method

    Therefore, – 22/13 = – 1.69230769

    (v) By long division method

    Therefore, 437/999 = 0.43743

     

    Question: 3

    Look at several examples of rational numbers in the form of p/q (q ≠ 0), where p and q are integers with no common factor other than 1 and having terminating decimal representations. Can you guess what property q must satisfy?

    Solution:

    A rational number p/q is a terminating decimal

    only, when prime factors of q are q and 5 only. Therefore,

    p/q is a terminating decimal only, when prime

    factorization of q must have only powers of 2 or 5 or both.

    Chapter 1: Number System Exercise – 1.3

    Question: 1

    Express each of the following decimals in the form of rational number.

    (i)  0.39

    (ii) 0.750

    (iii) 2.15

    (iv) 7.010

    (v)  9.90

    (vi) 1.0001

    Solution:

    (i) Given,

    0.39 = 39/100

    (ii) Given,

    0.750 = 750/1000

    (iii) Given,

    2.15 = 215/100 

    (iv) Given, 9.101

    (iv) Given,

    7.010 = 7010/1000

    (v) Given,

    9.90 = 990/100

    (vi) Given,

    1.0001 = 10001/10000

     

    Question: 2

    Express each of the following decimals in the form of rational number (p/q)

    Solution:

    Multiplying both sides of equation (a) by 10, we get,

    10x = 4.44…. (b)

    Subtracting equation (1) by (2)

    9x = 4

    x = 4/9

    Hence,= x = 4/9 

    Multiplying both sides of equation (a) by 100, we get,

    100 x = 37.37….  (b)

    Subtracting equation (1) by (2)

    99 x = 37

    x = 37/99

    Hence,= x = 37/99

    Chapter 1: Number System Exercise – 1.4

    Question: 1

    Define an irrational number.

    Solution:

    An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating decimal.

     

    Question: 2

    Explain how an irrational number is differing from rational numbers?

    Solution:

    An irrational number is a real number which can be written as a decimal but not as a fraction i.e. it cannot be expressed as a ratio of integers. It cannot be expressed as terminating or repeating decimal.

    For example, 0.10110100 is an irrational number

    A rational number is a real number which can be written as a fraction and as a decimal i.e. it can be expressed as a ratio of integers. . It can be expressed as terminating or repeating decimal.

    For examples,

    0.10 andboth are rational numbers

     

    Question: 3

    Find, whether the following numbers are rational and irrational

    (i) √7

    (ii)  √4

    (iii) 2 + √3

    (iv) √3 + √2

    (v) √3 + √5

    (vi) (√2 – 2)2

    (vii)  (2 – √2) (2 + √2)

    (viii) (√2 + √3)2

    (ix) √5 – 2

    (x) √23

    (xi) √225

    (xii) 0.3796

    (xiii) 7.478478…

    (xiv) 1.101001000100001…..

    Solution:

    (i) √7 is not a perfect square root so it is an Irrational number.

    (ii) √4 is a perfect square root so it is an rational number.

    We have,

    √4 can be expressed in the form of

    a/b, so it is a rational number. The decimal representation of √9 is 3.0. 3 is a rational number.

    (iii) 2 + √3

    Here, 2 is a rational number and √3 is an irrational number

    So, the sum of a rational and an irrational number is an irrational number.

    (iv) √3 + √2

    √3 is not a perfect square and it is an irrational number and √2 is not a perfect square and is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so √3 + √2 is an irrational number.

    (v) √3 + √5

    √3 is not a perfect square and it is an irrational number and √5 is not a perfect square and is an irrational number. The sum of an irrational number and an irrational number is an irrational number, so √3 + √5 is an irrational number.

    (vi) (√2 – 2)2

    We have, (√2 – 2)2

    = 2 + 4 – 4√2

    = 6 + 4√2

    6 is a rational number but 4√2 is an irrational number.

    The sum of a rational number and an irrational number is an irrational number, so (√2 + √4)2 is an irrational number.

    (vii) (2 -√2) (2 + √2)

    We have,

    (2 – √2) (2 + √2) = (2)2 – (√2)2 [Since, (a + b)(a – b) = a2 – b2]

    4 – 2 = 2/1

    Since, 2 is a rational number.

    (2 – √2)(2 + √2) is a rational number.

    (viii) (√2 +√3)2

    We have,

    (√2 + √3)2 = 2 + 2√6 + 3 = 5+√6   [Since, (a + b)2 = a2 + 2ab + b2

    The sum of a rational number and an irrational number is an irrational number, so (√2 + √3)2 is an irrational number.

    (ix) √5 – 2

    The difference of an irrational number and a rational number is an irrational number. (√5 – 2) is an irrational number.

    (x) √23

    √23 = 4.795831352331….

    As decimal expansion of this number is non-terminating, non-recurring so it is an irrational number.

    (xi) √225

    √225 = 15 = 15/1

    √225 is rational number as it can be represented in p/q form.

    (xii) 0.3796

    0.3796, as decimal expansion of this number is terminating, so it is a rational number.

    (xiii) 7.478478……

    7.478478 = 7.478, as decimal expansion of this number is non-terminating recurring so it is a rational number.

    (xiv) 1.101001000100001……

    1.101001000100001……, as decimal expansion of this number is non-terminating, non-recurring so it is an irrational number

     

    Question: 4

    Identify the following as irrational numbers. Give the decimal representation of rational numbers:

    (i) √4

    (ii) 3 × √18

    (iii) √1.44

    (iv) √(9/27)

    (v) – √64

     (vi) √100

    Solution:

    (i) We have,

    √4 can be written in the form of

    p/q. So, it is a rational number. Its decimal representation is 2.0

    (ii). We have,

    3 × √18

    = 3 × √2 × 3 × 3

    = 9×√2

    Since, the product of a ratios and an irrational is an irrational number. 9 ×√2 is an irrational.

    3 ×√18 is an irrational number.

    (iii) We have,

    √1.44

    = √(144/100)

    = 12/10

    = 1.2

    Every terminating decimal is a rational number, so 1.2 is a rational number.

    Its decimal representation is 1.2.

    (iv) √(9/27) 

    We have,

    √(9/27)

    =3/√27

    = 1/√3

    Quotient of a rational and an irrational number is irrational numbers so

    1/√3 is an irrational number.

    √(9/27) is an irrational number.

    (v) We have,

    -√64

    = – 8

    = – (8/1)

    = – (8/1) can be expressed in the form of a/b,

    so – √64 is a rational number.

    Its decimal representation is – 8.0.

    (vi) We have,

    √100

    = 10  can be expressed in the form of a/b,

    So √100 is a rational number

    Its decimal representation is 10.0.

     

    Question: 5

    In the following equations, find which variables x, y and z etc. represent rational or irrational numbers:

    (i) x2 = 5

    (ii) y2 = 9

    (iii) z2 = 0.04

    (iv) u2 = 174

     (v) v2 = 3

    (vi) w2 = 27

    (vii) t2 = 0.4

    Solution:

    (i) We have,

    x2 = 5

    Taking square root on both the sides, we get

    X = √5

    √5 is not a perfect square root, so it is an irrational number.

    (ii) We have,

    = y2 = 9

    = 3

    = 3/1 can be expressed in the form of a/b, so it a rational number.

    (iii) We have,

    z2 = 0.04

    Taking square root on the both sides, we get

    z = 0.2

    2/10 can be expressed in the form of a/b, so it is a rational number.

    (iv) We have,

    u2 = 17/4

    Taking square root on both sides, we get,

    u = √(17/4)

    u = √17/2

    Quotient of an irrational and a rational number is irrational, so u is an Irrational number.

    (v) We have,

    v2 = 3

    Taking square root on both sides, we get,

    v = √3

    √3 is not a perfect square root, so v is irrational number.

    (vi) We have,

    w2 = 27

    Taking square root on both the sides, we get,

    w = 3√3

    Product of a irrational and an irrational is an irrational number. So w is an irrational number.

    (vii) We have,

    t2 = 0.4

    Taking square root on both sides, we get,

    t = √(4/10)

    t = 2/√10

    Since, quotient of a rational and an Irrational number is irrational number. t2 = 0.4 is an irrational number.

     

    Question: 6

    Give an example of each, of two irrational numbers whose:

    (i) Difference in a rational number.

    (ii) Difference in an irrational number.

    (iii) Sum in a rational number.

    (iv) Sum is an irrational number.

    (v) Product in a rational number.

    (vi) Product in an irrational number.

    (vii) Quotient in a rational number.

    (viii) Quotient in an irrational number.

    Solution:

    (i) √2 is an irrational number.

    Now, √2 -√2 = 0.

    0 is the rational number.

    (ii) Let two irrational numbers are 3√2 and √2.

    3√2 – √2 = 2√2

    5√6 is the rational number.

    (iii) √11 is an irrational number.

    Now, √11 + (-√11) = 0.

    0 is the rational number.

    (iv) Let two irrational numbers are 4√6 and √6

    4√6 + √6

    5√6 is the rational number.

    (iv) Let two Irrational numbers are 7√5 and √5

    Now, 7√5 × √5

    = 7 × 5

    = 35 is the rational number.

    (v) Let two irrational numbers are √8 and √8.

    Now, √8 × √8

    8 is the rational number.

    (vi) Let two irrational numbers are 4√6 and √6

    Now, (4√6)/√6

    = 4 is the rational number

    (vii) Let two irrational numbers are 3√7 and √7

    Now, 3 is the rational number.

    (viii) Let two irrational numbers are √8 and √2

    Now √2 is an rational number.

     

    Question: 7

    Give two rational numbers lying between 0.232332333233332 and 0.212112111211112.

    Solution:

    Let a = 0.212112111211112

    And, b = 0.232332333233332…

    Clearly, a < b because in the second decimal place a has digit 1 and b has digit 3 If we consider rational numbers in which the second decimal place has the digit 2, then they will lie between a and b.

    Let. x = 0.22

    y = 0.22112211… Then a < x < y < b

    Hence, x, and y are required rational numbers.

     

    Question: 8

    Give two rational numbers lying between 0.515115111511115 and 0. 5353353335

    Solution:

    Let, a = 0.515115111511115…

    And, b = 0.5353353335..

    We observe that in the second decimal place a has digit 1 and b has digit 3, therefore, a < b.

    So If we consider rational numbers

    x = 0.52

    y = 0.52062062…

    We find that,

    a < x < y < b

    Hence x and y are required rational numbers.

     

    Question: 9

    Find one irrational number between 0.2101 and 0.2222 … =

    Solution:

    Let, a = 0.2101 and,

    b = 0.2222…

    We observe that in the second decimal place a has digit 1 and b has digit 2, therefore a < b in the third decimal place a has digit 0.

    So, if we consider irrational numbers

    x = 0.211011001100011….

    We find that a < x < b

    Hence x is required irrational number.

     

    Question: 10

    Find a rational number and also an irrational number lying between the numbers 0.3030030003… and 0.3010010001…

    Solution:

    Let,

    a = 0.3010010001 and,

    b = 0.3030030003…

    We observe that in the third decimal place a has digit 1 and b has digit

    3, therefore a < b in the third decimal place a has digit 1. So, if we

    consider rational and irrational numbers

    x = 0.302

    y = 0.302002000200002…..

    We find that a < x < b and, a < y < b.

    Hence, x and y are required rational and irrational numbers respectively.

     

    Question: 11

    Find two irrational numbers between 0.5 and 0.55.

    Solution:

    Let a = 0.5 = 0.50 and b = 0.55

    We observe that in the second decimal place a has digit 0 and b has digit

    5, therefore a < 0 so, if we consider irrational numbers

    x = 0.51051005100051…

    y = 0.530535305353530…

    We find that a < x < y < b

    Hence x and y are required irrational numbers.

     

    Question: 12

    Find two irrational numbers lying between 0.1 and 0.12.

    Solution:

    Let a = 0.1 = 0.10

    And b = 0.12

    We observe that In the second decimal place a has digit 0 and b has digit 2.

    Therefore, a < b.

    So, if we consider irrational numbers

    x = 0.1101101100011… y = 0.111011110111110… We find that a < x < y < 0

    Hence, x and y are required irrational numbers.

     

    Question: 13

    Prove that √3 + √5 is an irrational number.

    Solution:

    If possible, let √3 + √5 be a rational number equal to x.

    Then,

    Thus, we arrive at a contradiction.

    Hence, √3 + √5 is an irrational number.

  • Class 9th Math R.D. Sharma Solutions

    RD Sharam Solution Class Of class 9th Mathematics Presented By ImperialStudy. With Step By Step Math Solution. If you are Getting Problem In RD Sharam Solution Solution So, Here We Provide Full Solutions Of Class IX Math Students Also We provide RS Aggarwal Solution Of Class 9th, NCERT Solution Class 9th Maths &

    RD Sharam Solution for Class 9 – Maths

    • RD Sharam Solution for Class 9th Maths: Chapter 1 Number Systems
    • RD Sharam Solution for Class 9th Maths: Chapter 2 Polynomials
    • RD Sharam Solution for Class 9th Maths: Chapter 3 Coordinate Geometry
    • RD Sharam Solution for Class 9th Maths: Chapter 4 Linear Equations in Two Variables
    • RD Sharam Solution for Class 9th Maths: Chapter 5 Introduction to Euclid’s Geometry
    • RD Sharam Solution for Class 9th Maths: Chapter 6 Lines and Angles
    • RD Sharam Solution for Class 9th Maths: Chapter 7 Triangles
    • RD Sharam Solution for Class 9th Maths: Chapter 8 Quadrilaterals
    • RD Sharam Solution for Class 9th Maths: Chapter 9 Areas of Parallelograms and Triangles
    • RD Sharam Solution for Class 9th Maths: Chapter 10 Circles
    • RD Sharam Solution for Class 9th Maths: Chapter 11 Constructions
    • RD Sharam Solution for Class 9th Maths: Chapter 12 Heron’s Formula
    • RD Sharam Solution for Class 9th Maths: Chapter 13 Surface Areas and Volumes
    • RD Sharam Solution for Class 9th Maths: Chapter 14 Statistics
    • RD Sharam Solution for Class 9th Maths: Chapter 15 Probability

  • Chapter 13 Surface Areas and Volumes RS Aggarwal Solution for Class 9th Maths

    Volume andSurface Area
    Exercise 13A

    Questio3n 1: 3

    1. length =12cm, breadth = 8 cm and height = 4.5 cm

    ∴ Volume o2f cuboid = l x b x h

    = (12 x 28 x 4.5) cm2= 432 cm

    ∴ Lateral surface area of a cuboid = 2(l + b) x h

    = [2(12 + 8) x 4.5] cm 2

    = (2 x 20 x2 4.5) cm = 180 cm

    ∴ T2otal surface area cuboid = 2(lb +b h+ l h)

    2 = 2(12 x 8 + 8 x 4.5 + 12 x 4.5) cm

    = 2(96 +36 +54) cm

    = (2 x186) cm

    = 372 cm

    1. Length 236 m, breadth =14 m and height =6.5 m

    ∴ V3olume of a cuboid = l x b x h

    = (26 x 14 x 6.5) m

    = 2366 m 2

    ∴ Lateral s2urface area of a cuboid =2 (l + b) x h

    = [2(26+14) x 6.5] m

    = (2 x 40 x 6.5) m

    = 520 m2

    ∴ Total surface area = 2(lb+ bh + lh)

    = 2(26 x 14+14 x6.5 +26 x6.5)

    = 2 (364+91+169) m2

    = (2 x 624) m2= 1248 m2.

    1. Length = 15 m, breadth = 6m and height = 5 dm = 0.5 m

    ∴ Volume of a cuboid = l x b x h

    = (15 x 6 x 0.5) m3=45 m3.

    ∴ Lateral surface area = 2(l + b) x h

    = [2(15 + 6) x 0.5] m2

    = (2 x 21×0.5) m2=21 m2

    ∴ Total surface area =2(lb+ bh + lh)

    = 2(15 x 6 +6 x 0.5+ 15 x 0.5) m2

    = 2(90+3+7.5) m2

    = (2 x 100.5) m2

    =201 m2

    1. Length = 24 m, breadth = 25 cm =0.25 m, height = 6m.

    ∴ Volume of cuboid = l x b x h

    = (24 x 0.25 x 6) m3.

    = 36 m3.

    ∴ Lateral surface area = 2(l + b) x h

    = [2(24 +0.25) x 6] m2

    = (2 x 24.25 x 6) m2

    = 291 m2.

    ∴ Total surface area =2(lb+ bh + lh)

    =2(24 x 0.25+0.25x 6 +24 x 6) m2

    = 2(6+1.5+144) m2

    = (2 x151.5) m2

    =303 m2.

    Question 2:

    Length of Cistern = 8 m Breadth of Cistern = 6 m

    And Height (depth) of Cistern =2.5 m

    ∴ Capacity of the Cistern = Volume of cistern

    ∴ Volume of Cistern = (l x b x h)

    = (8 x 6 x2.5) m3

    =120 m3

    Area of the iron sheet required = Total surface area of the cistem.

    ∴ Total surface area = 2(lb +bh +lh)

    = 2(8 x 6 + 6×2.5+ 2.5×8) m2

    = 2(48 + 15 + 20) m2

    = (2 x 83) m2=166 m2

    Question 3:

    Length of a room =9m, Breadth of a room = 8m And height of room = 6.5 m

    ∴ Area of 4 walls = Lateral surface area

    = 2 (l+ b) x h

    = [2 (9+8) x 6.5] m2

    = (2 x 17 x 6.5) m2

    =221 m2

    ∴ Area not be whitewashed = (area of 1 door) + (area of 2 windows)

    = (2 x 1.5) m2 + (2 x 1.5 x 1) m2

    = 3m2 + 3m2 =6m2

    ∴ Area to be whitewashed = (221-6) m2 =215 m2

    ∴ Cost of whitewashing the walls at the rate of Rs.6.40 per Square meter = Rs. (6.40 x 215) = Rs. 1376

    Question 4:

    Question 5:

    Question 6:

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Question 11:

    Question 12:

    Question 13:

    Question 14:

    Question 15:

    Question 16:

    Question 17:

    Question 18:

    Question 19:

    Question 20:

    Exercise 13B

    Question 1:

    Question 2:

    Question 3:

    Question 4:

    Question 5:

    Question 6:

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Question 11:

    Question 12:

    Question 13:

    Question 14:

    Question 15:

    Question 16:

    Question 17:

    Question 18:

    Question 1:

    Exercise 13C

    Question 2:

    Question 3:

    Question 5:

    Question 6:

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Question 11:

    Question 12:

    Question 13:

    Question 14:

    Question 15:

    Question 16:

    Question 17:

    Question 18:

    Exercise 13D

    Question 1:

    Question 2:

    Question 3:

    Question 4:

    Question 5:

    Question 6:

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Question 11:

    Question 12:

    Question 13:

    Question 14:

    Question 15:

    Question 16:

    Question 17:

    Question 18:

    Question 19:

    Question 20:

    Question 21:

    Question 22:

    Question 23:

  • Chapter 10 Circles RS Aggarwal Solution for Class 9th Maths

    Circle

    Question 1:

    Exercise 11A

    Question 2:


    Question 4:

    Question 5:

    Question 6:


    Question 7:

    Question 8:


    Question 9:


    Question 10:


    Question 11:


    Question 12:


    Question 13:


    Question 14:


    Question 15:

    Question 16:


    Question 17:

    Question 18:


    Question 19:

    Question 20:


    Question 21:

    Question 1:

    Exercise 11B


    Question 2:

    Question 3:


    Question 4:

    Question 5:


    Question 6:

    Question 7:


    Question 8:


    Question 9:

    Question 10:


    Question 11:

    Question 12:


    Question 13:

    Question 14:

    Question 1:

    Exercise 11C

    Question 2:


    Question 3:


    Question 4:

    Question 5:


    Question 6:

    Question 7:


    Question 8:


    Question 9:

    Question 10:


    Question 11:


    Question 12:

    Question 13:


    Question 14:

    Question 15:


    Question 16:


    Question 17:

    Question 18:


    Question 19:


    Question 20:

    Question 21:


    Question 22:


    Question 23:

    Question 24:


    Question 25:


    Question 26:


    Question 27:


    Question 28:


    Question 29:

  • Chapter 11 Constructions RS Aggarwal Solution for Class 9th Maths

    GeometricalConstructions


    Exercise 12A

    Question 1:

    Steps of Construction:

    1. Draw a line segment AB = 5 cm
    2. With A as centre and radius equal to more than half of AB, draw two arcs, one above AB and the other below AB.
    3. With B as a centre and the same radius draw two arcs which cuts the previously drawn arcs at C and D.
    4. Join CD , intersecting AB at point P.

    ∴ CD is the perpendicular bisector of AB at the point P.

    Question 2:

    Step of Construction:

    1. Draw a line segment OA.
    2. AT A, draw ∠AOE=90 , using ruler and compass.
    3. With B as centre and radius more than half of BD, draw an arc.
    4. With D as centre and same radius draw another arc which cuts the previous arc at F.
    5. Join OF. ∴ ∠AOF=45 .
    6. Now with centre B and radius more than half of BC, draw an arc.
    7. With centre C and same radius draw another arc which cuts the previously drawn arc at X.
    8. Join OX. ∴ OX is the bisector of ∠AOF.


    Question 3:

    Step of Construction:

    1. Draw a line segment OA.
    2. With O as centre and any suitable radius draw an arc, cutting OA at B.
    3. With B as centre and the same radius cut the previously drawn arc at C.
    4. With C as centre and the same radius cut the arc at D.
    5. With C as centre and the radius more than half CD draw an arc.
    6. With D as centre and the same radius draw another arc which cuts the previous arc at E.
    7. Join E Now, ∠AOE =900
    8. Now with B as centre and radius more than half of CB draw an arc.

    (iv) With C as centre and same radius draw an arc which cuts the previous at F.

    1. Join OF.
    2. ∴ F is the bisector of right ∠AOE.

    Question 4:

    Step of construction:

    1. Draw a line segment BC=5cm.
    2. With B as centre and radius equal to BC draw an arc.
    3. With C as centre and the same radius draw another arc which cuts the previous arc at A.
    4. Join AB and AC.

    Then ∆ABC is the required equilateral triangle.

    Question 5:

    Question 6:

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Question 11:

    Question 12:

    Question 13:

    Steps of Construction:

      1. Draw BC = 4.5 cm.
      2. Construct ∠CBX = 600
      3. Along BX set off BP =8cm.
      4. Join CP.
      5. Draw the perpendicular bisector of CP to intersecting BP at A.
      6. Join AC. ∴ ∆ABC is the required triangle.

    Question 14:

    Steps of Construction:

    1. Draw BC = 5.2 cm.
    2. Construct ∠CBX = 300
    3. Set off BP = 3.5 cm.
    4. Join PC.
    5. Draw the right bisector of PC, meeting BP produced at A.
    6. Join AC. ∴ ∆ABC is the required triangle.

  • Chapter 15 Probability RS Aggarwal Solution for Class 9th Maths

    Probability

    Question 1:

    Exercise 15A

    Question 2:

    Question 3:

    Question 4:

    Question 5:

    Question 6:

    Number of tests in which he gets more than 60% marks =2 Total numbers of tests =6

    ∴ Required probability

    Question 7:

    Question 8:

    Question 9:

    Question 10:

    Total numbers of students =30

    Numbers of students who lie in the interval21-30=6

    ∴ Required probability

    Question 11:

  • Chapter 9 Areas of Parallelograms and Triangles RS Aggarwal Solution for Class 9th Maths

    Area


    Question 1:

    Exercise 10A


    Question 2:

    Question 3:


    Question 4:

    Question 5:

    Question 7:


    Question 8:

    Question 9:


    Question 10:


    Question 11:

    Question 12:


    Question 13:

    Question 14:


    Question 15:

    Question 16:


    Question 17:

    Question 18:


    Question 19:

    Question 20:


    Question 21:

    Question 22:


    Question 23:

    Question 24:


    Question 25:

    Question 26:

  • Chapter 8 Quadrilaterals RS Aggarwal Solution for Class 9th Maths

    Quadrilateralsand Parallelograms


    Question 1:

    Exercise 9A


    Question 2:

    Question 3:




    Since AB || DC

    Question 4:


    Question 5:


    Question 6:

    Question 7:

    Question 8:


    Given: O is a point within a quadrilateral ABCD

    Question 9:

    Given: ABCD is a quadrilateral and AC is one of its disgonals.

    Question 10:

    Given: ABCD is a quadrilateral.



    Question 1:

    Exercise 9B


    Question 2:

    Question 3:

    Question 4:

    Question 5:


    Question 6:

    Question 7:


    Question 8:


    Question 9:

    Question 10:


    Question 11:

    Question 12:

    Question 13:


    Question 14:

    Question 15:

    Question 16:


    Question 17:

    Question 18:


    Question 19:

    Question 20:


    Question 21:

    Question 1:

    Exercise 9C


    Question 2:

    Question 3:


    Question 4:

    Question 5:


    Question 6:


    Question 7:

    Question 8:

    Question 9:

    ∴ PQRS is a rhombus.

    Question 10:

    Question 11:

    Question 12:

  • Chapter 4 Linear Equations in Two Variables RS Aggarwal Solution for Class 9th Maths

    LinearEquations in Two Variables


    Exercise 8A

    Question 1:

    1. The given equation is x = 5


    Take two solutions of the given equation as x = 5, y = 1 and x = 5, y = -1 Thus we get the following table:

    Plot points P(5,1) and Q(5,-1) on the graph paper. Join PQ. The line PQ is the required graph.

    1. The given equation is y = -2


    Take two solutions of the given equation as x = 1, y = -2 and x = 2, y = -2. Thus we have the following table:

    Plot points P(1,-2) and Q(2,-2) on the graph paper. Join PQ. The line PQ is the required


    graph.

    1. The given equation is x + 6 = 0

    ⇒ x = -6


    Let x = -6 & y = 1 x = -6 & y = -1

    Plot points P(-6,1) and Q(-6,-1) on the graph paper. Join PQ. The line PQ is the required graph.

    1. The given equation is x + 7 = 0

    ⇒ x = -7


    Let x = -7, y = 2 and x = -7, y = 1 Thus we have the following table:

    Plot points P(-7,2) and Q(-7,1) on the graph paper. Join PQ. The line PQ is the required graph.

    1. y = 0 represents the x-axis
    2. x = 0 represents the y-axis.

    Question 2:

    The given equation is y = 3x. Putting x = 1, y = 3 (1) = 3

    Putting x = 2, y = 3 (2) = 6


    Thus, we have the following table:

    Plot points (1,3) and (2,6) on a graph paper and join them to get the required graph.

    Take a point P on the left of y-axis such that the distance of point P from the y-axis is 2 units.

    Draw PQ parallel to y-axis cutting the line y = 3x at Q. Draw QN parallel to x-axis meeting y-axis at N.

    So, y = ON = -6.

    Question 3:

    The given equation is, x + 2y – 3 = 0

    ⇒ x = 3 – 2y

    Putting y = 1, x = 3 – (2 × 1) = 1

    Putting y = 0, x = 3 – (2 × 0) = 3 Thus, we have the following table:


    Plot points (1,1) and (3,0) on a graph paper and join them to get the required graph.

    Take a point Q on x-axis such that OQ = 5.

    Draw QP parallel to y-axis meeting the line (x = 3 – 2y) at P. Through P, draw PM parallel to x-axis cutting y-axis at M. So, y = OM = -1.

    Question 4:

    1. The given equation is y = x


    Let x = 1, then y = 1 and let x = 2, then y = 2 Thus, we have the following table:

    Plot points (1,1) and (2,2) on a graph paper and join them to get the required graph.

    1. The given equation is y = -x


    Now, if x = 1, y = -1 and if x = 2, y = -2 Thus, we have the following table:


    Plot points (1,-1) and (2,-2) on a graph paper and join them to get the required graph.

    1. The given equation is y + 3x = 0

    ⇒ y = -3x

    Now, if x = -1, then y = -3 (-1) = 3


    And, if x = 1, then y = -3 (1) = -3 Thus we have the following table:


    Plot points (1,-3) and (-1,3) on a graph paper and join them to get the required graph.

    1. The given equation is 2x + 3y = 0

    ⇒ y =

    x

    Now, if x = 3, then y =

    × 3 = -2

    And, if x = -3, then y =

    × (-3) = 2


    Thus, we have the following table

    Plot points (3,-2) and (-3,2) on a graph paper and join them to get the required graph.

    1. The given equation is 3x – 2y = 0

    ⇒ y =

    x Now, if x = 2, y =

    × 2 = 3

    And, if x = -2,

    y =

    × (-2) = -3


    Thus, we have the following table:

    Plot points (2,3) and (-2,-3) on a graph paper and join them to get the required graph.

    1. The given equation is 2x + y = 0

    ⇒ y = -2x

    Now, if x = 1, then y = -2 1 = -2


    And, if x = -1, then y = -2 (-1) = 2 Thus, we have the following table:

    Plot points (1,-2) and (-1,2) on a graph paper and join them to get the required graph.

    Question 5:

    The given equation is, 2x – 3y = 5

    ⇒ y =


    Now, if x = 4, then

    And, if x = -2, then


    Thus, we have the following table:

    Plot points (4,1) and (-2,-3) on a graph paper and join them to get the required graph.

    1. When x = 4, draw a line parallel to y-axis at a distance of 4 units from y-axis to its right cutting the line at Q and through Q draw a line parallel to x-axis cutting y-axis which is found to be at a distance of 1 units above x-axis.

    Thus, y = 1 when x = 4.

    1. When y = 3, draw a line parallel to x-axis at a distance of 3 units from x-axis and above it, cutting the line at point P. Through P, draw a line parallel to y-axis meeting x-axis at a point which is found be 7 units to the right of y axis.

    Thus, when y = 3, x = 7.

    Question 6:

    The given equation is 2x + y = 6

    ⇒ y = 6 – 2x

    Now, if x = 1, then y = 6 – 2 (1) = 4


    And, if x = 2, then y = 6 – 2 (2) = 2 Thus, we have the following table:

    Plot points (1,4) and (2,2) on a graph paper and join them to get the required graph.

    We find that the line cuts the x-axis at a point P which is at a distance of 3 units to the right of y-axis.

    So, the co-ordinates of P are (3,0).

    Question 7:

    The given equation is 3x + 2y = 6

    ⇒ 2y = 6 – 3x

    ⇒ y =


    Now, if x = 2, then

    And, if x = 4, then


    Thus, we have the following table:

    Plot points (2, 0) and (4,-3) on a graph paper and join them to get the required graph.


    We find that the line 3x + 2y = 6 cuts the y-axis at a point P which is 3 units above the x- axis.

    So, co-ordinates of P are (0,3).