Category: CBSE Class 9

RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3

Question 1.
Find the cube of each of the following binomial expressions:

Solution:

Question 2.
If a + b = 10 and ab = 21, find the value of a³ + b³.
Solution:
a + b = 10, ab = 21
Cubing both sides,
(a + b)³ = (10)³
⇒ a³ + 6³ + 3ab (a + b) = 1000
⇒  a³ + b³ + 3 x 21 x 10 = 1000
⇒  a³ + b³ + 630 = 1000
⇒  a³ + b³ = 1000 – 630 = 370
∴ a³ + b³ = 370

Question 3.
If a – b = 4 and ab = 21, find the value of a³-b³.
Solution:
a – b = 4, ab= 21
Cubing both sides,
⇒ (a – A)³ = (4)³
⇒ a³ – b³ – 3ab (a – b) = 64
⇒ a³-i³-3×21 x4 = 64
⇒  a³ – 6³ – 252 = 64
⇒  a³ – 6³ = 64 + 252 =316
∴ a³ – b³ = 316

Question 4.

Solution:

Question 5.

Solution:

Question 6.

Solution:

Question 7.

Solution:

Question 8.

Solution:

Question 9.
If 2x + 3y = 13 and xy = 6, find the value of 8x³ + 21y³.
Solution:
2x + 3y = 13, xy = 6
Cubing both sides,
(2x + 3y)³ = (13)³
⇒ (2x)³ + (3y)³ + 3 x 2x x 3X2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18xy(2x + 3y) = 2197
⇒ 8x³ + 27y³ + 18 x 6 x 13 = 2197
⇒ 8X³ + 27y³ + 1404 = 2197
⇒  8x³ + 27y³ = 2197 – 1404 = 793
∴ 8x³ + 27y³ = 793

Question 10.
If 3x – 2y= 11 and xy = 12, find the value of 27x³ – 8y³.
Solution:
3x – 2y = 11 and xy = 12 Cubing both sides,
(3x – 2y)³ = (11)³
⇒  (3x)³ – (2y)³ – 3 x 3x x 2y(3x – 2y) =1331
⇒  27x³ – 8y³ – 18xy(3x -2y) =1331
⇒   27x³ – 8y³ – 18 x 12 x 11 = 1331
⇒  27x³ – 8y³ – 2376 = 1331
⇒  27X³ – 8y³ = 1331 + 2376 = 3707
∴ 2x³ – 8y³ = 3707

Question 11.
Evaluate each of the following:
(i)  (103)³
(ii) (98)³
(iii) (9.9)³
(iv) (10.4)³
(v) (598)³
(vi) (99)³
Solution:
We know that (a + bf = a³ + b³ + 3ab(a + b) and (a – b)³= a³ – b³ – 3 ab(a – b)
Therefore,
(i)  (103)³ = (100 + 3)³
= (100)³ + (3)³ + 3 x 100 x 3(100 + 3)    {∵ (a + b)³ = a³ + b³ + 3ab(a + b)}
= 1000000 + 27 + 900 x 103
= 1000000 + 27 + 92700
= 1092727
(ii) (98)³ = (100 – 2)³
= (100)³ – (2)³ – 3 x 100 x 2(100 – 2)
= 1000000 – 8 – 600 x 98
= 1000000 – 8 – 58800
= 1000000-58808
= 941192
(iii) (9.9)³ = (10 – 0.1)³
= (10)³ – (0.1)³ – 3 X 10 X 0.1(10 – 0.1)
= 1000 – 0.001 – 3 x 9.9
= 1000 – 0.001 – 29.7
= 1000 – 29.701
= 970.299
(iv) (10.4)³ = (10 + 0.4)³
= (10)³ + (0.4)³ + 3 x 10 x 0.4(10 + 0.4)
= 1000 + 0.064 + 12(10.4)
= 1000 + 0.064 + 124.8 = 1124.864
(v) (598)³ = (600 – 2)³
= (600)³ – (2)³ – 3 x 600 x 2 x (600 – 2)
= 216000000 – 8 – 3600 x 598
= 216000000 – 8 – 2152800
= 216000000 – 2152808
= 213847192
(vi) (99)³ = (100 – 1)³
= (100)³ – (1)³ – 3 x 100 x 1 x (100 – 1)
= 1000000 – 1 – 300 x 99
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299

Question 12.
Evaluate each of the following:
(i)  111³ – 89³
(ii) 46³ + 34³
(iii) 104³ + 96³
(iv) 93³ – 107³
Solution:
We know that a³ + b³ = (a + bf – 3ab(a + b) and a³ – b³ = (a – bf + 3 ab(a – b)
(i) 111³ – 89³
= (111 – 89)³ + 3 x ill x 89(111 – 89)
= (22)³ + 3 x 111 x 89 x 22
= 10648 + 652014 = 662662
(OR)
(a + b)³ – (a – b)³ = 2(b³ + 3a²b)
= 111³ – 89³ = (100 + 11)³ – (100 – 11)³
= 2(11³ + 3 x 100² x 11]
= 2(1331 + 330000]
= 331331 x 2 = 662662
(a + b)³ + (a- b)³ = 2(b³ + 3ab²)
(ii) 46³ + 34³ = (40 + 6)³ + (40 – 6)³
= 2[(40)³ + 3 x 40 x 6²]
= 2[64000 + 3 x 40 x 36]
= 2[64000 + 4320]
= 2 x 68320 = 136640
(iii) 104³ + 96³ = (100 + 4)³ + (100 – 96)³
= 2 [a³ + 3 ab²]
= 2[(100)³ + 3 x 100 x (4)²]
= 2[ 1000000 + 300 x 16]
= 2[ 1000000 + 4800]
= 1004800 x 2 = 2009600
(iv) 93³ – 107³ = -[(107)³ – (93)³]
= -[(100 + If – (100 – 7)³]
= -2[b³ + 3a²b)]
= -2[(7)³ + 3(100)² x 7]
= -2(343 + 3 x 10000 x 7]
= -2[343 + 210000]
= -2[210343] = -420686

Question 13.

Solution:

Question 14.
Find the value of 27X³ + 8y³ if
(i) 3x + 2y = 14 and xy = 8
(ii) 3x + 2y = 20 and xy = \(\frac { 14 }{ 9 }\)
Solution:

Question 15.
Find the value of 64x³ – 125z³, if 4x – 5z = 16 and xz = 12.
Solution:
4x – 5z = 16, xz = 12
Cubing both sides,
(4x – 5z)³ = (16)³
⇒ (4x)³ – (5y)³ – 3 x 4x x 5z(4x – 5z) = 4096
⇒ 64x³ – 125z³ – 3 x 4 x 5 x xz(4x – 5z) = 4096
⇒  64x³ – 125z³ – 60 x 12 x 16 = 4096
⇒ 64x³ – 125z³ – 11520 = 4096
⇒  64x³ – 125z³ = 4096 + 11520 = 15616

Question 16.

Solution:

Question 17.
Simplify each of the following:

Solution:

Question 18.

Solution:

Question 19.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 4 Algebraic Identities Ex 4.3 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS

Question 1.
Write the value of (2 + \(\sqrt { 3 } \) ) (2 – \(\sqrt { 3 } \)).
Solution:
(2+ \(\sqrt { 3 } \) )(2- \(\sqrt { 3 } \) ) = (2)²-(\(\sqrt { 3 } \) )²
{∵ (a + b) (a – b) = a² – b²}
= 4-3=1

Question 2.
Write the reciprocal of 5 + \(\sqrt { 2 } \).
Solution:

Question 3.
Write the rationalisation factor of 7 – 3\(\sqrt { 5 } \) .
Solution:
Rationalising factor of 7 – 3\(\sqrt { 5 } \) is 7 + 3\(\sqrt { 5 } \)
{∵ (\(\sqrt { a } \) + \(\sqrt { b } \)  ) (\(\sqrt { a } \) – \(\sqrt { b } \)) = a-b}

Question 4.

Solution:

Question 5.
If x =\(\sqrt { 2 } \) – 1 then write the value of \(\frac { 1 }{ x }\).
Solution:

Question 6.
If a = \(\sqrt { 2 } \) + h then find the value of a –\(\frac { 1 }{ a }\)
Solution:

Question 7.
If x = 2 + \(\sqrt { 3 } \), find the value of x + \(\frac { 1 }{ x }\).
Solution:

Question 8.
Write the rationalisation factor of \(\sqrt { 5 } \) – 2.
Solution:
Rationalisation factor of \(\sqrt { 5 } \) – 2 is \(\sqrt { 5 } \) + 2 as
(\(\sqrt { a } \) + \(\sqrt { b } \))(\(\sqrt { a } \) – \(\sqrt { b } \)) = a – b

Question 9.
Simplify : \(\sqrt { 3+2\sqrt { 2 } }\).
Solution:

Question 10.
Simplify : \(\sqrt { 3-2\sqrt { 2 } }\).
Solution:

Question 11.
If x = 3 + 2 \(\sqrt { 2 } \), then find the value of  \(\sqrt { x } \) – \(\frac { 1 }{ \sqrt { x } }\).
Solution:

Hope given RD Sharma Class 9 Solutions Chapter 3 Rationalisation VSAQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1

Question 1.
Simplify each of the following:

Solution:

Question 2.
Simplify the following expressions:
(i)  (4 + \(\sqrt { 7 } \)) (3 + \(\sqrt { 2 } \))
(ii) (3 + \(\sqrt { 3 } \) )(5- \(\sqrt { 2 } \))
(iii) (\(\sqrt { 5 } \) -2)(\(\sqrt { 3 } \) – \(\sqrt { 5 } \))
Solution:

Question 3.
Simplify the following expressions:
(i)  (11 + \(\sqrt { 11 } \)) (11 – \(\sqrt { 11 } \))
(ii) (5 + \(\sqrt { 7 } \) ) (5 – \(\sqrt { 7 } \) )
(iii) (\(\sqrt { 8 } \) – \(\sqrt { 2 } \)) (\(\sqrt { 8 } \)+ \(\sqrt { 2 } \))
Solution:

Question 4.
Simplify the following expressions:
(i) (\(\sqrt { 3 } \)+\(\sqrt { 7 } \))²
(ii) (\(\sqrt { 5 } \) – \(\sqrt { 3 } \) )²
(iii) (2\(\sqrt { 5 } \) + 3 \(\sqrt { 2} \))²
Solution:
(i)


Hope given RD Sharma Class 9 Solutions Chapter 3 Rationalisation Ex 3.1 are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers MCQS

These Solutions are part of RD Sharma Class 9 Solutions. Here we have given RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers MCQS

Mark the correct alternative in each of the following:
Question 1.
The value of {2 – 3 (2 – 3)³}³ is
(a) 5
(b) 125
(c) \(\frac { 1 }{ 5 }\)
(d) -125
Solution:
{2 – 3 (2 – 3)³}³ = {2 – 3 (-1)³}³
= {2 – 3 x (-1)}³
= (2 + 3)³ = (5)³
= 125    (b)

Question 2.
The value of x – y^x-y when x = 2 and y = -2 is
(a) 18
(b) -18
(c) 14
(d) -14
Solution:
x = 2, y = -2
x-y^x-y = 2 – (-2)^{2 – (-2)}
= 2 – (-2)^{2 + 2} = 2 – (-2)⁴
= 2 – (+16) = 2 – 16 = -14        (d)

Question 3.
The product of the square root of x with the cube root of x, is
(a) cube root of the square root of x
(b) sixth root of the fifth power of x
(c) fifth root of the sixth power of x
(d) sixth root of x
Solution:

Question 4.
The seventh root of x divided by the eighth root of x is

Solution:

Question 5.
The square root of 64 divided by the cube root of 64 is
(a) 64
(b) 2
(c) \(\frac { 1 }{ 2 }\)
(d) 64^{\(\frac { 2 }{ 3 }\)}
Solution:

Question 6.
Which of the following is (are) not equal to

Solution:

Question 7.
When simplified (x^–1 + y^–1)^–1 is equal to

Solution:

Question 8.
If 8^x+1 = 64, what is the value of 3 ^{2x +1}?
(a) 1
(b) 3
(c) 9
(d) 27
Solution:

Question 9.
If (2³)² = 4^x then   3^x =
(a) 3
(b) 6
(c) 9
(d) 27
Solution:

Question 10.
If x^-2= 64, then x^{\(\frac { 1 }{ 3 }\)} + x°=
(a) 2
(b) 3
(c) \(\frac { 3 }{ 2 }\)
(c) \(\frac { 2 }{ 3 }\)
Solution:

Question 11.
When simplified ( –\(\frac { 1 }{ 27 }\))^{\(\frac { -2 }{ 3 }\)}
(a) 9
(b) -9
(c) \(\frac { 1 }{ 9 }\)
(d) –\(\frac { 1 }{ 9 }\)
Solution:

Question 12.
Which one of the following is not equal to

Solution:

Question 13.
Which one of the following is not equal to

Solution:

Question 14.
If a, b, c are positive real numbers, then

Solution:

Question 15.

Solution:

Question 16.

Solution:

Question 17.

Solution:

Question 18.

Solution:

Question 19.

Solution:

Question 20.

Solution:

Question 21.
The value of {(23 + 2²)^2/3+ (150 -29)^1/2}²  is
(a) 196
(b) 289
(c) 324
(d) 400
Solution:
{(23 + 2²)^2/3 + (150 – 29)^1/2}²
= [(23×4)^{\(\frac { 2 }{ 3 }\)}  +(150 – 29)^{\(\frac { 1 }{ 2 }\)} ]²
= [(27)^{\(\frac { 2 }{ 3 }\)} + (121)^{\(\frac { 1 }{ 2 }\)} ]²
= [(3³)³ +(11²)^{\(\frac { 1 }{ 2 }\)}]² = (9 + 11)²
= (20)² = 400  (d)

Question 22.
(256)^0.16x (256)^0.09
(a) 4
(b) 16
(c) 64
(d) 256.25
Solution:

Question 23.
If 10^2y = 25, then 10^-y equals

Solution:

Question 24.
If 9^{X + 2} = 240 + 9^X. then x =
(a) 0.5
(b) 0.2
(c) 0.4
(d) 0.1
Solution:

Question 25.
If x is a positive real number and x² = 2, then x³ =
(a) \(\sqrt { 2 } \)
(b) 2\(\sqrt { 2 } \)
(c) 3\(\sqrt { 2 } \)
(d) 4
Solution:

Question 26.

Solution:

Question 27.

Solution:

Question 28.

Solution:

Question 29.

Solution:

Question 30.

Solution:

Question 31.

Solution:

Question 32.

Solution:

Question 33.
If (16)^{2x + 3} = (64)^{x + 3} , then 4^{2x – 2}  =
(a) 64
(b) 256
(c) 32
(d) 512
Solution:

Question 34.

Solution:

Question 35.

Solution:

Question 36.

Solution:

Question 37.

Solution:

Question 38.

Solution:

Question 39.

Solution:

Question 40.

Solution:

Hope given RD Sharma Class 9 Solutions Chapter 2 Exponents of Real Numbers MCQS are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.