Maharashtra Board Class 9 Math Solution Chapter 2 – Real Numbers
Balbharati Maharashtra Board Class 9 Math Solution Chapter 2: Real Numbers. Marathi or English Medium Students of Class 9 get here Real Numbers full Exercise Solution.
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Std |
Maharashtra Class 9 |
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Subject |
Math Solution |
| Chapter |
Real Numbers |
Practice set 2.1
Practice Set 2.2
(i) Show that 4√2 be irrational number.
Therefore, 4√2 can be expressed in the from p/q where p and q are integers and q ≠ o
Therefore, 4≠2 = p/q
=> √2 = p/4q
Therefore, p/4q is rational number
Therefore, √2 is also a rational number which is a contradiction due to the wrong assumption that 4√2 is rational
Therefore, 4√2 is irrational.
(2) Prove that 3 + √5 is an irrational number
A: Let, 3 + √5 be rational
Therefore, 3 + √5 can be expressed in the form p/q where p and q are integers and q ≠ o
Therefore, 3 + √5 = p/q
=> √5 = p/q – 3
Because, p/q – 3 is rational numbers
Therefore, √5 should also be rational which is contradiction due to the assumption that 3 + √5 is rational
Therefore, 3+√5 is irrational.
(3) Represent the numbers √5 and √10 on a number line
A: Because representation on a number line :
Because, √4 = 2 and √9 = 3 and √5 lies between √4 and √9, so, √5 lies 2 and 3.
We take OA = 2 units and draw AB = 1 unit perpendicular to the number line and join OB. With O as centre and OB as length we cut are on the number line. The point at which the fare cuts the numbers line gives √5 on the numbers line.
√10 representation on a number line
Because, √9 = 3 and √16 = 4 and √10 lies between √4 and √16 i.e., √10 lies between 3 and 4.
We take OA = 3 units and draw AB = 1 unit perpendicular to the number line and join OB. With O as centre and OB as length we cut are on the number line. The point at which the are cuts the number line gives √10 on the number line.
(4) Write any three rational numbers between the two numbers given below.
(i) 0.3 and -0.5
A: -0.3, -0.1, 0.1
(ii) -2.3 and -2.33
A: -2.31, -2.32, -2.312
(iii) 5.2 and 5.3
A: 5.21, 5.25, 5.29
(iv)- 4.5 and -4.6
A: – 4.52, -4.56, – 4.57
Practice set 2.3
(1) State the order of the surds given below.
(i) 3√7
(ii) 5√12
(iii) 4√10
(iv) √39
(v) 3√18
A: (i) 3
(ii) 2
(iii) 4
(iv) 2
(v) 3
(2) State which of the following are surds justify
(i) 3√51
A: Here a = 51, order of word, n = 4
But 3rd root of 51 is not a rational number.
Therefore, 3√51 is an irrational number
Therefore, 3√51 is a surd.
(ii) 4√16
A: Here a = 16, n = 4
Let, 16 = p4 and we know 4th power 2 gives 16
Therefore, 16 = 24 => 4√16 = 2 which is a rational number
Therefore, 4√16 is not surd.
(iii) 5√81
A: Here a = 81, order of surd, n = 5 but, 5th root of 81 is not a rational numbers
Therefore, 5√81 is an irrational numbers
Therefore, 5√81 a surd
(iv) √256
A: Here, a = 256 and n = 2
Let, 256 = p2 and we know that 256 = 162
Therefore, 256 = 162 => √256 = 16 which is a rational number
Therefore, √256 is not a surd
(v) 3√64
A: Here a = 64 and n = 3 but, 3rd root of 64 is 4
Let, 64 = p3 and we know that 64 = 43
Therefore, 64 = 43 => 3√64 = 4 which is a rational number
Therefore, 3√64 is not surd
(vi) √22/7
A: Here, a = 22/7 and n = 2
But square root of 22/7 is not a rational number
Therefore, √22/7 is an irrational number
Therefore, √22/7 is a surd
(3) Classify the given pair of surds into surds and unlike surds
(i) √52, 5√13
A: Here, √52 = √4 x 13 = √4 √13 = 2√13
Because, Both order and radicand of 2√13 and 5√13 are same
Therefore, √52 and 5√13 are like surds.
(8) Divided and write answer in simplest form:
(i) √98 ÷ √2
A: √98 ÷ √2
= √98/2
= √49
= 7
(ii) √125 ÷ √50
A √125 ÷ √50
= √125/50
= √5/2
= √5/√2
(iii) √54 ÷ √27
A: √54 ÷ √27
= √54/27
= √2
(iv) √310 ÷ √5
√310 ÷ √5
= √310/5
= √62
(9) Rationalize the denominator
(i) 3/√5
A: 3/√5 = 3/√5 x √5/√5 = 3√5/5
(ii) 1/√14
A: Multiplying and dividing √14
1/√14 = 1/√14 x √14/√14 = √14/14
(iii) 5/√7
= Multiplying and dividing by √7
5/√7 = 5/√7 x √7/√7
= 5√7/7
(iv) 6/9√3
A: Multiplying and dividing by √3
6/9√3 = 6/9√3 x √3/√3 = 6√3/9 x3 = 2√3/9
(v) 11/√3
A: Multiplying and dividing by √3
11/√3 = 11/√3 x √3/√3 = 11√3/3
Practice set 2.4
(i) Multiply
(i) √3(√7 – √3)
A: √3 (√7 – √3)
= √3 x √7 – √3 x √3
=21-3
= 3 + √21
(ii) (√5-√7) √2
A: (√5-√7) √2
= √5 x √2 – √7 x √2
= √10 – √14
(iii) (3√2 – √3) (4√3 – √2)
A: (3√2 – √3) (4√3 – √2) = 3√2(4√3 – √2) – √3 (4√3-√2)
= 3√2 x 4√3 – 3√2 x √2 – √3 x 4√3 + √3 + √3 x √2
= 12√6 – 3 x 2 – 4 x 3 + √6
= 16√6 -6 -12
= 13√6 – 18 = -18 + 13√6
(2) Rationalize the denominator
(i) 1/√7 + √2
A: Multiplying and dividing by √7 – √2
1/√7 + √2
= 1/√7 + √2 x √7 – √2/√7 – √2
= √7 – √2/√72 – √22
= √7 – √2/7-2
= √7 – √2/5
Practice set 2.5
(i) Find the value
(i) |15-2|
A: |15-2|
= |13|
= 13
(ii) |4-9|
A: |4-9|
= |-5|
= 5
(iii) 17| x |-4|
A: 17| x |-4|
= 7 x 4
= 28
(2) Solve:
(i) |3x – 5| = 1
A: |3x – 5| = 1
i.e., 3x – 5 = +- 1
Therefore, 3x – 5 = 1
=> 3x = 6
=> x = 2
Or, 3x – 5 = 1
Or, => 3x = 4
Or, => x = 4/3
(ii) |7 – 2x| = 5
A: |7-2x| = 5
i.e., 7-2x = +-5
Therefore, 7-2x = 5
=> 2x = 7-5
=> x = 2/2
=> x = 1
Or, 7 – 2x = -5
Or, => 2x = 7+5
Or, => x = 12/6
O=> x = 6
(iii) | 8 – x/2| = 5
A: |8 – x/2| = 5
i.e., 8 – x/2 = +- 5
Therefore, 8 – x/2 = 5
=> 8 – x = 10
=> x = 8-10
=> x = -2
Or, 8 – x/2 = -5
Or, => 8 – x = -10
Or => x = 8 + 10
Or, x = 18
(iv) |5 + x/4| = 5
A: |5 + x/4| = 5
i.e., 5 + x/4 = +-5
Therefore, 5 + x/4 = 5
=> x/4 = 5 – 5
=> x = 0
Or, 5 + x/4 = -5
Or, => x/4 = -5 -5
Or => x = -10 x 4 = -40
Problem set 2
(i) Choose the correct alternative answer for the questions given below.
(i) Which one of the following is an irrational number?
(A) √16/25
(B) √5
(C) 3/9
(D) √196
=A: (B) √5
(iii) Decimal expansion of which of the following is non-terminating recurring?
(A) 2/5
(B) 3/16
(C) 3/11
(D) 137/25
A: (C) 3/11
(iv) Every point on the number line represent, which of the following numbers?
(A) Natural numbers
(b) Irrational numbers
(C) Rational numbers
(D) Real numbers
A: (D) Real numbers.
Here is your solution of Maharashtra Board Class 9 Math Chapter 2 Real Numbers
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