Maharashtra Board solution 9th Class Maths Part 1

Problem Set 2 Algebra 9th Std Maths Part 1 Answers Chapter 2 Real Numbers

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Question 1.
Choose the correct alternative answer for the questions given below. [1 Mark each]

i. Which one of the following is an irrational number?

Answer:
√5

ii. Which of the following is an irrational number?
(A) 0.17
(B) 1.513¯¯¯¯¯¯¯¯
(C) 0.2746¯¯¯¯¯
(D) 0.101001000……..
Answer:
(D) 0.101001000……..

iii. Decimal expansion of which of the following is non-terminating recurring?

Answer:
(C) 311

iv. Every point on the number line represents which of the following numbers?
(A) Natural numbers
(B) Irrational numbers
(C) Rational numbers
(D) Real numbers
Answer:
(D) Real numbers

v. The number [/latex]0.\dot { 4 }[/latex] in pq form is ……

Answer:
(A) 49

vi. What is √n , if n is not a perfect square number ?
(A) Natural number
(B) Rational number
(C) Irrational number
(D) Options A, B, C all are correct.
Answer:
(C) Irrational number

vii. Which of the following is not a surd ?

Answer:
(C) 64−−√−−−−√3

viii. What is the order of the surd 5–√−−−√3 ?
(A) 3
(B) 2
(C) 6
(D) 5
Answer:
(C) 6

ix. Which one is the conjugate pair of 2√5 + √3 ?
(A) -2√5 + √3
(B) -2√5 – √3
(C) 2√3 – √5
(D) √3 + 2√5
Answer:
(A) -2√5 + √3

x. The value of |12 – (13 + 7) x 4| is ____ .
(A) – 68
(B) 68
(C) – 32
(D) 32
Answer:
(B) 68

Hints:
ii. Since the decimal expansion is neither terminating nor recurring, 0.101001000…. is an irrational number.

iii. 311
Denominator =11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal expansion of 311 will be non terminating recurring.

v. Let x = [/latex]0.\dot { 4 }[/latex]
∴10 x = [/latex]0.\dot { 4 }[/latex]
∴10 – x = [/latex]4.\dot { 4 }[/latex] – [/latex]0.\dot { 4 }[/latex]
∴9x = 4
∴ x = 49

vii. 61−−√3 = 4, which is not an irrational number.

viii. 5–√−−−√3=5–√3×2=5–√6
∴ Order = 6

ix. The conjugate of 2√5 + √3 is 2√5 – √3 or -2√5 + √3

x. |12 – (13+7) x 4| = |12 – 20 x 4|
= |12 – 80|
= |-68|
= 68

Question 2.
Write the following numbers in pq form.
i. 0.555
ii. 29.568¯¯¯¯¯¯¯¯
iii. 9.315315…..
iv. 357.417417…..
v . 30.219¯¯¯¯¯¯¯¯
Solution:

ii. Let x = 29.568¯¯¯¯¯¯¯¯ …(i)
x = 29.568568…
Since, three numbers i.e. 5, 6 and 8 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 29568.568568…
1000 x= 29568.568¯¯¯¯¯¯¯¯ …(ii)
Subtracting (i) from (ii),
1000x – x = 29568.568¯¯¯¯¯¯¯¯ – 29.568¯¯¯¯¯¯¯¯
∴ 999x = 29539

iii. Let x = 9.315315 … = 9.315¯¯¯¯¯¯¯¯ …(i)
Since, three numbers i.e. 3, 1 and 5 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 9315.315315…
∴1000x = 9315.315¯¯¯¯¯¯¯¯ …(ii)
Subtracting (i) from (ii),
1000x – x = 9315.315¯¯¯¯¯¯¯¯ – 9.315¯¯¯¯¯¯¯¯
∴ 999x = 9306

iv. Let x = 357.417417… = 357.417¯¯¯¯¯¯¯¯ …(i)
Since, three numbers i.e. 4, 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 357417.417417…
∴ 1000x = 357417.417 …(ii)
Subtracting (i) from (ii),
1000x – x = 357417.417¯¯¯¯¯¯¯¯ – 357.417¯¯¯¯¯¯¯¯
∴ 999x = 357060

v. Let x = 30.219¯¯¯¯¯¯¯¯ …(i)
∴ x = 30.219219
Since, three numbers i.e. 2, 1 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x= 30219.219219…
∴ 1000x = 30219.219¯¯¯¯¯¯¯¯ …(ii)
Subtracting (i) from (ii),
1000x – x = 30219.219¯¯¯¯¯¯¯¯ – 30.219¯¯¯¯¯¯¯¯
∴ 999x = 30189

Question 3.
Write the following numbers in its decimal form.

Solution:
i. −57

ii. 911

iii. √5

iv. 12113

v. 298

Question 4.
Show that 5 + √7 is an irrational number. [3 Marks]
Solution:
Let us assume that 5 + √7 is a rational number. So, we can find co-prime integers ‘a’ and ‘b’ (b ≠ 0) such that

Since, ‘a’ and ‘b’ are integers, b√a – 5 is a rational number and so √7 is a rational number.
∴ But this contradicts the fact that √7 is an irrational number.
Our assumption that 5 + √7 is a rational number is wrong.
∴ 5 + √7 is an irrational number.

Question 5.
Write the following surds in simplest form.

Solution:

Question 6.
Write the simplest form of rationalising factor for the given surds.

Solution:

Now, 4√2 x √2 = 4 x 2 = 8, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √32 .

Now, 5√2 x √2 = 5 x 2 = 10, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of √50 .

Now, 3√3 x √3 = 3 x 3 = 9, which is a rational number.
∴ √ 3 is the simplest form of the rationalising factor of √27 .

= 6, which is a rational number.
∴ √10 is the simplest form of the rationalising factor of 5–√3 √10 .

Now, 18√2 x √2 = 18 x 2 = 36, which is a rational number.
∴ √2 is the simplest form of the rationalising factor of 3√72.

vi. 4√11
4√11 x √11 = 4 x 11 = 44, which is a rational number.
∴ √11 is the simplest form of the rationalising factor of 4√11.

Question 7.
Simplify.

Solution:

Question 8.
Rationalize the denominator.

Solution:

Question 1.
Draw three or four circles of different radii on a card board. Cut these circles. Take a thread and measure the length of circumference and diameter of each of the circles. Note down the readings in the given table. (Textbook pg.no.23 )

Solution:
i. 14,44,3.1
ii. 16,50.3,3.1
iii. 11,34.6,3.1
From table, we observe that the ratio d−−√c is nearly 3.1 which is constant. This ratio is denoted by π (pi).

Question 2.
To find the approximate value of π, take the wire of length 11 cm, 22 cm and 33 cm each. Make a circle from the wire. Measure the diameter and complete the following table.

Verify that the ratio of circumference to the diameter of a circle is approximately 7–√22. (Textbook pg. no. 24)
Solution:
i. 3.5, 7–√22
ii. 7, 7–√22
iii. 10.5, 7–√22
∴ The ratio of circumference to the diameter of each circle is 7–√22.