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Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 2.1- Real Numbers

Class 9: Maths Chapter 2.1 solutions. Complete Class 9 Maths Chapter 2.1 Notes. Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 2.1- Real Numbers Maharashtra Board 9th Maths Chapter 2.1,...

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Class 9: Maths Chapter 2.1 solutions. Complete Class 9 Maths Chapter 2.1 Notes.

Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 2.1- Real Numbers

Maharashtra Board 9th Maths Chapter 2.1, Class 9 Maths Chapter 2.1 solutions

Question 1.
Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 1
Solution:

i. Denominator = 5 = 1 x 5
Since, 5 is the only prime factor denominator.
the decimal form of the rational number 135 will be terminating type.

ii. Denominator = 11 = 1 x 11
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number 211 will be non-terminating recurring type.

iii. Denominator = 16
= 2 x 2 x 2 x 2
Since, 2 is the only prime factor in the denominator.
∴ the decimal form of the rational number 2916 will be terminating type.

iv. Denominator = 125
= 5 x 5 x 5
Since, 5 is the only prime factor in the denominator.
the decimal form of the rational number 17125 will be terminating type.

v. Denominator = 6
= 2 x 3
Since, the denominator is other than prime factors 2 or 5.
∴ the decimal form of the rational number 116 will be non-terminating recurring type.

Question 2.
Write the following rational numbers in decimal form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 2
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 3
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 4
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 5
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 6
Solution:

i. −57
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 7

ii. 911
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 8

iii. √5
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 9

iv. 12113
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 10

v. 298
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 11

Question 3.
Write the following rational numbers in pq form.
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 12
Solution:

i. Let x = 0.6˙ …(i)
∴ x = 0.666…
Since, one number i.e. 6 is repeating after the decimal point.
Thus, multiplying both sides by 10,
10x = 6.666…
∴ 10 x 6.6 …(ii)
Subtracting (i) from (ii),
10x – x = 6.6 – 0.6
∴ 9x = 6
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 13

ii. Let x = 0.37¯¯¯¯¯
∴ x = 0.3737…
Since, two numbers i.e. 3 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 37.3737……
∴ 100x = 37.37¯¯¯¯¯ ……(ii)
Subtracting (i) from (ii),
100x – x = 37.37¯¯¯¯¯ – 0.37¯¯¯¯¯
∴ 99x = 37
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 14

iii. Letx = 3.17¯¯¯¯¯ …(i)
∴ x = 3.1717…
Since, two numbers i.e. 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x = 317.1717…
∴ 100x= 317.17 …(ii)
Subtracting (i) from (ii),
100x – x = 317.17¯¯¯¯¯ – 3.17¯¯¯¯¯
∴ 99x = 314
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 15

iv. Let x = 15.89¯¯¯¯¯ …….. (i)
∴ x = 15.8989…
Since, two numbers i.e. 8 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 100,
100x= 1589.8989…
∴ 100x = 1589.89¯¯¯¯¯ …(ii)
Subtracting (i) from (ii),
100x – x = 1589.89¯¯¯¯¯ – 15.89¯¯¯¯¯
∴ 99x = 1574
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 16

v. Let x = 2.514¯¯¯¯¯¯¯¯
∴ x = 2.514514…
Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
1000x = 2514.514514…
1000x = 2514.514¯¯¯¯¯¯¯¯ ….(ii)
Subtracting (i) from (ii),
1000x – x = 2514.514¯¯¯¯¯¯¯¯ – 2.514¯¯¯¯¯¯¯¯
∴ 999x = 2512
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 17

Question 1.
How to convert 2.43 in pq form ? (Textbook pg. no. 20)
Solution:

Let x = 2.43
In 2.43, the number 4 on the right side of the decimal point is not recurring.
So, in order to get only recurring digits on the right side of the decimal point, we will multiply 2.43 by 10.
∴ 10x = 24.3 …(i)
∴ 10x = 24.333…
Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, 100x = 243.333…
∴ 100x= 243.3 …(ii)
Subtracting (i) from (ii),
100x – 10x = 243.3 – 24.3
∴ 90x = 219
Maharashtra Board Class 9 Maths Solutions Chapter 2 Real Numbers Practice Set 2.1 18

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Maharashtra Board Solutions Class 9-Maths (Part 1): Chapter 2.1- Real Numbers

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