Maharashtra Board Class 9 Maths Chapter 2 Set 2.1 Algebra Standard Part 1 Real Numbers Solutions

Question 1. Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
(i) \( \frac{13}{5} \)
(ii) \( \frac{2}{11} \)
(iii) \( \frac{29}{16} \)
(iv) \( \frac{17}{125} \)
(v) \( \frac{11}{6} \)
Answer:
(i) \( \frac{13}{5} \) - Terminating type. The prime factor of the denominator is 5.
(ii) \( \frac{2}{11} \) - Non-terminating recurring type. The prime factor of the denominator is 11, which is other than 2 or 5.
(iii) \( \frac{29}{16} \) - Terminating type. The prime factors of the denominator 16 are only 2.
(iv) \( \frac{17}{125} \) - Terminating type. The prime factors of the denominator 125 are only 5.
(v) \( \frac{11}{6} \) - Non-terminating recurring type. The prime factors of the denominator 6 are 2 and 3, where 3 is other than 2 or 5.
In simple words: To find out if a fraction stops or repeats forever, look at the bottom number (denominator). If its prime factors are only 2s, only 5s, or both, the decimal stops; if there is any other prime number, it repeats forever.

🎯 Exam Tip: Always reduce the fraction to its simplest form before checking the prime factors of the denominator to avoid making mistakes.

Question 1. Classify the decimal form of the given rational numbers into terminating and non-terminating recurring type.
(i) \( \frac{13}{5} \)
(ii) \( \frac{2}{11} \)
(iii) \( \frac{29}{16} \)
(iv) \( \frac{17}{125} \)
(v) \( \frac{11}{6} \)
Answer:
(i) \( \frac{13}{5} \)
Denominator = \( 5 = 1 \times 5 \)
Since 5 is the only prime factor in the denominator, the decimal form of the rational number \( \frac{13}{5} \) will be of terminating type. This means the division ends after a finite number of decimal places.

(ii) \( \frac{2}{11} \)
Denominator = \( 11 = 1 \times 11 \)
Since the prime factor of the denominator is other than 2 or 5, \( \therefore \) the decimal form of the rational number \( \frac{2}{11} \) will be of non-terminating recurring type.

(iii) \( \frac{29}{16} \)
Denominator = \( 16 = 2 \times 2 \times 2 \times 2 \)
Since 2 is the only prime factor in the denominator, \( \dots \) the decimal form of the rational number \( \frac{29}{16} \) will be of terminating type.

(iv) \( \frac{17}{125} \)
Denominator = \( 125 = 5 \times 5 \times 5 \)
Since 5 is the only prime factor in the denominator, the decimal form of the rational number \( \frac{17}{125} \) will be of terminating type.

(v) \( \frac{11}{6} \)
Denominator = \( 6 = 2 \times 3 \)
Since the denominator has a prime factor (3) other than 2 or 5, \( \therefore \) the decimal form of the rational number \( \frac{11}{6} \) will be of non-terminating recurring type.
In simple words: If the denominator of a simplified fraction has only 2 or 5 (or both) as its prime factors, its decimal form stops (terminates). If it has any other prime factor like 3 or 11, the decimal form keeps repeating forever.

🎯 Exam Tip: To quickly identify the type, always simplify the fraction first and then check if the prime factors of the denominator contain only 2s, 5s, or both.

Question 2. Write the following rational numbers in decimal form.
(i) \( \frac{127}{200} \)
(ii) \( \frac{25}{99} \)
Answer:
(i) \( \frac{127}{200} \)
\( \frac{127}{200} = \frac{127 \times 5}{200 \times 5} = \frac{635}{1000} = 0.635 \)
Multiplying both numerator and denominator by 5 makes the denominator a power of 10, which simplifies the division process.

(ii) \( \frac{25}{99} \)
Dividing 25 by 99:
\( 25 \div 99 = 0.252525... = 0.\overline{25} \)
In simple words: To convert a fraction to a decimal, you divide the top number by the bottom number. If the denominator is 99, the digits in the numerator will repeat continuously after the decimal point.

🎯 Exam Tip: When the denominator is 9, 99, or 999, the numerator directly becomes the repeating block after the decimal point, which is a great shortcut to save time.

Question 1. Write the following rational numbers in decimal form:
(i) \( \frac{-5}{7} \)
(ii) \( \frac{9}{11} \)
(iii) \( \frac{23}{7} \)
(iv) \( \frac{4}{5} \)
(v) \( \frac{17}{8} \)
Answer:
(i) \( \frac{-5}{7} \)
To find the decimal form, we perform long division of 5 by 7:
- \( 5 \div 7 = 0 \) with remainder 5.
- Bring down 0 to make it 50: \( 50 \div 7 = 7 \) with remainder 1.
- Bring down 0 to make it 10: \( 10 \div 7 = 1 \) with remainder 3.
- Bring down 0 to make it 30: \( 30 \div 7 = 4 \) with remainder 2.
- Bring down 0 to make it 20: \( 20 \div 7 = 2 \) with remainder 6.
- Bring down 0 to make it 60: \( 60 \div 7 = 8 \) with remainder 4.
- Bring down 0 to make it 40: \( 40 \div 7 = 5 \) with remainder 5.
Since the remainder 5 repeats, the block of digits \( 714285 \) will recur continuously.
Therefore, \( \frac{-5}{7} = -0.\overline{714285} \)

(ii) \( \frac{9}{11} \)
To find the decimal form, we perform long division of 9 by 11:
- \( 9 \div 11 = 0 \) with remainder 9.
- Bring down 0 to make it 90: \( 90 \div 11 = 8 \) with remainder 2.
- Bring down 0 to make it 20: \( 20 \div 11 = 1 \) with remainder 9.
Since the remainder 9 repeats, the block of digits \( 81 \) will recur continuously.
Therefore, \( \frac{9}{11} = 0.\overline{81} \)
In simple words: To convert a fraction into a decimal, divide the numerator by the denominator. If the division doesn't end and the same numbers start repeating, put a bar over the repeating group of numbers to show they go on forever.

🎯 Exam Tip: Always remember to carry over the negative sign to your final decimal answer if the original fraction was negative, and place the recurring bar precisely over the repeating digits only.

Question 1. Find the decimal representation of the following:
(iii) \( \sqrt{5} \)
(iv) \( \frac{121}{13} \)
Answer:

(iii) \( \sqrt{5} \)
To find the square root of 5, we use the long division method as shown below:

\( \therefore \sqrt{5} = 2.2360679... \)

(iv) \( \frac{121}{13} \)
To find the decimal form of \( \frac{121}{13} \), we perform long division:

\( \therefore \frac{121}{13} = 9.307692... \)
In simple words: We can find the decimal values of square roots and fractions by using long division step-by-step. For square roots, we group digits in pairs, and for fractions, we divide the numerator by the denominator until we find a repeating pattern.

🎯 Exam Tip: When finding square roots or dividing fractions that do not terminate, carry out the division to at least 5 or 6 decimal places to clearly show the non-terminating or recurring nature of the number.

Question 2. (v) Express \( \frac{29}{8} \) in decimal form.
Answer:
By performing long division of 29 by 8:
\( 29 \div 8 = 3.625 \)
\( \therefore \frac{29}{8} = 3.625 \)
In simple words: To convert the fraction 29/8 into a decimal, we divide 29 by 8 to get 3.625.

🎯 Exam Tip: Always perform long division carefully until the remainder becomes zero to find the exact terminating decimal representation.

Question 3. Write the following rational numbers in \( \frac{p}{q} \) form.
(i) \( 0.\dot{6} \)
(ii) \( 0.\overline{37} \)
(iii) \( 3.\overline{17} \)
(iv) \( 15.\overline{89} \)
(v) \( 2.\overline{514} \)
Answer:
(i) \( 0.\dot{6} \)
Let \( x = 0.\dot{6} \) --- (1)
\( \therefore x = 0.666... \)
Since one digit (6) is repeating after the decimal point, multiply both sides by 10:
\( 10x = 6.666... \)
\( \therefore 10x = 6.\dot{6} \) --- (2)
Subtracting equation (1) from (2):
\( 10x - x = 6.\dot{6} - 0.\dot{6} \)
\( \implies 9x = 6 \)
\( \implies x = \frac{6}{9} \)
\( \implies x = \frac{2}{3} \)
\( \therefore 0.\dot{6} = \frac{2}{3} \)

(ii) \( 0.\overline{37} \)
Let \( x = 0.\overline{37} \) --- (1)
\( \therefore x = 0.3737... \)
Since two digits (3 and 7) are repeating after the decimal point, multiply both sides by 100:
\( 100x = 37.3737... \)
\( \therefore 100x = 37.\overline{37} \) --- (2)
Subtracting equation (1) from (2):
\( 100x - x = 37.\overline{37} - 0.\overline{37} \)
\( \implies 99x = 37 \)
\( \implies x = \frac{37}{99} \)
\( \therefore 0.\overline{37} = \frac{37}{99} \)

(iii) \( 3.\overline{17} \)
Let \( x = 3.\overline{17} \) --- (1)
\( \dots x = 3.1717... \)
Since two digits (1 and 7) are repeating after the decimal point, multiply both sides by 100:
\( 100x = 317.1717... \)
\( \therefore 100x = 317.\overline{17} \) --- (2)
Subtracting equation (1) from (2):
\( 100x - x = 317.\overline{17} - 3.\overline{17} \)
\( \implies 99x = 314 \)
\( \implies x = \frac{314}{99} \)
\( \therefore 3.\overline{17} = \frac{314}{99} \)

(iv) \( 15.\overline{89} \)
Let \( x = 15.\overline{89} \) --- (1)
\( \therefore x = 15.8989... \)
Since two digits (8 and 9) are repeating after the decimal point, multiply both sides by 100:
\( 100x = 1589.8989... \)
\( \therefore 100x = 1589.\overline{89} \) --- (2)
Subtracting equation (1) from (2):
\( 100x - x = 1589.\overline{89} - 15.\overline{89} \)
\( \implies 99x = 1574 \)
\( \implies x = \frac{1574}{99} \)
\( \therefore 15.\overline{89} = \frac{1574}{99} \)

(v) \( 2.\overline{514} \)
Let \( x = 2.\overline{514} \) --- (1)
\( \therefore x = 2.514514... \)
Since three digits (5, 1, and 4) are repeating after the decimal point, multiply both sides by 1000:
\( 1000x = 2514.514514... \)
\( \therefore 1000x = 2514.\overline{514} \) --- (2)
Subtracting equation (1) from (2):
\( 1000x - x = 2514.\overline{514} - 2.\overline{514} \)
\( \implies 999x = 2512 \)
\( \implies x = \frac{2512}{999} \)
\( \therefore 2.\overline{514} = \frac{2512}{999} \)
In simple words: To convert a repeating decimal to a fraction, we multiply it by 10, 100, or 1000 depending on how many digits repeat, and then subtract the original equation to cancel out the repeating decimal part.

🎯 Exam Tip: Remember that the number of 9s in the denominator of the final fraction corresponds directly to the number of repeating digits in the decimal.

Question 1. Express the following recurring decimals in \( \frac{p}{q} \) form:
(ii) \( 0.\overline{37} \)
(iii) \( 3.\overline{17} \)
(iv) \( 15.\overline{89} \)
Answer:
(ii) Let \( x = 0.\overline{37} \) ...(i)
\( \therefore x = 0.3737... \)
Since, two numbers i.e. 3 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
\( 100x = 37.3737... \)
\( \therefore 100x = 37.\overline{37} \) ...(ii)
Subtracting (i) from (ii),
\( 100x - x = 37.\overline{37} - 0.\overline{37} \)
\( \dots 99x = 37 \)
\( \therefore x = \frac{37}{99} \)
\( \therefore 0.\overline{37} = \frac{37}{99} \)

(iii) Let \( x = 3.\overline{17} \) ...(i)
\( \dots x = 3.1717... \)
Since, two numbers i.e. 1 and 7 are repeating after the decimal point.
Thus, multiplying both sides by 100,
\( 100x = 317.1717... \)
\( \therefore 100x = 317.\overline{17} \) ...(ii)
Subtracting (i) from (ii),
\( 100x - x = 317.\overline{17} - 3.\overline{17} \)
\( \dots 99x = 314 \)
\( \therefore x = \frac{314}{99} \)
\( \therefore 3.\overline{17} = \frac{314}{99} \)

(iv) Let \( x = 15.\overline{89} \) ...(i)
\( \therefore x = 15.8989... \)
Since, two numbers i.e. 8 and 9 are repeating after the decimal point.
Thus, multiplying both sides by 100,
\( 100x = 1589.8989... \)
\( \therefore 100x = 1589.\overline{89} \) ...(ii)
Subtracting (i) from (ii),
\( 100x - x = 1589.\overline{89} - 15.\overline{89} \)
\( \therefore 99x = 1574 \)
\( \therefore x = \frac{1574}{99} \)
\( \therefore 15.\overline{89} = \frac{1574}{99} \)
This systematic method of algebraic subtraction is highly reliable for converting any recurring decimal into its corresponding rational form.
In simple words: To convert a repeating decimal into a fraction, we multiply it by 100 because there are two repeating digits. This lets us subtract the original decimal and completely cancel out the infinite repeating part, leaving us with a simple fraction.

🎯 Exam Tip: Always count the number of repeating digits under the bar line to decide whether to multiply by 10, 100, or 1000. Clearly label your equations as (i) and (ii) to make your subtraction steps easy for the examiner to follow.

Question. Convert the recurring decimal \( 2.\overline{514} \) into \( \frac{p}{q} \) form.
Answer:
Let \( x = 2.\overline{514} \) ....(i)
\( \therefore x = 2.514514... \)
Since, three numbers i.e. 5, 1 and 4 are repeating after the decimal point.
Thus, multiplying both sides by 1000,
\( 1000x = 2514.514514... \)
\( 1000x = 2514.\overline{514} \) ....(ii)
Subtracting (i) from (ii),
\( 1000x - x = 2514.\overline{514} - 2.\overline{514} \)
\( \therefore 999x = 2512 \)
\( \therefore x = \frac{2512}{999} \)
\( \therefore 2.\overline{514} = \frac{2512}{999} \)
In simple words: Since three digits repeat, we multiply by 1000 to shift the decimal point past the repeating block, subtract the original value to cancel the decimals, and solve for x.

🎯 Exam Tip: Always count the number of repeating digits under the bar to decide whether to multiply by 10, 100, or 1000.

Question 1. How to convert \( 2.4\dot{3} \) in \( \frac{p}{q} \) form ? (Textbook pg. no. 20)
Answer:
Let \( x = 2.4\dot{3} \)
In \( 2.4\dot{3} \), 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 \( x \) by 10.
\( \therefore 10x = 24.\dot{3} \) ...(i)
\( \dots 10x = 24.333... \)
Here, digit 3 is the only recurring digit. Thus, by multiplying both sides by 10, \( 100x = 243.333... \)
\( \therefore 100x = 243.\dot{3} \) ...(ii)
Subtracting (i) from (ii),
\( 100x - 10x = 243.\dot{3} - 24.\dot{3} \)
\( \dots 90x = 219 \)
\( \therefore x = \frac{219}{90} = \frac{3 \times 73}{3 \times 30} = \frac{73}{30} \)
\( \therefore 2.4\dot{3} = \frac{73}{30} \)
In simple words: First, multiply by 10 to move the non-repeating digit before the decimal point. Then, multiply by another 10 to shift one repeating digit, subtract the two equations to eliminate the decimal, and simplify the fraction.

🎯 Exam Tip: Remember to simplify the final fraction to its lowest terms by dividing both the numerator and denominator by their greatest common divisor.