Samacheer Kalvi Class 9 Maths Solutions Chapter 2 Real Numbers Exercise 2.2

Question 1. Express the following rational numbers into decimal and state the kind of decimal expression.
(i) \( \frac{2}{7} \)
(ii) \( -5\frac{3}{11} \)
(iii) \( \frac{22}{3} \)
(iv) \( \frac{327}{200} \)
Answer:
(i) To convert \( \frac{2}{7} \) to a decimal, divide 2 by 7. The result is \( 0.285714285714... \). This decimal keeps repeating the block '285714'. So, we write it as \( 0.\overline{285714} \). This is a non-terminating and recurring (repeating) decimal expansion because the division never ends, and a pattern of digits repeats.
(ii) First, convert the fraction part \( \frac{3}{11} \) to a decimal. Dividing 3 by 11 gives \( 0.272727... \). This means the digits '27' repeat. So, \( -5\frac{3}{11} \) becomes \( -5 + 0.\overline{27} = -5.\overline{27} \). This is also a non-terminating and recurring decimal expansion.
(iii) To express \( \frac{22}{3} \) as a decimal, divide 22 by 3. The result is \( 7.333... \). Here, only the digit '3' repeats. So, we write it as \( 7.\overline{3} \). This is another example of a non-terminating and recurring decimal expansion.
(iv) To convert \( \frac{327}{200} \) to a decimal, we can divide 327 by 200. Alternatively, we can first divide 327 by 100 to get 3.27, and then divide 3.27 by 2, which gives 1.635. This decimal stops after a few digits, so it is a terminating decimal expansion.
In simple words: We convert each fraction to a decimal by dividing the numerator by the denominator. If the decimal stops, it's terminating. If digits repeat forever, it's non-terminating and recurring, shown with a bar.

🎯 Exam Tip: Always perform division carefully until a remainder repeats (for recurring) or becomes zero (for terminating).

Question 2. Express \( \frac{1}{13} \) in decimal form. Find the length of the period of decimals.
Answer: To express \( \frac{1}{13} \) as a decimal, we divide 1 by 13. The result is \( 0.076923076923... \). The repeating block of digits is '076923'. We write this as \( 0.\overline{076923} \). The length of the period, which is the number of digits in the repeating block, is 6.
In simple words: When you divide 1 by 13, the decimal part '076923' repeats. So, we put a bar over these 6 digits. The 'length of the period' means how many digits are in that repeating part, which is 6.

🎯 Exam Tip: The length of the period is exactly the count of digits in the repeating block, so count them precisely.

Question 3. Express the rational number \( \frac{1}{33} \) in recurring decimal form by using the recurring decimal expansion of \( \frac{1}{11} \). Hence write \( \frac{71}{33} \) in recurring decimal form.
Answer:
First, we find the decimal expansion of \( \frac{1}{11} \). Dividing 1 by 11 gives \( 0.0909... \), which is \( 0.\overline{09} \).
Next, we use this to find \( \frac{1}{33} \). We know that \( \frac{1}{33} = \frac{1}{3} \times \frac{1}{11} \).
So, \( \frac{1}{33} = \frac{1}{3} \times 0.\overline{09} = 0.\overline{03} \).
Now, to express \( \frac{71}{33} \) as a recurring decimal, we can write it as a mixed fraction: \( \frac{71}{33} = 2\frac{5}{33} \).
This can be written as \( 2 + \frac{5}{33} = 2 + 5 \times \frac{1}{33} \).
Substituting the value of \( \frac{1}{33} \), we get \( 2 + 5 \times 0.\overline{03} = 2 + 0.\overline{15} \).
Finally, this simplifies to \( 2.\overline{15} \).
In simple words: We first change \( \frac{1}{11} \) to a repeating decimal, \( 0.\overline{09} \). Then, we use this to find \( \frac{1}{33} \) by multiplying \( \frac{1}{3} \) by \( 0.\overline{09} \), which gives \( 0.\overline{03} \). Lastly, we write \( \frac{71}{33} \) as \( 2 + 5 \times \frac{1}{33} \). When we calculate it, the answer is \( 2.\overline{15} \).

🎯 Exam Tip: When asked to use a known expansion, break down the new fraction into parts that include the known one to simplify calculations.

Question 4. Express the following decimal expression into rational numbers.
(i) 0.24
(ii) 2.327
(iii) -5.132
(iv) 3.17
(v) 17.215
(vi) -21.2137
Answer:
(i) Let \( x = 0.242424... \) (Equation 1).
Since two digits are repeating, multiply Equation 1 by 100:
\( 100x = 24.2424... \) (Equation 2).
Subtract Equation 1 from Equation 2:
\( 100x - x = 24.2424... - 0.2424... \)
\( 99x = 24 \)
\( x = \frac{24}{99} \)
This fraction can be simplified by dividing both the numerator and denominator by 3, so \( x = \frac{8}{33} \).
(ii) Let \( x = 2.327327... \) (Equation 1).
Since three digits are repeating, we multiply Equation 1 by 1000:
\( 1000x = 2327.327327... \) (Equation 2).
Subtract Equation 1 from Equation 2:
\( 1000x - x = 2327.327327... - 2.327327... \)
\( 999x = 2325 \)
\( x = \frac{2325}{999} \)
This fraction can be simplified by dividing both the numerator and denominator by 3, resulting in \( x = \frac{775}{333} \).
(iii) We can express \( -5.132 \) by breaking it into its parts as shown in the source: \( -5 + \frac{1}{10} + \frac{3}{100} + \frac{2}{1000} \).
To combine these, we find a common denominator, which is 1000:
\( \frac{-5 \times 1000}{1000} + \frac{1 \times 100}{1000} + \frac{3 \times 10}{1000} + \frac{2}{1000} \)
\( = \frac{-5000 + 100 + 30 + 2}{1000} \)
\( = \frac{-4868}{1000} \)
This fraction can be simplified by dividing both the numerator and denominator by 4, which gives \( \frac{-1217}{250} \).
(iv) Let \( x = 3.1777... \) (Equation 1).
First, move the non-repeating digit '1' past the decimal. Multiply Equation 1 by 10:
\( 10x = 31.777... \) (Equation 2).
Now, to align the repeating part, multiply Equation 1 by 100:
\( 100x = 317.777... \) (Equation 3).
Subtract Equation 2 from Equation 3:
\( 100x - 10x = 317.777... - 31.777... \)
\( 90x = 286 \)
\( x = \frac{286}{90} \)
This fraction can be simplified by dividing both the numerator and denominator by 2, which gives \( x = \frac{143}{45} \).
(v) Let \( x = 17.2151515... \) (Equation 1).
First, move the non-repeating digit '2' past the decimal. Multiply Equation 1 by 10:
\( 10x = 172.151515... \) (Equation 2).
Next, to align the repeating part, multiply Equation 1 by 1000 (since two digits '15' repeat after '2'):
\( 1000x = 17215.151515... \) (Equation 3).
Subtract Equation 2 from Equation 3:
\( 1000x - 10x = 17215.151515... - 172.151515... \)
\( 990x = 17043 \)
\( x = \frac{17043}{990} \)
This fraction can be simplified by dividing both the numerator and denominator by 3, resulting in \( x = \frac{5681}{330} \).
(vi) Let \( x = -21.213777... \) (Equation 1).
First, shift the non-repeating digits '213' past the decimal. Multiply Equation 1 by 1000:
\( 1000x = -21213.777... \) (Equation 2).
Next, to align the repeating part, multiply Equation 1 by 10000 (since one digit '7' repeats after '213'):
\( 10000x = -212137.777... \) (Equation 3).
Subtract Equation 2 from Equation 3:
\( 10000x - 1000x = -212137.777... - (-21213.777...) \)
\( 9000x = -190924 \)
\( x = \frac{-190924}{9000} \)
This fraction can be simplified by dividing both the numerator and denominator by 4, which gives \( x = \frac{-47731}{2250} \).
In simple words: To change repeating decimals into fractions, we use algebra. We set the decimal as 'x', multiply it by powers of 10 to shift the digits, then subtract the equations to remove the repeating part. This leaves us with an equation like '99x = number', which we can solve for x as a fraction and then simplify it.

🎯 Exam Tip: Remember to multiply by powers of 10 that first shift the non-repeating part past the decimal, and then shift one full repeating block past the decimal, before subtracting.

Question 5. Without actual division, find which of the following rational numbers have terminating decimal expression.
(i) \( \frac{7}{128} \)
(ii) \( \frac{21}{15} \)
(iii) \( 4\frac{9}{35} \)
(iv) \( \frac{219}{2200} \)
Answer:
(i) The rational number is \( \frac{7}{128} \). The denominator is 128.
The prime factorization of 128 is \( 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^7 \).
Since the denominator 128 has only prime factors of 2, the decimal expansion of \( \frac{7}{128} \) will be terminating.
(ii) The rational number is \( \frac{21}{15} \). First, simplify the fraction by dividing both numerator and denominator by their greatest common divisor, 3: \( \frac{21}{15} = \frac{7}{5} \).
The denominator is now 5.
The prime factorization of 5 is \( 5^1 \).
Since the denominator has only a prime factor of 5, the decimal expansion of \( \frac{21}{15} \) will be terminating.
(iii) The rational number is \( 4\frac{9}{35} \). First, convert the mixed number to an improper fraction: \( 4\frac{9}{35} = \frac{(4 \times 35) + 9}{35} = \frac{140 + 9}{35} = \frac{149}{35} \).
The denominator is 35.
The prime factorization of 35 is \( 5 \times 7 \).
Since the denominator 35 contains a prime factor of 7 (which is not 2 or 5), the decimal expansion of \( 4\frac{9}{35} \) will be non-terminating and recurring.
(iv) The rational number is \( \frac{219}{2200} \). The denominator is 2200.
The prime factorization of 2200 is \( 2 \times 2 \times 2 \times 5 \times 5 \times 11 = 2^3 \times 5^2 \times 11 \).
Since the denominator 2200 contains a prime factor of 11 (which is not 2 or 5), the decimal expansion of \( \frac{219}{2200} \) will be non-terminating and recurring.
In simple words: To check if a fraction will have a decimal that stops (terminates) without dividing, first simplify the fraction. Then, look at the bottom number (denominator). If its prime factors are only 2s or only 5s (or both), the decimal will stop. If it has any other prime factors, like 7 or 11, the decimal will keep repeating.

🎯 Exam Tip: To check if a rational number has a terminating decimal, first simplify the fraction to its lowest terms, then find the prime factors of the denominator. Only powers of 2 and/or 5 mean it terminates.