OP Malhotra Class 9 Maths Solutions Chapter 1 Rational and Irrational Numbers Exercise 1 (A)

S Chand Class 9 ICSE Maths Solutions Chapter 1 Rational and Irrational Numbers Ex 1(A)

Question 1.
(i) Find a rational number between \( \frac { 1 }{ 2 } \) and \( \frac { 3 }{ 4 } \).
(ii) Find two rational numbers between 0.1 and 0.2.
(iii) How many rational numbers can you find between two given rational numbers?
Answer:
(i) One rational number between \( \frac { 1 }{ 2 } \) and \( \frac { 3 }{ 4 } \):
\( = \frac{1}{2}\left[\frac{1}{2}+\frac{3}{4}\right] \)
\( = \frac{1}{2}\left[\frac{2+3}{4}\right] \)
\( = \frac{1}{2} \times \frac{5}{4} = \frac{5}{8} \)
(ii) Two rational numbers between 0.1 and 0.2:
First number \( = \frac { 1 }{ 2 } [0.1 + 0.2] \)
\( = \frac { 1 }{ 2 } \times 0.3 = 0.15 \) or \( \frac { 15 }{ 100 } = \frac { 3 }{ 20 } \)
And second number \( = \frac{1}{2}\left[\frac{3}{20}+\frac{2}{10}\right] \)
\( = \frac{1}{2}\left[\frac{3+4}{20}\right]=\frac{1}{2} \times \frac{7}{20}=\frac{7}{40} \)
Therefore, two numbers are \( \frac { 3 }{ 20 } \) and \( \frac { 7 }{ 40 } \).
(iii) We can find infinite numbers of rational numbers between two given rational numbers. The concept of density of rational numbers means there's always another rational number between any two given ones.
In simple words: To find a rational number between two others, you can average them. If asked for multiple, you can keep averaging the new number with one of the originals. There are endless rational numbers between any two because you can always find a midpoint.

🎯 Exam Tip: Remember the formula for finding a rational number between 'a' and 'b' is \( \frac{a+b}{2} \). For multiple numbers, repeat the process using the newly found number and one of the previous endpoints.

Question 2.
Find two rational numbers between
(i) \( \frac { 4 }{ 5 } \) and \( \frac { 7 }{ 13 } \)
(ii) \( \frac { 3 }{ 4 } \) and \( 1\frac { 1 }{ 5 } \)
Answer:
(i) One rational number between \( \frac { 4 }{ 5 } \) and \( \frac { 7 }{ 13 } \):
\( = \frac{1}{2}\left[\frac{4}{5}+\frac{7}{13}\right] \)
\( = \frac{1}{2}\left(\frac{52+35}{65}\right)=\frac{87}{130} \)
And second rational number:
\( = \frac{1}{2}\left[\frac{87}{130}+\frac{7}{13}\right] \)
\( = \frac{1}{2}\left[\frac{87+70}{130}\right] \)
\( = \frac{1}{2}\left[\frac{157}{130}\right]=\frac{157}{260} \)
(ii) One rational number between \( \frac { 3 }{ 4 } \) and \( 1\frac { 1 }{ 5 } \) (or \( \frac { 3 }{ 4 } \) and \( \frac { 6 }{ 5 } \)):
\( = \frac{1}{2}\left[\frac{3}{4}+\frac{6}{5}\right] \)
\( = \frac{1}{2}\left[\frac{15+24}{20}\right] \)
\( = \frac{1}{2}\left[\frac{39}{20}\right]=\frac{39}{40} \)
And second rational number:
\( = \frac{1}{2}\left[\frac{39}{40}+\frac{6}{5}\right] \)
\( = \frac{1}{2}\left[\frac{39+48}{40}\right]=\frac{1}{2} \times \frac{87}{40}=\frac{87}{80} \)
In simple words: To find rational numbers, convert mixed fractions to improper fractions and then use the averaging method \( \frac{a+b}{2} \) repeatedly until you get the required number of rational numbers.

🎯 Exam Tip: Always simplify fractions and convert mixed numbers to improper fractions before performing calculations to avoid errors. Ensure your final answers are also in their simplest form.

Question 3.
Find three rational numbers between 0 and 0.2.
Answer:
First rational number between 0 and 0.2 \( = \frac { 1 }{ 2 } [0 + 0.2] = 0.1 \)
Second rational number between 0 and 0.1 \( = \frac { 1 }{ 2 } [0 + 0.1] \)
\( = \frac { 1 }{ 2 } [0.1] = 0.05 \)
And third rational number between 0.1 and 0.2 \( = \frac { 1 }{ 2 } [0.1 + 0.2] \)
\( = \frac { 1 }{ 2 } [0.3] \)
\( = 0.15 \)
Hence, three rational numbers are 0.05, 0.1 and 0.15. These numbers provide a good spread within the given interval.
In simple words: Start by finding the middle point (average) of the two given numbers. Then, find the middle point between the first number and the original starting number, and another middle point using the newly found number and the original ending number.

🎯 Exam Tip: When finding multiple rational numbers, systematically pick new intervals (e.g., between the first number and the new midpoint, or the new midpoint and the second number) to ensure distinct values.

Question 4.
Find three rational numbers between 3 and 4.
Answer:
First rational number between 3 and 4:
\( = \frac { 1 }{ 2 } [3 + 4] = \frac { 1 }{ 2 } \times 7 = \frac { 7 }{ 2 } \)
Second rational number between 3 and \( \frac { 7 }{ 2 } \):
\( = \frac{1}{2}\left[3+\frac{7}{2}\right] \)
\( = \frac{1}{2} \times \frac{6+7}{2}=\frac{13}{4} \)
And third number between \( \frac { 7 }{ 2 } \) and 4:
\( = \frac{1}{2}\left[\frac{7}{2}+4\right] \)
\( = \frac{1}{2}\left[\frac{7+8}{2}\right]=\frac{1}{2} \times \frac{15}{2}=\frac{15}{4} \)
Hence, three rational numbers are \( \frac { 13 }{ 4 }, \frac { 7 }{ 2 } \) and \( \frac { 15 }{ 4 } \). This systematic approach ensures all numbers lie within the original range.
In simple words: Find the midpoint of 3 and 4, which is 3.5 or \( \frac{7}{2} \). Then find the midpoint of 3 and 3.5, and finally the midpoint of 3.5 and 4.

🎯 Exam Tip: You can always find more rational numbers by continuing the averaging process. Make sure to clearly state which two numbers you are averaging at each step.

Question 5.
Find the rational number that is one seventh of the way from \( 1\frac { 3 }{ 4 } \) to \( 4\frac { 3 }{ 8 } \).
Answer:
First, convert mixed fractions to improper fractions:
\( 1\frac{3}{4} = \frac{7}{4} \) and \( 4\frac{3}{8} = \frac{35}{8} \)
To find numbers between them, make denominators common: \( \frac{7}{4} = \frac{14}{8} \).
So, we need to find a number \( \frac{1}{7} \)th of the way from \( \frac{14}{8} \) to \( \frac{35}{8} \).
The total difference is \( \frac{35}{8} - \frac{14}{8} = \frac{21}{8} \).
\( \frac{1}{7} \)th of this difference is \( \frac{1}{7} \times \frac{21}{8} = \frac{3}{8} \).
Add this to the starting number: \( \frac{14}{8} + \frac{3}{8} = \frac{17}{8} \).
Therefore, the rational number is \( \frac{17}{8} \), which can also be written as \( 2\frac{1}{8} \).
In simple words: Convert the mixed numbers into fractions with the same bottom number. Calculate the total distance between these two numbers. Then, find one-seventh of that distance and add it to the starting number to find the required point.

🎯 Exam Tip: For "x-th of the way from A to B" questions, the formula is \( A + \frac{1}{x}(B-A) \). Always convert to improper fractions and find a common denominator first.

Question 6.
Find four rational numbers between -1 and \( \frac { 1 }{ -2 } \).
Answer:
The two numbers are -1 and \( -\frac { 1 }{ 2 } \).
First rational number between -1 and \( -\frac { 1 }{ 2 } \):
\( = \frac{1}{2}\left[-1+\left(-\frac{1}{2}\right)\right] \)
\( = \frac{1}{2}\left[-1-\frac{1}{2}\right] \)
\( = \frac{1}{2}\left[-\frac{2+1}{2}\right] = \frac{1}{2} \times \left(-\frac{3}{2}\right) = -\frac{3}{4} \)
Second rational number between -1 and \( -\frac{3}{4} \):
\( = \frac{1}{2}\left[-1+\left(-\frac{3}{4}\right)\right] \)
\( = \frac{1}{2}\left[-\frac{4+3}{4}\right] \)
\( = \frac{1}{2}\left(-\frac{7}{4}\right) = -\frac{7}{8} \)
Third rational number between \( -\frac{3}{4} \) and \( -\frac{1}{2} \):
\( = \frac{1}{2}\left[-\frac{3}{4}+\left(-\frac{1}{2}\right)\right] \)
\( = \frac{1}{2}\left[-\frac{3}{4}-\frac{1}{2}\right] \)
\( = \frac{1}{2}\left[\frac{-3-2}{4}\right] \)
\( = \frac{1}{2}\left(-\frac{5}{4}\right) = -\frac{5}{8} \)
And fourth rational number between \( -\frac{5}{8} \) and \( -\frac{1}{2} \):
\( = \frac{1}{2}\left[-\frac{5}{8}+\left(-\frac{1}{2}\right)\right] \)
\( = \frac{1}{2}\left[-\frac{5}{8}-\frac{1}{2}\right] \)
\( = \frac{1}{2}\left[\frac{-5-4}{8}\right] \)
\( = \frac{1}{2}\left(-\frac{9}{8}\right) = -\frac{9}{16} \)
Hence, four rational numbers are \( -\frac{3}{4}, -\frac{7}{8}, -\frac{5}{8}, -\frac{9}{16} \). Always be careful with negative signs during calculation.
In simple words: Follow the same averaging method as before, but pay extra attention to the negative signs. Finding the average of two negative numbers results in another negative number between them.

🎯 Exam Tip: When working with negative numbers, use brackets to keep track of signs, especially when adding or subtracting fractions. A common error is incorrectly applying the sign rules.

Question 7.
Express \( \frac { 12 }{ 125 } \) as decimal fraction.
Answer:
(Dividing 12 by 125 by long division)
0.096 _________ 125 | 12.000 - 0 ---- 12 0 - 0 ---- 12 00 -11 25 ----- 750 -750 ---- X So, \( \frac { 12 }{ 125 } = 0.096 \). This is a terminating decimal because the prime factors of the denominator (125) are only powers of 5.
In simple words: To change a fraction into a decimal, divide the top number by the bottom number using long division.

🎯 Exam Tip: For fractions with denominators that are powers of 2 or 5, the decimal representation will always terminate. This knowledge can help you verify your long division results.

Question 8.
Find a vulgar fraction equivalent to 0.03.
Answer:
Let \( x = 0.033333... \)
Multiply by 10:
\( 10x = 0.33333... \) (i)
Multiply by 100:
\( 100x = 3.3333... \) (ii)
Subtracting (i) from (ii):
\( 100x - 10x = 3.3333... - 0.3333... \)
\( 90x = 3.00 \)
\( x = \frac { 3 }{ 90 } \)
Simplify the fraction:
\( x = \frac { 1 }{ 30 } \)
Therefore, the required vulgar fraction is \( \frac { 1 }{ 30 } \). This method effectively isolates the repeating part of the decimal.
In simple words: To turn a repeating decimal into a fraction, first set the decimal equal to 'x'. Then, multiply 'x' by powers of 10 until the repeating part aligns, and subtract the equations to remove the repeating part. Finally, simplify the resulting fraction.

🎯 Exam Tip: When converting repeating decimals to fractions, ensure you multiply by appropriate powers of 10 to align the repeating parts accurately before subtraction. The goal is to isolate the non-repeating part from the repeating part.

Question 9.
Express the following rational numbers in the form \( \frac { p }{ q } \), p, q are integers, \( q \ne 0 \).
(i) \( 6.\overline{46} \)
(ii) \( 0.1\overline{36} \)
(iii) \( 3.\overline{146} \)
(iv) \( - 5.\overline{12} \)
Answer:
(i) \( 6.\overline{46} = 6.464646... \)
Let \( x = 6.464646... \) (i)
Multiply by 100 (since 2 digits are repeating):
\( 100x = 646.464646... \) (ii)
Subtracting (i) from (ii):
\( 100x - x = 646.464646... - 6.464646... \)
\( 99x = 640 \)
\( x = \frac { 640 }{ 99 } \)

(ii) \( 0.1\overline{36} = 0.1363636... \)
Let \( x = 0.1363636... \)
First, move the non-repeating part past the decimal:
\( 10x = 1.363636... \) (i)
Now, multiply to move one full repeating block past the decimal (2 digits repeating, so \( 100 \times 10x = 1000x \)):
\( 1000x = 136.363636... \) (ii)
Subtracting (i) from (ii):
\( 1000x - 10x = 136.363636... - 1.363636... \)
\( 990x = 135 \)
\( x = \frac{135}{990} \)
Simplify the fraction by dividing by 5, then 9:
\( x = \frac{27}{198} = \frac{3}{22} \)
Therefore, the fraction is \( \frac { 3 }{ 22 } \).

(iii) \( 3.\overline{146} = 3.146146146... \)
Let \( x = 3.146146146... \) (i)
Multiply by 1000 (since 3 digits are repeating):
\( 1000x = 3146.146146146... \) (ii)
Subtracting (i) from (ii):
\( 1000x - x = 3146.146146146... - 3.146146146... \)
\( 999x = 3143 \)
\( x = \frac { 3143 }{ 999 } \)
Hence, the fraction is \( \frac { 3143 }{ 999 } \).

(iv) \( - 5.\overline{12} \)
Let \( x = -5.\overline{12} = -5.121212... \)
To handle the negative sign, let's work with the absolute value first, or keep it throughout carefully.
\( x = -5.121212... \) (i)
Multiply by 100 (since 2 digits are repeating):
\( 100x = -512.121212... \) (ii)
Subtracting (i) from (ii):
\( 100x - x = (-512.121212...) - (-5.121212...) \)
\( 99x = -512.121212... + 5.121212... \)
\( 99x = -507 \)
\( x = \frac{-507}{99} \)
Simplify the fraction by dividing by 3:
\( x = \frac{-169}{33} \)
Therefore, the fraction is \( \frac { -169 }{ 33 } \).
In simple words: For repeating decimals, set the number as 'x', then multiply by powers of 10 to shift the decimal so that repeating parts align. Subtract the original equation from the shifted one to eliminate the repeating part, leaving a simple equation to solve for 'x' as a fraction.

🎯 Exam Tip: Always identify the repeating block of digits. If there are non-repeating digits before the repeating block, multiply by 10 raised to the power of the number of non-repeating digits first, then proceed with the standard method for purely repeating decimals.

Question 10.
Write the terminating decimal numeral for the given rational number :
(i) \( \frac { 7 }{ 4 } \)
(ii) \( \frac { 29 }{ 50 } \)
(iii) \( \frac { 17 }{ 32 } \)
Answer:
(i) \( \frac { 7 }{ 4 } \):
\( 7 \div 4 = 1.75 \).
1.75 _______ 4 | 7.00 -4 ---- 3 0 -2 8 ---- 20 -20 --- X (ii) \( \frac { 29 }{ 50 } \):
\( 29 \div 50 = 0.58 \).
0.58 _______ 50 | 29.00 -25 0 ----- 4 00 -4 00 ----- X (iii) \( \frac { 17 }{ 32 } \):
\( 17 \div 32 = 0.53125 \).
0.53125 _________ 32 | 17.00000 -16 0 ------- 1 00 -96 ----- 40 -32 ---- 80 -64 ---- 160 -160 ---- X All these are terminating decimals because their denominators (4, 50, 32) can be expressed as products of only 2s and 5s.
In simple words: To convert a fraction to a decimal, simply divide the numerator (top number) by the denominator (bottom number). If the division ends with no remainder, it's a terminating decimal.

🎯 Exam Tip: Long division is a fundamental skill for converting fractions to decimals. Practice it carefully, especially with multiple decimal places, to avoid errors. Always double-check your subtraction steps.

Question 11.
Write the repeating decimal for each of the following and use a bar to show the repetend.
(i) \( \frac { 1 }{ 9 } \)
(ii) \( \frac { -4 }{ 3 } \)
(iii) \( \frac { 1 }{ 6 } \)
Answer:
(i) \( \frac { 1 }{ 9 } = 0.1111... \)
So, \( \frac { 1 }{ 9 } = 0.\overline{1} \).
0.1111... _________ 9 | 1.0000 -0 ---- 1 0 -9 ---- 10 -9 ---- 10 -9 ---- 1 (ii) \( \frac { -4 }{ 3 } = -1.3333... \)
So, \( \frac { -4 }{ 3 } = -1.\overline{3} \).
1.3333... _________ 3 | 4.0000 -3 ---- 1 0 -9 ---- 10 -9 ---- 10 -9 ---- 1 (iii) \( \frac { 1 }{ 6 } = 0.16666... \)
So, \( \frac { 1 }{ 6 } = 0.1\overline{6} \).
0.1666... _________ 6 | 1.0000 -0 ---- 1 0 -6 ---- 40 -36 ---- 40 -36 ---- 40 -36 ---- 4 For repeating decimals, the bar indicates the digit or block of digits that repeats infinitely. This is a compact way to represent non-terminating, repeating decimals.
In simple words: To write a fraction as a repeating decimal, perform long division. If the remainder keeps repeating, the digits in the quotient will also repeat. Place a bar over the repeating digits.

🎯 Exam Tip: Pay attention to where the repeating pattern starts. The bar should only cover the digits that truly repeat, not any non-repeating digits that precede them (as seen in \( 0.1\overline{6} \)).