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NCERT Class 9 Mathematics The World of Numbers: Exercise Solutions

Step-by-step working for every question of End-of-chapter exercises in NCERT's Class 9 Mathematics book, chapter “The World of Numbers”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

1. Exercises Q1

   By long division, write as terminating or repeating decimals: (i) 3/50, (ii) 2/9.

   1. (i) 3 ÷ 50: 50 does not go into 3 or into 30, so the decimal begins 0.0.
   2. (i) 50 goes into 300 exactly 6 times with remainder 0, so the division stops.
   3. (i) 3/50 = 0.06, a terminating decimal.
   4. (ii) 2 ÷ 9: 9 goes into 20 two times, remainder 2.
   5. (ii) The remainder 2 comes back at every step, so the digit 2 repeats forever.
   6. (ii) 2/9 = 0.222…, a non-terminating repeating decimal.

   Answer: (i) 0.06 (terminating); (ii) 0.222… (non-terminating, repeating)

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2. Exercises Q2

   Prove that √5 is irrational.

   1. Suppose, to the contrary, that √5 is rational. Then √5 = p/q with integers p, q, q ≠ 0, and p/q in lowest form (p and q have no common factor other than 1).
      why?

      Proof by contradiction: assume the opposite of what is to be proved and show that it leads to something impossible; then the assumption must be false.

   2. Squaring, 5 = p²/q², so p² = 5q². Hence p² is a multiple of 5.
   3. Then p itself is a multiple of 5: if it were not, p would be 5k + 1, 5k + 2, 5k + 3 or 5k + 4, and p² would leave the same remainder as 1², 2², 3² = 9 or 4² = 16, that is 1, 4, 4 or 1, on division by 5, never 0.
   4. Write p = 5m for an integer m. Then 25m² = 5q², so q² = 5m², and q² is a multiple of 5.
   5. By the same remainder argument, q is a multiple of 5.
   6. So 5 divides both p and q, which contradicts p/q being in lowest form.
      why?

      Proof by contradiction: assume the opposite of what is to be proved and show that it leads to something impossible; then the assumption must be false.

   7. The assumption is therefore false, and √5 is irrational.

   Answer: Assuming √5 = p/q in lowest form forces 5 to divide both p and q, a contradiction; so √5 is irrational.

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3. Exercises Q3

   Write as p/q: (i) 12.6, (ii) 0.0120, (iii) 3.0525252… (52 repeating), (iv) 1.2353535… (35 repeating), (v) 0.2323… (23 repeating), (vi) 2.0555… (5 repeating), (vii) 2.12555… (5 repeating), (viii) 3.12555… (5 repeating), (ix) 2.16251625… (1625 repeating).

   1. (i) 12.6 = 126/10 = 63/5
   2. (ii) 0.0120 = 120/10000 = 3/250
   3. (iii) Let x = 3.0525252…; 100x = 305.252525…; 100x − x = 302.2; 99x = 302.2; x = 3022/990 = 1511/495
   4. (iv) Let x = 1.2353535…; 100x = 123.53535…; 100x − x = 122.3; 99x = 122.3; x = 1223/990
   5. (v) Let x = 0.2323…; 100x = 23.2323…; 100x − x = 23; 99x = 23; x = 23/99
   6. (vi) Let x = 2.0555…; 10x = 20.555…; 10x − x = 18.5; 9x = 18.5; x = 185/90 = 37/18
   7. (vii) Let x = 2.12555…; 100x = 212.555…; 100x − x = 210.43; 99x = 210.43; x = 21043/9900 = 1913/900
   8. (viii) Let x = 3.12555…; 100x = 312.555…; 100x − x = 309.43; 99x = 309.43; x = 30943/9900 = 2813/900
   9. (ix) Let x = 2.16251625…; 10000x = 21625.1625…; 10000x − x = 21623; 9999x = 21623; x = 21623/9999

   Answer: (i) 63/5; (ii) 3/250; (iii) 1511/495; (iv) 1223/990; (v) 23/99; (vi) 37/18; (vii) 1913/900; (viii) 2813/900; (ix) 21623/9999

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4. Exercises Q4

   Locate on the number line: (i) 0.532, (ii) 1.1555… (5 repeating).

   1. (i) 0.532 lies between 0.53 and 0.54 (it is 0.530 + 0.002).
   2. (i) Magnify the part of the line from 0.53 to 0.54 and divide it into 10 equal parts of 0.001 each; 0.532 is the second mark after 0.53, so it is closer to 0.53.
   3. (ii) Let x = 1.1555…; then 100x = 115.555… and 10x = 11.555…, so 90x = 104 and x = 104/90 = 52/45.
   4. (ii) 1.1555… lies between 1.15 and 1.16; magnifying that part into 10 parts of 0.001, it lies between 1.155 and 1.156, a little past halfway from 1.15 to 1.16, so it is closer to 1.16.

   Answer: (i) 0.532: between 0.53 and 0.54, at the 2nd of 10 small divisions after 0.53; (ii) 1.1555… = 52/45: between 1.15 and 1.16 (between 1.155 and 1.156).

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5. Exercises Q5

   Find 6 rational numbers between 3 and 4.

   1. We need to find rational numbers between 3 and 4.
   2. Any number of the form p/q where p and q are integers and q is not 0 is a rational number.
   3. We can write 3 as 3/1 and 4 as 4/1.
   4. To find numbers between them, we can use decimals or fractions with a common denominator.
   5. Let us take the denominator as 10, so 3 = 30/10 and 4 = 40/10.
   6. The integers between 30 and 40 are 31, 32, 33, 34, 35, and 36.
   7. Therefore, the rational numbers are 31/10, 32/10, 33/10, 34/10, 35/10, and 36/10.
   8. Converting these to decimals gives 3.1, 3.2, 3.3, 3.4, 3.5, and 3.6.

   Answer: 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

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6. Exercises Q6

   Find 5 rational numbers between 2/5 and 3/5.

   1. To find rational numbers between 2/5 and 3/5, we first make their denominators equal by finding a common multiple.
   2. Since 30 = 5 × 6, use 30 as the common denominator by multiplying the top and bottom of each fraction by 6.
   3. Convert 2/5 to an equivalent fraction with denominator 30: (2 × 6)/(5 × 6) = 12/30.
   4. Convert 3/5 to an equivalent fraction with denominator 30: (3 × 6)/(5 × 6) = 18/30.
   5. Now, we need to find 5 rational numbers between 12/30 and 18/30.
   6. The integers between 12 and 18 are 13, 14, 15, 16, and 17.
   7. Therefore, the corresponding fractions are 13/30, 14/30, 15/30, 16/30, and 17/30.
   8. Simplify these fractions where possible: 14/30 = 7/15, 15/30 = 1/2, and 16/30 = 8/15.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   9. The five rational numbers are 13/30, 7/15, 1/2, 8/15, and 17/30.

   Answer: 13/30, 7/15, 1/2, 8/15, 17/30

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7. Exercises Q7

   Find 5 rational numbers between 1/6 and 2/5.

   1. Convert the given fractions to a common denominator of 30.
   2. 1/6 becomes 5/30 and 2/5 becomes 12/30.
   3. We need five rational numbers between 5/30 and 12/30.
   4. The integers between 5 and 12 are 6, 7, 8, 9, and 10.
   5. Write these as fractions with denominator 30: 6/30, 7/30, 8/30, 9/30, 10/30.
   6. Simplify 6/30 to 1/5.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   7. Simplify 8/30 to 4/15.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   8. Simplify 9/30 to 3/10.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   9. Simplify 10/30 to 1/3.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   10. The five rational numbers are 1/5, 7/30, 4/15, 3/10, and 1/3.

   Answer: 1/5, 7/30, 4/15, 3/10, 1/3

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8. Exercises Q8

   Find the rational number x if x/3 + x/5 = 16/15.

   1. Write the given equation: x/3 + x/5 = 16/15.
   2. Find the least common multiple of the denominators 3 and 5, which is 15.
   3. Rewrite the fractions with the common denominator: 5x/15 + 3x/15 = 16/15.
   4. Combine the numerators on the left side: (5x + 3x)/15 = 16/15.
   5. Simplify the numerator: 8x/15 = 16/15.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Since the denominators are the same, equate the numerators: 8x = 16.
   7. Divide both sides by 8 to solve for x: x = 16/8.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   8. Calculate the final value: x = 2.

   Answer: x = 2

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9. Exercises Q9

   a and b are non-zero rationals with a + 1/b = 0. Without using numbers, is ab positive or negative? Why?

   1. Given that a and b are non-zero rational numbers.
   2. The equation is a + 1/b = 0.
   3. Subtract 1/b from both sides to isolate a, which gives a = -1/b.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   4. Multiply both sides of the equation by b to find the product ab.
      why?

      Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

   5. This simplifies to ab = -1.
      why?

      Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

   6. Since -1 is less than 0, the product ab is negative.

   Answer: ab = −1, negative

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10. Exercises Q10

    A terminating decimal has its last non-zero digit in the 4th decimal place. Show it equals p/10⁴ with p an integer not divisible by 10. In lowest form, must its denominator be divisible by 2⁴ or 5⁴? Why?

    1. Let x be the number. Its last non-zero digit is in the 4th decimal place, so x × 10⁴ moves all its decimal digits to the left of the point: p = x × 10⁴ is an integer and x = p/10⁴.
    2. The last digit of p is that 4th decimal digit, which is not 0, so p is not divisible by 10.
    3. 10⁴ = 2⁴ × 5⁴. To reach lowest form, only factors 2 and 5 can cancel between p and 10⁴.
    4. Since p is not divisible by 10, p cannot be divisible by both 2 and 5.
    5. If p is not divisible by 5, no 5 cancels, so the lowest denominator still contains 5⁴. Example: 0.0002 = 2/10⁴ = 1/5000 = 1/(2³ × 5⁴), divisible by 5⁴ but not by 2⁴.
    6. If p is not divisible by 2, no 2 cancels, so the lowest denominator still contains 2⁴. Example: 0.0005 = 5/10⁴ = 1/2000 = 1/(2⁴ × 5³), divisible by 2⁴ but not by 5⁴.
    7. If p is divisible by neither (for example 0.0007 = 7/10⁴), the denominator stays 10⁴, divisible by both.

    Answer: x = p/10⁴ with p not divisible by 10. Yes: in lowest form the denominator is always divisible by 2⁴ or by 5⁴ (at least one of them), but not necessarily both, as 1/5000 and 1/2000 show.

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11. Exercises Q11

    Without dividing, does 18/125 terminate? If so, how many decimal places?

    1. Write the fraction as 18/125.
    2. Factorize the denominator 125 as 5 × 5 × 5, which is 5³.
    3. Since the denominator has only 5 as a prime factor, the decimal expansion is terminating.
    4. Multiply the numerator and denominator by 2³ to make the denominator 1000.
    5. The fraction becomes (18 × 8) / (125 × 8) = 144 / 1000.
    6. Divide 144 by 1000 to get 0.144.
    7. Count the digits after the decimal point, which is 3.

    Answer: terminating; 0.144; 3 decimal places

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12. Exercises Q12

    A rational number in lowest form has denominator 2³ × 5. How many decimal places does its decimal have? Why?

    1. A rational number in its lowest form is written as p/q, where p and q are co-prime integers.
    2. The denominator q is given as 2³ × 5.
    3. For a rational number to have a terminating decimal expansion, the denominator q must be expressible in the form 2ⁿ × 5ᵐ, where n and m are non-negative integers.
    4. Here, the denominator is already in the form 2³ × 5¹, so the decimal expansion will terminate.
    5. To convert the denominator to a power of 10, we need equal powers of 2 and 5.
    6. The current powers are 2³ and 5¹. To make them equal, we multiply the numerator and denominator by 5².
    7. This gives a new denominator of 2³ × 5¹ × 5² = 2³ × 5³ = (2 × 5)³ = 10³.
    8. Since the denominator is now 10³, the decimal expansion terminates after 3 places.
    9. Therefore, the decimal representation has exactly 3 decimal places.

    Answer: 3 decimal places

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13. Exercises Q13

    a = 7/12, b = 5/6. Write them as k₁/m and k₂/m with integers k₁, k₂, m and k₂ − k₁ > 6; list exactly five rationals between a and b with denominator m and integer numerators; explain the role of the condition k₂ − k₁ > n + 1 for finding n such numbers.

    1. Given a = 7/12 and b = 5/6.
    2. Find the least common multiple of the denominators 12 and 6, which is 12.
    3. Express a with denominator 12: a = 7/12, so k₁ = 7.
    4. Express b with denominator 12: b = 5/6 = 10/12, so k₂ = 10.
    5. Check the condition k₂ − k₁ > n + 1 for n = 5 numbers: 10 − 7 = 3, which is not greater than 5 + 1 = 6.
    6. Since the condition fails, multiply both numerator and denominator of a and b by 3 to get a common denominator m = 36.
    7. Now a = (7 × 3)/(12 × 3) = 21/36, so k₁ = 21.
    8. Now b = (10 × 3)/(12 × 3) = 30/36, so k₂ = 30.
    9. Check the condition again: k₂ − k₁ = 30 − 21 = 9, which is greater than 5 + 1 = 6.
    10. The integers strictly between k₁ and k₂ number k₂ − k₁ − 1, so n such numerators exist exactly when k₂ − k₁ − 1 ≥ n, that is k₂ − k₁ ≥ n + 1; the condition k₂ − k₁ > n + 1 guarantees this with room to spare (strictly, k₂ − k₁ = n + 1 already gives exactly n).
    11. The five integers between 21 and 30 are 22, 23, 24, 25, and 26.
    12. Thus, the five rational numbers between a and b with denominator 36 are 22/36, 23/36, 24/36, 25/36, and 26/36.

    Answer: With m = 36: a = 21/36 and b = 30/36 (k₂ − k₁ = 9 > 6); five rationals between them are 22/36, 23/36, 24/36, 25/36, 26/36. The numerators must be integers strictly between k₁ and k₂, and there are only k₂ − k₁ − 1 of those, so the gap must be large enough to hold n of them.

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14. Exercises Q14

    Rationals x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that x = y = z = 0.

    1. We are given two equations: x + y + z = 0 and xy + yz + zx = 0.
    2. Square the first equation to get (x + y + z)² = 0².
    3. Expand the left side to obtain x² + y² + z² + 2(xy + yz + zx) = 0.
       why?

       Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

    4. Substitute the value of xy + yz + zx from the second equation into this expanded form.
       why?

       Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.

    5. This simplifies to x² + y² + z² + 2(0) = 0, which means x² + y² + z² = 0.
       why?

       Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.

    6. Since x, y, and z are rational numbers, their squares x², y², and z² are non-negative real numbers.
    7. The sum of non-negative numbers is zero if and only if each individual number is zero.
    8. Therefore, x² = 0, y² = 0, and z² = 0.
    9. Taking the square root of each equation gives x = 0, y = 0, and z = 0.

    Answer: x = y = z = 0

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15. Exercises Q15

    Show that (a + b)/2 lies between the rational numbers a and b.

    1. Let a and b be two rational numbers such that a < b.
    2. Add a to both sides of the inequality a < b to get 2a < a + b.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    3. Add b to both sides of the inequality a < b to get a + b < 2b.
       why?

       Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.

    4. Combine the two inequalities to form the chain 2a < a + b < 2b.
    5. Divide every term in the inequality by 2, which is positive, to preserve the direction.
    6. This results in the inequality a < (a + b)/2 < b.
    7. Since a and b are rational, their sum is rational, and dividing by 2 keeps it rational.
    8. Thus, (a + b)/2 is a rational number strictly between a and b.

    Answer: Proved that (a + b)/2 lies between a and b.

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16. Exercises Q16

    Square root spiral (Fig. 3.14): find the hypotenuse of every right triangle in it.

    Uses a figure printed in the NCERT book — open the chapter above to see it.

    1. Fig. 3.14 has 10 right triangles: the first has both legs 1, and each next one has the previous hypotenuse and a new side of length 1 as its legs.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    2. Start with a right triangle having both legs of length 1.
    3. By Pythagoras theorem, the hypotenuse is √(1² + 1²) = √2.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    4. Draw a new leg of length 1 perpendicular to the previous hypotenuse √2.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    5. The new hypotenuse is √((√2)² + 1²) = √(2 + 1) = √3.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    6. Repeat the process: add a leg of length 1 perpendicular to the last hypotenuse.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    7. The next hypotenuse is √((√3)² + 1²) = √(3 + 1) = √4 = 2.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    8. Continue this pattern for the remaining triangles in the spiral.
    9. The hypotenuse after the 4th triangle is √((2)² + 1²) = √5.
       why?

       In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    10. The hypotenuse after the 5th triangle is √((√5)² + 1²) = √6.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    11. The hypotenuse after the 6th triangle is √((√6)² + 1²) = √7.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    12. The hypotenuse after the 7th triangle is √((√7)² + 1²) = √8 = 2√2.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    13. The hypotenuse after the 8th triangle is √((2√2)² + 1²) = √(8 + 1) = √9 = 3.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    14. The hypotenuse after the 9th triangle is √(3² + 1²) = √10.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    15. The hypotenuse after the 10th triangle is √((√10)² + 1²) = √11.
        why?

        In a right-angled triangle, (hypotenuse)² = (one side)² + (other side)² (Baudhāyana–Pythagoras theorem). The hypotenuse is the side opposite the right angle and is the longest side.

    Answer: The 10 hypotenuses are √2, √3, √4 = 2, √5, √6, √7, √8 = 2√2, √9 = 3, √10, √11.

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