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NCERT Class 9 Mathematics Exercise Set 3.5 Solutions: The World of Numbers

Step-by-step working for every question of Exercise Set 3.5 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. Ex 3.5 Q1

   Without dividing, decide which of 7/20, 4/15, 13/250 terminate and which repeat; then confirm by long division.

   1. Write the denominators of the given fractions: 20, 15, and 250.
   2. Factorize the denominators into prime factors: 20 = 2² × 5, 15 = 3 × 5, and 250 = 2 × 5³.
      why?

      Every whole number greater than 1 is a product of primes in exactly one way (Fundamental Theorem of Arithmetic). Dividing again and again by a prime that goes in exactly finds that product.

   3. Check if the denominators are of the form 2ⁿ × 5ᵐ where n and m are non-negative integers.
   4. The denominator 20 is of the form 2² × 5¹, so 7/20 is a terminating decimal.
   5. The denominator 15 includes the prime factor 3, so 4/15 is a non-terminating repeating decimal.
   6. The denominator 250 is of the form 2¹ × 5³, so 13/250 is a terminating decimal.
   7. Perform long division for 7 ÷ 20 to get 0.35.
   8. Perform long division for 4 ÷ 15 to get 0.2666…
   9. Perform long division for 13 ÷ 250 to get 0.052.

   Answer: 7/20 = 0.35 (terminating); 4/15 = 0.2666… (repeating); 13/250 = 0.052 (terminating)

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2. Ex 3.5 Q2

   Find the decimal of 1/13 and its repeating block. Is it cyclic when you work out 2/13? Then work out 3/13, 4/13 and so on: what do you see?

   1. Long division of 1 by 13: the remainders are 10, 9, 12, 3, 4, 1, giving the digits 0, 7, 6, 9, 2, 3, and then the remainder 1 returns.
   2. So 1/13 = 0.076923076923…, with the repeating block 076923 (6 digits).
   3. 2/13 = 0.153846153846…, with the block 153846.
   4. 153846 is not a rotation of 076923 (it does not even use the same digits), so 1/13 is not cyclic in the way 1/7 is: the multiples of 1/7 all reuse the block 142857.
   5. 3/13 = 0.230769230769…, block 230769, which is 076923 started from the digit 2 (a rotation).
   6. 4/13 = 0.307692307692…, block 307692, again a rotation of 076923.
   7. In fact the twelve fractions 1/13 to 12/13 fall into two families of six: 1, 3, 4, 9, 10, 12 thirteenths are rotations of 076923, and 2, 5, 6, 7, 8, 11 thirteenths are rotations of 153846.
   8. This happens because the block of 1/13 has only 6 digits, not 12 = 13 − 1, so it cannot hold all twelve multiples.

   Answer: 1/13 = 0.076923… (block 076923); 2/13 has block 153846, which is not a rotation, so 1/13 is not fully cyclic; 3/13 (block 230769) and 4/13 (block 307692) are rotations of 076923: the multiples split into two families of rotating blocks.

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3. Ex 3.5 Q3

   Rational or irrational: (i) √81, (ii) √12, (iii) 0.33333…, (iv) 0.123451234512345…, (v) 1.01001000100001…, (vi) 23.560185612239874790120? Give the fraction for each rational one.

   1. (i) √81 = 9 because 9 × 9 = 81; 9 = 9/1, so √81 is rational.
   2. (ii) 12 is not a perfect square (3² = 9 < 12 < 16 = 4²), so √12 is irrational; it simplifies to √(4 × 3) = 2√3.
      why?

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

   3. (iii) Let x = 0.333…; then 10x = 3.333…, and 10x − x = 3, so 9x = 3 and x = 1/3: rational.
   4. (iv) The block 12345 repeats. Let x = 0.1234512345…; then 100000x = 12345.12345…, so 99999x = 12345 and x = 12345/99999 = 4115/33333 (dividing by 3): rational.
   5. (v) In 1.01001000100001… the number of 0s between the 1s keeps growing, so no single block repeats and the decimal never ends: irrational.
   6. (vi) 23.560185612239874790120 terminates (21 decimal places, the last digit 0 can be dropped), so it is rational: 23.56018561223987479012 = 2356018561223987479012/10²⁰.
   7. (vi) The numerator ends in 12, so it is divisible by 4 but not by 8 and not by 5; dividing top and bottom by 4 gives 589004640305996869753/25000000000000000000.

   Answer: (i) √81 = 9, rational (9/1); (ii) √12 = 2√3, irrational; (iii) rational, 1/3; (iv) rational, 4115/33333; (v) irrational; (vi) rational, 589004640305996869753/25000000000000000000.

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4. Ex 3.5 Q4

   Using x = 0.999… (9 repeating), multiply by 10 and subtract to explain why 0.999… is exactly 1.

   1. Let x = 0.999…
   2. Multiply both sides by 10 to get 10x = 9.999…
      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. Subtract the original equation x = 0.999… from this new equation
   4. 10x − x = 9.999… − 0.999…
   5. 9x = 9
   6. Divide both sides by 9 to find x = 1
      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.

   Answer: 10x − x = 9.999… − 0.999… gives 9x = 9, so x = 1: 0.999… is exactly 1.

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5. Ex 3.5 Q5

   1/7 has a cyclic repeating block. Find more numbers n whose reciprocals 1/n have cyclic repeating blocks.

   1. 1/7 = 0.142857…, and 2/7, 3/7, 4/7, 5/7, 6/7 all have blocks that are rotations of 142857: the block has 6 = 7 − 1 digits, the longest possible.
   2. In the long division of 1 by n the remainders can only be 1, 2, …, n − 1, so the block has at most n − 1 digits.
   3. When the block has exactly n − 1 digits, every remainder 1 to n − 1 appears in one cycle, so each fraction k/n starts the same cycle at a different place: the block is cyclic.
   4. Such n must be prime (other than 2 and 5). Testing the primes below 100 by long division, the block of 1/n has n − 1 digits for n = 7, 17, 19, 23, 29, 47, 59, 61 and 97.
   5. Example: 1/17 = 0.0588235294117647…, block 0588235294117647 with 16 digits; 2/17 = 0.1176470588235294…, a rotation of it.
   6. Example: 1/19 = 0.052631578947368421…, block of 18 digits.
   7. By contrast 1/13 has a block of only 6 digits (not 12), so 13 does not work, and 1/11 = 0.0909… has a 2-digit block.

   Answer: n = 7, 17, 19, 23, 29, 47, 59, 61, 97 (below 100); for example 1/17 has the 16-digit cyclic block 0588235294117647. Larger such primes also exist.

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