EN हिं

NCERT Solutions for Class 9th Maths Chapter 3 The World of Numbers — End-of-Chapter Exercises

Book page 64 Updated on2026-09-08

Q1.
Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal, whichever the case may be, by the process of long division: (i) 3/50 (ii) 2/9
Answer
(i) 3/50. 50 = 2 × 5², only 2s and 5s → expect a terminating decimal
30 ÷ 50 = 0, remainder 30
300 ÷ 50 = 6, remainder 0 — stop
3/50 = 0.06, a terminating decimal with 2 places
(ii) 2/9. 9 = 3², a prime other than 2 or 5 → expect repetition
20 ÷ 9 = 2, remainder 2
20 ÷ 9 = 2, remainder 2 (the remainder returns at once)
2/9 = 0.2222… = 0.2
Why it happens: the predicted number of decimal places for 3/50 is max(1, 2) = 2, and indeed the answer stops after two places. In 2/9 the remainder can only be 1…8, and it lands back on 2 immediately, so the single digit 2 loops for ever.
Check it yourself: 3/50 = 6/100 = 0.06 without any division; and 0.2 = 2/9 by the pure-repeating rule, since 9x = 2.
Q2.
Prove that √5 is an irrational number.
Answer

Proof by contradiction, following the pattern of Section 3.5.1.

Step 1. Suppose √5 is rational.
Then √5 = p/q with p, q integers, q ≠ 0 and gcd(p, q) = 1.

Step 2. Square both sides: 5 = p²/q²

Step 3. Multiply by q²: 5q² = p²

Step 4. So 5 divides p². Since 5 is prime, 5 must divide p.
Write p = 5k with k an integer.

Step 5. Substitute: 5q² = (5k)² = 25k²

Step 6. Divide by 5: q² = 5k²

Step 7. So 5 divides q², and therefore 5 divides q.

Step 8. Now 5 divides both p and q, so they share the factor 5.
But Step 1 said gcd(p, q) = 1. Contradiction.

Every step after the assumption is forced, so the assumption itself must be false. Hence √5 cannot be written as p/q — it is irrational.

Why it happens: Step 4 is the heart of the argument and it needs 5 to be prime. If a prime divides a product, it must divide one of the factors; here the product is p × p, so the prime divides p. The same proof would collapse for √4, since 4 divides 6² without dividing 6 — and rightly so, because √4 = 2 is rational.
Tip: notice that Step 3 gives another quick argument. In p² every prime appears an even number of times; in 5q² the prime 5 appears an odd number of times. The two sides can never be equal.
Q3.
Convert the following decimal numbers in the form of p/q. (i) 12.6 (ii) 0.0120 (iii) 3.052 (iv) 1.235 (v) 0.23 (vi) 2.05 (vii) 2.125 (viii) 3.125 (ix) 2.1625
Answer

The first two are terminating; the rest use the shifting-and-subtracting method of Cases 2 and 3.

(i) 12.6 = 126/10 = 63/5
(ii) 0.0120 = 120/10000 = 12/1000 = 3/250
(iii) 3.052 — 1 non-repeating digit, 2 repeating
x = 3.0525252…
10x = 30.52 and 1000x = 3052.52
1000x − 10x = 3052.5252… − 30.5252… = 3022
990x = 3022 → x = 3022/990 = 1511/495
(iv) 1.235 — 1 non-repeating digit, 2 repeating
10x = 12.35, 1000x = 1235.35
990x = 1223 → x = 1223/990
(v) 0.23 — pure repeating, 2 digits
100x = 23.23, so 99x = 23 → x = 23/99
(vi) 2.05 — 1 non-repeating, 1 repeating
10x = 20.5, 100x = 205.5
90x = 185 → x = 185/90 = 37/18
(vii) 2.125 — 2 non-repeating, 1 repeating
100x = 212.5, 1000x = 2125.5
900x = 1913 → x = 1913/900
(viii) 3.125 — same shape as (vii)
1000x − 100x = 3125.5… − 312.5… = 2813
900x = 2813 → x = 2813/900
(ix) 2.1625 — pure repeating, 4 digits
10000x = 21625.1625
9999x = 21623 → x = 21623/9999
DecimalFractionDecimalFraction
12.663/52.0537/18
0.01203/2502.1251913/900
3.0521511/4953.1252813/900
1.2351223/9902.162521623/9999
0.2323/99
Why it happens: multiplying by 10^m pushes the m non-repeating digits to the left of the point; multiplying again by 10^n slides the decimal on by exactly one full repeating cycle. The two numbers then have identical infinite tails, so the subtraction wipes the tail out completely and leaves a whole number. That is the whole trick, and it is why the denominator always comes out as a string of 9s followed by a string of 0s: 990 = 99 × 10, 900 = 9 × 100, 9999 = 9999 × 1.
Check it yourself: 1511 ÷ 495 = 3.0525252…, 1223 ÷ 990 = 1.2353535…, 1913 ÷ 900 = 2.12555… — each returns the decimal we started from.
Q4.
Locate the following rational numbers on the number line. (i) 0.532 (ii) 1.15
Answer

(i) 0.532 is located by successive magnification: each decimal place cuts the interval into ten equal parts.

0.532 lies between 0.5 and 0.6  (first place is 5)
then between 0.53 and 0.54  (second place is 3)
then it is the 2nd of the ten parts of 0.53 to 0.54
01 0.50.6 0.530.54 0.532
Zooming in three times: 0.5–0.6, then 0.53–0.54, then the point 0.532.

(ii) 1.15 is a repeating decimal, so first turn it into a fraction and then place it exactly.

x = 1.1555…
10x = 11.5, 100x = 115.5
90x = 104 → x = 104/90 = 52/45

So 1.15 = 52/45 = 1 7/45. To mark it exactly, divide the interval from 1 to 2 into 45 equal parts and count 7 of them from 1. By magnification it lies between 1.15 and 1.16, just past one half of that gap.

Why it happens: a terminating decimal is located by repeated tenfold zooming, which is precisely reading off one decimal place at a time. A repeating decimal never finishes, so zooming alone can only ever approximate it — but its p/q form gives an exact position, since dividing a unit into q equal parts is an exact ruler-and-compass operation.
Q5.
Find 6 rational numbers between 3 and 4.
Answer

Write both endpoints over a denominator large enough to leave six whole numerators in between. Since we want 6 numbers, use 6 + 1 = 7.

3 = 21/7 and 4 = 28/7
the integers strictly between 21 and 28 are 22, 23, 24, 25, 26, 27 — exactly six

So six rational numbers between 3 and 4 are

22/7, 23/7, 24/7, 25/7, 26/7, 27/7
Why it happens: with a common denominator d, every number k/d with 3d < k < 4d lies between 3 and 4, and there are exactly d − 1 such k. Choosing d = 7 gives 6 of them — just enough. Choosing d = 10 would give nine candidates (3.1 to 3.9) and you could pick any six.
Tip: the answer is not unique. 3.1, 3.2, 3.3, 3.4, 3.5, 3.6 is equally correct, and so is repeated averaging.
Q6.
Find 5 rational numbers between 2/5 and 3/5.
Answer
the numerators 2 and 3 are consecutive — no room yet
multiply top and bottom of both by 6
2/5 = 12/30 and 3/5 = 18/30
integers strictly between 12 and 18: 13, 14, 15, 16, 17 — exactly five

Five rational numbers between 2/5 and 3/5:

13/30, 14/30 = 7/15, 15/30 = 1/2, 16/30 = 8/15, 17/30
Why it happens: the endpoints did not move — 12/30 and 18/30 are the same numbers as 2/5 and 3/5 — but the gap between the numerators grew from 1 to 6, which is what creates the five slots. To place n numbers this way the numerator gap must be at least n + 1.
Check it yourself: 15/30 = 1/2, and indeed 0.4 < 0.5 < 0.6. ✓
Q7.
Find 5 rational numbers between 1/6 and 2/5.
Answer
LCM(6, 5) = 30
1/6 = 5/30 and 2/5 = 12/30
integers strictly between 5 and 12: 6, 7, 8, 9, 10, 11 — six candidates, more than enough

Choosing five of them:

6/30 = 1/5, 7/30, 8/30 = 4/15, 9/30 = 3/10, 10/30 = 1/3

Check the order in decimals: 0.16 < 0.2 < 0.23 < 0.26 < 0.3 < 0.3 < 0.4 ✓

Why it happens: here the LCM alone already opens a numerator gap of 7, which is more than the 6 needed for five numbers. Unlike Q6, no extra enlargement was necessary. Whenever the gap is too small, simply multiply the common denominator by a suitable whole number.
Q8.
If x/3 + x/5 = 16/15, find the rational number x.
Answer
LCM(3, 5) = 15
x/3 + x/5 = 5x/15 + 3x/15 = 8x/15
8x/15 = 16/15
multiply both sides by 15: 8x = 16
x = 2
Why it happens: the two terms share the unknown x, so the distributive law lets us collect them: x(1/3 + 1/5) = x × 8/15. Since the denominators on the two sides are already the same, the equation reduces to a comparison of numerators alone.
Check it yourself: 2/3 + 2/5 = 10/15 + 6/15 = 16/15 ✓
Q9.
Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning any numerical values, determine whether ab is positive or negative. Justify your answer.
Answer

ab is negative — in fact ab = −1 exactly.

a + 1/b = 0 (b ≠ 0, so 1/b is defined)
a = −1/b
multiply both sides by b
ab = −1

Since −1 < 0, the product ab is negative for every such pair a, b.

Why it happens: the condition says a is the negative of the reciprocal of b. A number and its reciprocal always have the same sign — if b > 0 then 1/b > 0, and if b < 0 then 1/b < 0. Putting a minus sign in front therefore guarantees that a and b have opposite signs, and a product of a fortune and a debt is a debt. No numerical value was needed anywhere.
Check it yourself: b = 4 gives a = −1/4 and ab = −1. b = −2/3 gives a = 3/2 and ab = −1. Always −1.
Q10.
A rational number has a terminating decimal expansion whose last non-zero digit occurs in the 4th decimal place. Show that such a number can be written in the form p/10⁴, where p is an integer not divisible by 10. Is it necessary that the denominator of this rational number, when written in the lowest form, is divisible by 2⁴ or 5⁴? Give reasons.
Answer

Part 1: the form p/10⁴. A decimal that stops at the 4th place has the shape

N = d₀.d₁d₂d₃d₄
multiply by 10⁴ to clear the point
N × 10⁴ = an integer, call it p
so N = p/10⁴

Here d₄, the 4th decimal digit, is the last non-zero digit, and d₄ is exactly the units digit of p. Since d₄ ≠ 0, the integer p does not end in 0, so 10 does not divide p. ∎

Part 2: yes, it is necessary. Write N = p/10⁴ = p/(2⁴ · 5⁴) and reduce to lowest terms.

Because 10 ∤ p, the integer p cannot be divisible by 2 and 5 at the same time. So at least one of these two cases holds:

  • 2 ∤ p. Then no factor of 2 cancels, and the reduced denominator still contains the whole 2⁴ — so it is divisible by 2⁴.
  • 5 ∤ p. Then no factor of 5 cancels, and the reduced denominator still contains the whole 5⁴ — so it is divisible by 5⁴.

Hence the lowest-form denominator is always divisible by 2⁴ or by 5⁴ (possibly by both).

Numberp/10⁴Lowest formDenominator
0.00055/10⁴1/20002⁴ × 5³ — divisible by 2⁴
0.00022/10⁴1/50002³ × 5⁴ — divisible by 5⁴
0.12331233/10⁴1233/100002⁴ × 5⁴ — divisible by both
Why it happens: cancellation can only remove a prime from the denominator if the numerator supplies it. Since p is missing at least one of 2 and 5 entirely, the corresponding fourth power survives untouched. This is the converse side of the terminating-decimal rule: the number of decimal places equals the larger of the two exponents in the lowest-form denominator, and here that larger exponent must be 4.
Q11.
Without performing division, determine whether the decimal expansion of 18/125 is terminating or non-terminating. If it terminates, state the number of decimal places.
Answer
125 = 5³, and 18 = 2 × 3²
gcd(18, 125) = 1, so 18/125 is already in lowest terms
the only prime in the denominator is 5

The denominator has the form 2⁰ × 5³, so the decimal terminates. The number of places is the larger of the two exponents:

places = max(0, 3) = 3

Confirming without long division:

18/125 = (18 × 2³)/(5³ × 2³) = 144/1000 = 0.144 — 3 places ✓
Why it happens: to turn 5³ into a power of ten we must supply the missing 2³. That fixes the denominator at 10³, so the decimal cannot need more than three places — and it cannot need fewer either, because 144 does not end in 0.
Q12.
A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have? Explain your answer.
Answer

The expansion terminates after 3 decimal places.

q = 2³ × 5¹, so m = 3 and n = 1
k = max(m, n) = 3
multiply top and bottom by 5^(3−1) = 5² = 25
p/(2³ × 5) = 25p/(2³ × 5³) = 25p/1000

A whole number over 1000 stops after three places. For example, 3/40:

3/40 = (3 × 25)/1000 = 75/1000 = 0.075 — 3 places ✓
Why it happens: to make the denominator a power of 10 both primes must appear to the same exponent, and the exponent has to be at least as large as each of them — so it is the larger one, max(m, n). It is never more, because supplying extra factors of 10 only appends zeros. And it is never fewer, because in lowest terms the numerator shares no factor with q, so nothing can cancel and shorten the expansion.
Check it yourself: 1/40 = 0.025 and 7/40 = 0.175 — every fraction with this denominator uses exactly three places.
Q13.
Let a = 7/12 and b = 5/6. Express both a and b in the form k₁/m and k₂/m where k₁, k₂ and m are integers and k₂ – k₁ > 6. Using the same denominator m, write exactly five distinct rational numbers lying between a and b keeping an integer numerator. Explain why the condition k₂ – k₁ > n + 1 is necessary to find n such rational numbers between the two rational numbers a and b using this method.
Answer

Step 1: a common denominator that is large enough.

a = 7/12, b = 5/6 = 10/12
here k₂ − k₁ = 10 − 7 = 3, which is not > 6
so enlarge: multiply numerator and denominator by 3
a = 21/36 and b = 30/36
now m = 36, k₁ = 21, k₂ = 30 and k₂ − k₁ = 9 > 6 ✓

Step 2: five rational numbers between them. The integers strictly between 21 and 30 are 22, 23, 24, 25, 26, 27, 28, 29 — eight candidates. Choosing five:

22/36 = 11/18, 23/36, 24/36 = 2/3, 25/36, 26/36 = 13/18

Check: 21/36 < 22/36 < 23/36 < 24/36 < 25/36 < 26/36 < 30/36 ✓

Step 3: why the size condition is needed.

the numbers this method produces are k/m with k₁ < k < k₂
the number of such integers k is k₂ − k₁ − 1
to obtain n of them we need k₂ − k₁ − 1 ≥ n
that is, k₂ − k₁ ≥ n + 1

The book's condition k₂ − k₁ > n + 1 is the same requirement with a margin to spare: it guarantees k₂ − k₁ − 1 > n, so there are strictly more than n candidates and you can choose freely. With n = 5 it demands k₂ − k₁ > 6, and our m = 36 gives 9, comfortably enough.

Why it happens: once the denominator is fixed at m, the rationals available between a and b are locked to a grid of step 1/m. Widening the gap between the endpoints on that grid is the only way to create more grid points, and multiplying m by t multiplies the gap k₂ − k₁ by t while leaving a and b unmoved. So any number of rationals can be squeezed between two distinct rationals — the density of ℚ again, this time counted exactly.
Tip: strictly, k₂ − k₁ = n + 1 already yields exactly n numbers with no choice at all. The book's strict inequality is a safe sufficient condition rather than the sharpest one.
Q14.
Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that all the rational numbers x, y, z must be simultaneously zero.
Answer

Square the first condition and use the second.

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)
substitute x + y + z = 0 and xy + yz + zx = 0
0² = x² + y² + z² + 2 × 0
x² + y² + z² = 0

Now x², y², z² are each squares of rational numbers, so each is ≥ 0. Three non-negative numbers add to 0 only if every one of them is 0:

if, say, x² > 0 then x² + y² + z² ≥ x² > 0, a contradiction
so x² = 0, y² = 0, z² = 0
hence x = y = z = 0 ∎
Why it happens: the identity (x + y + z)² = x² + y² + z² + 2(xy + yz + zx) converts the two given conditions into one statement about a sum of squares, and squares of real numbers can never be negative — the very fact used in the Think and Reflect about √(−1). Note how essential that is: over the imaginary numbers the conclusion fails, since x = 1, y = ω, z = ω² (complex cube roots of 1) satisfy both conditions without being zero.
Check it yourself: try to find a non-zero example by hand. Take x = 1, y = −1; then z = 0 from the first condition, but xy + yz + zx = −1 ≠ 0. The second condition always blocks you.
Q15.
Show that the rational number (a + b)/2 lies between the rational numbers a and b.
Answer

Take a ≠ b and, without loss of generality, let a < b (otherwise swap the names).

Above a:
(a + b)/2 − a = (a + b − 2a)/2 = (b − a)/2
since a < b, b − a > 0, so (b − a)/2 > 0
therefore (a + b)/2 > a
Below b:
b − (a + b)/2 = (2b − a − b)/2 = (b − a)/2
again (b − a)/2 > 0
therefore (a + b)/2 < b

Combining, a < (a + b)/2 < b. ∎

It is also a rational number: a + b is rational because ℚ is closed under addition, and dividing by 2 ≠ 0 keeps it rational.

Why it happens: the two gaps computed above are identical, both equal to (b − a)/2. That is why the average is not merely somewhere in between but exactly at the midpoint — the same distance from each end. Geometrically it is the middle of the segment joining a and b on the number line.
Tip: the statement needs a ≠ b. If a = b then (a + b)/2 = a, and there is no number strictly between a and b to find.
Q16.
Find the lengths of the hypotenuses of all the right triangles in Fig. 3.14 which is referred to as the square root spiral.
Answer

Fig. 3.14 is built by the rule of Section 3.5.2: each new right triangle takes the previous hypotenuse as one leg and a fresh segment of length 1 as the other.

1st triangle: legs 1 and 1 → h₁² = 1 + 1 = 2 → h₁ = √2
2nd triangle: legs √2 and 1 → h₂² = 2 + 1 = 3 → h₂ = √3
3rd triangle: legs √3 and 1 → h₃² = 3 + 1 = 4 → h₃ = 2
in general hₙ = √(n + 1)

The figure prints ten right triangles, so their hypotenuses are:

TriangleLegsHypotenuse²Hypotenuse
11, 12√2 ≈ 1.414
2√2, 13√3 ≈ 1.732
3√3, 14√4 = 2
42, 15√5 ≈ 2.236
5√5, 16√6 ≈ 2.449
6√6, 17√7 ≈ 2.646
7√7, 18√8 = 2√2 ≈ 2.828
8√8, 19√9 = 3
93, 110√10 ≈ 3.162
10√10, 111√11 ≈ 3.317
O √2√32√5√6√7√83√10√11 1
Each blue segment has length 1; each grey spoke from O is the hypotenuse of the previous triangle, giving √2, √3, 2, √5, … , √11.

So the ten hypotenuses are √2, √3, 2, √5, √6, √7, 2√2, 3, √10 and √11.

Why it happens: every step adds exactly 1 to the square of the length, so the squares run through the whole numbers 2, 3, 4, 5, … in order. Three of these squares are perfect — 4, 9 — giving the rational lengths 2 and 3; all the others give irrational lengths. The spiral is therefore a picture of the fact that irrational and rational points sit side by side on the number line, and it also shows how to construct √n for every positive integer n with only a ruler and compass.
Did you know? The turn at each step is arctan(1/√n), which shrinks as n grows. After ten triangles the spiral has swept about 259°, and it takes seventeen triangles to complete a full turn.
Was this helpful? Report an error