Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises: The World of Numbers
Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises revise The World of Numbers through fractions, decimals, rational numbers, irrational numbers, real numbers, number-line questions, proof of √5 irrationality and square root spiral problems.
Chapter 3 ends by bringing together the full journey of numbers: counting numbers, zero, integers, rational numbers, irrational numbers, decimal expansions and real numbers. Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises test whether students can move between fractions, decimals and number-line positions, while also proving important results such as the irrationality of √5.
These questions from The World of Numbers include long division, conversion of repeating decimals into fractions, rational numbers between two numbers, terminating-decimal tests, proof-based questions and the square root spiral Class 9 figure. The solutions below follow the textbook order and keep calculations in a copy-friendly format for school notes and CBSE 2026 revision.
Class 9 Maths Ganita Manjari Chapter 3 End of Chapter Exercises
1. Convert the following rational numbers in the form of a terminating decimal or non-terminating and repeating decimal by long division.
(i) 3/50
Solution:
3/50 = 6/100
3/50 = 0.06
Answer: 0.06, terminating decimal.
(ii) 2/9
Solution:
On long division:
2 ÷ 9 = 0.2222…
So,
2/9 = 0.2̅
Answer: 0.2222… = 0.2̅, non-terminating repeating decimal.
2. Prove that √5 is an irrational number.
Solution:
Assume that √5 is rational.
Then it can be written in lowest form as:
√5 = p/q
where p and q are integers, q ≠ 0, and p and q have no common factor.
Squaring both sides:
5 = p²/q²
p² = 5q²
So, p² is divisible by 5. Therefore, p is also divisible by 5.
Let:
p = 5k
Substitute in p² = 5q².
(5k)² = 5q²
25k² = 5q²
q² = 5k²
So, q² is divisible by 5. Therefore, q is also divisible by 5.
This means both p and q are divisible by 5, which contradicts the assumption that p/q is in lowest form.
Answer: Therefore, √5 is irrational.
3. Convert the following decimal numbers in the form p/q.
(i) 12.6̅
Solution:
Let:
x = 12.6666…
Then:
10x = 126.6666…
Subtract:
10x − x = 126.6666… − 12.6666…
9x = 114
x = 114/9
x = 38/3
Answer: 12.6̅ = 38/3
(ii) 0.0120
Solution:
0.0120 = 120/10000
Simplify:
120/10000 = 3/250
Answer: 0.0120 = 3/250
(iii) 3.0\overline{52}
Solution:
Let:
x = 3.0525252…
There is 1 non-repeating digit after the decimal and 2 repeating digits.
First multiply by 10:
10x = 30.525252…
Then multiply by 1000:
1000x = 3052.525252…
Subtract:
1000x − 10x = 3052.525252… − 30.525252…
990x = 3022
x = 3022/990
x = 1511/495
Answer: 3.0\overline{52} = 1511/495
(iv) 1.\overline{235}
Solution:
Let:
x = 1.235235235…
Since 3 digits repeat, multiply by 1000.
1000x = 1235.235235…
Subtract:
1000x − x = 1235.235235… − 1.235235…
999x = 1234
x = 1234/999
Answer: 1.\overline{235} = 1234/999
(v) 0.\overline{23}
Solution:
Let:
x = 0.232323…
100x = 23.232323…
Subtract:
100x − x = 23.232323… − 0.232323…
99x = 23
x = 23/99
Answer: 0.\overline{23} = 23/99
(vi) 2.0\overline{5}
Solution:
Let:
x = 2.05555…
First multiply by 10:
10x = 20.5555…
Then multiply by 100:
100x = 205.5555…
Subtract:
100x − 10x = 205.5555… − 20.5555…
90x = 185
x = 185/90
x = 37/18
Answer: 2.0\overline{5} = 37/18
(vii) 2.12\overline{5}
Solution:
Let:
x = 2.125555…
There are 2 non-repeating decimal digits and 1 repeating digit.
100x = 212.5555…
1000x = 2125.5555…
Subtract:
1000x − 100x = 2125.5555… − 212.5555…
900x = 1913
x = 1913/900
Answer: 2.12\overline{5} = 1913/900
(viii) 3.\overline{125}
Solution:
Let:
x = 3.125125125…
1000x = 3125.125125…
Subtract:
1000x − x = 3125.125125… − 3.125125…
999x = 3122
x = 3122/999
Answer: 3.\overline{125} = 3122/999
(ix) 2.\overline{1625}
Solution:
Let:
x = 2.162516251625…
Since 4 digits repeat, multiply by 10000.
10000x = 21625.16251625…
Subtract:
10000x − x = 21625.16251625… − 2.16251625…
9999x = 21623
x = 21623/9999
Answer: 2.\overline{1625} = 21623/9999
4. Locate the following rational numbers on the number line.
(i) 0.532
Solution:
0.532 = 532/1000
So, it lies between 0 and 1, slightly more than 0.5.
Answer: Mark 0.532 between 0 and 1, at 532/1000 of the distance from 0 to 1.
(ii) 1.1\overline{5}
Solution:
1.1\overline{5} = 1.15555…
It lies between 1 and 2.
Convert to fraction:
Let:
x = 1.15555…
10x = 11.5555…
100x = 115.5555…
Subtract:
100x − 10x = 115.5555… − 11.5555…
90x = 104
x = 104/90
x = 52/45
So,
1.1\overline{5} = 52/45 = 1 + 7/45
Answer: Mark 1.1\overline{5} between 1 and 2, at 7/45 of the distance after 1.
5. Find 6 rational numbers between 3 and 4.
Solution:
Write 3 and 4 with denominator 10.
3 = 30/10
4 = 40/10
Six rational numbers between them are:
31/10, 32/10, 33/10, 34/10, 35/10, 36/10
Answer: 31/10, 32/10, 33/10, 34/10, 35/10 and 36/10
6. Find 5 rational numbers between 2/5 and 3/5.
Solution:
Convert both numbers to denominator 30.
2/5 = 12/30
3/5 = 18/30
Five rational numbers between them are:
13/30, 14/30, 15/30, 16/30, 17/30
Answer: 13/30, 14/30, 15/30, 16/30 and 17/30
7. Find 5 rational numbers between 1/6 and 2/5.
Solution:
Use denominator 60.
1/6 = 10/60
2/5 = 24/60
Five rational numbers between them are:
11/60, 12/60, 13/60, 14/60, 15/60
Answer: 11/60, 12/60, 13/60, 14/60 and 15/60
8. If x/3 + x/5 = 16/15, find the rational number x.
Solution:
Given:
x/3 + x/5 = 16/15
Take LCM 15.
5x/15 + 3x/15 = 16/15
8x/15 = 16/15
Multiply both sides by 15.
8x = 16
x = 2
Answer: x = 2
9. Let a and b be two non-zero rational numbers such that a + 1/b = 0. Without assigning values, determine whether ab is positive or negative.
Solution:
Given:
a + 1/b = 0
So,
a = −1/b
Now multiply both sides by b.
ab = −1
Since −1 is negative:
ab < 0
Answer: ab is negative.
10. 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 in lowest form is divisible by 2⁴ or 5⁴?
Solution:
If the last non-zero digit occurs in the 4th decimal place, the number can be written with exactly four decimal places.
So, it can be written as:
p/10000
Since:
10000 = 10⁴
the number is:
p/10⁴
Here, p is not divisible by 10, because the fourth decimal digit is the last non-zero digit. If p were divisible by 10, the decimal would end earlier.
Now:
10⁴ = 2⁴ × 5⁴
When the fraction is reduced to lowest form, factors common with p may cancel. Since p is not divisible by 10, it cannot contain both 2 and 5 as factors together. Therefore, after cancellation, at least one of 2⁴ or 5⁴ remains in the denominator.
Answer: Yes, the number can be written as p/10⁴, where p is not divisible by 10. In lowest form, the denominator is necessarily divisible by 2⁴ or 5⁴.
11. 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.
Solution:
Given rational number:
18/125
Denominator:
125 = 5³
Since the denominator has only 5 as a prime factor, the decimal expansion is terminating.
To make the denominator a power of 10:
125 × 8 = 1000
So, the decimal will have 3 decimal places.
Answer: 18/125 has a terminating decimal expansion with 3 decimal places.
12. A rational number in its lowest form has denominator 2³ × 5. How many decimal places will its decimal expansion have?
Solution:
Denominator:
2³ × 5
To make it a power of 10, the powers of 2 and 5 must be equal.
We have:
2³ × 5¹
Multiply by 5².
2³ × 5¹ × 5² = 2³ × 5³
2³ × 5³ = 10³
So, the decimal expansion will have 3 decimal places.
Answer: The decimal expansion will have 3 decimal places.
13. Let a = 5/6 and b = 7/12. Express both using a common denominator so that k₂ − k₁ > 6. Then write exactly five rational numbers between them.
Solution:
Given:
a = 5/6
b = 7/12
Since:
5/6 = 10/12
we can use a larger common denominator to create enough integer numerators between the two numbers.
Use denominator 72.
7/12 = 42/72
5/6 = 60/72
Here:
k₁ = 42
k₂ = 60
So,
k₂ − k₁ = 60 − 42 = 18
Since 18 > 6, we can choose five integer numerators between 42 and 60.
Five rational numbers are:
43/72, 44/72, 45/72, 46/72, 47/72
These all lie between:
42/72 and 60/72
So, they lie between 7/12 and 5/6.
Why is k₂ − k₁ > n + 1 needed?
If two rational numbers are written as:
k₁/m and k₂/m
then the integer numerators strictly between them are:
k₁ + 1, k₁ + 2, ..., k₂ − 1
The number of available numerators is:
k₂ − k₁ − 1
To find n rational numbers, we need enough integer numerators between the endpoints. So, the gap between k₂ and k₁ must be large enough.
Answer: Five rational numbers are 43/72, 44/72, 45/72, 46/72 and 47/72. The condition ensures enough integer numerators are available between the two endpoints.
14. Three rational numbers x, y, z satisfy x + y + z = 0 and xy + yz + zx = 0. Show that x, y, z must be simultaneously zero.
Solution:
Given:
x + y + z = 0
and
xy + yz + zx = 0
Square the first equation:
(x + y + z)² = 0²
Use identity:
x² + y² + z² + 2xy + 2yz + 2zx = 0
So,
x² + y² + z² + 2(xy + yz + zx) = 0
Since:
xy + yz + zx = 0
we get:
x² + y² + z² = 0
Now, squares of rational numbers are always non-negative. Therefore:
x² = 0, y² = 0, z² = 0
So:
x = 0, y = 0, z = 0
Answer: All three rational numbers must be 0.
15. Show that the rational number (a + b)/2 lies between the rational numbers a and b.
Solution:
Assume:
a < b
Then:
a + a < a + b < b + b
So:
2a < a + b < 2b
Divide by 2:
a < (a + b)/2 < b
Therefore, (a + b)/2 lies between a and b.
Answer: (a + b)/2 lies between a and b.
16. Find the lengths of the hypotenuses of all the right triangles in the square root spiral.
Solution:
In the square root spiral, each new right triangle has one side of length 1 and the previous hypotenuse as the other side.
For the first triangle:
Hypotenuse² = 1² + 1²
Hypotenuse² = 2
Hypotenuse = √2
For the second triangle:
Hypotenuse² = (√2)² + 1²
Hypotenuse² = 2 + 1 = 3
Hypotenuse = √3
Continuing in the same way, the hypotenuse lengths are:
√2, √3, √4, √5, √6, √7, √8, √9, √10, √11
Answer: The hypotenuse lengths are √2, √3, √4, √5, √6, √7, √8, √9, √10 and √11.
Related Links – Class 9 Maths Chapter 3