Mathematics is a subject that builds logical reasoning, enhances problem-solving abilities, and helps students interpret the world in a systematic way. The newly introduced Ganita Prakash textbook for Class 8 has been carefully designed to strengthen these skills through a wide range of chapters and exercises. Each chapter not only lays a strong foundation but also prepares learners for advanced concepts in higher classes.
In this blog, we bring you Class 8 Maths Ganita Prakash NCERT Solutions. These solutions are presented in a clear, step-by-step manner to make problem-solving easier, boost conceptual understanding, and help students practice effectively. Whether you’re revising for exams, preparing for a test, or simply looking for homework support, these solutions will serve as a reliable guide throughout the year.
Welcome to the complete Class 8 Maths Ganita Prakash NCERT Solutions. This blog provides chapter-wise step-by-step solutions designed to make learning easier and help you excel in exams. Whether you find some concepts tricky or want to practice effectively, these solutions cover all exercises and examples from your NCERT textbook. Each chapter is explained in a simple and clear manner to enhance your understanding and boost your confidence in Maths.
In this blog, we will go through the following chapters:-
In this chapter, we explore squares, cubes, and their properties. Our Class 8 Maths Ganita Prakash NCERT Solutions provide step-by-step guidance to calculate areas, volumes, and understand the relationship between numbers easily.
Textbook Page 1 – 3
Q. Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer?
Hint: Determine the number of times each locker is switched (toggled).
Solution:
He observed that the number of times a locker is toggled corresponds to the number of factors of its number. A locker toggled an odd number of times will remain open, while one toggled an even number of times will be closed. Since only perfect squares have an odd number of factors, the lockers numbered with perfect squares (1, 4, 9, 16, …) are the ones left open.
Q. Does every number have an even number of factors?
Solution:
Not every number has an even count of factors. Only perfect squares have an odd count, since one of their factors is repeated — the square root, which multiplied by itself gives the number.
Q. Can you use this insight to find more numbers with an odd number of factors?
Solution:
Q. Write the locker numbers that remain open.
Solution:
The lockers numbered with perfect squares — 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 — will remain open.
Q. Which are these five lockers?
Solution:
Lockers that are toggled exactly twice are the prime-numbered ones, as each prime has only two factors: 1 and itself. Hence, the code is 2-3-5-7-11.
Textbook Page 4
Q. Find the squares of the first 30 natural numbers and fill in the table below.
Solution:
Q. What patterns do you notice? Share your observations and make conjectures.
Solution:
Every perfect square ends with one of the digits 0, 1, 4, 5, 6, or 9, and never with 2, 3, 7, or 8.
Q. If a number ends in 0, 1, 4, 5, 6, or 9, is it always a square?
Solution:
A number cannot be confirmed as a perfect square simply by its unit digit. However, the unit digit can help us rule out some cases. If a number ends in 2, 3, 7, or 8, it is definitely not a perfect square. For instance, 26 ends with 6, yet it is not a perfect square.
Q. Write 5 numbers such that you can determine by looking at their unit digit that they are not squares.
Solution:
Here are 5 such numbers whose unit digit shows they are not perfect squares:
Textbook Page 5
Q. Which of the following numbers have the digit 6 in the units place?
(i) 382 (ii) 342 (iii) 462 (iv) 562 (v) 742 (vi) 822
Solution:
(i) 382
38 → Units digit → 8
8 × 8 = 64 (ends in 4)
So, 382 does not end in 6.
(ii) 342
34 → Units digit → 4
4 × 4 = 16 (ends in 6)
So, 342 ends in 6.
(iii) 462
46 → Units digit → 6
6 × 6 = 36 (ends in 6)
So, 462 ends in 6.
(iv) 562
56 → Units digit → 6
6 × 6 = 36 (ends in 6)
So, 562 ends in 6.
(v) 742
74 → Units digit → 4
4 × 4 = 16 (ends in 6)
So, 742 ends in 6.
(vi) 822
82 → Units digit → 2
2 × 2 = 4 (ends in 4)
So, 822 ends in 4.
Thus, if a number ends with 4 or 6, its square will have 6 in the units place.
Therefore, numbers like 342, 462, 562, and 742 will have 6 as the unit digit of their squares.
Q. Consider the following numbers and their squares.
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Solution:
The number of zeros at the end of a square is always twice the number of zeros at the end of the original number. Hence, if a number ends with 3 zeros, its square will end with 6 zeros.
Q. What do you notice about the number of zeros at the end of a number and the number of zeros at the end of its square? Will this always happen? Can we say that squares can only have an even number of zeros at the end?
Solution:
The square of a number always has twice as many zeros at the end as the original number. This rule holds true in every case, meaning a square can only end with an even number of zeros.
Q. What can you say about the parity of a number and its square?
Solution:
The square of an even number is always even, while the square of an odd number is always odd.
Textbook Page 7
Q. Find how many numbers lie between two consecutive perfect squares. Do you notice a pattern?
Solution:
There are exactly 2n numbers between n2 and (n+1)2.
For example, between 32 and 42, there are 2×3=6 numbers.
Q. How many square numbers are there between 1 and 100? How many are between 101 and 200? Using the table of squares you filled earlier, enter the values below, tabulating the number of squares in each block of 100. What is the largest square less than 1000?
Solution:
The largest square less than 1000 is given by 312 = 961.
Q. Can you see any relation between triangular numbers and square numbers? Extend the pattern shown and draw the next term.
Solution:
Textbook Page 9
Q. Find whether 1156 and 2800 are perfect squares using prime factorisation.
Solution:
(i) 1156 = (2 × 2) × (17 × 17)
= 22 × 172 = (2 × 17)2 = (34)2
∴ √1156 = 34. (ii) 2800 = (2 × 2) × (2 × 2) × (5 × 5) × 7
= 22 × 22 × 52 × 7
2800 is not a perfect square because the factors cannot be paired.
Textbook Page 10 – 11
Figure it Out
1. Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089
Solution:
A perfect square always ends with 0, 1, 4, 5, 6, or 9 in the unit’s place.
(i) 2032 ends with 2, so it is not a perfect square.
(ii) 2048 ends with 8, so it is not a perfect square.
(iii) 1027 ends with 7, so it is not a perfect square.
(iv) 1089 ends with 9, so it is a perfect square.
2. Which one among 642, 1082, 2922, 362 has the last digit 4?
Solution:
(i) 642
The unit digit of 64 is 4.
4×4=16 → last digit is 6.
(ii) 1082
The unit digit of 108 is 8.
8×8=64 → last digit is 4.
(iii) 2922
The unit digit of 292 is 2.
2×2=4→ last digit is 4.
(iv) 362
The unit digit of 36 is 6.
6×6=36→ last digit is 6.
Therefore, 1082 and 2922 have 4 as their last digits.
3. Given 1252 = 15625, what is the value of 1262?
(i) 15625 + 126
(ii) 15625 + 262
(iii) 15625 + 253
(iv) 15625 + 251
(v) 15625 + 252
Solution:
1252 = 15625
This means 15625 is the sum of the first 125 odd numbers.
1262 = 15625 + 127th odd number
127th odd number = (2 × 127) – 1 = 252 – 1 = 251.
∴ 1262 = 15625 + 251
Hence, the correct answer is (iv) 15625 + 251.
4. Find the length of the side of a square whose area is 441 m2.
Solution:
Area of the square = side × side = 441 m2
(side)2=441
side=441=±21 m
Since the side of a square cannot be negative, the length of the side is 21 m.
5. Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Solution:
The LCM of 4, 9, and 10 is 180.
Thus, the smallest number divisible by 4, 9, and 10 is 180.
Prime factorization:
180=2×2×3×3×5
180=22×32×5
Here, 5 is unpaired, so 180 is not a perfect square.
To make it a perfect square, we multiply by 5.
Hence, the required smallest perfect square is:
180×5=900.
6. Find the smallest number by which 9408 must be multiplied so that the product is a perfect square. Find the square root of the product.
Solution:
9408=2×2×2×2×2×2×3×7×7
9408=22×22×22×72×3
Here, 3 is unpaired.
So, 9408 is not a perfect square. To make it a perfect square, we multiply by 3.
9408×3=28224 which is a perfect square.
28224=22×22×22×32×72 so, square root of 28224=2×2×2×3×7=8×21=168
Thus, 28224=168.
7. How many numbers lie between the squares of the following numbers?
(i) 16 and 17 (ii) 99 and 100
Solution:
There are 2n numbers between n2 and (n + 1)2 .
(i) 162 and 172
Here, n = 16 and (n + 1) = 17
Therefore, the numbers between 162 and 172 = 2n = 2 × 16 = 32.
(ii) 992 and 1002
Here, n = 99 and (n + 1) = 100
Therefore, the numbers between the squares 992 and 1002 = 2n = 2 × 99 = 198.
8. In the following pattern, fill in the missing numbers:
12 × 22 × 22 = 32
22 × 32 × 62 = 72
32 × 42 × 122 = 132
42 × 52 × 202 = (___)2
92 × 102 × (___)2 = (___)2
Solution:
12 × 22 × 22 = 32
22 × 32 × 62 = 72
32 × 42 × 122 = 132
42 × 52 × 202 = (21)2
92 × 102 × (90)2 = (91)2
9. How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Solution:
Number of squares in each row = 9
Number of squares in each column = 9
Total squares in the figure = 9×9=81
Each square contains 5×5=25 tiny squares.
So, the total number of tiny squares in the figure = 25×81=2025.
Prime factorization of 2025:
2025 = (3 × 3) × (3 × 3) × (5 × 5)= 32 × 32 × 52
2= 34 × 52.
Textbook Page 12
Q. Complete the table below.
Solution:
Q. What patterns do you notice in the table above?
Solution:
(i) The cube of a number ending in 1 also ends in 1.
(ii) A number ending in 2 has a cube ending in 8, and a number ending in 8 has a cube ending in 2.
(iii) A number ending in 3 has a cube ending in 7, and a number ending in 7 has a cube ending in 3.
(iv) Numbers ending in 4, 5, 6, 9, or 0 retain the same unit digit when cubed.
Q. We know that 0, 1, 4, 5, 6, 9 are the only last digits possible for squares. What are the possible last digits of cubes?
Solution:
The last digits of cubes can be any digit from 0 to 9.
Textbook Page 13
Q. Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Solution:
1-digit cubes: 13=1, 23=8
Count: 2 cubes → (1, 8)
2-digit cubes: 33=27, 43=64
Count: 2 cubes → (27, 64)
3-digit cubes: 53=125, 63=216, 73=343, 83=512, 93=729
Count: 5 cubes → (125, 216, 343, 512, 729)
Thus, the number of perfect cubes increases with larger numbers.
Unlike squares, cubes grow faster, so fewer of them are found in smaller ranges.
Q. Can a cube end with exactly two zeroes (00)? Explain.
Solution:
No, a cube cannot end with exactly two zeros, because zeros in a cube always appear in multiples of three. If a number ends with one zero, its cube will end with three zeros.
Q. The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each of these can be expressed as the sum of two positive cubes.
Solution:
4104:
23 + 163 = 8 + 4096 = 4104
93 + 153 = 729 + 3375 = 4104 13832:
23 + 243 = 8 + 13824 = 13832
183 + 203 = 5832 + 8000 = 13832
Textbook Page 14
Q. Can you tell what this sum is without doing the calculation?
91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109.
Solution:
This series consists of 10 consecutive odd numbers, and their sum is 103=1000.
Textbook Page 15
Q. Find the cube roots of these numbers:
(i) ∛64 = (ii) ∛512 = (iii) ∛729 =
Solution:
(i) ∛64
64 = (2 × 2 × 2) × (2 × 2 × 2) = 23 × 23 = (2 × 2)3 = 43
∴ ∛64 = 4. (ii) ∛512
512 = (2 × 2 × 2) × (2 × 2 × 2) × (2 × 2 × 2) = 23 × 23 × 23 = (2 × 2 × 2)3 = 83
∴ ∛512 = 8. (iii) ∛729
729 = (3 × 3 × 3) × (3 × 3 × 3) = 33 × 33 = (3 × 3)3 = 93
∴ ∛729 = 9.
Textbook Page 16 – 17
Figure it Out
1. Find the cube roots of 27000 and 10648.
Solution:
(i) ∛27000
Prime factorization of 27000 = 2 × 2 × 2 × 3 × 3 × 3 × 5 × 5 × 5
27000 = 23 × 33 × 53 = (2 × 3 × 5)3 = 303
∴ ∛27000 = 30.
(ii) ∛10648
Prime factorisation of 10648 = 2 × 2 × 2 × 11 × 11 × 11
10648 = 23 × 113 = (2 × 11)3 = 223
∴ ∛10648 = 22.
2. What number will you multiply by 1323 to make it a cube number?
Solution:
Prime factorization of 1323=3×3×3×7×7.
Here, 7 does not form a complete triplet.
So, 1323 is not a perfect cube. To make it a cube, we multiply by 7:
1323×7=3×3×3×7×7×7=9261
which is a perfect cube.
Hence, the number that must be multiplied by 1323 to make it a cube is 7.
3. State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.
Solution:
| Statement | True/False | Reason |
|---|---|---|
| (i) | False | Cube of odd number is odd. |
| (ii) | False | Cubes of numbers ending in 2 or 8 can end with 8. |
| (iii) | False | Cube of 2-digit number ≥ 1000 → 4 digits. |
| (iv) | False | Cube of 2-digit number ≤ 970,299 → max 6 digits. |
| (v) | False | Only perfect squares have odd number of factors. |
4. You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.
Solution:
Yes! We can guess cube roots without full factorisation by looking at the last digit and the approximate size of the number.
So the cube roots are: 1331 → 11, 4913 → 17, 12167 → 23, 32768 → 32.
5. Which of the following is the greatest? Explain your reasoning.
(a) 673 – 663
(b) 433 – 423
(c) 672 – 662
(d) 432 – 422
Solution:
(a) 673−663 is the greatest.
Reason: For consecutive integers n and n−1,
Evaluating:
So the order is (a) > (b) > (c) > (d).
This chapter focuses on powers, exponents, and their applications. Using these Class 8 Maths Ganita Prakash NCERT Solutions, students can quickly grasp the rules of exponents and solve problems accurately.
Textbook Page 22 – 23
Figure it Out
1. Express the following in exponential form:
(i) 6 × 6 × 6 × 6
(ii) y × y
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
Solution:
(i) 6 × 6 × 6 × 6 = 64 (ii) y × y = y2 (iii) b × b × b × b = b4 (iv) 5 × 5 × 7 × 7 × 7 = 52 × 73 (v) 2 × 2 × a × a = 22 × a2 (vi) a × a × a × c × c × c × c × d = a3 × c4 × d
2. Express each of the following as a product of powers of their prime factors in exponential form.
(i) 648 (ii) 405 (iii) 540 (iv) 3600
Solution:
(i) 648 = 2 × 2 × 2 × 3 × 3 × 3 × 3 = 23 × 34
(ii) 405 = 3 × 3 × 3 × 3 × 5 = 34 × 5
(iii) 540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 × 33 × 5
(iv) 3600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5 = 24 × 32 × 52
3. Write the numerical value of each of the following:
(i) 2 × 103
(ii) 72 × 23
(iii) 3 × 44
(iv) (-3)2 × (-5)2
(v) 32 × 104
(vi) (-2)5 × (-10)6
Solution:
(i) 2 × 103
= 2 × (10 × 10 × 10) = 2 × 1000 = 2000
(ii) 72 × 23
= (7 × 7) × (2 × 2 × 2) = 49 × 8 = 392
(iii) 3 × 44
= 3 × (4 × 4 × 4 × 4) = 3 × (16 × 16) = 3 × 256 = 768
(iv) 32 × 104
= {(-3) × (-3)} × {(-5) × (-5)} = 9 × 25 = 225
(v) 32 × 104
= (3 × 3) × (10 × 10 × 10 × 10) = 9 × 10000 = 90000
(vi) (-2)5 × (-10)6
= {(-2) × (-2) × (-2) × (-2) × (-2)} × {(-10) × (-10) × (-10) × (-10) × (-10) × (-10)}
= {4 × 4 × (-2)} × {100 × 100 × 100} = (-32) × (1000000) = -32000000
Textbook Page 44 – 45
Figure it Out
1. Find out the units digit in the value of 2224 ÷ 432 ?
Solution:
2224 ÷ 432
= 2224 ÷ (22)32 = 2224 ÷ 22 × 32 = 2224 ÷ 264 = 2224 – 64 = 2160
21 = 2 (units digit 2)
22 = 4 (units digit 4)
23 = 8 (units digit 8)
24 = 16 (units digit 6)
25 = 32 (units digit 2)
26 = 64 (units digit 4)
27 = 128 (units digit 8)
The units digit repeats in a cycle of 4: 2, 4, 8, 6.
Since 2160÷4=540 (exact multiple of 4), the units digit of 22160 will be the same as that of 24, which is 6.
2. There are 5 bottles in a container. Every day, a new container is brought in. How many bottles would be there after 40 days?
Solution:
Containers added per day = 1
Containers after 40 days = 40
Bottles in each container = 5
Total bottles after 40 days = 40×5=200
∴ After 40 days, there will be 200 bottles.
3. Write the given number as the product of two or more powers in three different ways. The powers can be any integers.
(i) 643
(ii) 1928
(iii) 32-5
Solution: (i) 643
64 = 2 × 2 × 2 × 2 × 2 × 2 = 26
643 = (26)3 = 218
Three different ways are-
1. 643 = 218 = 29 × 29
2. 643 = 218 = 210 × 28
3. 643 = 218 = 26 × 26 × 26
(ii) 1928
192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 26 × 3
1928 = (26 × 3)8 = 26 × 8 × 38 = 248 × 38
Three different ways are-
1. 1928 = 248 × 38 = 240 × 28 × 38 = 240 × (2×3)8 = 240 × 68
2. 1928 = 248 × 38 = (224×34)×(224×34)
3. 1928 = 248 × 38 = (26)8 × 38
(iii) 32-5
32 = 2 × 2 × 2 × 2 × 2 = 25
32-5 = (25)-5 = 2-25
Three different ways are-
1. 32-5 = 2-25 = 2-15 × 2-10
2. 32-5 = 2-25 = 2−5 × 2−5 × 2−5 × 2−5 × 2−5
3. 32-5 = 2-25 = 2-5 × 2-20
4. Examine each statement below and find out if it is ‘Always True’, ‘Only Sometimes True’, or ‘Never True’. Explain your reasoning.
(i) Cube numbers are also square numbers.
(ii) Fourth powers are also square numbers.
(iii) The fifth power of a number is divisible by the cube of that number.
(iv) The product of two cube numbers is a cube number.
(v) q46 is both a 4th power and a 6th power (q is a prime number).
Solution:
(i) Cube numbers are also square numbers.
Answer: Only sometimes true.
Reason: A number n3 is a square exactly when n itself is a square, i.e. n=k2. Then n3=(k2)3=k6, a sixth power. So only those cubes that are sixth powers are also squares.
Example: 64=43=82 (true). 27=33 is not a square (false).
(ii) Fourth powers are also square numbers.
Answer: Always true.
Reason: n4=(n2)2, so every fourth power is a perfect square.
Example: 16=24=(22)2=42.
(iii) The fifth power of a number is divisible by the cube of that number.
Answer: Always true (for every nonzero integer).
Reason: n5=n3⋅n2, so n5 is divisible by n3. (Note: division by 0 is undefined, so the usual interpretation excludes n=0.)
Example: 55=53⋅52, so 55 is divisible by 53.
(iv) The product of two cube numbers is a cube number.
Answer: Always true.
example- 8 = 23, 27 = 33; 8 × 27 = 216, which is 63.
(v) q46 is both a 4th power and a 6th power ( q is prime).
Answer: Never true.
Reason: For q46 to be a 4th power, 4 must divide 46; for it to be a 6th power, 6 must divide 46. Neither divides 46. More generally, a number is both a 4th and a 6th power only if its exponent is a multiple of lcm(4,6)=12; 46 is not a multiple of 12. (Also q being prime rules out trivial exceptions like 1.)
5. Simplify and write these in the exponential form.
(i) 10–2 × 10–5
(ii) 57 ÷ 54
(iii) 9–7 ÷ 94
(iv) (13–2)–3
(v) m5n12(mn)9
Solution:
(i) 10–2 × 10–5 = 10-2 – 5 = 10-7 (ii) 57 ÷ 54 = 57 – 4 = 53 (iii) 9–7 ÷ 94 = 9-7 – 4 = 9–11 (iv) (13–2)–3 = 13-2 × -3 = 136 (v) m5n12(mn)9 = m5n12 × m9n9 = m5+9n12+9 = m14n21
6. If 122 = 144 what is
(i) (1.2)2
(ii) (0.12)2
(iii) (0.012)2
(iv) 1202
Solution:
7. Circle the numbers that are the same —
24 × 36, 64 × 32, 610, 182 × 62, 624
Solution:
(i) 24 × 36
(ii) 64 × 32 = (2 × 3)4 × 32 = 24 × 34 × 32 = 24 × 36
(iii) 610 = (2 × 3)10 = 210 × 310
(iv) 182 × 62 = (2 × 3 × 3)2 × (2 × 3)2 = 22 × 32 × 32 × 22 × 32 = 24 × 36
(v) 624 = (2 × 3)24 = 224 × 324
Same numbers are: 24 × 36 ,64 × 32 , 182 × 62
8. Identify the greater number in each of the following —
(i) 43 or 34 (ii) 28 or 82 (iii) 1002 or 2100
Solution: Let’s compare carefully step by step:
(i) 43 or 34
43 = 64 , 34 = 81 ; So, 34 is greater.
(ii) 28 or 82
28 = 256, 82 = 64; So, 28 is greater.
(iii) 1002 or 2100
1002 = 10,000; 2100 = (210)10 = 102410, which is far greater than 10,000.
So, 2100 is greater than 1002.
9. A dairy plans to produce 8.5 billion packets of milk in a year. They want a unique ID (identifier) code for each packet. If they choose to use the digits 0–9, how many digits should the code consist of?
Solution:
8.5 billion = 8,500,000,000
With digits 0–9 there are 10n possible codes of length n.
We need at least 8.5×10^9 codes.
10^9=1 000 000 000 (too few) and 1010=10 000 000 000 (enough).
So the minimum length is n=10.
10. 64 is a square number (82) and a cube number (43). Are there other numbers that are both squares and cubes? Is there a way to describe such numbers in general?
Solution: Yes. Any number that is both a perfect square and a perfect cube must be a sixth power.
Reason:
So such numbers are exactly n^6.
General Rule: The sixth power of any number (i.e., n6) is both a square and a cube. i.e.
16 = 1, 26 = 64, 36 = 729, 46 = 4096, 56 = 15,625
11. A digital locker has an alphanumeric (it can have both digits and letters) passcode of length 5. Some example codes are G89P0, 38098, BRJKW, and 003AZ. How many such codes are possible?
Solution:
Passcode length = 5
Choices for each position = 26 (letters) + 10 (digits) = 36
Total number of possible passcodes =
36×36×36×36×36=365
So, the total possible codes = 36⁵.
12. The worldwide population of sheep (2024) is about 109, and that of goats is also about the same. What is the total population of sheep and goats?
(i) 209 (ii) 1011 (iii) 1010
(iv) 1018 (v) 2 × 109 (vi) 109 + 109
Solution:
Population of sheep = 10⁹
Population of goats = 10⁹
Total population = 10⁹ + 10⁹ = 2 × 10⁹
Hence, both (v) 2 × 10⁹ and (vi) 10⁹ + 10⁹ represent the correct total population.
13. Calculate and write the answer in scientific notation:
(i) If each person in the world had 30 pieces of clothing, find the total number of pieces of clothing.
(ii) There are about 100 million bee colonies in the world. Find the number of honeybees if each colony has about 50,000 bees.
(iii) The human body has about 38 trillion bacterial cells. Find the bacterial population residing in all humans in the world.
(iv) Total time spent eating in a lifetime in seconds.
Solution:
(i) Clothing pieces
World population ≈ 8×109 people
Each person has 30 pieces
Total pieces=30×8×109=240×109=2.4×1011
Answer: 2.4×1011 pieces of clothing
(ii) Honeybees
Number of colonies ≈ 1×108
Bees per colony = 50,000 = 5×104
Total bees=(1×108)(5×104)=5×1012
Answer: 5×1012 honeybees
(iii) Bacteria in all humans
Bacteria per person ≈ 3.8×1013
World population ≈ 8×109
Total bacteria=(3.8×1013)(8×109)=30.4×1022=3.04×1023
Answer: 3.04×1023 bacterial cells
(iv) Lifetime eating time in seconds
Average eating time ≈ 1.5 hours/day = 1.5 × 3600 = 5400 seconds/day
Average lifespan ≈ 80 years = 80×365×24×3600≈2.52×109 seconds
Eating fraction = 540086400=116 of the day
Eating lifetime seconds=116×2.52×109≈1.575×108
Answer: 1.58×108 seconds
14. What was the date 1 arab/1 billion seconds ago?
Solution:
Step 1: Convert 1 billion seconds into years
1 billion seconds = 1,000,000,000 seconds.
Now divide:
1,000,000,00031,557,600≈31.69 years
So 1 billion seconds ≈ 31 years and 8 months.
Step 2: Subtract from today
Today is 24 August 2025.
Answer: Approximately 24 December 1993.
Discover the fascinating world of numbers, factors, and multiples in this chapter. The detailed Class 8 Maths Ganita Prakash NCERT Solutions help students understand divisibility, prime numbers, and other number concepts clearly.
Textbook Page 51 – 52
The Mechanism of Counting
Imagine that we are living in the Stone Age, say, around ten thousand years ago. Suppose we have a herd of cows. Here are some natural questions that we might ask about our herd —
Q1. How do we ensure that all cows have returned safely after grazing?
Q2. Do we have fewer cows than our neighbour?
Q3. If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour?
We need to tackle these questions without the use of the number names or written numbers of the Hindu number system. How do we do it?
Here are some possible methods.
Method 1: We could tackle the questions by using pebbles, sticks or any object that is available in abundance. Let us choose sticks. For every cow in the herd, we could keep a stick. The final collection of sticks tells us the number of cows, which can be used to check if any cows have gone missing.
This way of associating each cow with a stick, such that no two cows are associated or mapped to the same stick is called a one-to-one mapping. This mapping can then be used to come up with a way to represent numbers, as shown in the table.
Q. How will you use such sticks to answer the other two questions (Q2 and Q3)?
Solution:
Q2. Do we have fewer cows than our neighbour?
Answer:
Q3. If there are fewer, how many more cows would we need so that we have the same number of cows as our neighbour?
Answer:
Method 2: Instead of objects, we could use a standard sequence of sounds or names. For example, we could use the sounds of the letters of any language. While counting, we could make a one-to-one mapping between the objects and the letters: that is, associate each object to be counted with a letter, following the letter-order. This mapping can then be used to come up with a way of verbally representing numbers.
For example, we get the following number representation if we use English letters ‘a’ to ‘z’.
An obvious limitation of using only the letters of the English alphabet in this form is that it cannot be used to count collections having more than 26 objects.
Q. How many numbers can you represent in this way using the sounds of the letters of your language?
Solution: We can represent 52 numbers using the sounds of Hindi letters.
Method 3: We could use a sequence of written symbols as follows.
Q. Do you see a way of extending this method to represent bigger numbers as well? How?
Solution:
Yes, this method can be extended to represent much larger numbers by combining the basic symbols using additive and subtractive rules.
Rules to extend the system:
Textbook Page 54
Figure it out
Q1. Suppose you are using the number system that uses sticks to represent numbers, as in Method 1. Without using either the number names or the numerals of the Hindu number system, give a method for adding, subtracting, multiplying and dividing two numbers or two collections of sticks.
Solution:
1. Addition
2. Subtraction
3. Multiplication
4. Division
Q2. One way of extending the number system in Method 2 is by using strings with more than one letter — for example, we could use ‘aa’ for 27. How can you extend this system to represent all the numbers? There are many ways of doing it!
Solution:
One way to extend the number system in Method 2 is by using strings with more than one letter — for example, using “aa” for 27. You can extend this system to represent all numbers in many different ways!
Textbook Page 58
Q. What could be the difficulties with using a number system that counts only in groups of a single particular size? How would you represent a number like 1345 in a system that counts only by 5s?
Solution:
Difficulties with using a number system that counts only in groups of a single size:
Example:
1347÷5=269 remainder 2
So,
1347=5×269+2
In a system that only counts in groups of 5, you could show 269 groups of 5, but there would be 2 leftover, which the system cannot directly represent.
Textbook Page 59
Figure it Out
Q1. Represent the following numbers in the Roman system.
(i) 1222 (ii) 2999 (iii) 302 (iv) 715
Solution:
(i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1 = MCCXXII.
(ii) 2999 = 1000 + 1000 + 900 + 90 + 9
= 1000 + 1000 + (1000 – 100) + (100 – 10) + (10 – 1)
= MMCMXCIX.
(iii) 302 = 100 + 100 + 100 + 1 + 1 = CCCII.
(iv) 715 = 500 + 100 + 100 + 10 + 5 = DCCXV.
Example: Try adding the following numbers without converting them to Hindu numerals:
(a) CCXXXII + CCCCXIII
Let us find the total number of Is, Xs, and Cs, and group them starting from the largest landmark number.
Apparently, it looks like the largest landmark number is C, but note that 5 Cs (100s) make a D (500). So the sum is
Q. Do it yourself now:
(b) LXXXVII + LXXVIII
Solution:
Textbook Page 60 – 61
Q. How will you multiply two numbers given in Roman numerals, without converting them to Hindu numerals? Try to find the product of the following pairs of landmark numbers: V × L, L × D, V × D, VII × IX.
Solution:
Multiplying two numbers in Roman numerals without converting them to Hindu-Arabic numerals can be done using repeated addition or other techniques like doubling and halving.
Examples:
Figure it Out
Q1. A group of indigenous people in a Pacific island use different sequences of number names to count different objects. Why do you think they do this?
Solution:
Different number sequences are used because:
Q2. Consider the extension of the Gumulgal number system beyond 6 in the same way of counting by 2s. Come up with ways of performing the different arithmetic operations (+, –, ×, ÷) for numbers occurring in this system, without using Hindu numerals. Use this to evaluate the following:
(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar- ukasar-urapon)
(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar- ukasar)
(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)
(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)
Solution:
(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar- ukasar-urapon)
= (2 + 2 + 2 + 2 + 1) + (2 + 2 + 2 + 1) = 9 + 7 = 16.
(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar- ukasar)
= (2 + 2 + 2 + 2 + 1) – (2 + 2 + 2) = 9 – 6 = 3.
(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)
= (2 + 2 + 2 + 2 + 1) × (2 + 2) = 9 × 4 = 36.
(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)
= (2 + 2 + 2 + 2 + 2 + 2 + 2 + 2) ÷ (2 + 2) = 16 ÷ 4 = 4.
Q3. Identify the features of the Hindu number system that make it efficient when compared to the Roman number system.
Solution:
The Hindu number system is far more efficient than the Roman number system because:
Textbook Page 62
Q1. Represent the following numbers in the Egyptian system: 10458, 1023, 2660, 784, 1111, 70707.
Solution:
(i) 10458
= 10000 + 100 + 100 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =
(ii) 1023
= 1000 + 10 + 10 + 1 + 1 + 1 =
(iii) 2660
= 1000 + 1000 + 100 + 100 + 100 + 100 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 =
(iv) 784
= 100 + 100 + 100 + 100 + 100 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1 + 1 =
(v) 1111
= 1000 + 100 + 10 + 1 =
(vi) 70707
= 10000 + 10000 + 10000 + 10000 + 10000 + 10000 + 10000 + 100 + 100 + 100 + 100 + 100 + 100 + 100 + 1 + 1 + 1 + 1 + 1 + 1 + 1 =
Q2. What numbers do these numerals stand for?
Solution:
(I) =(100+100) +(10+10+10+10+10+10+10) +(1+1+1) +(1+1+1)= 200 + 70 + 6 = 276.
(II) =(1000+1000+1000 + 1000) + (100 +100+100)+(10+10) + (1+1) = 4000+300 +20 +2 = 4322.
Textbook Page 63
Figure it Out
Q1. Write the following numbers in the above base-5 system using the symbols in Table 2: 15, 50, 137, 293, 651.
Solution:
(i) 15 = 5 + 5 + 5 =
(ii) 50 = 25 + 25 =
(iii) 137 = 125 + 5 + 5 + 1 + 1 =
(iv) 293 = 125 + 125 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 5 + 1 + 1 + 1 =
(v) 651 = 625 + 5 + 5 + 5 + 5 + 5 + 1 =
Q2. Is there a number that cannot be represented in our base-5 system above? Why or why not?
Solution:
There is no number that cannot be represented in base-5. Every whole number can be represented in base-5, because the system uses digits 0 to 4 combined with increasing powers of 5.
Q3. Compute the landmark numbers of a base-7 system. In general, what are the landmark numbers of a base-n system?
Solution:
In a base-7 system, starting with the first landmark number as 1, each subsequent landmark number is 7 times the previous one. They are the powers of 7:
70=1, 71, 72, 73, 74,…
More generally, in a base‑n system, the landmark numbers are the powers of n starting from n0=1:
1, n, n2, n3,…
Textbook Page 65
Figure it Out
Q1. Add the following Egyptian numerals:
Solution:
Q2. Add the following numerals that are in the base-5 system that we created:
Solution:
Textbook Page 69
Figure it Out
Q1. Can there be a number whose representation in Egyptian numerals has one of the symbols occurring 10 or more times? Why not?
Solution:
No number in the Egyptian numeral system can have any symbol appear 10 or more times. This is because:
Q2. Create your own number system of base 4 and represent numbers from 1 to 16.
Solution:
I have created a number system called the Quad-Code System, which is based on base-4. Instead of regular digits, I use four special symbols:
In base-4, the place values increase as powers of 4: the rightmost place is 40=1, the next is 41=4, then 42=16, and so on. Using these symbols, any number can be written compactly.
Here’s how numbers from 1 to 16 are represented in the Quad-Code System:
Q3. Give a simple rule to multiply a given number by 5 in the base-5 system that we created.
Solution:
In my base-5 system using symbols (A = 0, B = 1, C = 2, D = 3, E = 4), multiplying a number by 5 is simple:
Rule: Add a zero (A) at the end of the number.
This works because in base-5, multiplying by 5 is equivalent to shifting all digits one place to the left, just like adding a zero in base-10 when multiplying by 10.
Example:
So, adding A at the right end multiplies the number by the base itself (5).
Textbook Page 73
Figure it Out
Q1. Represent the following numbers in the Mesopotamian system —
(i) 63 (ii) 132 (iii) 200 (iv) 60 (v) 3605
Answer:
Textbook Page 80
Figure it Out
Q1. Why do you think the Chinese alternated between the Zong and Heng symbols? If only the Zong symbols were to be used, how would 41 be represented? Could this numeral be interpreted in any other way if there is no significant space between two successive positions?
Solution:
Using Zong and Heng Symbols in the Chinese Number System:
Example:
By alternating between Zong and Heng, the Chinese system ensured place values were visually clear, making numbers easier to read and reducing the risk of mistakes in handwritten or closely packed texts.
Q2. Form a base-2 place value system using ‘ukasar’ and ‘urapon’ as the digits. Compare this system with that of the Gumulgal’s.
Solution:-
Base-2 System Using ‘ukasar’ and ‘urapon’:
We can create a base-2 place value system by assigning:
This system works like the binary system, where each position from right to left represents increasing powers of 2:
Examples:
Each position indicates how many of that power of 2 are present, using ukasar for 0 and urapon for 1.
Comparison with the Gumulgal System:
Q3. Where in your daily lives and in which professions, do the Hindu numerals and 0, play an important role? How might our lives have been different if our number system and 0 hadn’t been invented or conceived of?
Solution:
Importance of Hindu Numerals and Zero
Hindu numerals, including the digit 0, are used every day—for telling time, counting money, reading prices, doing math in school, and writing phone numbers. Many professions, such as banking, teaching, engineering, and science, depend heavily on this number system.
Zero is especially important for place value and calculations, making it easy to write and understand large numbers.
Without zero and the Hindu number system, life would be much harder. Calculations, trade, and even writing dates would be complicated. Modern technology like computers and calculators might not exist, slowing progress in all areas of life.
Q4. The ancient Indians likely used base 10 for the Hindu number system because humans have 10 fingers and so we can use our fingers to count. But what if we had only 8 fingers? How would we be writing numbers then? What would the Hindu numerals look like if we were using base 8 instead? Base 5? Try writing the base-10 Hindu numeral 25 as base-8 and base-5 Hindu numerals, respectively. Can you write it in base-2?
Solution:
If Humans Had Only 8 Fingers
If humans had only 8 fingers, we would likely have developed a base-8 (octal) number system instead of base-10. Just as we now count from 0 to 9 in base-10, we would count from 0 to 7 in base-8. All numerals, calculations, and mathematics would be based on powers of 8.
Conversion of the decimal number 25:
Learn about different types of quadrilaterals, their properties, and theorems. Our Class 8 Maths Ganita Prakash NCERT Solutions simplify geometry problems and make exam preparation easier.
Textbook Page 94
Figure it Out
1. Find all the other angles inside the following rectangles.
Solution:
(i)
Here, we have : ∠1 + ∠9 = 90° (All corner angles of a rectangle are 90°)
∠1 + 30° = 90° => ∠1 = 90° – 30° = 60° ∠1 = ∠5 = 60° and ∠9 = ∠4 = 30° (Alternate interior angles)
In △AOB, OA = OB, so, angles opposite them are equal
∴ ∠9 = ∠7 = 30° also, ∠7 = ∠3 = 30° (Alternate interior angles)
In △AOD, OA = OD, so, angles opposite them are equal
∴ ∠2 = ∠1 = 60° also, ∠2 = ∠6 = 60° (Alternate interior angles)
In △AOB, ∠9 + ∠7 + ∠AOB = 180° (Sum of angles of a triangle)
30° + 30° + ∠AOB = 180° => 60° + ∠AOB = 180°
∠AOB = 180° – 60° = 120°
∠AOB = ∠COD = 120° (Vertically opposite angles)
∠AOB + ∠AOD = 180° (Linear pair)
120° + ∠AOD = 180° , ∠AOD = 180° – 120°= 60°
∠AOD = ∠BOC = 60° (Vertically opposite angles)
Therefore, ∠1 = ∠5 = ∠2 = ∠6 = ∠AOD = ∠BOC = 60°
∠AOB = ∠COD = 120° and, ∠9 = ∠4 = ∠7 = ∠3 = 30°
(ii)
Here we have, ∠POS = ∠ROQ = 110° (Vertically opposite angles)
∠POS + ∠POQ = 180° (Linear Pair)
110° + ∠POQ = 180° so, ∠POQ = 180° – 110° = 70°
∠POQ = ∠SOR = 70° (Vertically opposite angles)
now, In △POS, OP = OS, then the angles opposite them are equal.
∴ ∠1 = ∠2 = a In △POS, ∠1 + ∠2 + ∠POS = 180° (Sum of angles of a triangle)
a + a + 110° = 180° so, a = 35°
∴ ∠1 = ∠2 = a = 35°
∠1 = ∠5 = 35° and ∠2 = ∠6 = 35° (Alternate interior angles)
Since ABCD is a rectangle, ∠P = 90°
∠9 = ∠1 + ∠8 so, 90° = 35° + ∠8 => ∠8 = 90° – 35° = 55°
also, ∠8 = ∠4 = 55° (Alternate interior angles)
Now, In △POQ, OP = OQ, then the angles opposite them are equal
so, ∠7 = ∠8 = 55° and ∠7 = ∠2 = 55° (Alternate interior angles)
Therefore, ∠POS = ∠ROQ = 110° and ∠POQ = ∠SOR = 70°
∠1 = ∠2 = ∠5 = ∠6 = 35° and ∠8 = ∠4 = ∠7 = ∠2 = 55°
2. Draw a quadrilateral whose diagonals have equal lengths of 8 cm that bisect each other, and intersect at an angle of
(i) 30° (ii) 40° (iii) 90° (iv) 140°
Solution:
(i) 30°
(ii) 40°
(iii) 90°
(iv) 140°
3. Consider a circle with centre O. Line segments PL and AM are two perpendicular diameters of the circle. What is the figure APML? Reason and/or experiment to figure this out.
Solution:
APML is a square.
Reason:
So the quadrilateral has all sides equal and all angles 90∘ ⇒ it is a square.
4. We have seen how to get 90° using paper folding. Now, suppose we do not have any paper but two sticks of equal length, and a thread. How do we make an exact 90° using these?
Solution:
Steps:
5. We saw that one of the properties of a rectangle is that its opposite sides are parallel. Can this be chosen as a definition of a rectangle? In other words, is every quadrilateral that has opposite sides parallel and equal a rectangle?
Solution:
Property being tested
“A rectangle is a quadrilateral whose opposite sides are parallel and equal.”
Step 1: Opposite sides parallel and equal
A quadrilateral with opposite sides parallel and equal is by definition a parallelogram.
Step 2: Does every parallelogram become a rectangle?
Conclusion-
No, we cannot define a rectangle only by “opposite sides parallel and equal.” That condition defines a parallelogram. To make it a rectangle, we must also require all interior angles are right angles (90°). So the correct definition of a rectangle is: A rectangle is a parallelogram with all angles equal to 90 degree.
Textbook Page 102
1. Find the remaining angles in the following quadrilaterals.
Solution:
(i) Here we have, PR ∥ EA, and PE ∥ RA
so, PEAR is a parallelogram.
∠P = ∠A = 40° (Opposite angles of parallelogram are equal)
∠P + ∠R = 180° (The sum of adjacent angles of a parallelogram is 180°)
40° + ∠R = 180° , ∠R = 180° – 40°= 140°.
∠R = ∠E = 140° (Opposite angles of parallelogram are equal)
(ii) Here we have, PQ ∥ SR, and PS ∥ QR
so PQRS is a parallelogram.
∠P = ∠R = 110° (Opposite angles of parallelogram are equal)
∠P + ∠S = 180° (sum of adjacent angles of parallelogram is 180°)
110° + ∠S = 180° , ∠S = 180° – 110° = 70°.
∠S = ∠Q = 70° (Opposite angles of parallelogram are equal)
(iii) Here we have, XWUV is a rhombus (all sides are equal).
now, In △VUX, UV = UX, therefore angles opposite them are equal.
∴ ∠UXV = ∠UVX = 30° => ∠UXV = ∠WXV = 30° (The diagonals of a rhombus bisect its angles)
Also we have, ∠UVX = ∠WVX = 30° (diagonals of a rhombus bisect its angles) now, ∠E = 2 × ∠UVX = 2 × 30° = 60°
∠V = ∠X = 60° (Opposite angles of rhombus are equal)
∠V + ∠U = 180° (sum of adjacent angles of rhombus is 180°)
60° + ∠U = 180°, ∠U = 180° – 60° = 120°
∠U = ∠W = 120° (Opposite angles of rhombus are equal)
(iv) Here we have, AEIO is a rhombus (all sides equal).
In △EAO we have , AE = AO, then angles opposite them are equal.
so, ∠AOE = ∠AEO = 20°
∠AEO = ∠IEO = 20° (diagonals of rhombus bisect its angles)
Also we have, ∠AOE = ∠IOE = 20° (diagonals of rhombus bisect its angles) ∠E = 2 × ∠AEO = 2 × 20° = 40°
∠E = ∠O = 40° (Opposite angles of rhombus are equal)
∠E + ∠A = 180° (sum of adjacent angles of rhombus is 180°)
40° + ∠A = 180°, ∠A = 180° – 40°= 140°
∠A = ∠I = 140° (Opposite angles of rhombus are equal)
2. Using the diagonal properties, construct a parallelogram whose diagonals are of lengths 7 cm and 5 cm, and intersect at an angle of 140°.
Solution:
Steps of Construction:
Thus, ABCD is the required parallelogram.
3. Using the diagonal properties, construct a rhombus whose diagonals are of lengths 4 cm and 5 cm.
Solution:
Steps of Construction:
Thus, ABCD is the required rhombus.
Textbook Page 107
1. Find all the sides and the angles of the quadrilateral obtained by joining two equilateral triangles with sides 4 cm.
Solution:
Since all sides of an equilateral triangle are equal, the sides of the quadrilateral formed are also equal.
PQ=QR=RS=SP=4 cm
Each angle of an equilateral triangle measures 60°.
∠P=∠R=60∘
At vertex S:
∠S=∠PSQ+∠RSQ=60∘+60∘=120∘
At vertex Q:
∠Q=∠PQR+∠RQS=60∘+60∘=120∘
Thus, the quadrilateral PQRS has all sides equal to 4 cm, with angles:
∠P=∠R=60∘,∠Q=∠S=120∘
2. Construct a kite whose diagonals are of lengths 6 cm and 8 cm.
Solution:
To construct kite ABCD:
Quadrilateral ABCD is the required kite.
3. Find the remaining angles in the following trapeziums —
Solution:
Since AB ∥ DC & AD is a tranversal line , we have
∠A + ∠D = 180° (Sum of angles on same side of transversal)
135° + ∠D = 180°, ∠D = 180° – 135°= 45°
now, since AB ∥ DC & BC is a tranversal line , we have
∠B + ∠C = 180° (Sum of angles on same side of transversal)
105° + ∠C = 180°, ∠C = 180° – 105°, ∠C = 75°
we have PQ ∥ SR & PS is a tranversal, we have
∠P + ∠S = 180° (Sum of angles on same side of transversal)
∠P + 100° = 180°, ∠P = 180° – 100° = 80°
∠S = ∠R = 100° (Angles opposite to equal sides are equal)
Also, we have PQ ∥ SR, and QR is a tranversal, so
∠Q + ∠R = 180° (Sum of angles on same side of transversal)
∠Q + 100° = 180°, ∠Q = 180° – 100° = 80°.
4. Draw a Venn diagram showing the set of parallelograms, kites, rhombuses, rectangles, and squares. Then, answer the following questions —
(i) What is the quadrilateral that is both a kite and a parallelogram?
(ii) Can there be a quadrilateral that is both a kite and a rectangle?
(iii) Is every kite a rhombus? If not, what is the correct relationship between these two types of quadrilaterals?
Solution:
(i) Quadrilateral that is both a kite and a parallelogram → Rhombus (since it has both properties).
(ii) Quadrilateral that is both a kite and a rectangle → Yes, the square (because it is a special kite and also a rectangle).
(iii) Is every kite a rhombus? → No.
5. If PAIR and RODS are two rectangles, find ∠IOD.
Solution:
Since PAIR and RODS are triangles, and
∠RIO=90∘ (corner angle of a rectangle).
In △RIO:
∠IRO+∠IOR+∠RIO=180∘30∘+∠IOR+90∘=180∘120∘+∠IOR=180∘ ⇒ ∠IOR=60∘
Therefore,
∠IOD=90∘−∠IOR=90∘−60∘=30∘
Hence, ∠IOD=30∘.
6. Construct a square with a diagonal 6 cm without using a protractor.
Solution:
Steps of Construction:
Thus, ABCD is the required square whose diagonal is 6 cm.
7. CASE is a square. The points U, V, W and X are the midpoints of the sides of the square. What type of quadrilateral is UVWX? Find this by using geometric reasoning, as well as by construction and measurement. Find other ways of constructing a square within a square such that the vertices of the inner square lie on the sides of the outer square, as shown in Figure (b).
Solution:
Other constructions for a square inside a square (vertices on sides):
8. If a quadrilateral has four equal sides and one angle of 90°, will it be a square? Find the answer using geometric reasoning as well as by construction and measurement.
Solution: Yes — it must be a square.
Geometric reasoning
Straightedge–compass construction (to verify)
Then AB=BC=CD=DA=s and ∠A=90∘. By the reasoning above, the other angles also measure 90∘. Measuring confirms a square ABCD.
Here’s the diagram. A quadrilateral with all sides equal and one right angle becomes a square, since the right angle forces all four angles to be 90°.
9. What type of quadrilateral is one in which the opposite sides are equal? Justify your answer.
Hint: Draw a diagonal and check for congruent triangles.
Solution:
IA quadrilateral in which the opposite sides are equal is a parallelogram.
Justification:
Here’s the diagram showing a quadrilateral with opposite sides equal.
Such a quadrilateral is a parallelogram, since equality of opposite sides guarantees parallelism and the parallelogram property.
10. Will the sum of the angles in a quadrilateral such as the following one also be 360°? Find the answer using geometric reasoning as well as by constructing this figure and measuring.
Solution:
Yes, the sum of the angles in any quadrilateral is always 360°.
Construction: Plot four non-collinear points A,B,C, and D, then join them in order to form quadrilateral ABCD.
Geometric reasoning:
In quadrilateral ABCD, draw diagonal BD, dividing it into two triangles.
In △BAD:
∠DBA+∠BAD+∠ADB=180∘ ……….(1)
In △BCD:
∠BCD+∠CDB+∠DBC=180∘ ……….(2)
Adding (1) and (2):
∠DBA+∠BAD+∠ADB+∠BCD+∠CDB+∠DBC=360∘
Rearranging:
(∠DBA+∠DBC)+(∠ADB+∠CDB)+∠BAD+∠BCD=360∘
∠ABC+∠ADC+∠BAD+∠BCD=360∘
Hence, the sum of the angles of a quadrilateral is always 360°.
11. State whether the following statements are true or false. Justify your answers.
(i) A quadrilateral whose diagonals are equal and bisect each other must be a square.
Solution: False.
A quadrilateral with equal diagonals that bisect each other is called a rectangle. When all its sides are also equal, it becomes a square, which is a special type of rectangle.
(ii) A quadrilateral having three right angles must be a rectangle.
Solution: True.
If a quadrilateral has three right angles, the fourth angle must also be a right angle. Hence, a quadrilateral with four right angles is a rectangle.
(iii) A quadrilateral whose diagonals bisect each other must be a parallelogram.
Solution: True.
If the diagonals of a quadrilateral bisect each other, they divide the figure into two congruent triangles. This implies that opposite sides are parallel, and therefore, the quadrilateral is a parallelogram.
(iv) A quadrilateral whose diagonals are perpendicular to each other must be a rhombus.
Solution: False.
Squares, kites, and certain other quadrilaterals also have perpendicular diagonals. Hence, perpendicular diagonals alone do not guarantee that the quadrilateral is a rhombus.
(v) A quadrilateral in which the opposite angles are equal must be a parallelogram.
Solution: True.
If both pairs of opposite angles are equal, then each pair of adjacent angles becomes supplementary, implying that opposite sides are parallel. Therefore, the quadrilateral is a parallelogram.
(vi) A quadrilateral in which all the angles are equal is a rectangle.
Solution: True
If all four angles are equal, then each angle = 360° ÷ 4 = 90°. A quadrilateral with four right angles is a rectangle.
(vii) Isosceles trapeziums are parallelograms.
Solution: False.
An isosceles trapezium has one pair of parallel sides with the non-parallel sides equal, whereas a parallelogram has two pairs of parallel sides. Hence, an isosceles trapezium is not a parallelogram.
Explore number patterns, puzzles, and interesting arithmetic tricks. These Class 8 Maths Ganita Prakash NCERT Solutions guide students to solve problems efficiently while enhancing number sense.
Figure it Out – Page 122
1. The sum of four consecutive numbers is 34. What are these numbers?
Solution:
Let the four consecutive numbers be x,x+1,x+2,x+3.
x+(x+1)+(x+2)+(x+3)=34 ⟹ 4x+6=34 ⟹ x=7
Hence, the numbers are 7, 8, 9, 10.
2. Suppose p is the greatest of five consecutive numbers. Describe the other four numbers in terms of p.
Solution:
If p is the largest of five consecutive numbers, the remaining four are (p−1),(p−2),(p−3), and (p−4).
3. For each statement below, determine whether it is always true, sometimes true, or never true. Explain your answer. Mention examples and non-examples as appropriate. Justify your claim using algebra.
(i) The sum of two even numbers is a multiple of 3.
Solution:
Let the two even numbers be 2a and 2b.
Sum=2a+2b=2(a+b)
For 2(a+b) to be divisible by 3, (a+b) must be a multiple of 3.
Examples:
Conclusion: The statement is sometimes true.
(ii) If a number is not divisible by 18, then it is also not divisible by 9.
Solution:
If a number is divisible by 18, then it must also be divisible by 9, since 9 is a factor of 18.
18a÷9=2a⇒divisible by 9.
However, if a number is divisible by 9, it is not necessarily divisible by 18.
9b÷18=b2⇒may not be an integer.
Examples:
Conclusion: The statement is sometimes true.
(iii) If two numbers are not divisible by 6, then their sum is not divisible by 6.
Solution:
Let the two numbers be a and b.
If neither is divisible by 6, their sum may still be divisible by 6.
Example: 2 and 4 are not divisible by 6, but 2+4=6 is divisible by 6.
Conclusion: Sometimes true.
(iv) The sum of a multiple of 6 and a multiple of 9 is a multiple of 3.
Solution:
Let a multiple of 6 be 6a and a multiple of 9 be 9b.
6a+9b=3(2a+3b) ⇒ always divisible by 3.
Examples:
Conclusion: Always true.
(v) The sum of a multiple of 6 and a multiple of 3 is a multiple of 9.
Solution:
Let a multiple of 6 be 6a and a multiple of 3 be 3b.
6a+3b=3(2a+b)
For this sum to be divisible by 9, (2a+b) must be a multiple of 3.
Examples:
Conclusion: Sometimes true.
4. Find a few numbers that leave a remainder of 2 when divided by 3 and a remainder of 2 when divided by 4. Write an algebraic expression to describe all such numbers.
Solution:
The L.C.M. of 3 and 4 is 12.
Such numbers can be expressed as:
12a+2
Examples:
(i) 12×1+2=14
(ii) 12×2+2=26
(iii) 12×3+2=38
5. “I hold some pebbles, not too many, When I group them in 3’s, one stays with me. Try pairing them up — it simply won’t do, A stubborn odd pebble remains in my view. Group them by 5, yet one’s still around,
But grouping by seven, perfection is found. More than one hundred would be far too bold, Can you tell me the number of pebbles I hold?”
Solution:
When grouped in 3’s, 2’s, or 5’s, one pebble is left each time.
When grouped in 7’s, none is left.
Also, the number is ≤ 100.
LCM of 2,3,5=30
So, the number of pebbles must be of the form 30k+1.
Hence, the number of pebbles is 91.
6. Tathagat has written several numbers that leave a remainder of 2 when divided by 6. He claims, “If you add any three such numbers, the sum will always be a multiple of 6.” Is Tathagat’s claim true?
Solution:
A number that leaves a remainder of 2 when divided by 6 can be expressed as 6k+2.
Let three such numbers be (6a+2),(6b+2),(6c+2).
(6a+2)+(6b+2)+(6c+2)=6(a+b+c)+6=6(a+b+c+1)
This is clearly divisible by 6.
So, Tathagat’s claim is always true.
Examples:
7. When divided by 7, the number 661 leaves a remainder of 3, and 4779 leaves a remainder of 5. Without calculating, can you say what remainders the following expressions will leave when divided by 7? Show the solution both algebraically and visually.
(i) 4779 + 661 (ii) 4779 – 661
Solution:
(i) 4779+661:
(ii) 4779−661:
8. Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?
Solution:
A number leaving remainder 2 when divided by 3 can be written as 3x+2.
A number leaving remainder 3 when divided by 4 is 4x+3.
A number leaving remainder 4 when divided by 5 is 5x+4.
Since the L.C.M. of 3, 4, and 5 is 60, and in each case the remainder is 1 less than the divisor, the required number must be 1 less than 60.
60−1=59
Thus, the smallest number satisfying all the conditions is 59.
Understand the concept of distributive properties and multiplication tricks. Using these Class 8 Maths Ganita Prakash NCERT Solutions, students can learn practical applications of multiplication easily.
Textbook Page 142
Q1. Observe the multiplication grid below. Each number inside the grid is formed by multiplying two numbers. If the middle number of a 3 × 3 frame is given by the expression pq, as shown in the figure, write the expressions for the other numbers in the grid.
Solution:
Q2. Expand the following products.
(i) (3 + u) (v – 3)
(ii) 2/3 (15 + 6a)
(iii) (10a+ b) (10c + d)
(iv) (3 – x) (x– 6)
(v) (–5a + b) (c+ d)
(vi) (5 + z) (y+ 9)
Solution:
(i) (3+u)(v–3)
=(3+u)v−(3+u)3=3v+uv−9−3u=uv+3v−3u−9
(ii) 23(15+6a)
=23×15+23×6a=10+4a
(iii) (10a+b)(10c+d)
=(10a+b)(10c)+(10a+b)d=100ac+10bc+10ad+bd
(iv) (3–x)(x–6)
=(3−x)x−(3−x)6=3x−x2−18+6x=−x2+9x−18
(v) (−5a+b)(c+d)
=(−5a+b)c+(−5a+b)d=−5ac+bc−5ad+bd=−5ac−5ad+bc+bd
(vi) (5+z)(y+9)
=(5+z)y+(5+z)9=5y+zy+45+9z=zy+5y+9z+45
Q3. Find 3 examples where the product of two numbers remains unchanged when one of them is increased by 2 and the other is decreased by 4.
Solution:
If the two numbers are x and y, then
x×y=(x+2)(y−4)
Expanding,
xy=xy+2y−4x−80=2y−4x−82y=4x+8⇒y=2x+4
Examples:
(i) For x=1, y=6:
Product =1×6=6
Check: (1+2)(6−4)=3×2=6
(ii) For x=2, y=8:
Product =2×8=16
Check: (2+2)(8−4)=4×4=16
(iii) For x=5, y=14:
Product =5×14=70
Check: (5+2)(14−4)=7×10=70
Therefore, (1,6),(2,8),(5,14) are valid solutions.
Q4. Expand (i) (a + ab – 3b2) (4 + b), and (ii) (4y + 7) (y + 11z – 3).
Solution:
(i) (a+ab–3b2)(4+b)
=(a+ab–3b2)⋅4+(a+ab–3b2)⋅b=4a+4ab−12b2+ab+ab2−3b3 =−3b3−12b2+ab2+5ab+4a
(ii) (4y+7)(y+11z−3)
=(4y+7)y+(4y+7)(11z)−(4y+7)3=4y2+7y+44yz+77z−12y−21=4y2−5y+44yz+77z−21
Q5. Expand (i) (a – b) (a + b), (ii) (a – b) (a2 + ab + b2), and (iii) (a – b)(a3 + a2b + ab2 + b3), Do you see a pattern? What would be the next identity in the pattern that you see? Can you check it by expanding?
Solution:
(i) (a−b)(a+b)
=a2−ab+ab−b2 =a2−b2
(ii) (a−b)(a2+ab+b2)
=a3−a2b+a2b−ab2+ab2−b3 =a3−b3
(iii) (a – b)(a3 + a2b + ab2 + b3)
=a4−a3b+a3b−a2b2+a2b2−ab3+ab3−b4=a4−b4
Thus, the next identity is:
(a−b)(a4+a3b+a2b2+ab3+b4)=a5−b5
This follows the general formula:
(a−b)(an−1+an−2b+⋯+abn−2+bn−1)=an−bn
This chapter focuses on ratios, proportions, and their real-life applications. Our Class 8 Maths Ganita Prakash NCERT Solutions help students solve proportion problems with clarity and confidence.
Textbook Page 165
Figure it Out
1. Circle the following statements of proportion that are true.
(i) 4 : 7 :: 12 : 21
(ii) 8 : 3 :: 24 : 6
(iii) 7 : 12 :: 12 : 7
(iv) 21 : 6 :: 35 : 10
(v) 12 : 18 :: 28 : 12
(vi) 24 : 8 :: 9 : 3
Solution:
(i) 4:7::12:21
(ii) 8:3::24:6
(iii) 7:12::12:7
(iv) 21:6::35:10
(v) 12:18::28:12
(vi) 24:8::9:3
2. Give 3 ratios that are proportional to 4 : 9.
Solution:
To find ratios proportional to 4 : 9, we multiply both terms by the same number:
Therefore, three ratios proportional to 4 : 9 are 8 : 18, 12 : 27, and 20 : 45.
3. Fill in the missing numbers for these ratios that are proportional to 18 : 24.
3 : ______, 12 : ______, 20 : ______, 27 : ______
Solution:
To simplify the ratio 18 : 24, we find the H.C.F = 6:
18÷6:24÷6=3:4
So any ratio proportional to 18 : 24 must equal 3 : 4.
(i) 3 : ____ = 3 : 4
(ii) 12 : ____ = 3 × 4 : 4 × 4 = 12 : 16
(iii) 20 : ____ = 3 × (20/3) : 4 × (20/3) = 20 : 80/3
(iv) 27 : ____ = 3 × 9 : 4 × 9 = 27 : 36
4. Look at the following rectangles. Which rectangles are similar to each other? You can verify this by measuring the width and height using a scale and comparing their ratios.
Solution:
| Rectangle | Width | Height | Ratio |
|---|---|---|---|
| A | 0.5 cm | 1.5 cm | 0.5 : 1.5 = 1 : 3 |
| B | 1.5 cm | 1 cm | 1.5 : 1 = 3 : 2 |
| C | 4.5 cm | 2 cm | 4.5 : 2 = 9 : 4 |
| D | 3.5 cm | 1 cm | 3.5 : 1 = 7 : 2 |
| E | 0.5 cm | 1.5 cm | 0.5 : 1.5 = 1 : 3 |
Rectangles A and E have the same simplified ratio of 1 : 3, which means they are similar to each other.
6. The following figure shows a small portion of a long brick wall with patterns made using coloured bricks. Each wall continues this pattern throughout the wall. What is the ratio of grey bricks to coloured bricks? Try to give the ratios in their simplest form.
Solution:
(a) we have ratio as Grey bricks : Coloured bricks = 18 : 33 = 6 : 11
(b) we have the ratio as Grey bricks : Coloured bricks = 48 : 71
7. Let us draw some human figures. Measure your friend’s body—the lengths of their head, torso, arms, and legs. Write the ratios as mentioned below—
Solution:
My friend’s body measurements are:
The ratios are:
Final ratios:
Congratulations! You have now completed all the chapters with our Class 8 Maths Ganita Prakash NCERT Solutions. These chapter-wise solutions are designed to make learning easier, strengthen your concepts, and improve problem-solving skills. By practicing regularly with these solutions, you can confidently tackle any question in your Class 8 Maths exams. Keep revising, solve exercises yourself, and make the most of these Class 8 Maths Ganita Prakash NCERT Solutions to achieve excellent results.
1. How do I start Class 8 Maths (CBSE)?
The best way to begin your preparation for Class 8 Maths is by using the Class 8 Maths Ganita Prakash NCERT Solutions. Cover all topics systematically so that you face no difficulties during the exams.
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4. Which mathematics book should I refer to in Class 8?
The recommended book for Class 8 Maths is the NCERT textbook along with its Class 8 Maths Ganita Prakash NCERT Solutions prescribed by the CBSE Board. These resources ensure comprehensive exam preparation.
5. What is the best way to study Class 8 Maths?
The most effective way to study Class 8 Maths is by practicing previous years’ papers and referring to the Class 8 Maths Ganita Prakash NCERT Solutions. Regular practice and revision will help you score better grades.
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