Class-8 Ganita prakash Worksheet Maths ch-1 Answer key
If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
Answer: No. These units digits are possible for squares but are not sufficient to guarantee a number is a square. (Example: 21 ends with 1 but is not a perfect square.) (CBQ — Number sense & reasoning)
Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Answer: Any numbers ending in 2, 3, 7 or 8 are never perfect squares. Examples: 12, 23, 37, 48, 102.
Which of the following numbers have the digit 6 in the unit's place?
(i) 38² (ii) 34² (iii) 46² (iv) 56² (v) 74² (vi) 82²
Answer: (ii) 34², (iii) 46², (iv) 56², (v) 74². (Because units digit depends only on last digit squared: 4²→6, 6²→6.)
If a number contains 3 zeros at the end, how many zeros will its square have at the end?
Answer: 6 zeros. (If number = N·10³ then square = N²·10⁶.)
What can you say about the parity of a number and its square?
Answer: Parity is preserved: an even number squared is even; an odd number squared is odd. (CBQ — Number properties)
Using the sum of successive odd numbers pattern, find 36², given that 35² = 1225.
Answer: 36² = 35² + (2·36 − 1) = 1225 + 71 = 1296. (CBQ — Pattern recognition)
What is the nth odd number?
Options: a) 2n b) 2n − 1 c) n + 1 d) n + 2
Answer: b) 2n − 1. (CBQ — Number sequence)
The 36th odd number is ____
Options: a)71 b)81 c)101 d)91
Answer: a) 71. (2·36 − 1 = 71) (CBQ)
Find how many numbers lie between two consecutive (m, m+1) perfect squares?
Options: a) 2m b) 3m c) 4m d) m
Answer: a) 2m. (Because (m+1)² − m² − 1 = 2m.) (CBQ — Algebraic manipulation)
How many square numbers are there between 1 and 100?
Answer: Squares 1²,2²,...,10² → 10 squares (1 to 100 inclusive).
How many are between 101 and 200?
Answer: 11²=121, 12²=144, 13²=169, 14²=196 → 4 squares.
What is the largest square less than 1000?
Answer: ⌊√1000⌋ = 31 → 31² = 961.
Extend the pattern shown and draw the next term.
Answer: (Cannot determine: pattern image not provided.) Please supply the pattern image so the next term can be drawn. (CBQ — Visual pattern extension)
The area of a square is 49 sq. cm. What is the length of its side?
Answer: Side = √49 = 7 cm.
If y = x² then x is the square root of ______.
Answer: x is the square root of y.
What is the square root of 64?
Answer: 8.
n2 = __________ a) ± n b) 0 c) 2 d) -n,-m
Answer: Likely the intended meaning is n² = n × n. None of the options is correct as written; the correct expression is n² = n·n. (If the question was mistyped, please clarify.) (CBQ — Algebra notation)
Find out if 576 or 327 is a perfect square? If it is a perfect square, find its square root? If not write the reason.
Answer: 576 = 24², so 576 is a perfect square; √576 = 24. 327 is between 18²=324 and 19²=361, so 327 is not a perfect square.
Is 324 a perfect square?
Answer: Yes. 324 = 18².
Is 156 a perfect square?
Answer: No. 12²=144 and 13²=169, so 156 is not a perfect square.
Find whether 1156 and 2800 are perfect squares using prime factorisation.
Answer:
1156 = 34² (1156 = 2²·17²) ⇒ perfect square (√1156 = 34).
2800 = 2⁴·5²·7¹. Since exponent of 7 is odd (1), it's not a perfect square.
Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Answer: 2032, 2048 and 1027 are not perfect squares. 1089 = 33² is a perfect square.
Which one among 642, 1082, 2922, 362 has last digit 4?
Answer: None — none of the listed numbers ends with digit 4. (Check the list; maybe a typo.) (CBQ — careful reading & number sense)
Given 125² = 15625, what is the value of 126²?
Options shown; use (n+1)² = n² + 2n + 1.
Answer: 126² = 125² + 2·125 + 1 = 15625 + 251 = 15876. (So choose the option equal to +251.)
Find the length of the side of a square whose area is 441 m².
Answer: Side = √441 = 21 m.
Find the smallest square number that is divisible by each of the following numbers: 4, 9, and 10.
Answer: LCM(4,9,10) = 180 = 2²·3²·5. To be a square all prime exponents must be even; multiply by 5 to make 5² → 180·5 = 900 = 30².
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 = 2⁶·3¹·7². To make exponents all even we need one more factor 3. Multiply by 3 → product = 9408·3 = 28224. √28224 = 168.
How many numbers lie between the squares of the following numbers?
(i) 16 and 17 (ii) 99 and 100
Answer:
(i) Between 16² and 17²: 17² − 16² − 1 = 2·16 = 32 numbers.
(ii) Between 99² and 100²: 2·99 = 198 numbers.
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
Answer & reasoning:
Observe the third term for the row with starting n is n(n+1). The right-hand side equals n² + n + 1.
For 4: third term = 4·5 = 20 and RHS = 4² + 4 + 1 = 21 → 4² + 5² + 20² = 21².
For the line with 9 & 10: n = 9 → third term = 9·10 = 90 and RHS = 9² + 9 +1 = 91 → 9² + 10² + 90² = 91². (CBQ — Pattern recognition & proof)
How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
Answer: Image not provided — cannot determine. Please upload the picture. (CBQ — spatial reasoning)
How many cubes of side 1 cm will make a cube of side 3 cm?
Answer: 3³ = 27 unit cubes.
How many cubes of side 1 cm make a cube of side 2 cm?
Answer: 2³ = 8 unit cubes.
Is 9 a cube?
Answer: No. Cubes near 9 are 2³ = 8 and 3³ = 27, so 9 is not a perfect cube.
Estimate the number of unit cubes in a cube with an edge length of 4 units.
Answer: 4³ = 64 unit cubes.
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?
Answer: All digits 0–9 are possible as last digits of cubes. (Check 0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9.) (CBQ — number properties)
Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Answer:
1-digit cubes: 1³=1 and 2³=8 → 2 cubes.
2-digit cubes: 3³=27 and 4³=64 → 2 cubes.
3-digit cubes: 5³=125, 6³=216, 7³=343, 8³=512, 9³=729 → 5 cubes.
Observation: Counts are small and not uniform; ranges for k‑digit cubes depend on where powers of 10 fall. (CBQ — data observation & explanation)
Can a cube end with exactly two zeroes (00)? Explain.
Answer: No. If a number ends with exactly two zeros it has factor 10² = 2²·5². For a number to be a perfect cube all prime exponents must be multiples of 3 — 2 and 5 exponents would need to be multiples of 3. Two (exponent 2) is not a multiple of 3, so a perfect cube cannot end with exactly two zeros. A cube may end with 0, 3, 6, ... zeros (multiples of 3). (CBQ — prime factor reasoning)
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.
Answer:
4104 = 2³ + 16³ = 9³ + 15³.
13832 = 2³ + 24³ = 18³ + 20³.
91 + 93 + 95 + ... + 109. What is this sum without doing the calculation?
Answer: This is the sum of 10 consecutive odd numbers whose middle average is 100, so sum = 100·10 = 1000. (Also equals 10³.) (CBQ — recognition of cube pattern)
Check if 3375 is a perfect cube.
Answer: 15³ = 3375 → Yes, cube root = 15.
Is 500 a perfect cube?
Answer: No. 7³ = 343, 8³ = 512, so 500 is not a perfect cube.
Find the cube roots of these numbers: (i) 364 (ii) 3512 (iii) 3729
Answer: None are perfect cubes. Approximate cube roots:
∛364 ≈ 7.140
∛3512 ≈ 15.200
∛3729 ≈ 15.507
(So no exact integer cube roots.)
Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice? Perfect Cubes: 1, 8, 27, 64, 125, 216 ...
Answer: First differences: 7, 19, 37, 61, 91.
Second differences: 12, 18, 24, 30.
Third differences: 6, 6, 6 → third differences are constant (6). (CBQ — finite differences & polynomial degree)
Find the cube roots of 27000 and 10648.
Answer: ∛27000 = 30 (30³ = 27000). ∛10648 = 22 (22³ = 10648).
What number will you multiply by 1323 to make it a cube number?
Answer: 1323 = 3³·7². Multiply by 7 to get 1323·7 = 9261 = 21³. So multiply by 7; cube root of the product = 21.
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.
Answers & explanations:
(i) False. Cube of odd → odd (odd·odd·odd is odd).
(ii) False. Example: 2³ = 8 ends with 8; many cubes can end with 8 (…2³ = 8, …12³ = 1728 ends with 8).
(iii) False. Smallest 2-digit number is 10 and 10³ = 1000 (4 digits) so cube of any 2-digit number is at least 4 digits.
(iv) False. Largest 2-digit number 99 → 99³ = 970299 (6 digits), so not 7 or more digits.
(v) False. Number of factors of a perfect cube need not be odd (example: 8 has 4 divisors). (A number has odd number of factors iff it is a perfect square.) (CBQ — factor & divisor properties)
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.
Answer: Recognise common cubes: 11³ = 1331, 17³ = 4913, 23³ = 12167, 32³ = 32768. So roots: 11, 17, 23, 32.
Square root is the inverse operation of ____________
Answer: Squaring.
A number obtained by multiplying a number by itself three times is called a ________.
Answer: A cube.
A number is a perfect cube if its prime factors can be split into _______ identical groups.
Answer: three identical groups (i.e., exponents multiples of 3).
Which of the following is the greatest? Explain your reasoning.
(i) 67³ – 66³ (ii) 43³ – 42³ (iii) 67² – 66² (iv) 43² – 42²
Answer: Compute formula for consecutive differences or compare numerically. The greatest is
(i) 67³ − 66³ = 13,267 (others: (ii)=5,419; (iii)=133; (iv)=85). So (i) is greatest — cube differences grow faster.
