Class-8 Ganita prakash Worksheet Maths ch-1 Answer key

Answer Key with full explanation

(1) If a number ends in 0, 1, 4, 5, 6 or 9, is it always a square?
No. These are the only possible last digits for squares, but having one of those last digits does not guarantee the number is a perfect square. Example: 41 ends in 1 but is not a square.

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(2) Write 5 numbers such that you can determine by looking at their units digit that they are not squares.
Pick any numbers ending in digits not in {0,1,4,5,6,9}. For example: 12, 23, 37, 48, 59. Each ends with 2,3,7,8,9 respectively (note: 9 is a possible last digit of a square — so change 59 → 52). Better list: 12, 23, 37, 48, 52 — units digits 2,3,7,8,2 are impossible for squares, so these numbers cannot be squares.

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(3) Which of the following numbers have digit 6 in the unit's place?
(i) $38^2$ — unit digit from $8^2=64$ ⇒ 4 → no
(ii) $34^2$ — $4^2=16$ ⇒ 6 → yes
(iii) $46^2$ — $6^2=36$ ⇒ 6 → yes
(iv) $56^2$ — $6^2=36$ ⇒ 6 → yes
(v) $74^2$ — $4^2=16$ ⇒ 6 → yes
(vi) $82^2$ — $2^2=4$ ⇒ 4 → no

So: (ii), (iii), (iv), (v) have unit digit 6.

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(4) If a number contains 3 zeros at the end, how many zeros will its square have at the end?
If $n$ ends with 3 zeros then $n$ is divisible by $10^3$. Then $n^2$ is divisible by $10^6$ → 6 zeros at the end.

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(5) What can you say about the parity of a number and its square?

• If a number is even, its square is even.

• If a number is odd, its square is odd.
  (Parity is preserved under squaring.)

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(6) Using the sum of successive odd numbers pattern, find $36^2$, given $35^2=1225$.
$(n+1)^2 = n^2 + 2n + 1$. With $n=35$: $36^2 = 1225 + 70 + 1 = 1296.$

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(7) What is the nth odd number?
The $n$th odd number is $2n-1$. (So option (b) 2n-1.)

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(8) The 36th odd number is ____
36th odd = $2\cdot36 -1 = 71$. So answer (a) 71.

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(9) Find how many numbers lie between two consecutive $(m,m+1)$ perfect squares?
Consecutive squares: $m^2$ and $(m+1)^2$. Difference = $(m+1)^2 - m^2 = 2m+1$. Numbers strictly between = that minus 1 = 2m. So option (a) 2m.

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(10) How many square numbers are there between 1 and 100?
Squares $\,1^2,2^2,\dots,10^2$ so 10 squares.

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(11) How many are between 101 and 200?
Squares greater than 100 and ≤200: $11^2=121,12^2=144,13^2=169,14^2=196$. So 4 squares.

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(12) What is the largest square less than 1000?
$\lfloor\sqrt{999}\rfloor =31$. $31^2=961$. So 961.

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(13) Extend the pattern shown and draw the next term.
(You referred to a pictured pattern — I don’t have the picture. Please upload the image or describe the pattern and I’ll draw/extend it.)

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(14) The area of a square is 49 sq. cm. What is the length of its side?
Side = $\sqrt{49} = 7$ cm.

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(15) If $y = x^2$ then $x$ is the square root of ______
$x$ is the square root of $y$.

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(16) What is the square root of 64?
Principal square root $= 8$. (Usually we give the positive root, 8. If asked for ±, $\pm8$.)

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(17) $n^2 =$ __________ (options a) ± n b) 0 c) 2 d) -n,-m )
If they mean “the square root of $n^2$ is …”, then $\sqrt{n^2}=|n|$, while the solutions of $x^2=n^2$ are $x=\pm n$. So option (a) ±n fits the usual multiple-choice intent.

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(18) find out if 576 or 327 is a perfect square; if so give √, if not give reason.

• $576 = 24^2$. So 576 is a perfect square, $\sqrt{576}=24$.

• $327$ lies between $18^2=324$ and $19^2=361$ and is not equal to either → not a perfect square.

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(19) Is 324 a perfect square?
Yes: $324=18^2$.

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(20) Is 156 a perfect square?
No. $12^2=144$, $13^2=169$; 156 is between them and not a square.

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(21) Find whether 1156 and 2800 are perfect squares using prime factorisation.

• $1156 = 34^2$ (since $34=2\cdot17$, so $1156 = 2^2\cdot17^2$). All prime exponents even ⇒ perfect square, $\sqrt{1156}=34$.

• $2800 = 2^4\cdot5^2\cdot7$. Exponent of 7 is 1 (odd) ⇒ not a perfect square.

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(22) Which of the following numbers are not perfect squares? (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089
Check: $1089=33^2$ (perfect square). The others — 2032, 2048, 1027 — are not perfect squares.

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(23) Which one among 642, 1082, 2922, 362 has last digit 4?
None of these numbers ends with digit 4 (they end with 2,2,2,2). So none.

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(24) Given $125^2=15625$, what is $126^2$?
$(n+1)^2 = n^2 + 2n +1$. So $126^2 = 15625 + 2\cdot125 +1 =15625 +251 = \mathbf{15876}$. (So the correct choice is the one that adds 251.)

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(25) Find the side of a square whose area is 441 m².
Side = $\sqrt{441} = 21$ m.

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(26) Find the smallest square number that is divisible by 4, 9, and 10.
LCM(4,9,10) = $2^2\cdot3^2\cdot5 = 180$. To be a perfect square we need every prime exponent even. 180 = $2^2\cdot3^2\cdot5^1$. Multiply by 5 to make exponent of 5 even → smallest square = $180\cdot5=900=30^2$.

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(27) 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.
Prime factorisation approach: $9408 = 2^6\cdot3\cdot7^2$. The exponent of 3 is odd (1). Multiply by $3$ to make all exponents even. So smallest multiplier = 3. Product = $9408\cdot3=28224$. $\sqrt{28224} = 2^3\cdot3\cdot7 = 8\cdot3\cdot7 = 168.$

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(28) How many numbers lie between the squares of the following numbers? (i) 16 and 17 (ii) 99 and 100
Count = $2m$ where $m$ is the smaller number. (i) $2\cdot16=32$. (ii) $2\cdot99=198.$

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(29) In the following pattern, fill in the missing numbers
Pattern shown:
$1^2+2^2+2^2 = 3^2$
$2^2+3^2+6^2 = 7^2$
$3^2+4^2+12^2 = 13^2$
$4^2+5^2+20^2 = (\_ )^2$
... Pattern: for each $n$, $n^2 + (n+1)^2 + [n(n+1)]^2 = [n(n+1)+1]^2.$

• For $n=4$: $n(n+1)=20$. So $4^2+5^2+20^2 = 21^2$. Fill 21.

• For the last line $9^2+10^2+( \_ )^2 = (\_ )^2$, with $n=9$: $n(n+1)=90$, RHS = $91^2$. So fill 90 and 91.

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(30) How many tiny squares are there in the following picture? Write the prime factorisation of the number of tiny squares.
I don’t have the picture. Please upload the image (or describe the grid size) and I’ll compute the count and prime factorisation.

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(31) How many cubes of side 1 cm will make a cube of side 3 cm?
Volume ratio: $3^3/1^3 = 27$. So 27 unit cubes.

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(32) How many cubes of side 1 cm make a cube of side 2 cm?
$2^3 = 8$ unit cubes.

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(33) Is 9 a cube?
No. $2^3=8$, $3^3=27$. So 9 is not a perfect cube.

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(34) Estimate the number of unit cubes in a cube with edge length 4 units.
Exactly $4^3=64$ unit cubes.

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(35) 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?
All digits 0–9 can appear as the last digit of a cube. Check $n^3$ for units digit of $n=0\dots9$:
0→0, 1→1, 2→8, 3→7, 4→4, 5→5, 6→6, 7→3, 8→2, 9→9. So possible last digits are 0,1,2,3,4,5,6,7,8,9 (every digit).

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(36) Similar to squares, can you find the number of cubes with 1 digit, 2 digits, and 3 digits? What do you observe?
Cubes:

• 1-digit cubes: $1^3=1,\;2^3=8$ → 2 numbers.

• 2-digit cubes: $3^3=27,\;4^3=64$ → 2 numbers.

• 3-digit cubes: $5^3=125,\;6^3=216,\;7^3=343,\;8^3=512,\;9^3=729$ → 5 numbers.

Observation: Cubes are sparse among small numbers; as the number of digits grows the count of cubes in each digit-length range changes slowly and irregularly (gaps between consecutive cubes grow).

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(37) Can a cube end with exactly two zeroes (00)? Explain.
No. If $n^3$ is divisible by $100=2^2\cdot5^2$, then $n$ must be divisible by $2^{\lceil 2/3\rceil}=2$ and $5^{\lceil 2/3\rceil}=5$ — i.e. by 10. But if $n=10k$ then $n^3=(10k)^3=1000k^3$, which ends in three zeros. So a cube cannot end with exactly two zeros; trailing zeros in a cube come in multiples of 3.

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(38) The next two taxicab numbers after 1729 are 4104 and 13832. Find the two ways in which each can be expressed as the sum of two positive cubes.

• $4104 = 16^3 + 2^3 = 4096 + 8$ and also $4104 = 15^3 + 9^3 = 3375 + 729.$

• $13832 = 24^3 + 2^3 = 13824 + 8$ and also $13832 = 20^3 + 18^3 = 8000 + 5832.$

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(39) The sums of consecutive odd numbers producing cubes — given the pattern … 91 + 93 + … +109. What is this sum without doing the calculation?
This is a sum of 10 consecutive odd numbers starting at 91 (since $(109-91)/2 + 1 = 10$). The pattern given earlier sums $n$ consecutive odd numbers to get $n^3$. Here $n=10$, so the sum = $10^3=\mathbf{1000}$.

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(40) check if 3375 is a perfect cube.
$15^3 = 3375$. So yes, $3375 = 15^3$.

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(41) Is 500 a perfect cube?
No. $7^3=343,\;8^3=512$, so 500 is between and not a cube.

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(42) Find the cube roots of these numbers: (i) 364 (ii) 3512 (iii) 3729
None are perfect cubes. Approximate cube roots:

• $\sqrt[3]{364}\approx 7.140$

• $\sqrt[3]{3512}\approx 15.200$

• $\sqrt[3]{3729}\approx 15.507$

(If you needed the nearest integer cube root: 7, 15, 16 respectively as floor/nearest.)

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(43) Compute successive differences over levels for perfect cubes until all the differences at a level are the same. What do you notice? Perfect Cubes $1,8,27,64,125,216,\dots$
Write first differences: $7,19,37,61,91,\dots$
Second differences: $12,18,24,30,\dots$ — these increase linearly by 6. Third differences: constant (=6). Observation: for cubes the third differences are constant (6). This matches the fact that cubes are generated by a cubic polynomial; for degree $k$ polynomials the $k$th finite differences are constant.

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(44) Find the cube roots of 27000 and 10648.

• $\sqrt[3]{27000}=30$ because $30^3=27000$.

• $\sqrt[3]{10648}=22$ because $22^3=10648$.

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(45) What number will you multiply by 1323 to make it a cube number?
Factor $1323 = 3^3\cdot7^2$. To make exponents multiples of 3, multiply by one more $7$. So multiply by 7. Product $= 1323\cdot7 = 9261 = 21^3$.

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(46) State true or false. Explain your reasoning.
(i) The cube of any odd number is even. — False. Odd cubed remains odd.
(ii) There is no perfect cube that ends with 8. — False. $2^3=8$ ends in 8; other cubes may also end in 8 (e.g. $12^3=1728$).
(iii) The cube of a 2-digit number may be a 3-digit number. — False. Smallest 2-digit is 10 and $10^3=1000$ (4 digits), so a 2-digit number’s cube is at least 4 digits.
(iv) The cube of a 2-digit number may have seven or more digits. — False. Max 2-digit is 99, $99^3 = 970299$ which has 6 digits. So not 7 or more digits.
(v) Cube numbers have an odd number of factors. — False (not generally). The number of factors depends on prime exponents: for a perfect cube each prime exponent is a multiple of 3, and number of divisors = product $(3k_i+1)$. These factors can be even or odd (example: $8=2^3$ has 4 divisors, which is even).

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(47) You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly for 4913, 12167, 32768.
Recall some cube memorization:

• $1331 = 11^3$ → cube root 11.

• $4913 = 17^3$ → 17.

• $12167 = 23^3$ → 23.

• $32768 = 32^3$ → 32.

(Recognizing ends/approx magnitudes helps: 10^3=1000, 12^3=1728, etc.)

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(48) Square root is the inverse operation of _____________
Squaring.

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(49) A number obtained by multiplying a number by itself three times is called a ________.
A cube (or a perfect cube when the base is integer).

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(50) A number is a perfect cube if its prime factors can be split into _______ identical groups
three identical groups.

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(51) Which of the following is the greatest? Explain your reasoning.
(i) $67^3 – 66^3$
(ii) $43^3 – 42^3$
(iii) $67^2 – 66^2$
(iv) $43^2 – 42^2$

Use formulas:

• $a^3-(a-1)^3 = 3a^2 - 3a +1$.

• $a^2-(a-1)^2 = 2a-1$.

Compute:

• (i) $=3\cdot67^2 -3\cdot67 +1 =13267$.

• (ii) $=3\cdot43^2 -3\cdot43 +1 =5419$.

• (iii) $=2\cdot67 -1 =133$.

• (iv) $=2\cdot43 -1 =85$.

Greatest is (i) $67^3-66^3$ (because cubic differences grow much faster for larger $a$).

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Items needing images / missing data

• “Extend the pattern shown and draw the next term.” — I need the pattern/picture.

• “How many tiny squares are there in the following picture? Write prime factorisation.” — I need the picture (or the grid dimensions).