[New Book] A SQUARE AND A CUBE NCERT Solutions Math Chapter 1, Ganita Prakash | Class 8

Solutions to questions under Figure it out page 10 and 11, Chapter 1: A SQUARE AND A CUBE, Ganita Prakash | Grade 8

Here we are with “A SQUARE AND A CUBE NCERT Solutions Math Chapter 1“. The solutions are expert-checked and are strictly based on the New NCERT Book Curiosity.

1.1 Square Numbers

1. Which of the following numbers are not perfect squares?
   (i) 2032 (ii) 2048 (iii) 1027 (iv) 1089

Solution:

Perfect squares can only end in digits 0,1,4,5,6,9.

• Check units digit:
  • 2032 → units digit 2 ⇒ cannot be a perfect square.
  • 2048 → units digit 8 ⇒ cannot be a perfect square.
  • 1027 → units digit 7 ⇒ cannot be a perfect square.
  • 1089 → units digit 9 ⇒ possibly a square — test further.
• For 1089 check known squares (page 8–9): $33^2 = 1089$. So 1089 is a perfect square.

2. Which one among $64^2,\;108^2,\;292^2,\;36^2$ has last digit 4?

The last digit of a square depends only on the last digit of the base:

• If base ends with 2 → $2^2=4$22=4 ⇒ square ends with 4.
• If base ends with 8 → $8^2=64$82=64 ⇒ square ends with 4.
• If base ends with 4 or 6 → square ends with 6 (since $4^2=16,6^2=36$).

Check endings:

• $64^2$— base ends 4 ⇒ square ends 6 (NOT 4).
• $108^2$ — base ends 8 ⇒ square ends 4.
• $292^2$ — base ends 2 ⇒ square ends 4.
• $36^2$ — base ends 6 ⇒ square ends 6 (NOT 4).

3. Given $125^2 = 15625$. What is $126^2$?

(i) 15625 + 126 (ii) 15625 + 26² (iii) 15625 + 253
(iv) 15625 + 251 (v) 15625 + 51²

Solution:

• Use $(n+1)^2 = n^2 + 2n + 1$
• Here $n=125$.
  • So $126^2 = 125^2 + 2\cdot125 + 1 = 15625 + 250 + 1 = 15625 + 251$

4. Find the length of the side of a square whose area is 441 m².

Solution :

441 is a known perfect square: $21 \times 21 = 441$. So side $= \sqrt{441}=21$ m.

5. Find the smallest square number that is divisible by each of the
   following numbers: 4, 9, and 10.

Solution:

1. Find $\mathrm{LCM}(4,9,10)$.
   • Prime factors:

• $4 = 2^2$
• $9 = 3^2$
• $10 = 2 \times 5$

2. LCM takes highest powers: $2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180$

So any number divisible by 4,9,10 must be a multiple of 180.

To be a perfect square, every prime exponent in factorisation must be even.

Factorise 180:
$$180 = 2^2 \times 3^2 \times 5^1.$$

Exponents are $2,2,1$— the 5 has an odd exponent 1.

4. Multiply by the smallest number that makes all exponents even:

Multiply by another $5$ to make the exponent of 5 equal 2.

Product $=180\times 5 = 900$.

Check: $900 = 30^2$

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:

Step-by-step factorisation (showing the divisions):

• $9408 \div 2 = 4704$
• $4704 \div 2 = 2352$
• $2352 \div 2 = 1176$
• $1176 \div 2 = 588$
• $588 \div 2 = 294$
• $294 \div 2 = 147$
  So we divided by $2$2 six times ⇒ $2^6$ factor.
  Now $147 = 3 \times 49 = 3 \times 7^2$

Therefore.$$9408 = 2^6 \times 3^1 \times 7^2.$$

Logic to make a square:

Exponents must be even. Current exponents: $2^6$ (even), $3^1$ (odd), $7^2$ (even).

Only $3$3 is unpaired. Multiply by one more $3$ to get $3^2$

• Smallest multiplier = 3.
• Product $= 9408 \times 3 = 28224$

Find the square root:

After multiplying by 3, prime powers become $2^6 \times 3^2 \times 7^2$26×32×72. Group into pairs of primes
(taking one of each pair for the root):$$\sqrt{28224} = 2^{3}\times 3^{1}\times 7^{1} = 8 \times 3 \times 7 = 168.$$

7. How many numbers lie between the squares of the following
   numbers?
   (i) 16 and 17 (ii) 99 and 100

Solution:

The number of integers strictly between $n^2$ and $(n+1)^2$equals $(n+1)^2 – n^2 – 1 = (2n+1) – 1 = 2n$

Calculate:

• For $n=16$n=16: numbers between = $2\times16 = 32$
• For $n=99$n=99: numbers between = $2\times99 = 198$

8. In the following pattern, fill in the missing numbers:

Solution:

Discover Pattern:

• For the first line $n=1$:
  • third term $=1\times2=2$. RHS $=1\times2+1=3$.
• For second line $n=2$:
  • third $=2\times3=6$ RHS $=7$
• So general identity: for natural $n$,$$n^2 + (n+1)^2 + [n(n+1)]^2 = [n(n+1)+1]^2.$$

Apply:

• For $n=4$
  • third term $=4\times5=20$
    RHS $=4\times5+1=21$.
    So ,$$4^2 + 5^2 + 20^2 = 21^2.$$
• For the last line with $n=9$
  • third term $=9\times10=90$ RHS $=9\times10+1=91$
    So,
    $$9^2 + 10^2 + 90^2 = 91^2.$$

9. How many tiny squares are there in the following picture? Write the
   prime factorisation of the number of tiny squares

Please refer to the image by clicking here

Solution:

Look at the picture, you will observe:

• There are 5 small blocks along one side.
• Each block contains 5 tiny squares along its side.

So total tiny squares per side:$$5 \times 5 = 25$$

Now the entire big square is:$$25 \times 25$$

Which gives:$$25^2 = 625$$

Prime factorisation: $$625 = 5 \times 5 \times 5 \times 5 = 5^4$$

1.2 Cubic Numbers

1. Find the cube roots of 27000 and 10648.

1. Cube root of 27000

First, we find the prime factorisation of 27000 by the long division method:

The prime factorisation of 27000 is:

Now, applying the rule:

If $y = x^3$, then $x$ is the cube root of $y$.

This is denoted by $x = \sqrt[3]{y}. \thickspace So, \\ \sqrt[3]{27000}\\ = \sqrt[3]{30^3}\\ = 30$

2. Cube root of 10648

First, we find the prime factorisation of 10648 by the long division method:

$\begin{array}{c|c} 2 & 10648 \\ \hline 2 & 5324 \\ \hline 2 & 2662 \\ \hline 11 & 1331 \\ \hline 11 & 121 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$

The prime factorisation of 10648 is:$$10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11$$$$10648 = 2^3 \times 11^3$$$$10648 = (2 \times 11)^3$$$$10648 = 22^3$$

Now, applying the rule:

If $y = x^3$, then $x = \sqrt[3]{y}$

Therefore,

2. What number will you multiply by 1323 to make it a cube number?

First, we find the prime factorisation of 1323 using the long division method:

So, we can write 1323 as the product of its prime factors:

To make a number a perfect cube, the powers (exponents) of all its prime factors must be multiples of 3.

Let’s check our factors:

• Power of 3 = 3 ✔
• Power of 7 = 2 ✘

We need one more 7.

So multiply by 7.
$$\mathrm{}$$

Hence, the required number 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.

Solutions:

(i) The cube of any odd number is even.

Example:$$3^3 = 27$$

27 is odd. Therefore, the statement is False.

(ii) There is no perfect cube that ends with 8.

$$2^3 = 8$$

So a cube can end with 8. Therefore, the statement is False.

(iii) The cube of a 2-digit number may be a 3-digit number.

Smallest 2-digit number = 10$$10^3 = 1000$$

1000 is a 4-digit number. Therefore, the statement is False.

(iv) The cube of a 2-digit number may have seven or more digits.

Largest 2-digit number = 99$$99^3 = 970299$$

970299 has 6 digits. Therefore, the statement is False.

(v) Cube numbers have an odd number of factors.

Example:$$8 = 2^3$$

Factors of 8 are: 1, 2, 4, 8
Total = 4 factors (even).

Therefore, the statement is False.

4. You are told that 1331 is a perfect cube. Can you guess without
   factorisation what is its cube root? Similarly, guess the cube roots of
   4913, 12167, and 32768.

Solution:

(i) Cube root of 1331

Step 1: Make groups of three from the right

1331 → 1 | 331

Step 2: Look at the last digit (1)

• Only $1^3 = 1$ ends in 1.

So, the unit digit of the cube root = 1

Step 3: Look at the first group (1)

Largest cube ≤ 1 is:$$1^3 = 1$$

So the tens digit = 1

ii) Cube root of 4913

4913 → 4 | 913

Step 2: Last digit is 3

From cube table:

• $7^3 = 343$ ends in 3.

So, the unit digit of the cube root = 7

Step 3: First group (4)

Largest cube ≤ 4 is:$$1^3 = 1$$

So tens digit = 1

(iii) Cube root of 12167

Step 1: Group digits

12167 → 12 | 167

Step 2: Last digit is 7

From cube table:

• $3^3 = 27$ ends in 7.

So unit digit = 3

Step 3: First group (12)

Largest cube ≤ 12 is:$$2^3 = 8$$

So tens digit = 2

(iv) Cube root of 32768

Step 1: Group digits

32768 → 32 | 768

Step 2: Last digit is 8

From cube table:

• $2^3 = 8$ ends in 8.

So unit digit = 2

Step 3: First group (32)

Largest cube ≤ 32 is:$$3^3 = 27$$

So tens digit = 3