Samacheer Kalvi Class 8 Maths Solutions Chapter 1 Numbers Exercise 1.5

 

Question 1. Fill in the blanks
(i) The ones digits in the cube of 73 is _______.
(ii) The maximum number of digits in the cube of a two digit number is _______.
(iii) The smallest number to be added to 3333 to make it a perfect cube is _______.
(iv) The cube root of \( 540 \times 50 \) is _______.
(v) The cube root of 0.000004913 is _______.
Answer:
(i) 7
(ii) 6
(iii) 42
(iv) 30
(v) 0.017
In simple words: The last digit of a cube depends only on the last digit of the original number. For a two-digit number, its cube can have up to six digits. Finding the smallest number to add or cube roots often involves prime factorization or recognition of perfect cubes.

🎯 Exam Tip: To find the ones digit of a cube, only consider the ones digit of the number being cubed. For instance, the ones digit of \( 73^3 \) is the same as the ones digit of \( 3^3 \), which is 7.

 

Question 2. Say True or False.
(i) The cube of 24 ends with the digit 4.
(ii) Subtracting 103 from 1729 gives 93.
(iii) The cube of 0.0012 is 0.000001728.
(iv) 79570 is not a perfect cube.
(v) The cube root of 250047 is 63.
Answer:
(i) True
(ii) True
(iii) False
(iv) True
(v) True
In simple words: We check these statements by doing the math or remembering properties of cubes and cube roots. For example, to find the last digit of \( 24^3 \), we only need to look at the last digit of \( 4^3 \), which is 64, so it ends in 4.

🎯 Exam Tip: Always double-check calculations for True/False questions. For cubes, remember the pattern of the last digits (e.g., numbers ending in 4 will have cubes ending in 4).

 

Question 3. Show that 1944 is not a perfect cube.
Answer: To show if 1944 is a perfect cube, we find its prime factors using repeated division.

From the prime factorization, we get:
\( 1944 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \)
We can group the factors into triplets:
\( 1944 = (2 \times 2 \times 2) \times (3 \times 3 \times 3) \times 3 \)
\( 1944 = 2^3 \times 3^3 \times 3 \)
Since there is an extra factor of 3 that cannot form a triplet, 1944 is not a perfect cube. A number is a perfect cube if all its prime factors can be grouped into sets of three.
In simple words: When we break 1944 down into its smallest parts (prime factors), we find that one of the number 3s is left alone. For a number to be a perfect cube, all its prime factors must appear in groups of three. Because there's a 3 by itself, 1944 is not a perfect cube.

🎯 Exam Tip: To prove a number is not a perfect cube, always use prime factorization. If any prime factor does not appear in a group of three, the number is not a perfect cube.

 

Question 4. Find the smallest number by which 10985 should be divided so that the quotient is a perfect cube.
Answer: To find the smallest number, we first determine the prime factors of 10985.

The prime factorization of 10985 is \( 5 \times 13 \times 13 \times 13 \).
We can see that 13 forms a triplet \( (13 \times 13 \times 13) \), but the factor 5 is left over.
To make 10985 a perfect cube, we need to remove the extra factor of 5. This means we should divide 10985 by 5.
\( \frac{10985}{5} = 2197 \)
And \( 2197 = 13 \times 13 \times 13 = 13^3 \), which is a perfect cube.
Thus, the smallest number to divide by is 5.
In simple words: We break 10985 into its prime factors. We see that the number 5 is left over, while 13 forms a group of three. To make the number a perfect cube, we need to get rid of the extra 5. So, we divide the original number by 5.

🎯 Exam Tip: To make a number a perfect cube by division, always identify the prime factors that are not part of a triplet. Dividing by the product of these 'leftover' factors will result in a perfect cube.

 

Question 5. Find the smallest number by which 200 should be multiplied to make it a perfect cube.
Answer: First, we find the prime factors of 200.

The prime factorization of 200 is \( 2 \times 2 \times 2 \times 5 \times 5 \).
When we group the prime factors into triplets, we have a triplet of 2s \( (2 \times 2 \times 2) \). However, the factors 5 and 5 \( (5 \times 5) \) are left over, meaning they only form a pair.
To make a triplet for the factor 5, we need one more 5. So, we multiply 200 by 5.
\( 200 \times 5 = 1000 \)
The prime factors of 1000 are \( (2 \times 2 \times 2) \times (5 \times 5 \times 5) = 2^3 \times 5^3 = (2 \times 5)^3 = 10^3 \).
Since 1000 is a perfect cube, the smallest number to multiply by is 5.
In simple words: We find the prime factors of 200. We see a group of three 2s, but only two 5s. To make a group of three 5s, we need one more 5. So, we multiply 200 by 5, which gives 1000, a perfect cube.

🎯 Exam Tip: To find the smallest number to multiply to make a perfect cube, always perform prime factorization and identify factors that are not part of a triplet. Multiply by the missing factors to complete the triplets.

 

Question 6. Find the cube root of \( 24 \times 36 \times 80 \times 25 \).
Answer: We need to find the prime factors of each number first:

Prime factorization of each number:
\( 24 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3 \)
\( 36 = 2 \times 2 \times 3 \times 3 = 2^2 \times 3^2 \)
\( 80 = 2 \times 2 \times 2 \times 2 \times 5 = 2^4 \times 5 \)
\( 25 = 5 \times 5 = 5^2 \)
Now, we multiply all these prime factors together:
\( 24 \times 36 \times 80 \times 25 = (2^3 \times 3) \times (2^2 \times 3^2) \times (2^4 \times 5) \times (5^2) \)
Combine the powers of the same prime factors:
\( = 2^{(3+2+4)} \times 3^{(1+2)} \times 5^{(1+2)} \)
\( = 2^9 \times 3^3 \times 5^3 \)
To find the cube root, we divide each exponent by 3:
\( \sqrt[3]{2^9 \times 3^3 \times 5^3} = 2^{(9/3)} \times 3^{(3/3)} \times 5^{(3/3)} \)
\( = 2^3 \times 3^1 \times 5^1 \)
\( = 8 \times 3 \times 5 \)
\( = 120 \)
So, the cube root of \( 24 \times 36 \times 80 \times 25 \) is 120.
In simple words: We first break down each number into its prime factors. Then, we multiply all these prime factors together, combining the same numbers by adding their powers. Finally, to find the cube root, we divide each power by three. This gives us 2 multiplied by 3 multiplied by 5, which equals 120.

🎯 Exam Tip: When finding the cube root of a product of numbers, it's often easiest to find the prime factorization of each individual number first, then combine all factors, and finally group them into sets of three to find the cube root.

 

Question 7. Find the cube root of 729 and 6859 prime factorisation.
Answer:
(i) Cube root of 729:

From the prime factorization, \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \).
The cube root of 729 is \( \sqrt[3]{729} = \sqrt[3]{3^6} = 3^{(6/3)} = 3^2 = 9 \).
(ii) Cube root of 6859:From the prime factorization, \( 6859 = 19 \times 19 \times 19 = 19^3 \).
The cube root of 6859 is \( \sqrt[3]{6859} = \sqrt[3]{19^3} = 19 \).
In simple words: To find the cube root of a number, we break it down into its smallest prime factors. Then, we look for groups of three identical factors. For 729, we find two groups of three 3s, which means the cube root is \( 3 \times 3 = 9 \). For 6859, we find one group of three 19s, making its cube root 19.

🎯 Exam Tip: When using prime factorization to find cube roots, ensure all prime factors appear in groups of three. If a factor's exponent is not a multiple of 3, you cannot directly find its integer cube root.

 

Question 8. What is the square root of cube root of 46656?
Answer: We need to find the cube root first, then its square root.

First, find the cube root of 46656 by prime factorization:
\( 46656 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \)
This can be written as \( 2^6 \times 3^6 \).
Now, calculate the cube root:
\( \sqrt[3]{46656} = \sqrt[3]{2^6 \times 3^6} = 2^{(6/3)} \times 3^{(6/3)} \)
\( = 2^2 \times 3^2 \)
\( = 4 \times 9 \)
\( = 36 \)
Next, we find the square root of this result (36):
\( \sqrt{36} = 6 \)
Therefore, the square root of the cube root of 46656 is 6.
In simple words: We first break down 46656 into its prime factors. This helps us find its cube root, which is 36. Then, we find the square root of 36, which is 6. So, we do the cube root first, and then the square root of that answer.

🎯 Exam Tip: Always follow the order of operations when dealing with nested roots. Calculate the innermost root first, then proceed outwards. Prime factorization is a reliable method for finding both square and cube roots of large numbers.

 

Question 9. If the cube of a squared number is 729, find the square root of that number.
Answer: Let the unknown number be \( x \).
The squared number is \( x^2 \).
The cube of the squared number is \( (x^2)^3 \).
We are given that \( (x^2)^3 = 729 \).
Using exponent rules, \( (x^2)^3 = x^{(2 \times 3)} = x^6 \).
So, \( x^6 = 729 \).
To find \( x \), we can find the sixth root of 729. We know that \( 729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 = 3^6 \).
So, \( x^6 = 3^6 \), which means \( x = 3 \).
The question asks for the square root of that number, which is \( x \).
So we need to find \( \sqrt{3} \).
Using long division method to find \( \sqrt{3} \):

The square root of 3 is approximately 1.732.
In simple words: We are told that taking a number, squaring it, and then cubing that result gives 729. We find that the original number must be 3. Then, we need to find the square root of this number 3, which is about 1.732.

🎯 Exam Tip: Always define your variables clearly (e.g., "Let the number be \( x \)"). When dealing with powers of powers, multiply the exponents, i.e., \( (a^m)^n = a^{mn} \). Remember to simplify expressions like \( \sqrt{3} \) to their decimal approximations if requested.

 

Question 10. Find two smallest perfect square numbers which when multiplied together gives a perfect cube number.
Answer: We need to find two perfect square numbers, say \( a^2 \) and \( b^2 \), such that their product \( a^2 \times b^2 \) is a perfect cube.
Let's consider small perfect square numbers. The smallest perfect squares are 1, 4, 9, 16, 25, 36, 49, 64, etc.
We are looking for two such numbers that when multiplied, the result is a perfect cube.
Consider the numbers \( 2^2 = 4 \) and \( 4^2 = 16 \). Both are perfect squares.
Now, let's multiply them:
\( 4 \times 16 = 64 \)
We know that 64 is a perfect cube, because \( 4^3 = 4 \times 4 \times 4 = 64 \).
Also, \( 64 = 2^6 \), which is \( (2^2)^3 = 4^3 \) and also \( (2^3)^2 = 8^2 \).
Therefore, the two smallest perfect square numbers that fit this condition are 4 and 16.
In simple words: We are looking for two numbers that are themselves perfect squares (like 4 or 9) and whose product is a perfect cube (like 8 or 27). The numbers 4 and 16 are perfect squares, and when we multiply them, we get 64. Since 64 is a perfect cube (because \( 4 \times 4 \times 4 = 64 \)), these two numbers work.

🎯 Exam Tip: When solving such problems, start by listing small perfect squares and perfect cubes to quickly identify potential combinations. Look for numbers whose prime factor exponents are multiples of both 2 (for squares) and 3 (for cubes), implying multiples of 6.