RD Sharma Solutions Class 8 Chapter 4 Exercise 4.1
Q1. Find the cubes of the following numbers:
(i) 7 (ii) 12 (iii) 16 (iv) 21 (v) 40 (vi) 55 (vii) 100 (viii) 302 (ix) 301
Solution:
To find the cube of the given numbers
Cube of a number is defined as the number raised to the power 3.
(i) Cube of 7 = 7^{3 }
= 7 x 7 x 7
Therefore, 7^{3}= 343
(ii) Cube of 12 = 12^{3}
= 12 x 12 x 12
Therefore, 12^{3}= 1728
(iii) Cube of 16 = 16^{3}
= 16 x 16 x 16
Therefore, 16^{3} = 4096
(iv) Cube of 21 = 21^{3}
= 21 x 21 x 21
Therefore, 21^{3}= 9261
(v) Cube of 40 = 40^{3}
= 40 x 40 x 40
Therefore, 40^{3}= 64000
(vi) Cube of 55 = 55^{3}
= 55 x 55 x 55
Therefore, 55^{3}= 166375
(vii) Cube of 100 = 100^{3}
= 100 x 100 x 100
Therefore, 100^{3} = 1000000
(viii) Cube of 302 = 302^{3}
= 302 x 302 x 302
Therefore, 302^{3}= 27543608
(ix) Cube of 301 = 301^{3}
= 301 x 301 x 301
Therefore, 301^{3}= 27270901
Q2. Write the cubes of all natural numbers between 1 and 0 and verify the following statements:
(i) Cubes of all odd natural numbers are odd.
(ii) Cubes of all even natural numbers are even,
Solution:
The natural numbers between from 1 to 10 along with the cubes are listed and classified in the following table.
In order to check whether the cubes of the given number is even or odd, it is enough to check its divisibility by 2.
If the number is divided by 2, and the remainder will be zero, then it is an even number, otherwise, it will an odd number.
(i) From the given below table, it is evident that cubes of all the natural odd numbers are odd.
(ii) From the given below table, it is evident that cubes of all the natural even numbers are even.
Number | Cube | Classification |
1 | 1 | Odd |
2 | 8 | Even (Last digit is even, i.e., 0, 2, 4, 6, 8) |
3 | 27 | Odd (Not an even number) |
4 | 64 | Even (Last digit is even, i.e., 0, 2, 4, 6, 8) |
5 | 125 | Odd (Not an even number) |
6 | 216 | Even (Last digit is even, i.e., 0, 2, 4, 6, 8) |
7 | 343 | Odd (Not an even number) |
8 | 512 | Even (Last digit is even, i.e., 0, 2, 4, 6, 8) |
9 | 729 | Odd (Not an even number) |
10 | 1000 | Even (Last digit is even, i.e., 0, 2, 4, 6, 8) |
Q3. Observe the following pattern:
Write the next three rows and calculate the value of 1^{3} + 2^{3} + 3^{3} + ……. + 9^{3} + 10^{3} by the above pattern.
Solution:
Expand the given pattern of numbers are as follows:
Therefore, the next three rows of the given number pattern is given below
Also, the required value of the given pattern is,
1^{3} + 2^{3} + 3^{3} + 4^{3} + 5^{3} + 6^{3} + 7^{3} + 8^{3} + 9^{3} + 10^{3} = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)^{ 2}
= 55^{2} = 3025
Hence the, the required value is 3025
Q4. Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings: “The cube of a natural number which is a multiple of 3 is a multiple of 27”.
Solution:
We know that the five natural numbers, that are multiples of 3 are: 3, 6, 9, 12 and 15.
So, the cubes of these five numbers are given as follows:
3^{3} = 3 x 3 x 3 = 27
6^{3} = 6 x 6 x 6 = 216
9^{3} = 9 x 9 x 9 = 729
12^{3} = 12 x 12 x 12 = 1728
15^{3} = 15 x 15 x 15 = 3375
Now,write the cubes that are the multiples of 27. so, we have
27 = 27 x 1
216 = 27 x 8
729 = 27 x 27
1728 = 27 x 64
3375 = 27 x 125
It is observed that the cubes of the multiples of 3 also be written as the multiples of 27.
Hence, the cube of a natural number, which is a multiple of 3, is a multiple of 27 is verified.
Q5. Write the cubes of 5 natural numbers which are of the form 3n + 1 (e.g. 4, 7, 10, …) and verify the following:
‘The cube of a natural number of the form 3n + I is a natural number of the same form i.e. when divided by 3 it leaves the remainder 1’.
Solution:
Five natural numbers of the form (3n + 1) could be written by choosing n = 1, 2, 3, … etc.
Let five such numbers be 4, 7, 10, 13, and 16.
The cubes of these five numbers are: 4^{3} = 64, 7^{3} = 343, 10^{3} = 1000, 13^{3 }= 2197 and l6^{3} = 4096
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as:
64 = 3 x 21 + 1,
It is of the form (3n + 1) for n = 21
343 = 3 x 114 + 1,
It is of the form (3n + 1) for n = 114
1000 = 3 x 333 + 1,
It is of the form (3n + 1) for n = 333
2197 = 3 x 732 + 1,
It is of the form (3n + 1) for n = 732
4096 = 3 x 1365 + 1,
It is of the form (3n + 1) for n = 1365
The cubes of the numbers 4, 7, 10, 13, and 16 could be expressed as the natural numbers of the form (3n + 1) for some natural number n; therefore, the statement is verified.
Q6. Write the cubes of 5 natural numbers of the form 3n + 2 (i.e. 5, 8, 11,…) and verify the following:
“The cube of a natural number of the form 3n + 2 is a natural number of the same form i.e. when it is divided by 3 the remainder is 2”.
Solution:
We know that the five natural numbers are of the form (3n + 2) and it can be written by choosing n = 1, 2, 3… etc.
So, the five such numbers are 5, 8, 11, 14, and 17.
When you the cubes of these five numbers, it becomes
5^{3} = 125,
8^{3} = 512,
11^{3} = 1331,
14^{3} = 2744,
17^{3} = 4913.
Therefore, the cubes of the numbers 5, 8, 11, 14 and 17 are expressed as:
125 = 3 x 41 + 2,
It is of the form (3n + 2) for n = 41
512 = 3 x 170 + 2,
It is of the form (3n + 2) for n = 170
1331 = 3 x 443 + 2,
It is of the form (3n + 2) for n = 443
2744 = 3 x 914 + 2,
It is of the form (3n + 2) for n = 914
4913 = 3 x 1637 + 2,
It is of the form (3n + 2) for n = 1637
Therefore, form the above obsevations, we can say that the cubes of the numbers 5, 8, 11, 14, and 17 can also be expressed as the natural numbers of the form (3n + 2) for some natural number n.
Hence, the given statement is verified.
Q7. Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following: “The cube of a multiple of 7 is a multiple of 7^{3}”
Solution:
We know that the first five multiples of 7 can be written by choosing various values of a natural number “n” using the expression 7n.
So. the five multiples be 7, 14, 21, 28 and 35.
Thus, the cubes of the above five numbers are:
7^{3} = 343,
14^{3} = 2744,
21^{3} = 9261,
28^{3} = 21952,
35^{3} = 42875
Write the above obtained cubes numbers as a multiple of 7^{3} as follows:
343 = 7^{3} x l
2744 = 14^{3} = 14 x 14 x 14
= (7 x 2) x (7 x 2) x (7 x 2)
= (7 x7 x 7) x (2 x 2 x 2)
= 7^{3} x 2^{3}
9261 = 21^{3} = 21 x 21 x 21
=(7 x 3) x (7 x 3) x (7 x 3)
= 7^{3} x 3^{3}
21952 = 28^{3} = 28 x 28 x 28
= (7 x 4) x (7 x 4) x (7 x 4)
= (7 x 7 x 7) x (4 x 4 x 4)
=7^{3} x 4^{3}
42875 = 35^{3} = 35 x 35 x 35
= (7 x 5) x (7 x 5) x (7 x 5)
= (7 x 7 x 7) x (5 x 5 x 5)
= 7^{3} x 5^{3 }
It is observed that the cube of multiple of 7 is a multiple of 7^{3}.
Hence, the given statement is verified.
Q8. Which of the following are perfect cubes?
(i) 64 (ii) 216 (iii) 243 (iv) 1000 (v) 1728 (vi) 3087 (vii) 4608 (viii) 106480 (ix) 166375 (x) 456533
Solution:
(i) Factorise the number 64 into prime factors,
We get,
64 = 2 x 2 x 2 x 2 x 2 x 2
Now, group the factors in triples of equal factors:
64 = {2 x 2 x 2} x {2 x 2 x 2}
So, it is observed that the prime factors of 64 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 64 is a perfect cube.
(ii) Factorise the number 216 into prime factors,
We get,
216 = 2 x 2 x 2 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
216 = {2 x 2 x 2} x {3 x 3 x 3}
So, it is observed that the prime factors of 216 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 216 is a perfect cube.
(iii) Factorise the number 243 into prime factors,
We get,
243 = 3 x 3 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
243 = {3 x 3 x 3} x 3 x3
So, it is observed that the prime factors of 243 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 243 is not a perfect cube.
(iv) Factorise the number 1000 into prime factors,
We get,
1000 = 2 x 2 x 2 x 5 x 5 x 5
Now, group the factors in triples of equal factors:
1000 = {2 x 2 x 2} x {5 x 5 x 5}
So, it is observed that the prime factors of 1000 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 1000 is a perfect cube
(v) Factorise the number 1728 into prime factors,
We get,
1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
1728 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3}
So, it is observed that the prime factors of 1728 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 1728 is a perfect cube
(vi) Factorise the number 3087 into prime factors,
We get,
3087 = 3 x 3 x 7 x 7 x 7
Now, group the factors in triples of equal factors:
3087 = 3 x 3 x {7 x 7 x 7}
So, it is observed that the prime factors of 3087 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 3087 is not a perfect cube
(vii) Factorise the number4608 into prime factors,
We get,
4608 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3
Now, group the factors in triples of equal factors:
4608 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x 3 x 3
So, it is observed that the prime factors of 4608 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 4608 is not a perfect cube
(viii) Factorise the number106480 into prime factors,
We get,
106480 = 2 x 2 x 2 x 2 x 5 x 11 x 11 x 11
Now, group the factors in triples of equal factors:
106480 = {2 x 2 x 2} x 2 x 5 x {11 x 11 x 11}
So, it is observed that the prime factors of 106480 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 106480 is not a perfect cube
(ix) Factorise the number166375 into prime factors,
We get,
166375 = 5 x 5 x 5 x 11 x 11 x 11
Now, group the factors in triples of equal factors:
166375 = {5 x 5 x 5} x {11 x 11 x 11}
So, it is observed that the prime factors of 166375 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 166375 is a perfect cube
(x) Factorise the number 456533 into prime factors,
We get,
456533 = 7 x 7 x 7 x 11 x 11 x11
Now, group the factors in triples of equal factors:
456533={7 x 7 x 7} x {11 x 11 x 11}
So, it is observed that the prime factors of 456533 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 456533 is a perfect cube
Q9. Which of the following are cubes of even natural numbers?
216, 512, 729, 1000, 3375, 13824
Solution:
Given:
We know that,
the cubes of all even natural numbers are even.
The numbers which are the cubes of even natural numbers are: 216, 512, 1000 and 13824
These numbers are even and it can ve verified using the divisibility rule test of 2.
It means that if a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8.
Hence, the cubes of even natural numbers are 216, 512, 1000 and 13824.
Q10. Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859
Solution:
Given:
We know that,
the cubes of all odd natural numbers are odd.
The numbers which are the cubes of odd natural numbers are: 125, 343, and 6859
These numbers are odd and it can ve verified using the divisibility rule test of 2.
It means that if a number is divisible by 2, it is even number. i.e., if the number ends with 0, 2, 4, 6 or 8. Otherwise, it is an odd number.
Hence, the cubes of odd natural numbers are 125, 343, and 6859
Q11. What is the smallest number by which the following numbers must be multiplied so that the products are perfect cubes?
(i) 675 (ii) 1323 (iii) 2560 (iv) 7803 (v) 107811 (vi) 35721
Solution:
(i) Factorise the number 675 into prime factors,
We get,
675 = 3 x 3 x 3 x 5 x 5
Now, group the factors in triples of equal factors:
675 = {3 x 3 x 3} x 5 x 5
So, it is observed that the prime factors of 675 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 675 is not a perfect cube
However, if the number 675 is multiplied by 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 5 should be multiplied with 675 to make it a perfect cube.
(ii) Factorise the number 1323 into prime factors,
We get,
1323 = 3 x 3 x 3 x 7 x 7
Now, group the factors in triples of equal factors:
1323 = {3 x 3 x 3} x 7 x 7
So, it is observed that the prime factors of 1323 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence,1323 is not a perfect cube
However, if the number 1323 is multiplied by 7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 7 should be multiplied with 1323 to make it a perfect cube.
(iii) Factorise the number 2560 into prime factors,
We get,
2560 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2 x 5
Now, group the factors in triples of equal factors:
2560 = {2 x 2 x 2} x {2 x 2 x 2} x {2 x 2 x 2} x 5
So, it is observed that the prime factors of 2560 can be grouped into triples of equal factors
And, also, there is 1 factor left over.
Hence, 2560 is not a perfect cube
However, if the number 2560 is multiplied by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 25 should be multiplied with 2560 to make it a perfect cube.
(iv) Factorise the number 7803 into prime factors,
We get,
7803 = 3 x 3 x 3 x 17 x 17
Now, group the factors in triples of equal factors:
7803 = {3 x 3 x 3} x 17 x 17
So, it is observed that the prime factors of 7803 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 7803 is not a perfect cube
However, if the number 7803 is multiplied by 17, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 17 should be multiplied with 7803 to make it a perfect cube.
(v) Factorise the number 107811 into prime factors,
We get,
107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11
Now, group the factors in triples of equal factors:
107811 ={3 x 3 x 3} x 3 x{11 x 11 x 11}
So, it is observed that the prime factors of 107811 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 107811 is not a perfect cube
However, if the number 107811 is multiplied by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 9 should be multiplied with 107811 to make it a perfect cube.
(vi) Factorise the number 35721 into prime factors,
We get,
35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7
Now, group the factors in triples of equal factors:
35721 = {3 x 3 x 3} x {3 x 3 x 3} x 7 x 7
So, it is observed that the prime factors of 35721 can be grouped into triples of equal factors
And, also, there is 1 factor left over.
Hence, 35721 is not a perfect cube
However, if the number 35721 is multiplied by 7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 7 should be multiplied with 35721 to make it a perfect cube.
Q12. By which smallest number must the following numbers be divided so that the quotient is a perfect cube?
(i) 675 (ii) 8640 (iii) 1600 (iv) 8788 (v) 7803 (vi) 107811 (vii) 35721 (viii) 243000
Solution:
(i) Factorise the number 675 into prime factors,
We get,
675 = 3 x 3 x 3 x 5 x 5
Now, group the factors in triples of equal factors:
675 = {3 x 3 x 3} x {5 x 5}
So, it is observed that the prime factors of 675 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 675 is not a perfect cube
However, if the number 675 is divided by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 675 should be divided by 25 to make it a perfect cube.
(ii) Factorise the number 8640 into prime factors,
We get,
8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5
Now, group the factors in triples of equal factors:
8640 = {2 x 2 x 2} x {2 x 2 x 2 } x {3 x 3 x 3} x 5
So, it is observed that the prime factors of 8640 can be grouped into triples of equal factors
And, also, there is 1 factor left over.
Hence, 8640 is not a perfect cube
However, if the number 8640 is divided by 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 8640 should be divided by 5 to make it a perfect cube.
(iii) Factorise the number 1600 into prime factors,
We get,
1600 = 2 x 2 x 2 x 2 x 2 x 2 x 5 x 5
Now, group the factors in triples of equal factors:
1600 = {2 x 2 x 2} x {2 x 2 x 2} x 5 x 5
So, it is observed that the prime factors of 1600 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 1600 is not a perfect cube
However, if the number 1600 is divided by 5 x 5, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 1600 should be divided by 25 to make it a perfect cube.
(iv) Factorise the number 8788 into prime factors,
We get,
8788 = 2 x 2 x 13 x 13 x 13
Now, group the factors in triples of equal factors:
8788 = 2 x 2 x {13 x 13 x 13}
So, it is observed that the prime factors of 8788 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 8788 is not a perfect cube
However, if the number 8788 is divided by 2 x 2, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 8788 should be divided by 4 to make it a perfect cube.
(v) Factorise the number 7803 into prime factors,
We get,
7803 = 3 x 3 x 3 x 17 x 17
Now, group the factors in triples of equal factors:
7803 = {3 x 3 x 3} x 17 x 17
So, it is observed that the prime factors of 7803 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 7803 is not a perfect cube
However, if the number 7803 is divided by 17 x 17, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 7803 should be divided by 289 to make it a perfect cube.
(vi) Factorise the number 107811 into prime factors,
We get,
107811 = 3 x 3 x 3 x 3 x 11 x 11 x 11
Now, group the factors in triples of equal factors:
{3 x 3 x 3} x 3 x {11 x 11 x 11}
So, it is observed that the prime factors of 107811 can be grouped into triples of equal factors
And, also, there is 1 factor left over.
Hence, 107811 is not a perfect cube
However, if the number 107811 is divided by 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 107811 should be divided by 3 to make it a perfect cube.
(vii) Factorise the number 35721 into prime factors,
We get,
35721 = 3 x 3 x 3 x 3 x 3 x 3 x 7 x 7
Now, group the factors in triples of equal factors:
35721 = {3 x 3 x 3} x {3 x 3 x 3} x 7 x7
So, it is observed that the prime factors of 35721 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 35721 is not a perfect cube
However, if the number 35721 is divided by 7 x7, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 35721 should be divided by 49 to make it a perfect cube.
(viii) Factorise the number 243000 into prime factors,
We get,
243000 = 2 x 2 x 2 x 3 x 3 x 3 x 3 x 3 x 5 x 5 x 5
Now, group the factors in triples of equal factors:
243000 = {2 x 2 x 2} x {3 x 3 x 3} x 3 x 3 x {5 x 5 x 5}
So, it is observed that the prime factors of 243000 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 243000 is not a perfect cube
However, if the number 243000 is divided by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 243000 should be divided by 9 to make it a perfect cube.
Q13. Prove that if a number is trebled they its cube is 27 times the cube of the given number.
Solution:
Let us assume a number be “n”.
So, the cube of a number is n^{3}.
When the number n is trebled,
i.e., 3n, its cube becomes
(3n)^{3} = 3^{n} x n^{3 }
= 27 n^{3}
It is observed that the cube of 3n is 27 times the cube of the number “n”.
Hence, the given statement is proved.
Q14. What happens to the cube of a number if the number is multiplied by
(i) 3? (ii) 4? (iii) 5?
Solution:
(i) Let us assume a number be “n”.
So, the cube of a number is n^{3}.
When the number n is multiplied by 3,
i.e., 3n, its cube becomes
(3n)^{3} = 3^{n} x n^{3 }
= 27 n^{3}
Therefore, it is observed that the cube of 3n is 27 times the cube of the number “n”.
Hence, if a number is multiplied by 3, then the cube is 27 times the cube of that number.
(ii) Let us assume a number be “n”.
So, the cube of a number is n^{3}.
When the number n is multiplied by 4,
i.e., 4n, its cube becomes
(4n)^{3} = 3^{n} x n^{3 }
= 64 n^{3}
Therefore, it is observed that the cube of 4n is 46 times the cube of the number “n”.
Hence, if a number is multiplied by 4, then the cube is 64 times the cube of that number.
(iii) Let us assume a number be “n”.
So, the cube of a number is n^{3}.
When the number n is multiplied by 5,
i.e., 5n, its cube becomes
(5n)^{3} = 3^{n} x n^{3 }
= 125 n^{3}
Therefore, it is observed that the cube of 5n is 125 times the cube of the number “n”.
Hence, if a number is multiplied by 5, then the cube is 125 times the cube of that number.
Q15. Find the volume of a cube, one face of which has an area of 64 m^{2}.
Solution:
Given that. the area of a cube face is 64 m^{2}
We know that A = a^{2} Square units
where “a” = Side of the cube
Also, the volume of a cube is given by:
V = a^{3 } cubic units
To find the side of a cube, substitute the given values in the area formula,
s^{2} = 64
=> s = √64= 8 m
Therefore, the side of the cube = 8m
Now, the volume of a cube is given by:
V = s^{3} = 8^{3}
V = 8 x 8 x 8 = 512 m^{3}
Hence, the volume of the cube is 512 m^{3}.
Q16. Find the volume of a cube whose surface area is 384 m^{2}.
Solution:
Given that,
The surface area of a cube, SA = 384 m^{3}.
where a = Side of the cube
Also, the volume of a cube is given by:
V = a^{3 } cubic units
We know that the surface area of a cube, SA = 6a^{2 }square units
Substitute the given values in the above formula, we get,
6a^{2 }=384
a^{2}= 384 /6
Therefore, a = 8 m
Now, substitute the value of “a” in volume formula, we get,
V = 8^{3 } cubic units
V = 8 x 8 x 8
V =512
Therefore, the volume of a cone = 512 m^{2}
Q17. Evaluate the following:
(i) {(5^{2} + 12^{2})^{1/2}}3 (ii) {(6^{2} + 8^{2})^{1/2}}^{3}
Solution:
(i) Given expression: {(5^{2} + 12^{2})^{1/2}}^{3}
T0 evaluate the expression, proceed with the steps,
= {(25 + 144)^{1/2}}^{3}
^{ }= {√169}^{3}
= (√(13 x13))^{3}
Cancel square and square root, it becomes (13)^{3}
= 13 x13 x13 = 2197
Therefore, {(5^{2} + 12^{2})^{1/2}}^{3}= 2197
(ii) Given expression:{(6^{2} + 8^{2})^{1/2}}^{3}
T0 evaluate the expression, proceed with the steps,
= {(36 + 64)^{1/2}}^{3}
= {√100}^{3}
= (√(10 x10))^{3}
Cancel square and square root, it becomes (10)^{3}
= 10 x10 x10 = 1000
Therefore, {(6^{2} + 8^{2})^{1/2}}^{3} = 1000
Q18. Write the units digit of the cube of each of the following numbers: 31, 109, 388, 833, 4276, 5922, 77774, 44447, 125125125
Solution:
We know that the propertie of cube numbers are:
When numbers end with digits 1, 4, 5, 6 or 9, then its cube will end with the same digit.
When a number ends with 2, then its cube will end with 8.
When a number ends with 8, then its cube will end with 2.
When a number ends with 3, then its cube will end with 7.
When a number ends with 7, then its cube will end with 3.
From the above-given proper properties, we can say that,
31 = Cube of the number 31 ends with 1.
109 = Cube of the number 109 ends with 9.
388 = Cube of the number 388 ends with 2.
833 = Cube of the number 833 ends with 7.
4276 = Cube of the number 4276 ends with 6.
5922 = Cube of the number 5922 ends with 8.
77774 = Cube of the number 77774 ends with 4.
44447 = Cube of the number 44447 ends with 3.
125125125 = Cube of the number 125125125 ends with 5.
Q19. Find the cubes of the following numbers by column method:
(i) 35 (ii) 56 (iii) 72
Solution:
(i) To find the cube of a number 35 using the column method, proceed with the steps
Let a = 3 and b = 5
Column I
a^{3} |
Column II
3 x a^{2} x b |
Column III
3 x a x b^{2} |
Column IV
b^{3} |
3^{3} = 27 | 3 x a^{2} x b = 3 x 3^{2} x 5 = 135 | 3 x a x b^{2} = 3 x 3 x 5^{2} = 225 | 5^{3} = 125 |
+ 15 | + 23 | + 12 | 125 |
42 | 158 | 237 | |
42 | 8 | 7 | 5 |
Therefore, the cube of 35 is 42875.
(ii) To find the cube of a number 56 using the column method, proceed with the steps
Let a = 5 and b = 6
Column I
a^{3} |
Column II
3 x a^{2} x b |
Column III
3 x a x b^{2} |
Column IV
b^{3} |
5^{3} = 125 | 3 x a^{2} x b = 3 x 5^{2} x 6 = 450 | 3 x a x b^{2} = 3 x 5 x 6^{2} = 540 | 6^{3} = 216 |
+ 50 | + 56 | + 21 | 216 |
175 | 506 | 561 | |
175 | 6 | 1 | 6 |
Therefore, the cube of 56 is 175616
(iii) To find the cube of a number 72 using the column method, proceed with the steps
Let a = 7 and b = 2
Column I
a^{3} |
Column II
3 x a^{2} x b |
Column III
3 x a x b^{2} |
Column IV
b^{3} |
7^{3} | 3 x a^{2} x b = 3 x 7^{2} x 2 = 294 | 3 x a x b^{2} = 3 x 7 x 2^{2} = 84 | 2^{3} = 8 |
+ 30 | + 8 | + 0 | 8 |
373 | 302 | 84 | |
373 | 2 | 4 | 8 |
Therefore, the cube of 72 is 373248
Q20. Which of the following numbers are not perfect cubes?
(i) 64 (ii) 216 (iii) 243 (iv) 1728
Solution:
(i) Factorise the number 64 into prime factors,
We get,
64 = 2 x 2 x 2 x 2 x 2 x 2
Now, group the factors in triples of equal factors:
{2 x 2 x 2} x {2 x 2 x 2}
So, it is observed that the prime factors of 64 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 64 is a perfect cube
(ii) Factorise the number 216 into prime factors,
We get,
216 = 2 x 2 x 2 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
216 = {2 x 2 x 2} x {3 x 3 x 3}
So, it is observed that the prime factors of 216 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 216 is a perfect cube
(iii) Factorise the number 243 into prime factors,
We get,
243 = 3 x 3 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
243 = {3 x 3 x 3} x 3 x 3
So, it is observed that the prime factors of 243 can be grouped into triples of equal factors
And, also there are 2 factors left over.
Hence, 243 is not a perfect cube
(iv) Factorise the number 1728 into prime factors,
We get,
1728 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
1728 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3}
So, it is observed that the prime factors of 1728 can be grouped into triples of equal factors
And, also, there is no factor left over.
Hence, 1728 is a perfect cube
Therefore, (iii) 243 is the required number, which is not a perfect cube.
Q21. For each of the non-perfect cubes in Q. No. 20 find the smallest number by which it must be:
(a) multiplied so that the product is a perfect cube.
(b) divided so that the quotient is a perfect cube.
Solution:
The non-perfect cube in Q, No. 20 is 243.
Therefore, the smallest number which is obtained by multiplying and diving to get a perfect cube as follows:
(a) Factorise the number 243 into prime factors,
We get,
243 = 3 x 3 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
243 ={3 x 3 x 3}x 3 x 3
So, it is observed that the prime factors of 243 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 243 is not a perfect cube
However, if the number 243 is multiplied by 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 3 should be multiplied with 243 to make it a perfect cube.
(b) Factorise the number 243 into prime factors,
We get,
243 = 3 x 3 x 3 x 3 x 3
Now, group the factors in triples of equal factors:
243 = {3 x 3 x 3} x 3 x 3
So, it is observed that the prime factors of 243 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 243 is not a perfect cube
However, if the number 243 is divided by 3 x 3, the obtained factors are grouped into triples of equal factors and no factor will be leftover.
Therefore, the number 243 should be divided by 9 to make it a perfect cube.
Q.22: By taking three different values of n, verify the truth of the following statements:
(i) If n is even, then n^{3 }is also even.
(ii) if n is odd, then n^{3} is also odd.
(iii) If n leaves remainder 1 when divided by 3, then n^{3} also leaves 1 as the remainder when divided by 3.
(iv) If a natural number n is of the form 3p + 2 then n^{3} also a number of the same type.
Solution:
(i) consider three even natural numbers: 2, 4 and 8.
The cubes of the numbers are:
2^{3} = 8,
4^{3 }= 64,
8^{3 }= 512
With the help of the divisibility test, it is observed that 8, 64 and 512 are divisible by 2.
Therefore, the numbers given are even.
Hence, it verifies the statement.
(ii) consider the three odd natural numbers: 3, 9 and 27.
The cubes of the above-given numbers are:
3^{3} = 27,
9^{3} = 729,
27^{3} = 19683
With the help of the divisibility test, it is observed that 27, 729 and 19683 are divisible by 3.
Thus, the numbers are odd.
Hence, it verifies the statement.
(iii) Given that, the three natural numbers of the form (3n + 1) is written by choosing n = 1,2,3… etc.
Now, consider the three numbers such as 4,7 and 10.
The cubes of the three chosen numbers are:
4^{3} = 64,
7^{3} = 343
10^{3} = 1000
So, the cubes of 4,7 and 10 can written as:
64 = 3 x 21 + 1,
It is of the form (3n + 1) for n = 21
343 = 3 x 114 + 1,
It is of the form (3n + 1) for n = 114
1000 = 3 x 333 + 1,
It is of the form (3n + 1) for n = 333
Therefore, the cubes of 4, 7, and 10 can be written as the natural numbers of the form (3n + 1) for some natural number n.
Hence, the given statement is verified.
(iv) Given that, the three natural numbers of the form (3p + 2) can be expressed by choosing p = 1,2,3… etc.
Now, consider the three numbers such as 5, 8 and 11.
Therefore, the cubes of the three chosen numbers are:
5^{3} = 125,
8^{3} = 512
11^{3} = 1331
So, the cubes of 5, 8, and 11 can be expressed as:
125 = 3 x 41 + 2,
It is of the form (3p + 2) for p = 41
512 = 3 x 170 + 2,
It is of the form (3p + 2) for p = 170
1331 = 3 x 443 + 2,
It is of the form (3p + 2) for p = 443
Therefore, the cubes of 5, 8, and 11 could be written as the natural numbers of the form (3p + 2) for some natural number p.
Hence, the given statement is verified.
Q23. Write true (T) or false (F) for the following statements:
(i) 392 is a perfect cube.
(ii) 8640 is not a perfect cube.
(iii) No cube can end with exactly two zeros.
(iv) There is no perfect cube which ends in 4.
(v) For an integer a, a^{3} is always greater than a^{2}.
(vi) If a and b are integers such that a^{2} > b^{2}, then a^{3} > b^{3}.
(vii) if a divides b, then a^{3} divides b^{3}
(viii) If a^{2} ends in 9, then a^{3} ends in 7.
(ix) If a^{2} ends in 5, then a^{3} ends in 25.
(x) I a^{2} ends in an even number of zeros, then a^{3} ends in an odd number of zeros.
Solution:
(i) False
Factorise the number 392 into prime factors,
We get,
392 = 2 x 2 x 2 x 7 x 7
Now, group the factors in triples of equal factors:
392 = {2 x 2 x 2} x 7 x 7
So, it is observed that the prime factors of 392 can be grouped into triples of equal factors
And, also, there are 2 factors left over.
Hence, 392 is not a perfect cube
(ii) True
Factorise the number 8640 into prime factors,
We get,
8640 = 2 x 2 x 2 x 2 x 2 x 2 x 3 x 3 x 3 x 5
Now, group the factors in triples of equal factors:
8640 = {2 x 2 x 2} x {2 x 2 x 2} x {3 x 3 x 3} x 5
So, it is observed that the prime factors of 8640 can be grouped into triples of equal factors
And, also, there is 1 factor left over.
Hence, 8640 is not a perfect cube
(iii) True
It is noted that perfect cube always ends with multiples of 3 zeros,
It means that, it will always ends with 3 zeros, 6 zeros etc.
(iv) False.
Since the unit digit of 64 is 4, the number 64 is a perfect cube
(v) False
The given statement is not true for a negative integer
For Example: (-3)^{2 }= 9; (-3)^{3} = – 27
=> (-3)^{3} < (-3)^{2}
(vi) False
The given statement is not true for negative integers.
For Example: (-6)^{2} > (-5)^{2} but (-6)^{3} < (-5)^{3}
(vii) True
For the given condition, a divides b
\(\frac{b^{3}}{a^{3}} = \frac{b \times b \times b}{a \times a \times a} = \frac{(ax) \times (ax) \times (ax)}{a \times a \times a}\)
a divides b
b = ax for some x
\(\frac{b^{3}}{a^{3}} = \frac{(ax) \times (ax) \times (ax)}{a \times a \times a} = x^{3} \\ => b^{3} = a^{3}(x^{3}) a^{3} \; divides \; b^{3}\)
(viii) False
It is known that, a^{3} ends in 7 if “a” ends with 3.
So for every a^{2} ending in 9, it is not necessary that a is 3.
For example, if 49 is a square of 7, then the cube of 7 is 343.
(ix) False
Example : 35^{2} = 1225 but 35^{3} = 42875
(x) False
For example: 100^{2} = 10000 and 100^{3} = 100000