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Maths 5 Mark Question

Cubes and Cube Roots · Maharashtra Board STD 8 (MSBSHSE - Semi English Medium). 13 questions.

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Question 15 Marks
Making use of the cube root table, find the cube root 7532
Answer
We have:7500 < 7532 < 7600
$\Rightarrow\sqrt[3]{7500}<\sqrt[3]{7532}<\sqrt[3]{7600}$
From the cube root table, we have:
$\sqrt[3]{7500}=19.57$ and $\sqrt[3]{7600}=19.66$
For the difference (7600 - 7500), i.e., 100, the difference in values
= 19.66 - 19.57 = 0.09
$\therefore$ For the difference of (7532 - 7500), i.e., 32, the difference in values
$=\frac{0.09}{100}\times32=0.0288=0.029$ (upto three decimal places)
$\therefore\sqrt[3]{7532}$
$=19.57+0.029$
$=19.599$
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Question 25 Marks
Write the cubes of all natural numbers between 1 and 10 and verify the following statements:
  1. Cubes of all odd natural numbers are odd.
  2. Cubes of all even natural numbers are even.
Answer
The cubes of natural numbers between 1 and 10 are listed and classified in the following table.
We can classify all natural numbers as even or odd number; therefore, to check whether the cube of a natural number is even or odd, it is sufficient to check its divisibility by 2.
If the number is divisible by 2, it is an even number, otherwise it will an odd number.
  1. From the above table, it is evident that cubes of all odd natural numbers are odd.
  2. From the above table, it is evident that cubes of all even natural 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)
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Question 35 Marks
What is the smallest number by which 8192 must be divided so that quotient is a perfect cube? Also, find the cube root of the quotient so obtained.
Answer
First we find out the prime factors of 8192 ,
$8192=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=2^3 \times 2^3 \times 2^3 \times 2^3$
$\because$ one triples remained incomplete, hence 8192 is not a perfect cube.
So, we divide 8192 by 2 to make its quotient a perfect cube.
$\frac{8192}{2}=4096=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=2^3 \times 2^3 \times 2^3 \times 2^3$
Cube root of product:
$=3 \sqrt{=2^3 \times 2^3 \times 2^3 \times 2^3}$
$=3_{\sqrt{2 \times 2 \times 2 \times 2}}$
$=16$
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Question 45 Marks
The volume of a cubical box is 474.552 cubic metres. Find the length of each side of the box.
Answer
Volume of a cube is given by:
$V = s ^3$, where $s =$ side of the cube
Now
$S ^3=474.552$ cubic is given by:
$\Rightarrow S =\sqrt[3]{474.552}$
$=\sqrt[3]{\frac{474552}{1000}}$
$=\frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}$
To find the cube root of 474552 , we need to proceed as follows:
On factorising 474552 into prime factors, we get:
$474552=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 13 \times 13 \times 13$
On grouping the factors in triples of equal factors, we get:
$474552=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{13 \times 13 \times 13\}$
Now, taking one factor from each triple, we get:
$\sqrt[3]{474552}=\sqrt[3]{\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{13 \times 13 \times 13\}}$
$=2 \times 3 \times 13=78$
Also
$\sqrt[3]{1000}=10$
$\therefore s=\frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}$
$=\frac{78}{10}$
$=7.8$
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Question 55 Marks
Find the cube root of the following rational numbers:
$\frac{-39304}{-42875}$
Answer
Let us consider the following rational number:
$\frac{-39304}{-42875}$
Now
$\sqrt[3]{\frac{-39304}{-42875}}$
$=\frac{\sqrt[3]{-393049}}{\sqrt[3]{-42875}}$ $\Big(\because\sqrt[3]{\frac{\text{a}}{\text{}b}}=\sqrt[3]{\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}}\Big)$
$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}}$ $\Big(\because\sqrt[3]{\text{-a}}=-\sqrt[3]{\text{a}} \Big)$
Cube root by factors:
On factorising 39304 into prime factors, we get:
39304 = 2 × 2 × 2 × 17 × 17 × 17
On grouping the factors in triples of equal factors, we get:
39304 = {2 × 2 × 2} × {17 × 17 × 17}
Now, taking one factor from each triple, we get:
$\sqrt[3]{39304 }$ = 2 × 17 = 27
Also
On grouping the factors 42875 in triples of equal factors, we get:
42875 = 5 × 5 × 5 × 7 × 7 × 7
Now, taking one factor from the triple, we get:
42875 = {5 × 5 × 5} × {7 × 7 × 7}
Now, taking one factor from each triple, we get:
$\sqrt[3]{42875}=5\times7=35$
Now
$\sqrt[3]\frac{-39304}{-42875}$
$=\frac{\sqrt[3]{-39304}}{\sqrt[3]{-42875}}$
$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}}$
$=\frac{-34}{-35}$
$=​​​​​​\frac{34}{35}$
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Question 65 Marks
Find the smallest number which when multiplied with 3600 will make the product a perfect cube. Further, find the cube root of the product.
Answer
First we find out the prime factors of 3600 ,
$3600=2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5=2^3 \times 3^2 \times 5^2 \times 2$
$\because$ Only one triples is formed and three factors remained ungrouped in triples.
Hence, 3600 is not a perfect cube.
To make it a perfect cube we have to multiply it by:
$(2 \times 2 \times 3 \times 5)=60$
$3600 \times 60=216000(\text { which is a perfect cube of } 60)$
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Question 75 Marks
Making use of the cube root table, find the cube root 34.2
Answer
The number 34.2 could be written as $\frac{342}{10}.$
Now
$\sqrt[3]{34.2}$
$=\sqrt[3]{\frac{342}{10}}$
$=\frac{\sqrt[3]{342}}{\sqrt[3]{10}}$
Also
340 < 342 < 350
$\Rightarrow\sqrt[3]{340}<\sqrt[3]{342}<\sqrt[3]{350}$
From the cube root table, we have:
$\sqrt[3]{340}=6.980$ and $\sqrt[3]{350}=7.047$
$\therefore$ For the difference (350 - 340), i.e., 10, the difference in values
= 7.047 - 6.980 = 0.067.
$\therefore$ For the difference (342 - 340), i.e., 2, the difference in values
$=\frac{0.067}{10}\times2=0.013$ (upto three decimal places)
$\therefore\sqrt[3]{342}$
$=3.980+0.0134=6.993$ (upto three decimal places)
From the cube root table, we also have:
$\sqrt[3]{10}=2.154$
$\therefore\sqrt[3]{34.2}$
$=\frac{\sqrt[3]{342}}{\sqrt[3]{10}}$
$=3.246$
Thus, the required cube root is 3.246.
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Question 85 Marks
Find the cube root of the following rational numbers:
$\frac{-19683}{24689}$
Answer
Let us consider the following rational number:
$\frac{10648}{12167}$
Now
$\sqrt[3]{\frac{-19689}{24389}}$
$=\frac{\sqrt[3]{-19689}}{\sqrt[3]{24689}}$ $\Big(\because\sqrt[3]{\frac{\text{a}}{\text{}b}}=\sqrt[3]{\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}}\Big)$
$=\frac{-\sqrt[3]{19689}}{\sqrt[3]{24689}}$ $\Big(\because\sqrt[3]{\text{-a}}=-\sqrt[3]{\text{a}} \Big)$
Cube root by factors:
On factorising 19683 into prime factors, we get:
19683 = 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
19683 = {3 × 3 × 3} × {3 × 3 × 3} × {3 × 3 × 3}
Now, taking one factor from each triple, we get:
$\sqrt[3]{19683 }$ = 3 × 3 × 3 = 27
Also
On grouping the factors in triples of equal factors, we get:
24389 = 29 × 29 × 29
Now, taking one factor from the triple, we get:
24389 = {29 × 29 × 29}
Now, taking one factor from each triple, we get:
$\sqrt[3]{24389}=29$
Now
$\sqrt[3]\frac{-19683}{24389}$
$=\frac{\sqrt[3]{-19689}}{\sqrt[3]{24689}}$
$=\frac{-\sqrt[3]{19689}}{\sqrt[3]{24689}}$
$=\frac{-27}{29}$
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Question 95 Marks
Find the cube root of the following rational numbers:
$\frac{10648}{12167}$
Answer
Let us consider the following rational number:
$\frac{10648}{12167}$
Now
$\sqrt[3]\frac{10648}{12167}$
$=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}}$ $\Big(\because\sqrt[3]{\frac{\text{a}}{\text{}b}}=\sqrt[3]{\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}}\Big)$
Cube root by factors:
On factorising 10648 into prime factors, we get:
10648 = 2 × 2 × 2 × 11 × 11 × 11
On grouping the factors in triples of equal factors, we get:
10648 = {2 × 2 × 2} × {11 × 11 × 11}
Now, taking one factor from each triple, we get:
12167 = 23 × 23 × 23
On grouping the factors in triples of equal factors, we get:
12167 = {23 × 23 × 23}
Now, taking one factor from the triple, we get:
$\sqrt[3]{12167}=23$
Now
$\sqrt[3]\frac{10648}{12167}$
$=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}}$
$=\frac{22}{23}$
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Question 105 Marks
Making use of the cube root table, find the cube root 8.65
Answer
The number 8.65 can be written as $\frac{865}{100}.$
Now
$\sqrt[3]{8.65}$
$=\sqrt[3]{\frac{865}{100}}$
$={\frac{\sqrt[3]{865}}{\sqrt[3]{100}}}$
Also
860 < 865 < 870
$\Rightarrow\sqrt[3]{860}<\sqrt[3]{865}<\sqrt[3]{870}$
From the cube root table, we have:
$\sqrt[3]{860}=9.510$ and $\sqrt[3]{870}=9.546$
For the difference (870 - 860), i.e., 10, the difference in values
= 9.546 - 9.510 = 0.036
$\therefore$ For the difference of (865 - 860), i.e., 5, the difference in values
$=\frac{0.036}{10}\times5=0.018$ (upto three decimal places)
$\therefore\sqrt[3]{865}$
$=9.510+0.018$
$=9.528$ (upto three decimal places)
From the cube root table, we also have:
$\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{8.65}$
$=\frac{\sqrt[3]{865}}{\sqrt[3]{100}}$
$=\frac{9.528}{4.642}$
$=2.053$ (upto three decimal places)
Thus, the required cube root is 2.053.
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Question 115 Marks
Find the cube root of the following rational numbers:
0.001728
Answer
We have:
$0.001728=\frac{1728}{1000000}$
$\therefore\sqrt[3]{0.0001728}$
$=\sqrt[]{\frac{1828}{1000000}}$
$={\frac{\sqrt[3]{1828}}{\sqrt[3]{1000000}}}$
Now
On factorising 1728 into prime factors, we get:
1728 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
1728 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3}
Now, taking one factor from each triple, we get:
$\sqrt[3]{1728}=2\times2\times3=12$
Also
$\sqrt[3]{1000000}$
$\sqrt[3]{100\times100\times100}=100$
$\therefore\sqrt[3]{0.001728}$
$={\frac{\sqrt[3]{1828}}{\sqrt[3]{1000000}}}$
$\frac{12}{100}=0.12$
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Question 125 Marks
Find the smallest number that must be subtracted from those of the numbers which are not perfect cubes, to make them perfect cubes. What are the corresponding cube roots?
  1. 130
  2. 345
  3. 792
  4. 1331
Answer
In previous question there are three numbers which are not perfect cubes.
  1. 130
Apply subtraction method,

130 - 1 = 129

129 - 7 = 122

122 - 19 = 103

103 - 37 = 66

66 - 61 = 5

$\because$ Next number to be subtracted is 91, which is greter than 5

Hence, 130 is not a perfect cube. So, to make it perfect cube we subtract 5 from it.

130 - 5 = 125 (which is a perfect cube of 5)
  1. 345
Apply subtraction method,

345 - 1 = 344

344 - 7 = 337

337 - 19 = 318

318 - 37 = 281

281 - 61 = 220

220 - 91 = 129

129 - 127 = 2

$\because$ Next number to be subtracted is 169, which is greter than 2

Hence, 345 is not a perfect cube. So, to make it a perfect cube we subtract 2 from it.

345 - 2 = 343 (which is a perfect cube of 7)
  1. 792
Apply subtraction method,

792 - 1 = 791

791 - 7 = 784

784 - 19 = 765

765 - 37 = 728

728 - 61 = 667

667 - 91 = 576

576 - 127 = 449

449 - 169 = 280

280 - 217 = 63

$\because$ Next number to be subtracted is 271, which is greter than 63

Hence, 792 is not a perfect cube. So, to make it a perfect cube we subtract 63 from it.

792 - 63 = 729 (which is a perfect cube of 9)
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Question 135 Marks
Find the tens digit of the cube root of each of the numbers:
  1. 226981
  2. 13824
  3. 571787
  4. 175616
Answer
i. Let us consider the number 226981.
The unit digit is 1 ; therefore, the unit digit of the cube root of 226981 is 1 .
After striking out the units, tens and hundreds digits of the given number, we are left with 226.
Now, 6 is the largest number, whose cube is less than or equal to $226\left(6^3<226<7^3\right)$.
Therefore, the tens digit of the cube root of 226981 is 6 .
ii. Let us consider the number 13824 .
The unit digit is 4 ; therefore, the unit digit of the cube root of 13824 is 4 .
After striking out the units, tens and hundreds digits of the given number, we are left with 13.
Now, 2 is the largest number, whose cube is less than or equal to $13\left(2^3<13<3^3\right)$.
Therefore, the tens digit of the cube root of 13824 is 2 .
iii. Let us consider the number 571787 .
The unit digit is 7; therefore, the unit digit of the cube root of 571787 is 3 .
After striking out the units, tens and hundreds digits of the given number, we are left with 571.
Now, 8 is the largest number, whose cube is less than or equal to $571\left(8^3<571<9^3\right)$.
Therefore, the tens digit of the cube root of 571787 is 8.
iii. Let us consider the number 175616 .
The unit digit is 6 ; therefore, the unit digit of the cube root of 175616 is 6 .
After striking out the units, tens and hundreds digits of the given number, we are left with 175.
Now, 5 is the largest number, whose cube is less than or equal to $175\left(5^3<175<6^3\right)$.
Therefore, the tens digit of the cube root of 175616 is 5 .
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