Question 15 Marks
Making use of the cube root table, find the cube root $7532$
AnswerWe 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$
View full question & answer→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.
AnswerThe 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)$
|
View full question & answer→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.
AnswerFirst 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\times2\times2\times2}$
$=16$
View full question & answer→Question 45 Marks
The volume of a cubical box is $474.552$ cubic metres. Find the length of each side of the box.
AnswerVolume of a cube is given by:
$\text{V}=\text{s}^3,$ where $s$ = side of the cube
Now
$S^3=474.552$ cubic is given by:
$\Rightarrow\text{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 \times3\times13=78$
Also
$\sqrt[3]{1000}=10$
$\therefore\text{s}=\frac{\sqrt[3]{474552}}{\sqrt[3]{1000}}$
$=\frac{78}{10}$
$=7.8$
View full question & answer→Question 55 Marks
Find the cube root of the following rational numbers: $\frac{-39304}{-42875}$
AnswerLet 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 \times 2 \times 2 \times 17 \times 17 \times 17$ On grouping the factors in triples of equal factors,
we get: $39304 = {2 \times 2 \times 2} \times {17 \times 17 \times 17}$ Now, taking one factor from each triple,
we get: $\sqrt[3]{39304 }$$ = 2 \times 17 = 27$ Also On grouping the factors $42875$ in triples of equal factors,
we get: $42875 = 5 \times 5 \times 5 \times 7 \times 7 \times 7$ Now, taking one factor from the triple,
we get: $42875 = {5 \times 5 \times 5} \times {7 \times 7 \times 7}$ Now, taking one factor from each triple,
we get: $\sqrt[3]{42875}=5\times7=35$ Now $\sqrt[3]\frac{-39304}{-42875}$
$=\frac{\sqrt[3]{-39304}}{\sqrt[3]{-42875}}$
$=\frac{-\sqrt[3]{39304}}{-\sqrt[3]{42875}}$
$=\frac{-34}{-35}$
$=\frac{34}{35}$
View full question & answer→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.
AnswerFirst 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:
$( 2 \times 2 \times 3 \times 5) = 60$
$3600 \times 60 = 216000$ ( which is a perfect cube of $60$)
View full question & answer→Question 75 Marks
Making use of the cube root table, find the cube root $34.2$
AnswerThe 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 $3.246.$
View full question & answer→Question 85 Marks
Find the cube root of the following rational numbers: $\frac{-19683}{24689}$
AnswerLet 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: $19683$ $= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors,
we get: $19683 = {3 \times 3 \times 3} \times {3 \times 3 \times 3} \times {3 \times 3 \times 3}$
Now, taking one factor from each triple,
we get: $\sqrt[3]{19683 }$ $= 3 \times 3 \times 3 = 27$
Also On grouping the factors in triples of equal factors,
we get: $24389 = 29 \times 29 \times 29$ Now, taking one factor from the triple,
we get: $24389 = {29 \times 29 \times 29}$ Now, taking one factor from each triple,
we get: $\sqrt[3]{24389}=29$ Now $\sqrt[3]\frac{-19683}{24389}$
$=\frac{\sqrt[3]{-19689}}{\sqrt[3]{24689}}$
$=\frac{-\sqrt[3]{19689}}{\sqrt[3]{24689}}$
$=\frac{-27}{29}$
View full question & answer→Question 95 Marks
Find the cube root of the following rational numbers: $\frac{10648}{12167}$
AnswerLet 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: $10648 = 2 \times 2 \times 2 \times 11 \times 11 \times 11$ On grouping the factors in triples of equal factors,
we get: $10648 = {2 \times 2 \times 2} \times {11 \times 11 \times 11}$ Now, taking one factor from each triple,
we get: $12167 = 23 \times 23 \times 23$ On grouping the factors in triples of equal factors,
we get: $12167 = {23 \times 23 \times 23}$ Now, taking one factor from the triple,
we get: $\sqrt[3]{12167}=23$ Now $\sqrt[3]\frac{10648}{12167}$
$=\frac{\sqrt[3]{10648}}{\sqrt[3]{12167}}$
$=\frac{22}{23}$
View full question & answer→Question 105 Marks
Making use of the cube root table, find the cube root $8.65$
AnswerThe 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 $2.053$.
View full question & answer→Question 115 Marks
Find the cube root of the following rational numbers:
$0.001728$
AnswerWe 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:
$1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors, we get:
$1728 = {2 \times 2 \times 2} \times {2 \times 2 \times 2} \times {3 \times 3 \times 3}$
Now, taking one factor from each triple, we get:
$\sqrt[3]{1728}=2\times2\times3=12$
Also
$\sqrt[3]{1000000}$
$\sqrt[3]{100\times100\times100}=100$
$\therefore\sqrt[3]{0.001728}$
$={\frac{\sqrt[3]{1828}}{\sqrt[3]{1000000}}}$
$\frac{12}{100}=0.12$
View full question & answer→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?
$i. 130$
$ii. 345$
$iii. 792$
$iv. 1331$
AnswerIn previous question there are three numbers which are not perfect cubes.
$i. 130$
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 $5)$
$ii. 345$
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 $7)$
$iii. 792$
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 $9)$
View full question & answer→Question 135 Marks
Find the tens digit of the cube root of each of the numbers:
$i. 226981$
$ii. 13824$
$iii. 571787$
$iv. 175616$
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 $226 (6^3 < 226 < 7^3).$
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 $13 (2^3 < 13 < 3^3).$
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 $571 (8^3 < 571 < 9^3).$
Therefore, the tens digit of the cube root of $571787$ is $8$.
$iv.$ 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 $175 (5^3 < 175 < 6^3).$
Therefore, the tens digit of the cube root of $175616$ is $5$.
View full question & answer→Question 145 Marks
Find the cube root of the following numbers: $-1728 \times 216$
AnswerProperty:
For any two integers $a$ and $b$, $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$ From the above property,
we have: $\sqrt[3]{-1728\times216}$ $=\sqrt[3]{-1728}\times\sqrt[3]{216}$ $=\sqrt[3]{1728}\times\sqrt[3]{216}$ (For any positive integer $\text{x,}\sqrt{\text{-x}}-=\sqrt{\text{x}}$ )
Cube root using units digit: Let us consider the number $1728$ .$0000000$
The unit digit is $8$; therefore, the unit digit in the cube root of $1728$ will be $2$.
After striking out the units, tens and hundreds digits of the given number, we are left with $1$.
Now, $1$ is the largest number whose cube is less than or equal to $1$.
Therefore, the tens digit of the cube root of $1728$ is $1$.
$\therefore\sqrt[3]{1728}=12$ On factorising $216$ into prime factors,
we get: $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors,
we get: $216 = {2 \times 2 \times 2} \times {3 \times 3 \times 3}$ Now, taking one factor from each triple,
we get: $=\sqrt[3]{216}$ $=2\times3=6$
Thus $\sqrt[3]{-1728\times216}$
$=-\sqrt[3]{1728}\times\sqrt[3]{216}$
$=-12\times6=-72$
View full question & answer→Question 155 Marks
Making use of the cube root table, find the cube root $732$
AnswerWe have: $730 < 732 < 740$ $\Rightarrow\sqrt[3]{730}<\sqrt[3]{732}<\sqrt[3]{740}$
From cube root table, we have: $\sqrt[3]{730}=9.004 $ and $\sqrt[3]{740}=9.045$
For the difference $(740 -730)$, i.e., $10$, the difference in values $= 9.045 - 9.004 = 0.041$
$\therefore$ For the difference of $(732 - 730),$ i.e.,$ 2,$ the difference in the values $= 0.0082$
$\therefore\sqrt[3]{732}$
$=9.004 +0.008$
$=9.012$
View full question & answer→Question 165 Marks
Making use of the cube root table, find the cube root $1346$
AnswerBy prime factorisation, we have:$1346 = 2 \times 673$
$\Rightarrow\sqrt[3]{1346}$ $=\sqrt[3]{2}\times\sqrt[3]{673}$ Also $670 < 673 < 680$
$\Rightarrow\sqrt[3]{670}<\sqrt[3]{673}<\sqrt[3]{680}$ For the difference $(680 - 670)$, i.e., $10$,
the difference in the values $= 8.794 - 8.750 = 0.044$
$\therefore$ For the difference of $(673 - 670),$ i.e., $3$,
the difference in the values $=\frac{0.004}{10}\times3= 0.013$ (upto three decimal places)
$\therefore\sqrt[3]{673}$ $=8.750 +0.013$ $=8.763$ Now $\sqrt[3]{1346}$
$=\sqrt[3]{2}\times=\sqrt[3]{673}$ $=1.260\times8.763$
$=11.041$ (Up to three decimal places) Thus, the answer is $11.041$.
View full question & answer→Question 175 Marks
Making use of the cube root table, find the cube root $7342$
AnswerWe have:
From cube root table, we have:
$7300 < 7342 < 7400$
$\Rightarrow\sqrt[3]{7000}<\sqrt[3]{7342}<\sqrt[3]{7400}$
From the cube root table, we have:
$\sqrt[3]{7300}=19.39 $ and $\sqrt[3]{7400}=19.48$
$\therefore$ For the difference of $(7400 - 7300)$, i.e., $100$, the difference in the values
$= 19.48 - 19 - 39 = 0.09$
$\therefore$ For the difference of $(7342 - 7300)$, i.e., $42$, the difference in the values
$=\frac{0.09}{100}\times42= 0.0378$
$= 0.037$
$\therefore\sqrt[3]{7342}$
$=19.39 +0.037$
$=19.427$
View full question & answer→Question 185 Marks
Making use of the cube root table, find the cube root $37800$
Answer$37800=2^3 \times 3^3 \times 175$
$\Rightarrow \sqrt[3]{37800}$
$=\sqrt[3]{2^3\times3^3\times175}$
$=6\times\sqrt[3]{175}$
Also
$170 < 175 < 180$
$ \Rightarrow\sqrt[3]{170}<\sqrt[3]{175}<\sqrt[3]{180}$
From cube root table, we have:
$\sqrt[3]{170}=5.540 $ and $\sqrt[3]{180}=5.646$
$\therefore$ For the difference $(180 - 170)$, i.e., $10$,
the difference in values $= 5.646 - 5.540 = 0.106$
$\therefore$ For the difference of $(175 - 170)$, i.e., $5$,
the difference in values
$=\frac{0.106}{10}\times5=0.053$
$\therefore\sqrt[3]{175}$ $=5.540+0.053$
$=5.593$
Now
$37800$
$= 6\times\sqrt[3]{175}=6\times5.593=33.558$
Thus, the required cube root is $33.558$.
View full question & answer→Question 195 Marks
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 dividend by $3$ the remainder is $2'$.
AnswerFive natural numbers of the form $(3n + 2)$ could be written by choosing $n = 1, 2, 3 ...$ etc.
Let five such numbers be $5, 8, 11, 14,$ and $17$.
The cubes of these five numbers are: $5^3=125,8^3=512,11^3=1331,14^3=2744$ and $17^3=4913.$
The cubes of the numbers $5, 8, 11, 14,$ and $17$ could expressed as:
$125 \times 3 \times 41 + 2,$ which is of the form $(3n + 2)$ for $n = 41$
$512 = 3 \times 170 + 2,$ which is of the form $(3n + 2)$ for $n = 170$
$1331 = 3 \times 443 + 2$, which is of the form $(3n + 2)$ for $n = 443$
$2744 = 3 \times 914 + 2$, which is of the form $(3n + 2)$ for $n = 914 4$
$913 = 3 \times 1637 + 2$, which is of the form $(3n + 2)$ for $n = 1637$
The cubes of the numbers $5, 8, 11, 14,$ and $17$ could be expressed as the natural numbers of the form $(3n + 2)$ for some natural number $n$; therefore, the statement is verified.
View full question & answer→Question 205 Marks
Which of the following are cubes of odd natural numbers? $125, 343, 1728, 4096, 32768, 6859$
AnswerWe know that the cubes of all odd natural numbers are odd.
The numbers $125, 343,$ and $6859$ are cubes of odd natural numbers.
Any natural numbers could be either even or odd.
Therefore, if a natural number is not even, it is odd
. Now, the numbers $125, 343$ and $6859$ are odd (It could be verified by divisibility test of $2$, i.e., a number is divisible by $2$ if it ends with $0, 2, 4, 6$ or $8$).
None of the three numbers $125, 343$ and $6859$ are divisible by $2$.
Therefore, they are not even, they are odd.
The numbers $1728, 4096$ and $32768$ are even.
Thus, cubes of odd natural numbers are $125, 343$ and $6859$.
View full question & answer→Question 215 Marks
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 $3n + 1$ is a natural number of the same form i.e. when divided by 3 it leaves the remainder $1'$.
AnswerFive 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 $16^3=4096$
The cubes of the numbers $4, 7, 10, 13,$ and 16 could expressed as:
$64 \times 3 \times 21 + 1$, which is of the form $(3n + 1)$ for $n = 21$
$343 = 3 \times 114 + 1$, which is of the form $(3n + 1)$ for $n = 114$
$1000 = 3 \times 333 + 1$, which is of the form $(3n + 1)$ for $n = 333$
$2197 = 3 \times 732 + 1$, which is of the form $(3n + 1)$ for $n = 732$
$4096 = 3 \times 1365 + 1$, which 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.
View full question & answer→Question 225 Marks
Find the cubes of the following number by column method: $56$
AnswerThus, cube of $35$ is $42875$. We have to find the cube of $56$ using column method. We have: $a = 5$ and $b = 6$
|
Column I
|
Column II
|
Column III
|
Column IV
|
|
$a^3$
|
$3 \times a^2 \times b$
|
$3 \times a \times b^2$
|
$B^3$
|
|
$5^3= 125$
|
$3 \times a^2\times b = 3 \times 5^2\times 6 = 450$
|
$3 \times a \times b^2= 3 \times 5 \times 6^2= 540$
|
$6^3= 216$
|
|
$+50$
|
$+56$
|
$+21$
|
$216$
|
|
$175$
|
$506$
|
$561$
|
|
|
$175$
|
$6$
|
$1$
|
$6$
|
View full question & answer→Question 235 Marks
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 $27''$.
AnswerFive natural numbers, which are multiples of $3$, are $3, 6, 9, 12$ and $15$.
Cubes of these five numbers are:
$ 3^3=3 \times 3 \times 3=27 $
$ 6^3=6 \times 6 \times 6=216 $
$ 9^3=9 \times 9 \times 9=729 $
$ 12^3=12 \times 12 \times 12=1728 $
$ 15^3=15 \times 15 \times 15=3375 $
Now, let us write the cubes as a multiple of $27$.We have:
$ 27 = 27 \times 1$
$ 216 = 27 \times 8 $
$729 = 27 \times 27 $
$1728 = 27 \times 64 $
$3375 = 27 \times 125 $
It is evident that the cubes of the above multiples of $3$ could be written as multiples of $27$.
Thus, it is verified that the cube of a natural number, which is a multiple of $3$, is a multiple of $27$
View full question & answer→Question 245 Marks
Three numbers are to one another $2 : 3 : 4$. The sum of their cubes is $0.334125$. Find the numbers.
AnswerLet the numbers be $2x, 3x$ and $4x$.
According to the question:
$ (2 x)^3+(3 x)^3+(4 x)^3=0.334125 $
$ \Rightarrow 8 x^3 27 x^3+64 x^3=0.334125 $
$ \Rightarrow 8 x^3 27 x^3+64 x^3=0.334125 $
$ \Rightarrow 99 x^3=0.334125 $
$\Rightarrow\text{x}^3=\frac{3341125}{1000000\times99}$
$\Rightarrow\text{x}=\sqrt[3]{\frac{3375}{1000000}}$
$\Rightarrow\text{x}=\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}$
$\Rightarrow\text{x}=\frac{15}{100}$ $=0.15.$
View full question & answer→Question 255 Marks
What is the length of the side of a cube whose volume is $275\ cm^3$. Make use of the table for the cube root.
AnswerVolume of a cube is given by
$\text{V}=\text{a}^3,$ where a = side of the cube
$\therefore$ Side of a cube $= \text{a}=\sqrt[3]{\text{v}}$
If the volume of a cube is $275\ cm^3$, the side of the cube will be $\sqrt[3]{275}.$
We have:
$270 < 275 < 280 $
$\Rightarrow\sqrt[3]{270}<\sqrt[3]{275}<\sqrt[3]{280}$
From the cube root table, we have:
$\sqrt[3]{270}=6.463$ and $\sqrt[3]{280}=6.542$
$\therefore$ For the difference $(280 - 270)$, i.e., $10$, the difference in values
$= 6.542 - 6.463 = 0.079$
$\therefore$ For the difference $(275 - 270)$, i.e., $5$, the difference in values
$=\frac{0.079}{10}\times5=0.0395\simeq0.04$ (upto three decimal places)
$\therefore\sqrt[3]{275}=6.463+0.04=6.503$ (upto three decimal places)
Thus, the length of the side of the cube is $6.503\ cm$.
View full question & answer→Question 265 Marks
Find the cubes of the following number by column method: $72$
AnswerThus, cube of $56$ is $175616$. We have to find the cube of $72$ using column method.
We have: $a = 7$ and $b = 2$
|
Column $I$
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Column $II$
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Column $III$
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Column $IV$
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| $a^3$ |
$3 \times a^2 \times b$ |
$3 \times a \times b^2$ |
$B^3$ |
| $7^3=343$ |
$3 \times a^2 \times b=3 \times 7^2 \times 2=294$ |
$3 \times a \times b^2=3 \times 7 \times 2^2=84$ |
$2^3=8$ |
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$+30$
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$+8$
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$+0$
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$8$
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$373$
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$302$
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$84$
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|
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$373$
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$2$
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$4$
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$8$
|
View full question & answer→Question 275 Marks
Making use of the cube root table, find the cube root $833$
AnswerWe have:$830 < 833 < 840$
$\Rightarrow\sqrt[3]{830}<\sqrt[3]{833}<\sqrt[3]{840}$
From the cube root table, we have:
$\sqrt[3]{830}=9.398$ and $\sqrt[3]{840}=9.435$
For the difference $(840 - 830),$ i.e., $10$, the difference in values
$= 9.435 - 9.398 = 0.037$
$\therefore$ For the difference of $(833 - 830)$, i.e., $3$, the difference in values
$=\frac{0.037}{10}\times3=0.0111=0.011$ (upto three decimal places)
$\therefore\sqrt[3]{833}$
$=9.398+0.011$
$=9.409$
View full question & answer→Question 285 Marks
Find the cube root of the following rational numbers: $0.003375$
AnswerWe have: $0.003375=\frac{3375}{1000000}$
$\therefore\sqrt[3]{0.0003375}$
$=\sqrt[]{\frac{3375}{1000000}}$
$={\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}}$
Now On factorising $1728$ into prime factors,
we get: $3375$ $= 3 \times 3 \times 3 \times 5 \times 5 \times 5$ On grouping the factors in triples of equal factors,
we get: $3375$ $= {3 \times 3 \times 3} \times {5 \times 5 \times 5}$ Now, taking one factor from each triple,
we get: $\sqrt[3]{3375}=3\times5=15$ Also $\sqrt[3]{1000000}$
$\sqrt[3]{100\times100\times100}=100$
$\therefore\sqrt[3]{0.003375}$
$={\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}}$
$\frac{15}{100}=0.15$
View full question & answer→Question 295 Marks
Multiply $210125$ by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.
AnswerFirst we find out the prime factors of $210125$,
$210125 = 5 \times 5 \times 5 \times 41 \times 41$
$\because$ one triples remained incomplete, hence $210125$ is not a perfect cube.
We see that if we multiply the factors by $41$, we will get $2$ triples as $2^3$ and $41^3$.
And the product become:
$210125 \times 41 = 8615125 = 5 \times 5 \times 5 \times 41 \times 41 \times 41$
Cube root of product:
$=3_\sqrt{8615125}$
$=3_\sqrt{5^3\times41^3}$
$=5\times41$
$=205$
View full question & answer→Question 305 Marks
Find the cubes of the following numbers by column method: $35$
AnswerWe have to find the cube of 35 using column method. We have: $a = 3 $and $b = 5$
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Column $I$
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Column $II$
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Column $III$
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Column $IV$
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$a^3$
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$3 \times a^2 \times b$
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$3 \times a \times b^2$
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$B^3$
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$3^3= 27$
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$3 \times a^2\times b = 3 \times 3^2\times 5 = 135$
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$3 \times a \times b^2= 3 \times 3 \times 5^2= 255$
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$5^3= 125$
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$+15$
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$+23$
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$+12$
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$125$
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$42$
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$158$
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$237$
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|
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$42$
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$8$
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$7$
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$5$
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View full question & answer→Question 315 Marks
The volume of a cube is $9261000\ m3$. Find the side of the cube.
AnswerVolume of a cube is given by:
$\text{V}=\text{s}^3,$ where s = Side of the cube
It is given that the volume of the cube is $9261000\ m^3$; therefore, we have:
$S^3= 9261000$
Let us find the cube root of $9261000$ using prime factorisation:
$9261000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$
$= {2 \times 2 \times 2} \times {3 \times 3 \times 3} \times {5 \times 5 \times 5} \times {7 \times 7 \times 7}$
9261000 could be written as a triples of equal factors; therefore, we get:
Cube root $= 2 \times 3 \times 5 \times 7 = 210$
$S^2= 9261000$
$\Rightarrow\text{s} =3_\sqrt{9261000} = 210$
Hence, the length of the side of cube is $210\ m$.
View full question & answer→Question 325 Marks
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 $7^3$.
AnswerFive multiples of $7$ can be written by choosing different values of a natural number $n$ in the expression $7n$.
Let the five multiples be $7, 14, 21, 28$ and $35.$
The cubes of these numbers are: $7^3=343,14^3=2744,21^3=9261,28^3=21952$, and $35^3=42875$
Now, write the above cubes as a multiple of $7^3$. Proceed as follows:
$ 343=7^3 \times 1 $
$ 2744=14^3=14 \times 14 \times 14=(7 \times 2) \times(7 \times 2) \times(7 \times 2)=(7 \times 7 \times 7) \times(2 \times 2 \times 2)=7^3 \times 2^3 $
$ 9261=21^3=21 \times 21 \times 21=(7 \times 3) \times(7 \times 3) \times(7 \times 3)=(7 \times 7 \times 7) \times(3 \times 3 \times 3)=7^3 \times 3^3 $
$ 21952=28^3=28 \times 28 \times 28=(7 \times 4) \times(7 \times 4) \times(7 \times 4)=(7 \times 7 \times 7) \times(4 \times 4 \times 4)=7^3 \times 4^3 $
$ 42875=35^3=35 \times 35 \times 35=(7 \times 5) \times(7 \times 5) \times(7 \times 5)=(7 \times 7 \times 7) \times(5 \times 5 \times 5)=7^3 \times 5^3$
Hence, the cube of multiple of $7$ is a multiple of $7^3$.
View full question & answer→Question 335 Marks
Find the cube root of the following numbers: $-729 \times 15625$
AnswerProperty:
For any two integers $a$ and $b$, $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}},$
From the above property, we have: $\sqrt[3]{-729\times-15625}$
$=\sqrt[3]{-729}\times\sqrt[3]{-15625}$
$-=\sqrt[3]{729}\times-\sqrt[3]{15625}$ (For any positive integer $\text{x},\sqrt[3]{-\text{x}}=-\sqrt[3]{\text{x}}$)
Cube root using units digit:
Let us consider the number $15625$.
The unit digit is $4$; therefore, the unit digit in the cube root of $15625$ will be $5$.
After striking out the units, tens, and hundreds digits of the given number, we are left with $15$.
Now, $2$ is the largest number whose cube is less than or equal to $15(2^3 < 15 < 3^3)$
Therefore, the tens digit of the cube root of $15625$ is $2$.
$\therefore\sqrt[3]{15625}=25$
Also
$\sqrt[3]{-729\times-15625}$
$=-\sqrt[3]{729}\times-\sqrt[3]{15625}$
$=-9\times6=-225$
View full question & answer→Question 345 Marks
Find the volume of a cube, one face of which has an area of $64m^2$.
AnswerArea of a face of cube is given by:
$\text{A} = \text{s}^2$ , where $s$ = Side of the cube
Further, volume of a cube is given by:
$\text{V} = \text{s}^3$, where $s$ = Side of the cube
It is given that the area of one face of the cube is $64\ m^2$.
Therefore we have: $\text{s}^2=64\Rightarrow\text{s}=\sqrt{64}=8\text{m}$
Now, volume is given by:
$\text{V}=\text{s}^3=8^3$
$\Rightarrow\text{V}=8\times8\times8$
$=512\text{m}^3$
Thus, the volume of the cube is $512\ m^3$.
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