Question 12 Marks
Which of the following are perfect cube? $456533$
AnswerOn factorising $456533$ into prime factors, we get $456533 = 7 \times 7 \times 7 \times 11 \times 11 \times 11$ Group the factors in triples of equal factors as: $456533 = \{7 \times 7 \times 7\} \times \{11 \times 11 \times 11\}$ It is evident that the prime factors of $456533$ can be grouped into triples of equal factors and no factor is left over. Therefore, $456533$ is a perfect cube.
View full question & answer→Question 22 Marks
Find the cube root of each of the following natural numbers: $74088000$
AnswerBy prime factorization method, $=3_\sqrt{2\times2\times2\times2\times2\times2\times3\times3\times3\times7\times7\times7}$
$=3_\sqrt{2^3\times2^3\times3^3\times5^3\times7^3}$
$ = 2 \times 2 \times 3 \times 5 \times 7 = 420$
View full question & answer→Question 32 Marks
What is the smallest number by which the following number must be multiplied, so that the product are perfects cube$? 7803$
AnswerOn factorising $7803$ into prime factors, we get: $7803 = 3 \times 3 \times 3 \times 17 \times 17$ On grouping the factors in triples of equal factors, we get: $7803 = \{3 \times 3 \times 3\} \times 17 \times 17$ It is evident that the prime factors of $7803$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $7803$ is a not perfect cube. However, if the number is multiplied by $17,$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $7803$ should be multiplied by $17$ to make it a perfect cube.
View full question & answer→Question 42 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$? 1600$
AnswerOn factorising $1600$ into prime factors, we get: $1600 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5 \times 5$ On grouping the factors in triples of equal factors, we get: $1600 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 5 \times 5$ It is evident that the prime factors of $1600$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $1600$ is a not perfect cube. However, if the number is divided by $(5 \times 5 = 255 \times 5 = 25),$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $1600$ should be divided by 25 to make it a perfect cube.
View full question & answer→Question 52 Marks
Find the cube roots of the following integers: $-5832$
AnswerWe have, $=\sqrt[3]{-5832}$
$=-\sqrt[3]{5832}$ To find the cube root of $5832,$ we use the method of unit digits. Let us consider the number $5832.$ The unit digit is $2;$ therefore the unit digit in the cube root of $5832$ will be $8.$ After striking out the units, tens and hundreds digits of the given number, we are left with $5.$ Now, $1$ is the largest number whose cube is less than or $\therefore\sqrt[3]{5832}=18$
$=\sqrt[3]{-5832}$
$= -\sqrt[3]{5832}$
$=-18$
View full question & answer→Question 62 Marks
Find the cube root of each of the following natural numbers: $343$
AnswerBy prime factorization method, $3_\sqrt{343}$
$=3_\sqrt{7\times7\times7}$
$=7$
View full question & answer→Question 72 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 8640$
AnswerOn factorising $8640$ into prime factors, we get: $8640 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5$ On grouping the factors in triples of equal factors, we get: $8640 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\} \times 5$ It is evident that the prime factors of $8640$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $8640$ is a not perfect cube. However, if the number is divided by $5,$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $8640$ should be divided by $5$ to make it a perfect cube.
View full question & answer→Question 82 Marks
Show that:
$\sqrt[3]{-125-1000}=\sqrt[3]{-125}\times\sqrt[3]{-1000}$
Answer$\text{L.H.S}=\sqrt[3]{-125\times-1000}$
$=\sqrt[3]{-5\times-5\times-5\times-10\times-10\times-10}$
$=\sqrt[3]{\{-5\times-5\times-5\times\}\times\{-10\times-10\times-10\}}$
$=-5\times-10=-50$
$\text{R.H.S}=\sqrt[3]{-125}\times\sqrt[3]{-1000}$
$=\sqrt[3]{-5\times-5\times-5}\times\sqrt[3]{{\{-10\times-10\times-10\}}}$
$=-5\times-10=50$
Because $LHS$ is equal to $RHS,$ the equation is true.
View full question & answer→Question 92 Marks
Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
$1331$
AnswerApplying subtraction method, We have,
$1331 - 1 = 1330$
$1330 - 7 = 1323$
$1323 - 19 = 1304$
$1304 - 37 = 1267$
$1267 - 61 = 1206$
$1206 - 91 = 1115$
$1115 - 127 = 988$
$988 - 169 = 819$
$819 - 217 = 602$
$602 - 271 = 331$
$331 - 331 = 0$
$\because$ Subtraction is performed $11$ times.
Hence, $1331$ is a perfect cube
View full question & answer→Question 102 Marks
Find the volume of a cube whose surface area is $384m^2$.
AnswerSurface area of a cube is given by:
$\text{SA}=6\text{s}^2,$ 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 surface area of the cube is $384m^2$. Therefore, we have:
$6\text{s}^2=384$
$\Rightarrow\text{s}=\sqrt{\frac{384}{6}}$
$=\sqrt{64}=8\text{m}$
Now, volume is given by:
$\text{V}=\text{s}^3=8^3$
$\Rightarrow \text{V}=8\times8\times8=512\text{m}^3$
Thus, the required volume is $512m^3$.
View full question & answer→Question 112 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 243000$
AnswerOn factorising $243000$ into prime factors, we get: $243000 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5$ On grouping the factors in triples of equal factors, we get: $243000 = \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\} \times 3 \times 3 \times \{5 \times 5 \times 5\}$ It is evident that the prime factors of $243000$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $243000$ is a not perfect cube. However, if the number is divided by $(3 \times 3 = 9),$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $243000$ should be divided by 9 to make it a perfect cube.
View full question & answer→Question 122 Marks
Find the cube root of the following rational numbers: $\frac{-125}{729}$
AnswerLet us consider the following rational number: $\frac{-125}{729}$ Now $\sqrt[3]\frac{-125}{729}$
$=\frac{\sqrt[3]{-125}}{\sqrt[3]{729}}$
$\Big(\because\sqrt[3]{\frac{\text{a}}{\text{}b}}=\sqrt[3]{\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}}\Big)$
$=\frac{-\sqrt[3]{125}}{\sqrt[3]{729}}$
$\Big(\because\sqrt[3]{\text{-a}}=-\sqrt[3]{a}\Big)$
$=-\frac{5}{9}$
$\Big(\because 9 \times 9 \times 9$ and $125 = 5 \times 5 \times 5\Big)$
View full question & answer→Question 132 Marks
Which of the following number are not perfect cube$?\ 1728$
AnswerOn 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: $1728 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\}$ It is evident that the prime factors of $1728$ can be grouped into triples of equal factors and no factor is left over. Therefore, $1728$ is a perfect cube. Thus, $(iii) 243$ is the required number, which is not a perfect cube.
View full question & answer→Question 142 Marks
What is the smallest number by which the following number must be multiplied, so that the products are perfect cube$?\ 1323$
AnswerOn factorising $675$ into prime factors, we get: $1323 = 3 \times 3 \times 3 \times 7 \times 7$ On grouping the factors in triples of equal factors, we get: $1323 = \{3 \times 3 \times 3\} \times 5 \times 5$ It is evident that the prime factors of $1323$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $1323$ is a not perfect cube. However, if the number is multiplied by $7,$ the factors can be grouped into triples of equal factors and no factor will be left over. Thus, $1323$ should be multiplied by $7$ to make it a perfect cube.
View full question & answer→Question 152 Marks
Making use of the cube root table, find the cube root $0.27$
AnswerThe number $0.27$ can be written as $\frac{27}{100}.$
Now
$\sqrt[3]{0.27}$
$=\sqrt[3]{\frac{27}{100}}$
$={\frac{\sqrt[3]{27}}{\sqrt[3]{100}}}$
$=\frac{3}{\sqrt[3]{100}}$
By cube root table, we have:
$\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{0.27}$
$=\frac{3}{\sqrt[3]{100}}$
$=\frac{3}{4.642}$
$=0.646$
Thus, the required cube root is $0.646.$
View full question & answer→Question 162 Marks
Making use of the cube root table, find the cube root $250$
AnswerWe have:
$250 = 25 × 100$
$\therefore$ Cube root of $250$ would be in the column of $\sqrt[3]{10\text{x}}$ against $25.$
By the cube root table, we have:
$\sqrt[3]{250}=6.3$
Thus, the required cube root is $6.3.$
View full question & answer→Question 172 Marks
Which of the following are perfect cube$?$
$243$
AnswerOn factorising $243$ into prime factors, we get
$243 = 3 × 3 × 3 × 3 × 3$
Group the factors in triples of equal factors as:
$243 = \{3 × 3 × 3\} × 3 × 3$
It is evident that the prime factors of $243$ can be grouped into triples of equal factors and no factor is left over.
Therefore, $243$ is a perfect cube.
View full question & answer→Question 182 Marks
What happens to the cube of a number if the number is multiplied by. $5?$
AnswerLet us consider a number $n.$ Its cube would be $n^3$.
If n is multiplied by $5,$ it becomes $5n.$ Let us now find the cube of $4n,$ we get:
$(5 n)^3=5^3 \times n^3=125 n^3$
Therefore, the cube of $5n$ is $125$ times of the cube of $n.$
Thus, if a number is multiplied by $5,$ its cube is $125$ times of the cube of that number.
View full question & answer→Question 192 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 8788$
AnswerOn factorising 8788 into prime factors, we get: $8788 = 2 \times 2 \times 13 \times 13 \times 13$ On grouping the factors in triples of equal factors, we get: $8788 = 2 \times 2 \times \{13 \times 13 \times 13\}$ It is evident that the prime factors of $8788$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $8788$ is a not perfect cube. However, if the number is divided by $(2 \times 2 = 42 \times 2 = 4),$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $8788$ should be divided by $4$ to make it a perfect cube.
View full question & answer→Question 202 Marks
Making use of the cube root table, find the cube root $8.6$
AnswerThe number $8.6$ can be written as $\frac{86}{10}.$
Now $\sqrt[3]{8.6}$
$=\sqrt[3]{\frac{86}{100}}$
$={\frac{\sqrt[3]{86}}{\sqrt[3]{10}}}$ By cube root table,
we have: $\sqrt[3]{86}=4.414$ and $\sqrt[3]{10}=2.154$
$\therefore\sqrt[3]{8.6}$
$=\frac{\sqrt[3]{86}}{\sqrt[3]{10}}$
$=\frac{4.414}{2.154}$
$=2.049$ Thus, the required cube root is $2.049.$
View full question & answer→Question 212 Marks
Find the cube roots of the following numbers by successive subtraction of numbers: $1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ... 64$
AnswerWe have, $64 - 1 = 63 63 - 7 = 56 56 - 19 =37 37 - 37 = 0$
$\because$ Subtraction is performed $4$ times.
Hence, cube root of $64$ is $4.$
View full question & answer→Question 222 Marks
Show that: $\sqrt[3]{27}\times\sqrt[3]{64}=\sqrt[3]{27\times64}$
Answer$\text{L.H.S}=\sqrt[3]{27}\times\sqrt[3]{64}$
$\sqrt[3]{27\times64}$
$=\sqrt[3]{3\times3\times3}\times\sqrt[3]{4\times4\times4}$
$=3\times4=12$
$\text{R.H.S}=\sqrt[3]{27}\times\sqrt[3]{64}$
$=\sqrt[3]{3\times3\times3\times4\times4\times4}$
$=\sqrt[3]{\{3\times3\times3\times\}\{4\times4\times4\}}$
$=3\times4=12$ Because $LHS$ is equal to $RHS,$ the equation is true.
View full question & answer→Question 232 Marks
Making use of the cube root table, find the cube root $70$
AnswerBecause $70$ lies between $1$ and $100,$ we will look at the row containing $70$ in the column of $x.$ By the cube root table, we have: $\sqrt[3]{70}=4.121$
View full question & answer→Question 242 Marks
Making use of the cube root table, find the cube root $133100$
AnswerWe have: $133100 = 1331 \times 100 $
$\Rightarrow\sqrt[3]{1331100}$
$=\sqrt[3]{1331\times100} $
$=11\times\sqrt[3]{100}$ By cube root table, we have: $\sqrt[3]{100}=4.642 $
$\therefore\sqrt[3]{133100}$
$=11\times\sqrt[3]{100}$
$=11\times4.642$
$=51.062$
View full question & answer→Question 252 Marks
What is the smallest number by which the following number must be multiplied, so that the products are perfect cube$?\ 675$
AnswerOn factorising 675 into prime factors, we get: $675 = 3 × 3 × 3 × 5 × 5$ On grouping the factors in triples of equal factors, we get: $675 = \{3 × 3 × 3\} × 5 × 5$ It is evident that the prime factors of $675$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $675$ is a not perfect cube. However, if the number is multiplied by $5,$ the factors can be grouped into triples of equal factors and no factor will be left over. Thus, $675$ should be multiplied by $5$ to make it a perfect cube.
View full question & answer→Question 262 Marks
Evaluate the following: $\sqrt[3]{0.1\times0.1\times0.1\times13\times13\times13}$
AnswerTo evaluate the value of the expression, we need to proceed as follows: $\sqrt[3]{0.1\times0.1\times0.1\times13\times13\times13}$ $=\sqrt[3]{\frac{1}{10}\times\frac{1}{10}\times\frac{1}{10}\times13\times13\times13}$ $=\sqrt[3]{\frac{13\times13\times13}{10\times10\times10}}$ $=\frac{\sqrt[3]{13\times13\times13}}{\sqrt[3]{10\times10\times10}}$ $=\frac{13}{10}=1.3$ Thus, the answer is $1.3.$
View full question & answer→Question 272 Marks
Which of the following are perfect cube$?$
$3087$
AnswerOn factorising $3087$ into prime factors, we get 3087 $= 3 \times 3 \times 7 \times 7 \times 7$ Group the factors in triples of equal factors as: $3087 = 3 \times 3 \times \{7 \times 7 \times 7\}$ It is evident that the prime factors of $3087$ can be grouped into triples of equal factors and no factor is left over. Therefore, $243$ is a perfect cube.
View full question & answer→Question 282 Marks
What is the smallest number by which the following number must be multiplied, so that the products are perfect cube$?\ 2560$
AnswerOn factorising $2560$ into prime factors, we get: $2560 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$ On grouping the factors in triples of equal factors, we get: $2560 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 5$ It is evident that the prime factors of $2560$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $2560$ is a not perfect cube. However, if the number is multiplied by $5 \times 5 = 255 \times 5 = 25,$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $2560$ should be multiplied by $25$ to make it a perfect cube.
View full question & answer→Question 292 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?$
$675$
AnswerOn factorising 675 into prime factors, we get:
$675 = 3 \times 3 \times 3 \times 5 \times 5$
On grouping the factors in triples of equal factors, we get:
$675 = \{3 \times 3 \times 3\} \times 5 \times 5$
It is evident that the prime factors of $675$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $675$ is a not perfect cube. However, if the number is divided by $5 \times 5 = 255 \times 5 = 25,$ the factors can be grouped into triples of equal factors such that no factor is left over.
Thus, $675$ should be divided by $25$ to make it a perfect cube.
View full question & answer→Question 302 Marks
Using the method of successive subtraction examine whether or not the following numbers are perfect cubes: $792$
AnswerApplying subtraction method, We have, $792 - 1 = 791$
$ 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
View full question & answer→Question 312 Marks
Evaluate: $\sqrt[3]{700\times2\times49\times5}$
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: $\sqrt[3]{700\times2\times49\times5}$
$=\sqrt[3]{2\times2\times5\times5\times7\times2\times7\times7\times5}$
$\therefore$ $700 = 2 \times 2 \times 5 \times 5 \times 7$ and $49 = 7 \times 7$ $=\sqrt[3]{2^3\times5^3\times7^3}$
$=\sqrt[3]{2^3}\times\sqrt[3]5^3\times\sqrt[3]{7^3}$
$=2\times5\times7=70$
View full question & answer→Question 322 Marks
Making use of the cube root table, find the cube root $0.86$
AnswerThe number $0.86$ can be written as $\frac{86}{100}.$
Now $\sqrt[3]{0.86}$
$=\sqrt[3]{\frac{86}{100}}$
$={\frac{\sqrt[3]{86}}{\sqrt[3]{100}}}$ By cube root table,
we have: $\sqrt[3]{86}=4.414$ and $\sqrt[3]{100}=4.642$
$\therefore\sqrt[3]{0.86}$
$=\frac{\sqrt[3]{86}}{\sqrt[3]{100}}$
$=\frac{4.414}{4.642}$
$=0.951$ (upto three decimal places) Thus, the required cube root is $ 0.951.$
View full question & answer→Question 332 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube $?\ 35721$
AnswerOn factorising $35721$ into prime factors, we get: $35721 = 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 7 \times 7$ On grouping the factors in triples of equal factors, we get: $35721 = \{3 \times 3 \times 3\} \times \{3 \times 3 \times 3\} \times 7 \times 7$ It is evident that the prime factors of $35721$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $35721$ is a not perfect cube. However, if the number is divided by $(7 \times 7 = 497),$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $35721$ should be divided by $49$ to make it a perfect cube.
View full question & answer→Question 342 Marks
Evaluate the following: $\sqrt[3]{\frac{0.027}{0.008}}\div\sqrt\frac{0.09}{0.04}-1$
AnswerTo evaluate the value of the given expression, we need to proceed as follows: $\sqrt[3]{\frac{0.027}{0.008}}\div\sqrt\frac{0.09}{0.04}-1$
$=\sqrt[3]{\frac{\frac{27}{1000}}{\frac{8}{1000}}}\div\sqrt[3]{\frac{\frac{9}{100}}{\frac{4}{100}}}-1$
$=\sqrt[3]{\frac{27}{8}}\div\sqrt[3]{\frac{9}{4}}-1$
$={\frac{\sqrt[3]{27}}{\sqrt[3]{8}}}\div{\frac{\sqrt[3]{9}}{\sqrt[3]{4}}}-1$$=\frac{3}{2}\div\frac{3}{2}-1$
$= \frac{3^1}{2}\times\frac{2^1}{3}-1$
$=1-1=0$
Thus, the answer is $0.$
View full question & answer→Question 352 Marks
Show that: $\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}=\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}$
Answer$\text{LHS}=\frac{\sqrt[3]{729}}{\sqrt[3]{1000}}$
$=\frac{\sqrt[3]{9\times9\times9}}{\sqrt[3]{10\times10\times10}}$
$=\frac{9}{10}$
$\text{RHS}=\sqrt[3]\frac{{729}}{{1000}}$
$=\sqrt[3]\frac{{9\times9\times9}}{{10\times10\times10}}$
$=\sqrt[3]{\frac{9}{10}\times\frac{9}{10}\times\frac{9}{10}}$
$=\sqrt[3]{(\frac{9}{10})^3}$
$=\frac{9}{10}$ Because $LHS$ is equal to $RHS$, the equation is true.
View full question & answer→Question 362 Marks
Show that: $\sqrt[3]{64}\times\sqrt[3]{729}=\sqrt[3]{64\times729}$
Answer$\text{L.H.S}=\sqrt[3]{64\times729}$
$=\sqrt[3]{4\times4\times4\times9\times9\times9}$
$=\sqrt[3]{\{4\times4\times4\}\times\{9\times9\times9\}}$
$=4\times9=36$
$\text{R.H.S}=\sqrt[3]{64}\times\sqrt[3]{729}$
$=\sqrt[3]{4\times4\times4}\times\sqrt[3]{9\times9\times9}$
$=4\times9=36$ Because $LHS$ is equal to $RHS,$ the equation is true.
View full question & answer→Question 372 Marks
Using the method of successive subtraction examine whether or not the following numbers are perfect cubes$: 130$
AnswerApplying subtraction method,
We have, $130 - 1 = 129$
$129 - 7 = 122$
$122 - 19 = 103$
$103 - 37 = 66$
$66 - 61 = 5$
$\because$ Next number to be subtracted is $91,$ which is greter than $5$ Hence, $130$ is not a perfect cube.
View full question & answer→Question 382 Marks
Which of the following are perfect cube$?\ 1728$
AnswerOn factorising $1728$ into prime factors, we get $1728 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 3$ Group the factors in triples of equal factors as: $1728 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{3 \times 3 \times 3\}$ It is evident that the prime factors of $1728$ can be grouped into triples of equal factors and no factor is left over. Therefore, $1728$ is a perfect cube.
View full question & answer→Question 392 Marks
What happens to the cube of a number if the number is multiplied by. $3?$
AnswerLet us consider a number $n.$ Its cube would be $n^3$.
If $n$ is multiplied by $3,$ it becomes $3n.$ Let us now find the cube of $3n,$ we get:
$(3 n)^3=3^3 \times n^3=27 n^3$
Therefore, the cube of $3n$ is $27$ times of the cube of $n$.
Thus, if a number is multiplied by $3,$ its cube is $27$ times of the cube of that number.
View full question & answer→Question 402 Marks
Which of the following are perfect cube? $216$
AnswerOn factorising $216$ into prime factors, we get $216 = 2 × 2 × 2 × 3 × 3 × 3$ Group the factors in triples of equal factors as: $216 = \{2 × 2 × 2\} × \{3 × 3 × 3\}$ It is evident that the prime factors of $216$ can be grouped into triples of equal factors and no factor is left over. Therefore, $216$ is a perfect cube.
View full question & answer→Question 412 Marks
Which of the following number are not perfect cube$?\ 243$
AnswerOn factorising $243$ into prime factors, we get: $243 = 3 \times 3 \times 3 \times 3 \times 3$ On grouping the factors in triples of equal factors, we get: $243 = \{3 \times 3 \times 3\} \times 3 \times 3$ It is evident that the prime factors of $243$ can be grouped into triples of equal factors and no factor is left over. Therefore, $243$ is a perfect cube.
View full question & answer→Question 422 Marks
What is the smallest number by which the following number must be multiplied, so that the product are perfects cube$?\ 357221$
AnswerOn factorising $35721$ into prime factors, we get: $35721 = 3 × 3 × 3 × 3 × 3 × 3 × 7 × 7$ On grouping the factors in triples of equal factors, we get: $35721 = \{3 × 3 × 3\} × \{3 × 3 × 3\} × 7 × 7$ It is evident that the prime factors of $35721$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $35721$ is a not perfect cube. However, if the number is multiplied by $7,$ the factors be grouped into triples of equal factors such that no factor is left over. Thus, $35721$ should be multiplied by $7$ to make it a perfect cube.
View full question & answer→Question 432 Marks
Which of the following number are not perfect cube$?$
$64$
AnswerOn factorising 64 into prime factors, we get:
$64 = 2 × 2 × 2 × 2 × 2 × 2$
On grouping the factors in triples of equal factors, we get:
$64 = \{2 × 2 × 2\} × \{2 × 2 × 2\}$
It is evident that the prime factors of $64$ can be grouped into triples of equal factors and no factor is left over. Therefore, $64$ is a perfect cube.
View full question & answer→Question 442 Marks
Which of the following are perfect cube$?\ 166375$
AnswerOn factorising $166375$ into prime factors, we get $166375 = 5 × 5 × 5 × 11 × 11 × 11$ Group the factors in triples of equal factors as: $166375 = \{5 × 5 × 5\} × \{11 × 11 × 11\}$ It is evident that the prime factors of $166375$ can be grouped into triples of equal factors and no factor is left over. Therefore, $166375$ is a perfect cube.
View full question & answer→Question 452 Marks
Making use of the cube root table, find the cube root $700$
AnswerWe have: $700 = 70 \times 10$
$\therefore$ Cube root of $700$ will be in the column of $\sqrt[3]{10\text{x}}$ against $70.$ By the cube root table, we have: $\sqrt[3]{700}=8.879$ Thus, the answer is $8.879.$
View full question & answer→Question 462 Marks
What happens to the cube of a number if the number is multiplied by. $4?$
AnswerLet us consider a number $n.$ Its cube would be $n^3$.
If n is multiplied by $4,$ it becomes $4n.$ Let us now find the cube of $4n,$ we get:
$(4 n)^3=4^3 \times n^3=64 n^3$
Therefore, the cube of $4n$ is $64$ times of the cube of $n.$ Thus, if a number is multiplied by $4,$ its cube is $64$ times of the cube of that number.
View full question & answer→Question 472 Marks
Which of the following are perfect cube$?\ 64$
AnswerOn factorising $64$ into prime factors, we get $64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2$ Group the factors in triples of equal factors as: $64 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\}$ It is evident that the prime factors of $64$ can be grouped into triples of equal factors and no factor is left over. Therefore, $64$ is a perfect cube.
View full question & answer→Question 482 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 107811$
AnswerOn factorising $107811$ into prime factors, we get: $107811 = 3 \times 3 \times 3 \times 3 \times 11 \times 11 \times 11$ On grouping the factors in triples of equal factors, we get: $107811 = \{3 \times 3 \times 3\} \times 3 \times \{11 \times 11 \times 11\}$ It is evident that the prime factors of $107811$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $107811$ is a not perfect cube. However, if the number is divided by $3,$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $107811$ should be divided by $3$ to make it a perfect cube.
View full question & answer→Question 492 Marks
Using the method of successive subtraction examine whether or not the following numbers are perfect cubes:
$345$
AnswerApplying subtraction method, We have,
$345 - 1 = 344$
$344 - 7 = 337$
$337 - 19 = 318$
$318 - 37 = 281$
$281 - 61 = 220$
$220 - 91 = 129$
$129 - 127 = 2$
$\because$ Next number to be subtracted is $169,$ which is greter than $2$
Hence, $345$ is not a perfect cube
View full question & answer→Question 502 Marks
What is the smallest number by which the following number must be multiplied, so that the product are perfects cube$?$
$107811$
AnswerOn factorising 107811 into prime factors, we get:
$107811 = 3 \times 3 \times 3 \times 3 \times 11 \times 11 \times 11$
On grouping the factors in triples of equal factors, we get:
$107811 = \{3 \times 3 \times 3\} \times 3 \times \{11 \times 11 \times 11\}$
It is evident that the prime factors of $107811$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $107811$ is a not perfect cube. However, if the number is multiplied by $3 \times 3 = 93 \times 3 = 9,$ the factors be grouped into triples of equal factors such that no factor is left over.
Thus, $107811$ should be multiplied by $9$ to make it a perfect cube.
View full question & answer→Question 512 Marks
Show that: $\sqrt[3]{-125\times216}=\sqrt[3]{-125}\times\sqrt[3]{216}$
Answer$\text{L.H.S}=\sqrt[3]{-125\times216}$ $=\sqrt[3]{-5\times-5\times-5\times\{2\times2\times2\times3\times3\times3}\}$ $=\sqrt[3]{\{-5\times-5\times-5\times\}\times\{2\times2\times2\}\times\{\times3\times3\times3}\}$ $=-5\times2\times3=-30$ $\text{R.H.S}=\sqrt[3]{-125}\times\sqrt[3]{216}$ $=\sqrt[3]{-5\times-5\times-5}\times\sqrt[3]{\{2\times2\times2\}\times{\{3\times3\times3\}}}$ $=-5\times(2\times3)=-30$ Because LHS is equal to RHS, the equation is true.
View full question & answer→Question 522 Marks
Making use of the cube root table, find the cube root $780$
AnswerWe have: $780 = 78 \times 10$
$\therefore$ Cube root of $780$ would be in the column of $\sqrt[3]{10\text{x}}$ against $78.$
By the cube root table,
we have: $=\sqrt[3]{780}=9.205$
Thus, the answer is $9.205.$
View full question & answer→Question 532 Marks
Fill in the blanks: $\sqrt[3]{125\times27}=3\times...$
Answer$\because\sqrt[3]{125\times27}$
$=\sqrt[3]{125}\times\sqrt[3]{27}$
$=\sqrt[3]{5\times5\times5}\times\sqrt[3]{3\times3\times3}$
$=5\times3$ (Commutative law)
View full question & answer→Question 542 Marks
Making use of the cube root table, find the cube roots $7$
AnswerBecause $7$ lies between $1$ and $100,$ we will look at the row containing $7$ in the column of $x.$ By the cube root table, we have: $\sqrt[3]{7}=1.913$ Thus, the answer is $1.913.$
View full question & answer→Question 552 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 7803$
AnswerOn factorising $7803$ into prime factors, we get: $7803 = 3 \times 3 \times 3 \times 17 \times 17$ On grouping the factors in triples of equal factors, we get: $7803 = \{3 \times 3 \times 3\} \times 17 \times 17$ It is evident that the prime factors of $7803$ cannot be grouped into triples of equal factors such that no factor is left over. Therefore, $7803$ is a not perfect cube. However, if the number is divided by $(17 \times 17 = 289),$ the factors can be grouped into triples of equal factors such that no factor is left over. Thus, $7803$ should be divided by $289$ to make it a perfect cube.
View full question & answer→Question 562 Marks
Which of the following are perfect cube$?\ 1000$
AnswerOn factorising $1000$ into prime factors, we get $1000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5$ Group the factors in triples of equal factors as: $1000 = \{2 \times 2 \times 2\} \times \{5 \times 5 \times 5\}$ It is evident that the prime factors of $1000$ can be grouped into triples of equal factors and no factor is left over. Therefore, $1000$ is a perfect cube.
View full question & answer→Question 572 Marks
Which of the following number are not perfect cube$?\ 216$
AnswerOn factorising $216$ into prime factors, we get: $216 = 2 × 2 × 2 × 3 × 3 × 3$ On grouping the factors in triples of equal factors, we get: $216 = \{2 × 2 × 2\} × \{3 × 3 × 3\}$ It is evident that the prime factors of $216$ can be grouped into triples of equal factors and no factor is left over. Therefore, $216$ is a perfect cube.
View full question & answer→Question 582 Marks
Find the cube roots of the following numbers by successive subtraction of numbers: $1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ...1728$
AnswerWe have, $1728 - 1 = 1727$
$1727 - 7 = 1720$
$1720 - 19 = 1701$
$1701 - 37 = 1664$
$1664 - 91 = 1512$
$1512 - 127 = 1385$
$1385 - 169 = 1216$
$1216 - 217 = 999$
$999 - 271 = 728$
$728 - 331 = 397$
$397 - 397 = 0$
$\because$ Subtraction is performed $12$ times. Hence, cube root of $1728$ is $12.$
View full question & answer→Question 592 Marks
By taking three different values of $n$ verify the truth of the following statement: If $n$ is even , then $n^3$ is also even.
AnswerLet the three even natural numbers be $2, 4$ and $8.$
Cubes of these numbers are:
$2^3=8,4^3=64,8^3=512$
By divisibility test, it is evident that $8, 64$ and $512$ are divisible by $2.$
Thus, they are even. This verifies the statement.
View full question & answer→Question 602 Marks
By taking three different values of $n$ verify the truth of the following statement: If $n$ is even , then $n^3$ is also odd.
AnswerLet the three odd natural numbers be $3, 9$ and $27.$
Cubes of these numbers are:
$3^3=27,9^3=729,27^3=19683$
By divisibility test, it is evident that $27, 729$ and $19683$ are divisible by $3.$
Thus, they are odd.
This verifies the statement.
View full question & answer→Question 612 Marks
Prove that if a number is trebled then its cube is $27$ times the cube of the given number.
AnswerLet us consider a number $n.$ Then its cube would be $n^3$.
If the number $n$ is trebled, i.e., $3n,$ we get:
$(3 n)^3=3^3 \times n^3=27 n^3$
It is evident that the cube of $3n$ is $27$ times of the cube of $n$.
Hence, the statement is proved.
View full question & answer→Question 622 Marks
Which of the following are perfect cube$?\ 4608$
AnswerOn factorising $3087$ into prime factors, we get $4608 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3$ Group the factors in triples of equal factors as: $4608 = \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times \{2 \times 2 \times 2\} \times 3 \times 3$ It is evident that the prime factors of $4608$ can be grouped into triples of equal factors and no factor is left over. Therefore, $4608$ is a perfect cube.
View full question & answer→Question 632 Marks
Which of the following are perfect cube$?\ 106480$
AnswerOn factorising $106480$ into prime factors, we get $106480 = 2 \times 2 \times 2 \times 2 \times 5 \times 11 \times 11 \times 11$ Group the factors in triples of equal factors as: $106480 = \{2 \times 2 \times 2\} \times 2 \times 5 \times \{11 \times 11 \times 11\}$ It is evident that the prime factors of $106480$ can be grouped into triples of equal factors and no factor is left over. Therefore, $106480$ is a perfect cube.
View full question & answer→Question 642 Marks
Find the cube roots of the following numbers by successive subtraction of numbers: $1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ... 512$
AnswerWe have, $512 - 1 = 511$
$511 - 7 = 504$
$504 - 19 = 485$
$485 - 37 = 448$
$448 - 61 = 387$
$387 - 91 = 296$
$296 - 127 = 169$
$169 - 169 = 0$
$\therefore$ Subtraction is performed $8$ times.Hence, cube root of $512$ is $8.$
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