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MATHS 2 Marks Questions

Cubes and Cube Roots · Bihar Board (BSEB) STD 8 (BSEB - English Medium). 64 questions.

Questions · Page 2 of 2

2 Marks Questions

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.
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Question 522 Marks
Making use of the cube root table, find the cube root $780$
Answer
We 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 $9.205.$
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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)
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Question 542 Marks
Making use of the cube root table, find the cube roots $7$
Answer
Because $7$ 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 $1.913.$
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Question 552 Marks
By which smallest number must the following numbers be divided so that the quotient is a perfect cube$?\ 7803$
Answer
On 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.
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Question 562 Marks
Which of the following are perfect cube$?\ 1000$
Answer
On 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.
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Question 572 Marks
Which of the following number are not perfect cube$?\ 216$
Answer
On 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.
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Question 582 Marks
Find the cube roots of the following numbers by successive subtraction of numbers: $1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ...1728$
Answer
We 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 $12.$
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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.
Answer
Let 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.
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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.
Answer
Let 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.
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Question 612 Marks
Prove that if a number is trebled then its cube is $27$ times the cube of the given number.
Answer
Let 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.
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Question 622 Marks
Which of the following are perfect cube$?\ 4608$
Answer
On 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.
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Question 632 Marks
Which of the following are perfect cube$?\ 106480$
Answer
On 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.
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Question 642 Marks
Find the cube roots of the following numbers by successive subtraction of numbers: $1, 7, 19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ... 512$
Answer
We 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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