2 Mark Question

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 × 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

$\sqrt[3]{125\times27}=3\times\underline5$
Solution:
$\because\sqrt[3]{125\times27}$
$=\sqrt[3]{125}\times\sqrt[3]{27}$
$=\sqrt[3]{5\times5\times5}\times\sqrt[3]{3\times3\times3}$
$=5\times3$ (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 × 3 × 3 × 17 × 17
On grouping the factors in triples of equal factors, we get:
7803 = {3 × 3 × 3} × 17 × 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 × 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 × 2 × 2 × 5 × 5 × 5
Group the factors in triples of equal factors as:
1000 = {2 × 2 × 2} × {5 × 5 × 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., $3 n$, we get:
$(3 n)^3=3^3 \times n^3=27 n^3$
It is evident that the cube of $3 n$ 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 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3
Group the factors in triples of equal factors as:
4608 = {2 × 2 × 2} × {2 × 2 × 2} × {2 × 2 × 2} × 3 × 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 × 2 × 2 × 2 × 5 × 11 × 11 × 11
Group the factors in triples of equal factors as:
106480 = {2 × 2 × 2} × 2 × 5 × {11 × 11 × 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 = 448448 - 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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Maths 2 Mark Question