TRUE-FLASE [1 Marks Each]

Question 11 Mark

There is no perfect cube which ends in 4.

Answer

False.
Solution:
64 is a perfect cube, and it ends with 4.

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Question 21 Mark

No cube can end with exactly two zeros.

Answer

True.
Solution:
Because a perfect cube always ends with multiples of 3 zeros, e.g., 3 zeros, 6 zeros etc.

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Question 31 Mark

If a² ends in 5, then a³ ends in 25.

Answer

False.
Solution:
$\because$ 35²= 1225 but 53³ = 42875

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Question 41 Mark

If a²ends in 9, then a³ ends in 7.

Answer

False.
Solution:
a³ ends in 7 if a ends with 3.
But for every a² ending in 9, it is not necessary that a is 3.
E.g., 49 is a square of 7 and cube of 7 is 343.

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Question 51 Mark

If a² ends in an even number of zeros, then a³ ends in an odd number of zeros.

Answer

False.
Solution:
$\because$ 100² = 10000 but 100³ = 100000

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Question 61 Mark

If a divides b, then a³ divides b³.

Answer

True.
Solution:
$\because$ a divides b
$\therefore\frac{\text{b}^3}{\text{a}^3}=\frac{\text{b}\times\text{b}\times\text{b}}{\text{a}\times\text{a}\times\text{a}}=\frac{\text{(ak)}\times\text{(ak)}\times\text{(ak)}}{\text{a}\times\text{a}\times\text{a}}$
$\because$ a divides b
$\therefore$ b = ak fore some k
$\therefore\frac{\text{b}^3}{\text{a}^3}=\frac{\text{(ak)}\times\text{(ak)}\times\text{(ak)}}{\text{a}\times\text{a}\times\text{a}}=\text{k}^3$
$\Rightarrow\text{k}^3=\text{b}^3=\text{a}^3(\text{k}^3)$
$\therefore$ a³ divides b³

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Question 71 Mark

If a and b are integers such that a² > b², then a³ > b³.

Answer

False.
Solution:
It is not true for negative integers.
Example:
$(-5)^2 > (-4)^2$ but $(-5)^3<(-4)^3$

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Question 81 Mark

For an integer a, a³ is always greater than a².

Answer

False.
Solution:
It is not true for a negative integer.
Example:
$(-5)^2 = 25;(-5)^3$
$= -125$
$\Rightarrow(-5)^3<(-5)^2$

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Question 91 Mark

8640 is not a perfect cube.

Answer

True.
Solution:
On factorising 8640 into prime factors, we got:
8640 = 2 × 2 × 2 × 2 × 2 × 2 × 3 × 3 × 3 × 5
On grouping the factors in triples of equal factors, we get:
8640 = {2 × 2 × 2} × {2 × 2 × 2} × {3 × 3 × 3} × 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 not a perfect cube.

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Question 101 Mark

392 is a perfect cube.

Answer

False.
Solution:
On factorising 392 into prime factors, we got:
392 = 2 × 2 × 2 × 7 × 7
On grouping the factors in triples of equal factors, we get:
392 = {2 × 2 × 2} × 7 × 7
It is evident that the prime factors of 392 cannot be grouped into triples of equal factors such that no factor is left over. Therefore, 392 is not a perfect cube.

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MATHS TRUE-FLASE [1 Marks Each]

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