Cubes and Cube Roots Test

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Cubes and Cube Roots Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

Find the cube root of the following number:
$$274625000$$

A
$$650$$

B
$$625$$

C
$$750$$

D
$$450$$

Solution

On prime factorising, we get,
$$274625000=(5 \times 5\times 5)\times (5\times5\times5)\times(2\times2\times2)\times(13\times13\times13)$$
$$= 5^3 \times 5^3 \times 2^3 \times 13^3$$.
Then, cube root of $$274625000$$ is:
$$\sqrt[3]{274625000}=\sqrt[3]{5^3 \times 5^3 \times 2^3 \times 13^3}= 5\times 5\times 2\times 13=650$$.
Therefore, option $$A$$ is correct.
Question 2
1 / -0

$$8640$$ is not a perfect cube.

A
True

B
False

C
Ambiguous

D
Insufficient information

Solution

Prime factorising of $$8640$$ gives:
$$8640 = 3 \times 3 \times 3 \times 2 \times 2 \times 2\times 2\times 2\times 2 \times 5$$
$$= 2^6 \times 5 ^1 \times 3^3 $$.....(1)
$$\Rightarrow$$ We know, a perfect cube has multiples of $$3$$ as powers of prime factors.
$$\Rightarrow$$ $$8640$$ is not a perfect cube as the power of $$5$$ in (1) is $$1$$
That is, the given statement is true and option $$A$$ is correct.
Question 3
1 / -0

Directions For Questions

Find the smallest number that must be subtracted from the given number which is not a perfect cube so as to make them perfect cubes.

...view full instructions

$$130$$

A
$$6$$

B
$$5$$

C
$$4$$

D
$$3$$

Solution

$$130-1=129, 129-7=122, 122-19=103, 103-37=66, 66-61=5.$$
$$\displaystyle \Rightarrow $$ The remainder got is $$5.$$
$$\displaystyle \Rightarrow $$ $$5$$ is the number to be subtracted from $$130$$ to make it a perfect cube.
$$\displaystyle \Rightarrow $$ $$130-5=125$$ is the perfect cube.
$$\displaystyle \therefore $$ The corresponding cube $$\displaystyle \sqrt [ 3 ]{ 125 } =5.$$
Question 4
1 / -0

Find the cube root of $$389017$$ by finding their units and ten digits.

A
$$63$$

B
$$67$$

C
$$73$$

D
$$77$$

Solution

Given, $$389017$$.
Here, the unit digit is $$7.$$
$$\therefore$$ The unit digit of its cube root is $$3$$.....$$[\because 3^3=27]$$.
After grouping the last three digits from the right, the number left is $$389.$$
Now, $$7^3=343<389$$ and $$8^3=512>389.$$
$$\therefore $$The tens digit of the cube root is $$7.$$
Therefore, $$\sqrt[3]{389017}=73$$.

Hence, option $$C$$ is correct.
Question 5
1 / -0

If $$\displaystyle { a }^{ 2 }$$ ends in an even number of zeros, then $$\displaystyle { a }^{ 3 }$$ ends in an odd number of zeros.

A
True

B
False

C
Ambiguous

D
Insufficient information

Solution

False, $$100 \times 100$$ $$= 10000$$ ends in even number of zeros.

$$100 \times 100 \times 100 = 1000000$$ also ends in even number of zeros.

So the statement if $$\displaystyle { a }^{ 2 }$$ ends in an even number of zeros, then $$\displaystyle { a }^{ 3 }$$ ends in an odd number of zeros is not true for all cases.
Question 6
1 / -0

Find the cube root of the following number by finding their units and ten digits:
$$46656$$

A
$$36$$

B
$$31$$

C
$$32$$

D
$$35$$

Solution

Here the unit digit is $$6.$$
$$\therefore$$The unit digit of the cube root is $$6.$$
$$\because 6^3=36$$

After striking out the last three digits from the right, the number left is $$46.$$
Now,
$$3^3=27<46$$ and $$4^3=64>46$$

The ten-digit of the cube root is $$3.$$
$$\therefore \sqrt[3]{46656}=36$$
Question 7
1 / -0

Find the smallest number by which $$135$$ must be divided, so that the quotient is a perfect cube.

A
$$3$$

B
$$5$$

C
$$9$$

D
$$15$$

Solution

$$\displaystyle 135=5\times 3\times 3\times 3$$
$$\displaystyle =\left( 5\times { 3 }^{ 3 } \right) $$
$$\displaystyle ={ \left( 3 \right) }^{ 3 }\times 5$$
In the above factorization there is triplet for $$5$$.
So, $$135$$ is not a perfect cube.
$$\displaystyle \therefore $$ $$135$$ must be divided by $$5$$ to make the quotient a perfect cube.
Question 8
1 / -0

Find the smallest number by which $$192$$ must be divided, so that the quotient is a perfect cube.

A
$$2$$

B
$$3$$

C
$$4$$

D
$$7$$

Solution

By prime factorisation method, we have
$$\displaystyle 192=2\times 2\times 2\times 2\times 2\times 2\times 3$$
$$\displaystyle =\left( { 2 }^{ 3 }\times { 2 }^{ 3 }\times 3 \right) $$
$$\displaystyle ={ \left( 2\times 2 \right) }^{ 3 }\times 3$$
In the above factorization there is no triplet for $$3$$.
So, $$192$$ is not a perfect cube.
Therefore, $$192$$ must be divided by $$3$$ to make the quotient a perfect cube.
Question 9
1 / -0

Find the cube root $$64$$ by successive subtraction of numbers:
$$(1, 7 ,19, 37, 61, 91, 127, 169, 217, 271, 331, 397, ...)$$

A
$$6$$

B
$$12$$

C
$$8$$

D
$$4$$

Solution

By consecutive subtraction of numbers $$1,7,19,37,61,91,127,...$$ from $$64$$, we get,
$$64-1=63, 63-7=56, 56-19=37, 37-37=0.$$
That is, we obtain the difference is $$0$$ after $$4$$ successive subtractions.
$$\displaystyle \therefore $$ The cube root of $$64=4$$.
Hence, option $$D$$ is correct.
Question 10
1 / -0

Find the cube root of $$614125$$ using prime factorization:

A
$$65$$

B
$$75$$

C
$$85$$

D
$$95$$

Solution

On prime factorising, we get,
$$614125 = 5\times5\times5\times17\times17\times17$$
$$= 5^3\times17^3$$.

Then, cube root of $$614125$$ is:
$$\sqrt[3]{614125}=\sqrt[3]{5^3 \times 17^3}= 5\times17=85$$.

Therefore, option $$C$$ is correct.

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Cubes and Cube Roots Test - 23

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Cubes and Cube Roots Test - 23

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SHARING IS CARING

Question 1

$$650$$

$$625$$

$$750$$

$$450$$

Solution

Question 2

True

False

Ambiguous

Insufficient information

Solution

Question 3

$$6$$

$$5$$

$$4$$

$$3$$

Solution

Question 4

$$63$$

$$67$$

$$73$$

$$77$$

Solution

Question 5

True

False

Ambiguous

Insufficient information

Solution

Question 6

$$36$$

$$31$$

$$32$$

$$35$$

Solution

Question 7

$$3$$

$$5$$

$$9$$

$$15$$

Solution

Question 8

$$2$$

$$3$$

$$4$$

$$7$$

Solution

Question 9

$$6$$

$$12$$

$$8$$

$$4$$

Solution

Question 10

$$65$$

$$75$$

$$85$$

$$95$$

Solution

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Question Analysis

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