3 Marks Question

Question 13 Marks

Find the smallest number by which $8788$ must be divided so that the quotient is a perfect cube.

Answer

$\begin{array}{c|c}2&1600\\\hline2&800\\\hline2&400\\\hline2&200\\\hline2&100\\\hline2&50\\\hline2&25\\\hline5&5\\\hline&1\end{array}$
$8788$ can be expressed as the product of prime factors as $2 \times 2 \times 13 \times 13 \times 13.$
Therefore, $8788$ should be divided by $4,$ i.e. $(2 \times 2),$ so that the quotient is a perfect cube.

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Question 23 Marks

Find the value of using the short-cut method: $(68)^3$

Answer

$(68)^3$We know that short-cut method for finding the cube of any two digit numbers is as given.
$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3$
$=a^2 \times a+a^2 \times 3 b+b^2 \times 3 a+b^2 \times b$
Here,
$a=6$
$b=8$

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Question 33 Marks

Find the value of using the short-cut method: $(84)^3$

Answer

$(84)^3$We know that short-cut method for finding the cube of any two digit numbers is as given.
$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3$
$=a^2 \times a+a^2 \times 3 b+b^2 \times 3 a+b^2 \times b$
Here,
$a = 8$
$b = 4$

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Question 43 Marks

Evaluate $\sqrt[3]{216\times343}$

Answer

By prime factorisation method $\sqrt[3]{216\times343}$
$=\sqrt[3]{216}\times\sqrt[3]{343}$
$=\sqrt[3]{2\times2\times2\times3\times3\times3}\times\sqrt[3]{7\times7\times7}$
$=\sqrt[3]{(2)^3\times(3)^3}\times\sqrt[3]{(7)^3}$
$\sqrt[3]{216\times343}=(2)\times(3)\times(7)$
$\sqrt[3]{216\times343}=42$
$\therefore\sqrt[3]{216\times343}=42$

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Question 53 Marks

Evaluate $\sqrt[3]{4096}$

Answer

By prime factorisation method

$\sqrt[3]{4096}$

$=\sqrt[3]{2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2\times2}$

$=\sqrt[3]{(2)^3\times(2)^3\times(2)^3\times(2)^3}$

$\sqrt[3]{4096}=(2)\times(2)\times(2)\times(2)$

$\sqrt[3]{4096}=16$

$\therefore\sqrt[3]{4096}=16$

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Question 63 Marks

Find the smallest number by which $2560$ must be multiplied so that the product is a perfect cube.

Answer

$\begin{array}{c|c}2&2560\\\hline2&1280\\\hline2&640\\\hline2&320\\\hline2&160\\\hline2&80\\\hline2&40\\\hline2&20\\\hline2&10\\\hline5&5\\\hline&1\end{array}$
$2560 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 5$
To make this a perfect square, we have to multiply it by $5 \times 5.$
Therefore, $2560$ should be multiplied by $25$ so that the product is a perfect cube.

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Question 73 Marks

Which of the following are the cubes of odd numbers$?$
$i. 125$
$ii. 343$
$iii. 1728$
$iv. 4096$
$v. 9261$

Answer

The cubes of an odd numbers is an odd number.
Therefore, $125,343$ and $9261$ are the cubes of odd numbers.
$125=5 \times 5 \times 5$
$=5^3$
$343=7 \times 7 \times 7$
$=7^3$
$9261=3 \times 3 \times 3 \times 7 \times 7 \times 7$
$=3^3 \times 7^3$
$=21^3$

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Question 83 Marks

Which of the following are the cubes of even numbers$?$
$i. 216$
$ii. 729$
$iii. 512$
$iv. 3375$
$v. 1000$

Answer

The cubes of even numbers are always even.
Therefore, $216, 512$ and $1000$ are the cubes of even numbers.
$216=2 \times 2 \times 2 \times 3 \times 3 \times 3$
$=2^3 \times 3^3$
$=6^3$
$512=2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$=2^3 \times 2^3 \times 2^3$
$=8^3$
$1000=2 \times 2 \times 2 \times 5 \times 5 \times 5$
$=2^3 \times 5^3$
$=10^3$

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Question 93 Marks

What is the smallest number by which $1600$ must be divided so that the quotient is a perfect cube$?$

Answer

$1600$
$1600$ can be expressed as the product of prime factors in the following manner
$\begin{array}{c|c}2&1600\\\hline2&800\\\hline2&400\\\hline2&200\\\hline2&100\\\hline2&50\\\hline2&25\\\hline5&5\\\hline&1\end{array}$
$1600 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5$
Therefore, to make the quotient a perfect cube, we have to divide $1600$ by: $5 × 5 = 25$

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Question 103 Marks

Find the value of using the short-cut method: $(47)^3$

Answer

$(47)^3$We know that short-cut method for finding the cube of any two digit numbers is as given.
$(a+b)^3=a^3+3 a^2 b+3 a b^2+b^3$
$=a^2 \times a+a^2 \times 3 b+b^2 \times 3 a+b^2 \times b$
Here,
a = 4
b = 7

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Question 113 Marks

Evaluate $\sqrt[3]{\frac{-64}{125}}$

Answer

$\sqrt[3]{\frac{-64}{125}}$ By prime factorisation method $\sqrt[3]{\frac{-64}{125}}$ $=\frac{\sqrt[3]{-64}}{\sqrt[3]{125}}$ $=\frac{\sqrt[3]{(-4)\times(-4)\times(-4)}}{\sqrt[3]{5\times5\times5}}$ $=\frac{\sqrt[3]{(-4)^3}}{\sqrt[3]{(5)^3}}$ $\sqrt[3]{\frac{-64}{125}}=\frac{-4}{5}$ $\therefore\sqrt[3]{\frac{-64}{125}}=\frac{-4}{5}$

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Maths 3 Marks Question

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