3 Mark 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
8788 can be expressed as the product of prime factors as 2 × 2 × 13 × 13 × 13.
Therefore, 8788 should be divided by 4, i.e. (2 × 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 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 5
To make this a perfect square, we have to multiply it by 5 × 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?

125
343
1728
4096
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?

216
729
512
3375
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 Mark Question

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