5 Marks Questions

Question 15 Marks

You are told that 1331 is a perfect cube. Can you guess without factorisation what its cube root is? Similarly, guess the cube roots of 4913, 12167, and 32768.

Answer

(i) Yes, the cube root of 1331 can be guessed without factorization.
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{1331}<20$
Since 1331 ends in 1, its cube root ends in 1.
The number between 10 and 20 that ends in 1 is 11 .
$\therefore \sqrt[3]{1331}=11 .$

(ii) $\sqrt[3]{4913}$
$10^3=1000 \text { and } 20^3=8000$
So, $10<\sqrt[3]{4913}<20$
Since 4913 ends in 3 , its cube root ends in 7.
The number between 10 and 20 that ends in 7 is 17.
$\therefore \sqrt[3]{4913}=17 .$

(iii) $\sqrt[3]{12167}$
$20^3=8000 \text { and } 30^3=27000$
$\text { So, } 20<\sqrt[3]{12167}<30$
Since 12167 ends in 7 , its cube root ends in 3.
The number between 20 and 30 that ends in 3 is 23 .
$\therefore \sqrt[3]{12167}=23 .$

(iv) $\sqrt[3]{32768} $
$30^3=27000 \text { and } 40^3=64000$
$\text { So, } 30<\sqrt[3]{32768}<40$
Since 32768 ends in 8 , its cube root ends in 2.
The number between 30 and 40 that ends in 2 is 32 .
$\sqrt[3]{32768} =32 .$

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Question 25 Marks

State true or false. Explain your reasoning.
(i) The cube of any odd number is even.
(ii) There is no perfect cube that ends with 8.
(iii) The cube of a 2-digit number may be a 3-digit number.
(iv) The cube of a 2-digit number may have seven or more digits.
(v) Cube numbers have an odd number of factors.

Answer

(i) False because the cube of an odd number is always odd.
For example: $3^3=27,5^3=125,7^3=343$, all are odd.
(ii) False because some cubes do end with 8.
For example: $2^3=8,12^3=1728$, both end with 8 .
(iii) False, because the cube of a 2-digit number can range from 4-digit to 6-digit numbers.
For example: $10^3=1000,99^3=970299$.
(iv) False because the largest 2-digit number ' $99^{\prime}$ has a cube that is a 6 -digit number, i.e. $99^3=$ 970299.
So, a 2-digit whole number will always have a cube with at most 6 digits.
(v) False because only perfect squares have an odd number of factors.
For example :
Factors of $8=1,2,4,8 \rightarrow 4$ factors (even).
Factors of $27=1,3,9,27 \rightarrow 4$ factors (even).

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