4 Mark Question

Question 14 Marks

Which of the following is the greatest? Explain your reasoning.
(a) $67^3-66^3$
(b) $43^3-42^3$
(c) $67^2-66^2$
(d) $43^2-42^2$

Answer

Using, $n^3-(n-1)^3=3 n^2-3 n+1 ;$
$n^2-(n-1)^2=2 n-1$.
$
\begin{array}{l}
\text { (a) } 67^3-66^3=3 \times 67^2-3 \times 67+1 \\
=3 \times 4489-201+1 \\
=13467-200=13267 .
\end{array}
$
$
\begin{array}{l}
\text { (b) } 43^3-42^3=3 \times 43^2-3 \times 43+1 \\
=3 \times 1848-129+1 \\
=5547-128=5419
\end{array}
$
(c) $67^2-66^2=2 \times 67-1$
$
=134-1=133 .
$
(d) $43^2-42^2=2 \times 43-1$
$
=86-1=85 .
$
Thus, (i) $67^3-66^3$ is the greatest.

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Question 24 Marks

Find the cube roots of 27000 and 10648.

Answer

(i) $\sqrt[3]{2} 27000$
Prime factorization of $27000=\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}$
$\begin{array}{l}27000=2^3 \times 3^3 \times 5^3 \\27000=(2 \times 3 \times 5)^3 \\27000=30^3 \\\therefore \sqrt[3]{27000}=30\end{array}$

(ii) $\sqrt[3]{10648}$
Prime factorisation of $10648=\underline{2 \times 2 \times 2} \times \underline{11 \times 11 \times 11}$
$\begin{array}{l}10648=2^3 \times 11^3 \\10648=(2 \times 11)^3 \\10648=22^3 \\\therefore \sqrt[3]{10648}=22\end{array}$

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Question 34 Marks

Which one among $64^2, 108^2, 292^2, 36^2$ has the last digit $4 ?$

Answer

(i) $64^2$
The unit's digit of 64 is 4
$4 \times 4=16$ (last digit 6)

(ii) $108^2$
The unit's digit of 108 is 8
$8 \times 8=64$ (last digit 4)

(iii) $292^2$
The unit's digit of 292 is 2
$2 \times 2=4$ (last digit 4)

(iv) $36^2$
The unit's digit of 36 is 6
$6 \times 6=36$ (last digit 6)

Therefore, $108^2$ and $292^2$ have 4 as their last digits.

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Question 44 Marks

Which of the following numbers are not perfect squares?
(i) 2032
(ii) 2048
(iii) 1027
(iv) 1089

Answer

A perfect square ends in 0, 1, 4, 5, 6, or 9 at its unit’s place.
(i) 2032 ends in 2 at the unit’s place. So, it is not a perfect square.
(ii) 2048 ends in 8 at the unit’s place. So, it is not a perfect square.
(iii) 1027 ends in 7 at the unit’s place. So, it is not a perfect square.
(iv) 1089 ends in 9 at the unit’s place. So, it is a perfect square.

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Question 54 Marks

Before the process begins, Khoisnam realises that he already knows which lockers will be open at the end. How did he figure out the answer?

Answer

He noticed that the number of toggles a locker receives equals the number of factors of that locker’s number. A locker toggled an odd number of times will be open at the end; a locker toggled an even number of times will be closed. Only perfect squares have an odd number of factors, so exactly the lockers numbered with perfect squares (1, 4, 9, 16, …) remain open.

Locker Number	Factors	Number of factors/Number of locker toggles
1	1(o)	1 (odd)
2	1(o), 2(c)	2 (even)
3	1(o), 3(c)	2 (even)
4	1(o), 2(c), 4(o)	3 (odd)
5	1(o), 5(c)	2 (even)
6	1(o), 2(c), 3(o), 6(c)	4 (even)
7	1(o), 7(c)	2 (even)
8	1(o), 2(c), 4(o), 8(c)	4 (even)
9	1(o), 3(c), 9(o)	3 (odd)
10	1(o), 2(c), 5(o), 10(c)	4 (even)

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