2 Marks Questions

Question 12 Marks

Express the following numbers as the sum of odd numbers using the pattern given below
(i) $6^3$$\quad$(ii) $8^3$$\quad$(iii) $7^3$
$\begin{array}{r}1=1=1^3 \\ 3+5=8=2^3 \\ 7+9+11=27=3^3 \\ 13+15+17+19=64=4^3 \\ 21+23+25+27+29=125=5^3\end{array}$

Answer

From the given pattern, we observe that it follows the relation
$\begin{aligned} \text{n}^3=[\text{n}(\text{n}-1)+1]+ {[\text{n}(\text{n}-1)+3] } +[\text{n}(\text{n}-1)+5]+\ldots+\text{n} \text { terms }\end{aligned}$
(i) 216
(ii) 512
(iii) 343

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Question 22 Marks

Find the cube root of each of the following number by prime factorisation method:
10648

Answer

We have, 10648
Resolving 10648 into prime factors, we get
$10648=\underline{2\times2\times2}\times\underline{11\times11\times11}$
$=2^{3} \times 11^{3}$
$\therefore \sqrt[3]{10648}=2\times11=22$
Hence, the cube root of 10648 is 22.

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Question 32 Marks

Parikshit makes a cuboid of plasticine of sides 5 cm, 2 cm and 5 cm. How many such cubolds will he need to form a cube?

Answer

Given, sides of the cuboid are 5 cm, 2 cm and 5 cm.
$\begin{aligned} \therefore \text { Volume of the cuboid } & =l \times b \times h \\ & =5 \text{ cm} \times 2 \text{ cm} \times 5 \text{ cm}=50 cm^3\end{aligned}$
To form it as a cube, its dimension should be in the group of triples. Here, the prime factors of 2 and 5 are not in group of three (triples).
So, we have to multiply 50 by $2\times2\times5$ i.e. by 20 to make it a perfect cube.
Hence, Parikshit needed 20 cuboids to form a cube.

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Question 42 Marks

Find the smallest number by which each of the following number must be divided to obtain a perfect cube :
81

Answer

We have, 81
Resolving 81 into prime factors, we get

3	81
3	27
3	9
3	3
	1

$81=\underline{3 \times 3 \times 3} \times 3$
The prime factor 3 does not appear in a group of three.
So, 81 is not a perfect cube. Thus, we will divide the number by 3.
$\therefore \quad 81 \div 3=27=\underline{3 \times 3 \times 3}$
Hence, the smallest number by which 81 should be divided to make it a perfect cube is 3.

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Question 52 Marks

Hardy-Ramanujan number 1729 is the smallest Hardy-Ramanujan number. There are infinitely many such numbers. Few are 4104 (2, 16; 9, 15), 13832 (18, 20 ; 2, 24). Check it with the numbers given in the brackets.

Answer

Verification
$\begin{aligned} 4104 & =4096+8=16^3+2^3, \\ 4104 & =3375+729=15^3+9^3, \\ 13832 & =5832+8000=18^3+20^3\end{aligned}$
and $13832=13824+8=24^3+2^3$
Hence, it is the required verification.

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Question 62 Marks

Express $\left(6^3\right)$ as the sum of odd numbers.

Answer

Given, number is $6^3$.
So, $\text{n}=6$ and $(\text{n}-1)=(6-1)=5$
We will start with $(6 \times 5)+1=30+1=31$
So, odd numbers from 31 are $31,33,35,37,39,41$.
Now, the sum of odd numbers
$=31+33+35+37+39+41=216$

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Question 72 Marks

Is 150 a perfect cube?

Answer

Given, number is 150.
Now, prime factorisation of 150
$150=2 \times 3 \times 5 \times 5$

2	150
5	75
5	15
3	3
	1

The prime factors of 150 do not appear in group of triples.
Hence, 150 is not a perfect cube.

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