5 Marks Questions

Question 15 Marks

Find the length of each side of a cube, if its volume is $512 \text{ cm}^3.$

Answer

Given, the volume of cube $=512\text{ cm}^3$
We know that the volume of a cube $=(\text {Side})^3$

2	512
2	256
2	128
2	64
2	32
2	16
2	8
2	4
2	2
	1

Let the length of each side be a.
So, $\quad\text{(a)}^3=512$
$\Rightarrow \quad a=\sqrt[3]{512}$
$\begin{aligned}\text{a} & =\sqrt[3]{\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}} \\ & =2 \times 2 \times 2 \\ & =8\text{ cm}\end{aligned}$
Hence, the length of the each side of a cube is 8 cm.

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Question 25 Marks

Evaluate $\left(5^2+\left(12^2\right)^{1 / 2}\right)^3.$

Answer

Given, $\left[5^2+\left(12^2\right)^{1 / 2}\right]^3$
Now, $(5)^2=25$
and $\quad\left(12^2\right)^{1 / 2}=(12)^{2 \times \frac{1}{2}}$
$=(12)^1=12$
$\begin{aligned}\text{So, }(25+12)^3 & =(37)^3 \\ & =50653\end{aligned}$

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Question 35 Marks

Difference of two numbers which are perfect cubes is 189. If the cube root of the smaller of the two numbers is 3, then find the cube root of the larger number.

Answer

Let the cube root of the larger number be a and smaller number be b.
Then, according to the question,
$a^3-b^3=189$
$\Rightarrow \quad a^3-(3)^3=189$
$\Rightarrow \quad a^3-27=189$
$\Rightarrow \quad a^3=189+27$
$\Rightarrow \quad a^3=216$
$\Rightarrow \quad a^3=6^3$
$\therefore~ \quad a=6\qquad$ [by taking cube root on both sides]
Hence, the cube root of the larger number is 6.

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Question 45 Marks

To collect rain water, Aditya made a cubical tank which can hold $91125\text{ m}^3$ water. He uses this water for watering the plants of his garden.
(i) What is the height of the tank?
(ii) What value is depicted here?

Answer

(i) Volume of the cubical tank $=91125\text{ m}^3$
Let the length of each side of cubical tank be $\text{x m.}$
$\therefore$ Volume of a cube $=(\text {Side})^3$
So, $(\text{x})^3=91125$
$\Rightarrow\text{x}=\sqrt[3]{91125}$
$\Rightarrow x=\sqrt[3]{\underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3} \times \underline{5 \times 5 \times 5}}$
$\Rightarrow \quad\text{x}=3 \times 3 \times 5$
$\Rightarrow \quad\text{x}=45$

3	91125
3	30375
3	10125
3	3375
3	1125
3	375
5	125
5	25
5	5
	1

Hence, the height of the tank will be 45 m.
(ii) The value depicted here is that Aditya is conscious for environment. He saves water by reserving rain water.

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Question 55 Marks

Evaluate $\sqrt[3]{27}+\sqrt[3]{8}+\sqrt[3]{64}$

Answer

Given, $\sqrt[3]{27}+\sqrt[3]{8}+\sqrt[3]{64}$
$\because \quad \sqrt[3]{27}=\sqrt[3]{3 \times 3 \times 3}=3,$
$\sqrt[3]{8}=\sqrt[3]{8}$
$=\sqrt[3]{2 \times 2 \times 2}=2$

3	27
3	9
3	3
	1

$\begin{aligned}\text{and }\sqrt[3]{64} & =\sqrt[1]{64} \\& =\sqrt[3]{2 \times 2 \times 2 \times 2 \times 2 \times 2} \\& =4\end{aligned}$
$\begin{aligned} \therefore \quad \sqrt[3]{27} & +\sqrt[3]{8}+\sqrt[3]{64} \\ & =3+2+4 \\ & =9\end{aligned}$

2	8
2	4
2	2
	1

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Question 65 Marks

Three numbers are in the ratio $1: 2: 3$ and the sum of their cubes is 4500 . Find the numbers.

Answer

Given, three numbers are in the ratio $1: 2: 3$.
Let the numbers be $\text{x}, 2 \text{x}$ and $3\text{x}.$
If the sum of their cubes is 4500, the
$(\text{x})^3+(2\text{x})^3+(3\text{x})^3=4500$
$\Rightarrow x^3+8 x^3+27 x^3=4500$
$\Rightarrow 36 x^3=4500$
$\Rightarrow x^3=\frac{4500}{36}$
$\Rightarrow x^3=125$
$\Rightarrow x=\sqrt[3]{125}$
$\Rightarrow x=\sqrt[3]{5 \times 5 \times 5}$
$\Rightarrow x=5$
So, the numbers are $5 \times 1,5 \times 2,5 \times 3$
i.e. 5,10 and 15.

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Question 75 Marks

Is 9720 a perfect cube? If not, find the smallest number by which, it should be divided to get a perfect cube.

Answer

Resolving 9720 into prime factors, we get
$9720=\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times 3 \times 3 \times 5$

2	9720
2	4860
2	2430
3	1215
3	405
3	135
3	45
3	15
5	5
	1

The prime factors 3 and 5 are not forming the group of three (triples).
So, 9720 is not a perfect cube.
In prime factorisation of 9720, two factors 3 and 5 remain ungrouped.
So, if we divide the number by $3 \times 3 \times 5$, then the prime factorisation of the quotient will not contain 45.
$\begin{aligned} \therefore 9720 \div 45 & =216=(6)^3 \\ & =\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\end{aligned}$
Hence, the smallest number by which 9720 should be divided to get a perfect cube is 45.

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MATHS 5 Marks Questions

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