3 Marks Question - MATHS STD 8 Questions - Vidyadip

Questions

3 Marks Question

🎯

Test yourself on this topic

16 questions · timed · auto-graded

Question 13 Marks

Consider the following pattern
$\begin{array}{l}2^3-1^3=1+2 \times 1 \times 3,3^3-2^3=1+3 \times 2 \times 3 \\4^3-3^3=1+4 \times 3 \times 3\end{array}$
Using the above pattern, find the value of the following:
(i) $7^3-6^3$$\qquad\quad$(ii) $12^3-11^3$
(iii) $20^3-19^3$$\quad~~$(iv) $51^3-50^3$

Answer

Using the given pattern, we can write
$\begin{aligned}\text{(i)} ~7^3-6^3 & =1+7 \times 6 \times 3 \\& =1+7 \times 18=1+126=127\end{aligned}$
$\begin{aligned}\text{(ii)} ~12^3-11^3 & =1+12 \times 11 \times 3 \\& =1+12 \times 33=1+396=397\end{aligned}$
$\begin{aligned}\text{(iii)}~ 20^3-19^3 & =1+20 \times 19 \times 3=1+20 \times 57 \\& =1+1140=1141\end{aligned}$
$\begin{aligned}\text{(iv)}~ 51^3-50^3 & =1+51 \times 50 \times 3=1+51 \times 150 \\& =1+7650=7651\end{aligned}$

View full question & answer→

Question 23 Marks

Check which of the following are perfect cubes :
2700

Answer

We have, 2700
Resolving 2700 into prime factors,
we get

2	2700
2	1350
3	675
3	225
3	75
5	25
5	5
	1

$2700=2 \times 2 \times \underline{3 \times 3 \times 3} \times 5 \times 5$
Clearly, the prime factors of 2 and 5 do not appear in groups of three (triples).
So, 2700 is not a perfect cube.

View full question & answer→

Question 33 Marks

Find the cube root of each of the following number by prime factorisation method:
512

Answer

We have, 512
Resolving 512 into prime factors, we get

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

$512=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{2\times2\times2}$
$=2^{3} \times 2^{3} \times 2^{3}$
$\therefore \sqrt{512}=2 \times 2 \times 2=8$
Hence, the cube root of 512 is 8.

View full question & answer→

Question 43 Marks

Find the cube root of each of the following number by prime factorisation method:
64

Answer

We have, 64
Resolving 64 into prime factors, we get

2	64
2	32
2	16
2	8
2	4
	2
	1

$\begin{aligned} 64 & =\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \\ & =2^3 \times 2^3 \\ \therefore \sqrt[3]{64} & =2 \times 2=4\end{aligned}$
Hence, the cube root of 64 is 4.

View full question & answer→

Question 53 Marks

Find the smallest number by which each of the following number must be divided to obtain a perfect cube :
128

Answer

We have, 128
Resolving 128 into prime factors, we get

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

$128=\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times 2$
The prime factor 2 does not appear in a group of three (triples).
So, 128 is not a perfect cube,
Thus, we will divide the number by 2.
$\therefore 128 \div 2=64=\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2}$
Hence, the smallest number by which 128 should be divided to make it a perfect cube is 2.

View full question & answer→

Question 63 Marks

Find the cube of the following numbers.
(i) 16 $\quad$ (ii) 17 $\quad$ (iii) 23.

Answer

(i) Given, number is 16.
Cube of $16=16 \times 16 \times 16=4096 \quad\left[\because a^3=a \times a \times a\right]$
(ii) Given, number is 17.
Cube of $17=17 \times 17 \times 17=4913$
(iii) Given, number is 23.
Cube of $23=23 \times 23 \times 23=12167$

View full question & answer→

Question 73 Marks

Complete the following cubes of numbers 1 to 10.

Number	Cube	Number	Cube
1	$1^3= 1$	6	$6^3=$
2	$2^3= 8$	7	$7^3=$
3	$3^3= 27$	8	$8^3=$
4	$4^3=$	9	$9^3=$
5	$5^3=$	10	$10^3=$

Answer

Number	Cube	Number	Cube
1	$1^3= 1$	6	$6^3=216$
2	$2^3= 8$	7	$7^3=343$
3	$3^3= 27$	8	$8^3=512$
4	$4^3=64$	9	$9^3=729$
5	$5^3=125$	10	$10^3=1000$

View full question & answer→

Question 83 Marks

What is the value of $\sqrt[3]{64}-\frac{1}{\sqrt[3]{64}}?$

Answer

$\begin{array}{l}\sqrt[3]{64}-\frac{1}{\sqrt[3]{64}} \\ \because \text { Cube root of } 64=\sqrt[3]{64}=\overline{4 \times 4 \times 4}=4 \\ \therefore 4-\frac{1}{4}=\frac{15}{4}~ \text { or }~ 3 \frac{3}{4}\end{array}$

View full question & answer→

Question 93 Marks

Magic Game Board
A magic game board is an array of numbers having the same number of row and columns arranged, so that the sum of the numbers in each row and column is the same.
Complete the magic game board below.

View full question & answer→

Question 103 Marks

Find the value of $17^3-16^3.$

Answer

We have, $17^3-16^3$
We know that $\text{b}^3-\text{a}^3=1+3 \text{ab,} \text{b}>\text{a},$
where $\text{a}$ and $\text{b}$ are consecutive numbers.
On putting $\text{a}=16$ and $\text{b}=17,$ we get
$17^3-16^3=1+3 \times 16 \times 17=817$

View full question & answer→

Question 113 Marks

The volume of a cubical box is $4913\text{ cm}^3.$ Find the length of side of the cubical box.

Answer

Given the volume of cubical box
$=4913\text{ cm}^3$
By prime factorisation,
$4913=17 \times 17 \times 17$
$\Rightarrow \quad \sqrt[3]{4913}=17$
$\therefore$ The length of side of the cubical box is 17$\text{ cm.}$

View full question & answer→

Question 123 Marks

Using prime factorisation, find the cube root of 5832 .

Answer

The prime factorisation of 5832
$\therefore 5832=\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3}$

2	5832
2	2916
2	1458
3	729
3	243
3	81
3	27
3	9
3	3
	1

Therefore, $\sqrt[3]{5832}$
$\begin{array}{l}=\sqrt[3]{\underline{2 \times 2 \times 2}} \times \underline{3 \times 3 \times 3} \times \underline{3 \times 3 \times 3} \\=2 \times 3 \times 3=18\end{array}$

View full question & answer→

Question 133 Marks

Find the cube root of 614125 through estimation.

Answer

Given, number is 614125.
So, groups of 614125 are

In the first group, the number 125 ends with 5.
We know that 5 comes at the unit place of a number only when its cube end in 5.
So, 5 will come at unit place.
Now, in the second group, the number is 614.
$\begin{array}{l}\because(8)^3=512 \text { and }(9)^3=729 \\ \text { and } 512<614<729\end{array}$
So, tens place of required cube root is 8.
Hence, $\sqrt[3]{614125}=85$

View full question & answer→

Question 143 Marks

Is 68600 a perfect cube? If not, then find the smallest number by which 68600 must be multiplied to get a perfect cube.

Answer

Given number is 68600.
Now, prime factorisation of 68600
$68600=\underline{2 \times 2 \times 2} \times 5 \times 5 \times \underline{7 \times 7 \times 7}$

2	68600
2	34300
2	17150
5	8575
5	1715
7	343
7	49
7	7
	1

The prime factor 5 does not have triples.
$\therefore 68600$ is not a perfect cube.
So, if we multiply the number 68600 by 5, then we get a perfect cube.
i.e. $68600 \times 5=343000$
$\begin{array}{l}=2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7 \\ =2 \times 5 \times 7=70\end{array}$

View full question & answer→

Question 153 Marks

Check whether 21600 is a perfect cube or not.

Answer

Given number is 21600.
Now, prime factorisation of 21600
$\therefore 21600=\underline{2 \times 2 \times 2} \times \underline{3 \times 3 \times 3}\times 2 \times 2 \times 5 \times 5$

2	21600
2	10800
2	5400
3	2700
3	900
3	300
2	100
2	50
5	25
5	5
	1

Prime factors 2 and 5 do not have triples.
So, 21600 is not a perfect cube.

View full question & answer→

Question 163 Marks

Find the smallest number by which 256 must be multiplied to obtain a perfect cube.

Answer

Given, number іs 256.
Now, prime factorisation of 256
$256=\underline{2 \times 2 \times 2} \times \underline{2 \times 2 \times 2} \times 2 \times 2$

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

Here, the prime factor 2 does not appear in a group of triples.
So, to make 256 a perfect cube, we have to multiply 256 by 2.
$\begin{array}{lrl}\therefore & 256 \times 2 & =512 \\ \Rightarrow & (8)^3 & =512\end{array}$

View full question & answer→

Generate Paper Free All Cubes and Cube Roots topics

3 Marks Question - MATHS STD 8 Questions - Vidyadip