4 Mark Question - Maths STD 8 Questions - Vidyadip

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21 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

Find the cube root of the following numbers:
-1728 × 216

Answer

Property:

For any two integers a and b, $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}}$
From the above property, we have:
$\sqrt[3]{-1728\times216}$
$=\sqrt[3]{-1728}\times\sqrt[3]{216}$
$=\sqrt[3]{1728}\times\sqrt[3]{216}$ (For any positive integer $\text{x,}\sqrt{\text{-x}}-=\sqrt{\text{x}}$ )
Cube root using units digit:
Let us consider the number 1728.
The unit digit is 8; therefore, the unit digit in the cube root of 1728 will be 2.
After striking out the units, tens and hundreds digits of the given number, we are left with 1.
Now, 1 is the largest number whose cube is less than or equal to 1.
Therefore, the tens digit of the cube root of 1728 is 1.
$\therefore\sqrt[3]{1728}=12$
On factorising 216 into prime factors, we get:
216 = 2 × 2 × 2 × 3 × 3 × 3
On grouping the factors in triples of equal factors, we get:
216 = {2 × 2 × 2} × {3 × 3 × 3}
Now, taking one factor from each triple, we get:
$=\sqrt[3]{216}$
$=2\times3=6$
Thus
$\sqrt[3]{-1728\times216}$
$=-\sqrt[3]{1728}\times\sqrt[3]{216}$
$=-12\times6=-72$

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Question 24 Marks

Making use of the cube root table, find the cube root 732

Answer

We have:
730 < 732 < 740
$\Rightarrow\sqrt[3]{730}<\sqrt[3]{732}<\sqrt[3]{740}$
From cube root table, we have:
$\sqrt[3]{730}=9.004 $ and $\sqrt[3]{740}=9.045$
For the difference (740 -730), i.e., 10, the difference in values
= 9.045 - 9.004 = 0.041
$\therefore$ For the difference of (732 - 730), i.e., 2, the difference in the values
$= 0.0082$
$\therefore\sqrt[3]{732}$
$=9.004 +0.008$
$=9.012$

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Question 34 Marks

Making use of the cube root table, find the cube root 1346

Answer

By prime factorisation, we have:1346 = 2 × 673
$\Rightarrow\sqrt[3]{1346}$ $=\sqrt[3]{2}\times\sqrt[3]{673}$ Also 670 < 673 < 680 $\Rightarrow\sqrt[3]{670}<\sqrt[3]{673}<\sqrt[3]{680}$ For the difference (680 - 670), i.e., 10, the difference in the values = 8.794 - 8.750 = 0.044 $\therefore$ For the difference of (673 - 670), i.e., 3, the difference in the values $=\frac{0.004}{10}\times3= 0.013$ (upto three decimal places) $\therefore\sqrt[3]{673}$ $=8.750 +0.013$ $=8.763$ Now $\sqrt[3]{1346}$ $=\sqrt[3]{2}\times=\sqrt[3]{673}$ $=1.260\times8.763$ $=11.041$ (Up to three decimal places) Thus, the answer is 11.041.

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Question 44 Marks

Making use of the cube root table, find the cube root 7342

Answer

We have:
From cube root table, we have:
7300 < 7342 < 7400
$\Rightarrow\sqrt[3]{7000}<\sqrt[3]{7342}<\sqrt[3]{7400}$
From the cube root table, we have:
$\sqrt[3]{7300}=19.39 $ and $\sqrt[3]{7400}=19.48$
$\therefore$ For the difference of (7400 - 7300), i.e., 100, the difference in the values
= 19.48 - 19 - 39 = 0.09
$\therefore$ For the difference of (7342 - 7300), i.e., 42, the difference in the values
$=\frac{0.09}{100}\times42= 0.0378$
$= 0.037$
$\therefore\sqrt[3]{7342}$
$=19.39 +0.037$
$=19.427$

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Question 54 Marks

Making use of the cube root table, find the cube root 37800

Answer

$37800 = 2^3 × 3^3 × 175$
$\Rightarrow \sqrt[3]{37800}$
$=\sqrt[3]{2^3\times3^3\times175}$
$=6\times\sqrt[3]{175}$
Also
170 < 175 < 180
$\Rightarrow\sqrt[3]{170}<\sqrt[3]{175}<\sqrt[3]{180}$
From cube root table, we have:
$\sqrt[3]{170}=5.540 $ and $\sqrt[3]{180}=5.646$
$\therefore$ For the difference (180 - 170), i.e., 10, the difference in values
= 5.646 - 5.540 = 0.106
$\therefore$ For the difference of (175 - 170), i.e., 5, the difference in values
$=\frac{0.106}{10}\times5=0.053$
$\therefore\sqrt[3]{175}$
$=5.540+0.053$
$=5.593$
Now
37800
$= 6\times\sqrt[3]{175}=6\times5.593=33.558$
Thus, the required cube root is 33.558.

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Question 64 Marks

Write the cubes of 5 natural numbers of the form $3 n+2$ (i.e. $5,8,11, \ldots)$ and verify the following:
'The cube of a natural number of the form $3 n+2$ is a natural number of the same form i.e. when it is dividend by 3 the remainder is $2 '$.

Answer

Five natural numbers of the form $(3 n+2)$ could be written by choosing $n=1,2,3 \ldots$ etc.
Let five such numbers be $5,8,11,14$, and 17 .
The cubes of these five numbers are: $5^3=125,8^3=512,11^3=1331,14^3=2744$ and $17^3=4913$.
The cubes of the numbers $5,8,11,14$, and 17 could expressed as:
$125 \times 3 \times 41+2$, which is of the form $(3 n+2)$ for $n=41$
$512=3 \times 170+2$, which is of the form $(3 n+2)$ for $n=170$
$1331=3 \times 443+2$, which is of the form $(3 n+2)$ for $n=443$
$2744=3 \times 914+2$, which is of the form $(3 n+2)$ for $n=914$
$4913=3 \times 1637+2$, which is of the form $(3 n+2)$ for $n=1637$
The cubes of the numbers $5,8,11,14$, and 17 could be expressed as the natural numbers of the form $(3 n+2)$ for some natural number n ; therefore, the statement is verified.

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Question 74 Marks

Which of the following are cubes of odd natural numbers?
125, 343, 1728, 4096, 32768, 6859

Answer

We know that the cubes of all odd natural numbers are odd. The numbers 125, 343, and 6859 are cubes of odd natural numbers.
Any natural numbers could be either even or odd. Therefore, if a natural number is not even, it is odd. Now, the numbers 125, 343 and 6859 are odd (It could be verified by divisibility test of 2, i.e., a number is divisible by 2 if it ends with 0, 2, 4, 6 or 8). None of the three numbers 125, 343 and 6859 are divisible by 2. Therefore, they are not even, they are odd. The numbers 1728, 4096 and 32768 are even.
Thus, cubes of odd natural numbers are 125, 343 and 6859.

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Question 84 Marks

Write the cubes of 5 natural numbers which are of the form $3 n+1$ (e.g. $4,7,10, \ldots$ ) and verify the following:
'The cube of a natural number of the form $3 n+1$ is a natural number of the same form i.e. when divided by 3 it leaves the remainder $1^{\prime}$.

Answer

Five natural numbers of the form $(3 n+1)$ could be written by choosing $n=1,2,3 \ldots$ etc.
Let five such numbers be $4,7,10,13$, and 16 .
The cubes of these five numbers are: $4^3=64,7^3=343,10^3=1000,13^3=2197$ and $16^3=4096$
The cubes of the numbers $4,7,10,13$, and 16 could expressed as:
$64 \times 3 \times 21+1$, which is of the form $(3 n+1)$ for $n=21$
$343=3 \times 114+1$, which is of the form $(3 n+1)$ for $n=114$
$1000=3 \times 333+1$, which is of the form $(3 n+1)$ for $n=333$
$2197=3 \times 732+1$, which is of the form $(3 n+1)$ for $n=732$
$4096=3 \times 1365+1$, which is of the form $(3 n+1)$ for $n=1365$
The cubes of the numbers $4,7,10,13$, and 16 could be expressed as the natural numbers of the form ( $3 n+1$ ) for some natural number n; therefore, the statement is verified.

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Question 94 Marks

Find the cubes of the following number by column method:
$56$

Answer

Thus, cube of 35 is 42875 .
We have to find the cube of 56 using column method.
We have: $a=5$ and $b=6$

Column I	Column II	Column III	Column IV
$a ^3$	$3 \times a ^2 \times b$	$3 \times a ^{-1} b^2$	$B^3$
$5^3=125$	$3 \times a ^2 \times b =3 \times 5^2 \times 6=450$	$3 \times a ^2 \times b ^2=3 \times 5 \times 6^2=540$	$6^3=216$
$+50$	+56	$+21$	$216$
$175$	506	$561$
$175$	6$	$1$	$6$

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Question 104 Marks

Write the cubes of 5 natural numbers which are multiples of 3 and verify the followings: 'The cube of a natural number which is a multiple of 3 is a multiple of $27^{\prime \prime}$.

Answer

Five natural numbers, which are multiples of 3 , are $3,6,9,12$ and 15 .
Cubes of these five numbers are:
$3^3=3 \times 3 \times 3=27$
$6^3=6 \times 6 \times 6=216$
$9^3=9 \times 9 \times 9=729$
$12^3=12 \times 12 \times 12=1728$
$15^3=15 \times 15 \times 15=3375$
Now, let us write the cubes as a multiple of 27 . We have:
$27=27 \times 1$
$216=27 \times 8$
$729=27 \times 27$
$1728=27 \times 64$
$3375=27 \times 125$
It is evident that the cubes of the above multiples of 3 could be written as multiples of 27 . Thus, it is verified that the cube of a natural number, which is a multiple of 3 , is a multiple of 27

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Question 114 Marks

Three numbers are to one another $2: 3: 4$. The sum of their cubes is 0.334125 . Find the numbers.

Answer

Let the numbers be $2 x, 3 x$ and $4 x$.
According to the question:
$(2 x)^3+(3 x)^3+(4 x)^3=0.334125$
$\Rightarrow 8 x^3 27 x^3+64 x^3=0.334125$
$\Rightarrow 8 x^3 27 x^3+64 x^3=0.334125$
$\Rightarrow 99 x^3=0.334125$
$\Rightarrow x^3=\frac{3341125}{100000 \times 99}$
$\Rightarrow x=\sqrt[3]{\frac{3375}{1000000}}$
$\Rightarrow x=\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}$
$\Rightarrow x=\frac{15}{100}$
$= 0 . 1 5 .$

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Question 124 Marks

What is the length of the side of a cube whose volume is $275 cm^3$. Make use of the table for the cube root.

Answer

Volume of a cube is given by
$\text{V}=\text{a}^3,$ where a = side of the cube
$\therefore$ Side of a cube $= \text{a}=\sqrt[3]{\text{v}}$
If the volume of a cube is $275 cm^3$, the side of the cube will be $\sqrt[3]{275}$.
We have:
270 < 275 < 280
$\Rightarrow\sqrt[3]{270}<\sqrt[3]{275}<\sqrt[3]{280}$
From the cube root table, we have:
$\sqrt[3]{270}=6.463$ and $\sqrt[3]{280}=6.542$
$\therefore$ For the difference (280 - 270), i.e., 10, the difference in values
= 6.542 - 6.463 = 0.079
$\therefore$ For the difference (275 - 270), i.e., 5, the difference in values
$=\frac{0.079}{10}\times5=0.0395\simeq0.04$ (upto three decimal places)
$\therefore\sqrt[3]{275}=6.463+0.04=6.503$ (upto three decimal places)
Thus, the length of the side of the cube is 6.503cm.

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Question 134 Marks

Find the cubes of the following number by column method:
$72$

Answer

Thus, cube of 56 is 175616 .
We have to find the cube of 72 using column method.
We have: $a =7$ and $b =2$

Column I	Column II	Column III	Column IV
$a ^3$	$3 \times a ^2 \times b$	$3 \times a \times b ^2$	$B^3$
$7^3=343$	$3 \times a ^2 \times b =3 \times 7^2 \times 2=294$	$3 \times a \times b ^2=3 \times 7 \times 2^2=84$	$2^3=8$
+30	+8	+0	8
373	302	84
373	2	4	8

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Question 144 Marks

Making use of the cube root table, find the cube root 833

Answer

We have:830 < 833 < 840
$\Rightarrow\sqrt[3]{830}<\sqrt[3]{833}<\sqrt[3]{840}$
From the cube root table, we have:
$\sqrt[3]{830}=9.398$ and $\sqrt[3]{840}=9.435$
For the difference (840 - 830), i.e., 10, the difference in values
= 9.435 - 9.398 = 0.037
$\therefore$ For the difference of (833 - 830), i.e., 3, the difference in values
$=\frac{0.037}{10}\times3=0.0111=0.011$ (upto three decimal places)
$\therefore\sqrt[3]{833}$
$=9.398+0.011$
$=9.409$

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Question 154 Marks

Find the cube root of the following rational numbers:
0.003375

Answer

We have:
$0.003375=\frac{3375}{1000000}$
$\therefore\sqrt[3]{0.0003375}$
$=\sqrt[]{\frac{3375}{1000000}}$
$={\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}}$
Now
On factorising 1728 into prime factors, we get:
3375 = 3 × 3 × 3 × 5 × 5 × 5
On grouping the factors in triples of equal factors, we get:
3375 = {3 × 3 × 3} × {5 × 5 × 5}
Now, taking one factor from each triple, we get:
$\sqrt[3]{3375}=3\times5=15$
Also
$\sqrt[3]{1000000}$
$\sqrt[3]{100\times100\times100}=100$
$\therefore\sqrt[3]{0.003375}$
$={\frac{\sqrt[3]{3375}}{\sqrt[3]{1000000}}}$
$\frac{15}{100}=0.15$

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Question 164 Marks

Multiply 210125 by the smallest number so that the product is a perfect cube. Also, find out the cube root of the product.

Answer

First we find out the prime factors of 210125 ,
$210125=5 \times 5 \times 5 \times 41 \times 41$
$\because$ one triples remained incomplete, hence 210125 is not a perfect cube.
We see that if we multiply the factors by 41 , we will get 2 triples as $2^3$ and $41^3$.
And the product become:
$210125 \times 41=8615125=5 \times 5 \times 5 \times 41 \times 41 \times 41$
$\text { Cube root of product: }$
$=3_{\sqrt{8615125}}$
$=3 \sqrt{5^3 \times 41^3}$
$=5 \times 41$
$=205$

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Question 174 Marks

Find the cubes of the following numbers by column method:
$35$

Answer

We have to find the cube of 35 using column method.
We have: $a=3$ and $b=5$

Column I	Column II	Column III	Column IV
$a ^3$	$3 \times a ^2 \times b$	$3 \times a ^{-1} b^2$	$B^3$
$3^3=27$	$3 \times a ^2 \times b =3 \times 3^2 \times 5=135$	$3 \times a \times b ^2=3 \times 3 \times 5^2=255$	$5^3=125$
+15	+23	+12	125
42	158	237
42	8	7	5

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Question 184 Marks

The volume of a cube is $9261000 m3$. Find the side of the cube.

Answer

Volume of a cube is given by:
$V = s ^3$, where $s=$ Side of the cube
It is given that the volume of the cube is $9261000 m^3$; therefore, we have:
$S^3=9261000$
Let us find the cube root of 9261000 using prime factorisation:
$9261000=2 \times 2 \times 2 \times 3 \times 3 \times 3 \times 5 \times 5 \times 5 \times 7 \times 7 \times 7$
$=\{2 \times 2 \times 2\} \times\{3 \times 3 \times 3\} \times\{5 \times 5 \times 5\} \times\{7 \times 7 \times 7\}$
9261000 could be written as a triples of equal factors; therefore, we get:
Cube root $=2 \times 3 \times 5 \times 7=210$
$s^2=9261000$
$\Rightarrow s=3_{\sqrt{9261000}}=210$
Hence, the length of the side of cube is 210 m .

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Question 194 Marks

Write the cubes of 5 natural numbers of which are multiples of 7 and verify the following:
'The cube of a multiple of 7 is a multiple of $7^{3+}$.

Answer

Five multiples of 7 can be written by choosing different values of a natural number n in the expression 7 n . Let the five multiples be $7,14,21,28$ and 35 .
The cubes of these numbers are: $7^3=343,14^3=2744,21^3=9261,28^3=21952$, and $35^3=42875$
Now, write the above cubes as a multiple of $7^3$. Proceed as follows:
$343=7^3 \times 1$
$2744=14^3=14 \times 14 \times 14=(7 \times 2) \times(7 \times 2) \times(7 \times 2)=(7 \times 7 \times 7) \times(2 \times 2 \times 2)=7^3 \times 2^3$
$9261=21^3=21 \times 21 \times 21=(7 \times 3) \times(7 \times 3) \times(7 \times 3)=(7 \times 7 \times 7) \times(3 \times 3 \times 3)=7^3 \times 3^3$
$21952=28^3=28 \times 28 \times 28=(7 \times 4) \times(7 \times 4) \times(7 \times 4)=(7 \times 7 \times 7) \times(4 \times 4 \times 4)=7^3 \times 4^3$
$42875=35^3=35 \times 35 \times 35=(7 \times 5) \times(7 \times 5) \times(7 \times 5)=(7 \times 7 \times 7) \times(5 \times 5 \times 5)=7^3 \times 5^3$
Hence, the cube of multiple of 7 is a multiple of $7^3$.

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Question 204 Marks

Find the cube root of the following numbers:
$-729 \times 15625$

Answer

Property:
For any two integers a and b, $\sqrt[3]{\text{ab}}=\sqrt[3]{\text{a}}\times\sqrt[3]{\text{b}},$
From the above property, we have:
$\sqrt[3]{-729\times-15625}$
$=\sqrt[3]{-729}\times\sqrt[3]{-15625}$
$-=\sqrt[3]{729}\times-\sqrt[3]{15625}$ (For any positive integer $\text{x},\sqrt[3]{-\text{x}}=-\sqrt[3]{\text{x}}$)
Cube root using units digit:
Let us consider the number 15625.
The unit digit is 4; therefore, the unit digit in the cube root of 15625 will be 5.
After striking out the units, tens, and hundreds digits of the given number, we are left with 15.
Now, 2 is the largest number whose cube is less than or equal to $15\left(2^3<15<3^3\right)$
Therefore, the tens digit of the cube root of 15625 is 2 .
$\therefore\sqrt[3]{15625}=25$
Also
$\sqrt[3]{-729\times-15625}$
$=-\sqrt[3]{729}\times-\sqrt[3]{15625}$
$=-9\times6=-225$

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Question 214 Marks

Find the volume of a cube, one face of which has an area of $64m^2.$

Answer

Area of a face of cube is given by:
$\text{A} = \text{s}^2$ , where s = Side of the cube
Further, volume of a cube is given by:
$\text{V} = \text{s}^3$, where s = Side of the cube
It is given that the area of one face of the cube is $64m^2.$ Therefore we have:
$\text{s}^2=64\Rightarrow\text{s}=\sqrt{64}=8\text{m}$
Now, volume is given by:
$\text{V}=\text{s}^3=8^3$
$\Rightarrow\text{V}=8\times8\times8$
$=512\text{m}^3$
Thus, the volume of the cube is $512m^3.$

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