3 Marks Question - MATHS STD 8 Questions - Vidyadip

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19 questions · timed · auto-graded

Question 13 Marks

Find the decimal fraction which when multiplied by itself gives $84.64.$

Answer

Let the decimal fraction be $x$.
When a decimal fraction is multiplied by itself, then product $= x × x = x^2$
But product = 84.64 [given]
$\therefore\text{x}^2=84.64$
$\Rightarrow\text{x}=\sqrt{84.64}$
Now, place the over the numbers, then square root is given below.
$\begin{array}{c|c} & 9.2 \\ \hline9 & \overline{84}\ {\overline{64}\\81}\\\hline182&{364\\364}\\\hline&0\end{array}$
$\therefore\text{x}=\sqrt{84.64}=9.2$
Hence, the required decimal is $9.2.$

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Question 23 Marks

Evaluate: $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$

Answer

We have, $\sqrt[3]{27}+\sqrt[3]{0.008}+\sqrt[3]{0.064}$
$=(27)^\frac{1}{3}+(0.008)^\frac{1}{3}+(0.064)^\frac{1}{3}$
$=(3\times3\times3)^\frac{1}{3}+(0.2\times0.2\times0.2)^\frac{1}{3}$
$+(0.4\times0.4\times0.4)^\frac{1}{3}$
$=(3)^{3\times\frac{1}{3}}+(0.2)^{3\times\frac{1}{3}}+(0.4)^{3\times\frac{1}{3}}$
$=3+0.2+0.4=3.6$

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Question 33 Marks

The area of a square plot is $101\frac{1}{400}\text{m}^2.$ Find the length of one side of the plot.

Answer

Let length of the square plot be a. Then, the area of square $= a^2$
According to the question,
Area $=101\frac{1}{400}\text{m}^2$ [given]
$\therefore\text{a}^2=101\frac{1}{400}\Rightarrow\text{a}^2=\frac{40401}{400}$
$\Rightarrow\text{a}=\sqrt{\frac{40401}{400}}\Rightarrow\text{x}=\sqrt{\frac{201\times201}{20\times20}}$
$\therefore\text{a}=\frac{201}{20}=10\frac{1}{20}\text{m}$ [taking square root on both sides]
Hence, the length of one side of the plot is $10\frac{1}{20}$ m.

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Question 43 Marks

Find the length of the side of a square if the length of its diagonal is $10\ cm.$

Answer

Given, length of diagonal $= 10cm$
Suppose, the length of side of a square is $x cm.$
By using Pythagoras theorem, $(10)^2=\text{x}^2+\text{x}^2$
$\Rightarrow100=2\text{x}^2$
$\Rightarrow\text{x}^2=50$
$\Rightarrow\text{x}=\sqrt{50}$ [taking square root on both sides] $\therefore\text{x}=5\sqrt{2}\text{cm}$
Hence, the length of the side of square is $\sqrt{50}$ or $5\sqrt{2}\text{cm}.$

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Question 53 Marks

Find three numbers in the ratio $2 : 3 : 5$, the sum of whose squares is $608.$

Answer

Let the three numbers be $2x, 3x$ and $5x$, respectively.
According to the question,
$(2x)^2+ (3x)^2+ (5x)^2= 608$[given]
$\Rightarrow 4x^2+ 9x^2+ 25x^2= 608$
$\Rightarrow 38x^2= 608$
$\Rightarrow\text{x}^2=\frac{608}{38}$
$\Rightarrow x^2= 16 = (4)^2$
$\therefore x = 4$
Hence, the numbers are $2 \times 4, 3 \times 4, 5 \times 4, i.e. 8, 12$ and $20.$

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Question 63 Marks

Write the Pythagorean triplet whose one of the numbers is $4.$

Answer

We know that, for any natural number greater than $1,\left(2 m, m^2-1, m^2+1\right)$ is a pythagorean triplet.
So, one number is $2m$, then other two numbers are $m^2+1$ and $m^2-1$
Hence, one number is $4$, then pythagorean triplet,
$2m = 4 $
$\Rightarrow m = 2$
$\therefore m^2+1=2^2+1=4+1=5$
and $m^2-1=2^2-1=4-1=3$
Now, $3^2+4^2=5^2$
$\Rightarrow 9 + 16 = 25 $
$\Rightarrow 25 = 25$
So, $3, 4$ and $5$ are pythagorean triplets.

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Question 73 Marks

8649 students were sitting in a lecture room in such a manner that there were as many students in the row as there were rows in the lecture room. How many students were there in each row of the lecture room?

Answer

$\begin{array}{c|c} & 93 \\ \hline9 & {\overline{86}\\81}\ {\overline{49}}\\\hline183&{549\\549}\\\hline&0\end{array}$
Let number of students in each row of the lecture room be $x.$
Then, number of row $= x$
$\therefore$ Total students $= x \times x = x^2$
According to the question,
$\text{x}=8649$
$\Rightarrow\text{x}=\sqrt{8649}$
$\therefore\text{x}=93$

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Question 83 Marks

Find the least square number which is exactly divisible by $3, 4, 5, 6$ and $8.$

Answer

The least square numbers divisible by each of $3, 4, 5, 6$ and $8$, is equal to the
LCM of $3, 4, 5, 6$ and $8$.
$\begin{array}{c|c}2 & 3,4,5,6,8 \\ \hline2 & 3,2,5,3,4\\\hline2&3,1,5,3,2\\\hline3&3,1,5,3,1\\\hline5&1,1,5,1,1\end{array}$
$\therefore LCM$ of $3, 4, 5, 6$ and $8 = 2 \times 2 \times 2 \times 3 \times 5 = 120$
The prime factorisation of $120 = (2 \times 2) \times 2 \times 3 \times 5$
Here, prime factors $2, 3$ and $5$ are unpaire Clearly, to make it a perfect square, it must be multiplied by $2 \times 3 \times 5$
, i.e. $30$. Therefore, required number $= 120 \times 30 = 3600$
Hence, the least square number is $3600$

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Question 93 Marks

A king wanted to reward his advisor, a wise man of the kingdom. So he asked the wiseman to name his own reward. The wiseman thanked the king but said that he would ask only for some gold coins each day for a month. The coins were to be counted out in a pattern of one coin for the first day, $3$ coins for the second day, $5$ coins for the third day and so on for $30$ days. Without making calculations, find how many coins will the advisor get in that month?

Answer

Let the advisor get $x$ coins in that month.
According to the question, advisor get coins upto $30$ days as pattern
$1 + 3 + 5 + ...$
Here, $n = 30$
We know that, sum of frist n odd natural numbers $= n^2= (30)^2= 30 \times 30 = 900$
Hence, the advisor get $900$ coins in that month.

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Question 103 Marks

The dimensions of a rectangular field are $80\ m$ and $18\ m$. Find the length of its diagonal.

Answer

Here, length of a rectangular field $(l) = 80\ m$ and breadth of a rectangular field $(b) = 18\ m$

$\therefore$ Length of diagonal $=\sqrt{\text{l}^2+\text{b}^2}$
$=\sqrt{(80)^2+(18)^2}$
$=\sqrt{6400+324}$
$=\sqrt{6724}=82\text{m}$

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Question 113 Marks

A $5.5\ m$ long ladder is leaned against a wall. The ladder reaches the wall to a height of $4.4\ m$. Find the distance between the wall and the foot of the ladder.

Answer

Let the distance between the wall and foot of the ladder be $x m$

In right angled $\triangle\text{ABC},$ by using Pythagoras theorem, we get
$\begin{array}{c|c} & 3.3 \\ \hline3 & {\overline{10}\\9}\ \overline{89}\\\hline63&{189\\189}\\\hline&0\end{array}$
$ BC^2=A B^2+A C^2 $
$ \Rightarrow(5.5)^2=x^2+(4.4)^2 $
$ \Rightarrow x^2=(5.5)^2-(4.4)^2 $
$ \Rightarrow x^2=30.25-19.36 $
$ \Rightarrow x^2=10.89 $
$\Rightarrow\text{x}=\sqrt{10.89}$
$\therefore x = 3.3m$
Hence, the between the wall and the foot of the ladder is $3.3\ m.$

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Question 123 Marks

Find the smallest square number divisible by each one of the numbers $8, 9$ and $10.$

Answer

The least number divisible by each of the numbers $8, 9$ and $10$ is equal to the $LCM$ of $8, 9$ and $10.$
$\begin{array}{c|c}2 & 8,2,10 \\ \hline2 & 4,9,5\\\hline2&2,9,5\\\hline3&1,9,5\\\hline3&1,3,5\\\hline5&1,1,5\\\hline&1,1,1\end{array}$
$\therefore LCM$ of $8, 9$ and $10 = 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 360$
Prime factors of$ 360 = (2 \times 2) \times 2 \times (3 \times 3) \times 5$
Here, prime factors $2$ and $5$ are unpaired.
Clearly, to make it a perfect square, it must be multiplied by $2 \times 5,$ i.e. $10$.
Therefore, required number $= 360 \times 10 = 3600$
Hence, the smallest square number divisible by each of the numbers $8, 9$ and $10$ is $3600.$

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Question 133 Marks

A decimal number is multiplied by itself. If the product is $51.84$, find the number.

Answer

Let the number be $x$. Then, product $= x \times x = x^2$
But product $= 51.84$
$\therefore\text{x}^2=51.84$ [given]
$\Rightarrow\text{x}=\sqrt{51.84}$
Now, place the bar over the numbers, then square root is given below.
$\therefore\text{x}=\sqrt{51.84}=7.2$
$\begin{array}{c|c} & 7.2 \\ \hline 7 & \overline{51}\ {\overline{84}\\49}\\\hline142&{284\\284}\\\hline&0\end{array}$
Hence, the required number is $7.2.$

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Question 143 Marks

By what smallest number should $216$ be divided so that the quotient is a perfect square. Also find the square root of the quotient.

Answer

Prime factors of $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
Grouping the factors into pairs of equal factors,
we get $216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3$
We find that there is no prime factor to form a pair with $2$ and $3.$
Therefore, we must divide the number by $6$, so that the quotient becomes a perfect square.
If we divide the given number by $2 \times 2$
i.e. $6,$ then New number $=\frac{216}{6}=36$
Taking one factor from each, we square root of new number (quotient) $= 2 \times 3 = 6$

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Question 153 Marks

Find the side of a square whose area is equal to the area of a rectangle with sides $6.4\ m$ and $2.5\ m.$

Answer

Given dimensions of a rectangle are $6.4m$ and $2.5m.$
Area of a rectangle = Length $\times $ Breadth $= 6.4 \times 2.5 = 16m^2$
Let the side of square be $x$ m.
According to the question,
Area of square = Area of rectangle
$\Rightarrow\text{x}^2=16$
$\Rightarrow\text{x}=\sqrt{16}$
$\Rightarrow\text{x}=\sqrt{4\times4}$
$\therefore\text{ x}=4\text{m}$
Hence, the side of a square is $4m.$

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Question 163 Marks

Rahul walks $12\ m$ north from his house and turns west to walk $35\ m$ to reach his friend’s house. While returning, he walks diagonally from his friend’s house to reach back to his house. What distance did he walk while returning?

Answer

Let Rahul walked $x \ m$, while returning home.

In $\triangle\text{ABC},$ by using Pythagoras theorem,
we get $\begin{array}{c|c} & 37 \\ \hline3 & {\overline{13}\\9}\ {\overline{69}}\\\hline67&{469\\469}\\\hline&0\end{array}$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(12)^2+(35)^2$
$\Rightarrow\text{AC}^2=144+1225=1369$
$\Rightarrow\text{AC}=\sqrt{1369}$
$\Rightarrow\text{AC}=37\text{m}$
Hence, Rahul walked $37\ m$ distance for returning to his house diagonally.

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Question 173 Marks

Find the number of plants in each row if $1024$ plants are arranged so that number of plants in a row is the same as the number of rows.

Answer

$\begin{array}{c|c} & 32 \\ \hline3 & {\overline{10}\\9}\ {\overline{24}}\\\hline62&{124\\124}\\\hline&0\end{array}$
Let the number of plants in each row be $x.$
Then, number of rows $=$ Number of plants in each row $= x$
$\therefore$ Total plants $= x \times x = x^2$
According to the question,
$\text{x}^3=1024$
$\Rightarrow\text{x}=\sqrt{1024}=\sqrt{32\times32}$
$\therefore\text{x}=32$
Hence, there are $32$ plants in each row.

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Question 183 Marks

A hall has a capacity of $2704$ seats. If the number of rows is equal to the number of seats in each row, then find the number of seats in each row.

Answer

$\begin{array}{c|c} & 52 \\ \hline5 &{\overline{27}\\25}\ {\overline{04}}\\\hline102&{204\\204}\\\hline&0\end{array}$
Let the number of seets in each row be $x.$
Then, numbers = Number of seets in each row $= x$
$\therefore$ total seets $= x \times x = x^2$
According to the question,
$\text{x}^2=2704$
$\Rightarrow\text{x}=\sqrt{2704}=\sqrt{50\times52}$
$\Rightarrow\text{x}=52$
Hence, there are $52$ seets in each row.

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Question 193 Marks

Find the square root of $324$ by the method of repeated subtraction.

Answer

Given different of two perfect cubes $= 189$
and cube root of the smaller number $= 3$
$\therefore$ Cube of smaller number $= (3)^3= 27$
Let cube root of the larger number be $x.$
Then, cube of larger number $= x^3$
According to the question,
$\text{x}^3-27=189$
$\Rightarrow\text{x}^3=189+27$
$\Rightarrow\text{x}^3=216$
$\Rightarrow\text{x}=\sqrt[3]{216}=\sqrt[3]{6\times6\times6}$
$\therefore\text{ x}=65$
Hence, the cube root of the larger number is $6.$

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3 Marks Question - MATHS STD 8 Questions - Vidyadip