5 Marks Questions

Question 15 Marks

The area of a square plot is $101\frac{1}{100}\text{m}^2$ Find the length of one side of the plot.

Answer

Let the numbers be $2x, 3x$ and $4x$, respectively.
$\because$ Sum of their cubes $= 0.334125$ According to the question,
$\Rightarrow(2\text{x})^3+(3\text{x})^3+(4\text{x})^3=0.334125$
$\Rightarrow8\text{x}^3+27\text{x}^3+64\text{x}^3=0.334125$
$\Rightarrow99\text{x}^3=0.334125$
$\Rightarrow\text{x}^3=\frac{0.334125}{99}$
$\Rightarrow\text{x}^3=0.0003375$
$\Rightarrow\text{x}^3=\frac{3375}{1000000}$
$\Rightarrow\text{x}=\sqrt[3]{\frac{15\times15\times15}{10\times10\times10\times10\times10\times10}}$ [taking cube root on both sides]
$\Rightarrow\text{x}=\frac{15}{10\times10\times10}$
Hence, the required numbers are $2 \times 0.015, 3 \times 0.015$ and $4 \times 0.015$, i.e. $0.03, 0.45$ and $0.06$.

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Question 25 Marks

The perimeters of two squares are $40$ and $96$ metres respectively. Find the perimeter of another square equal in area to the sum of the first two squares.

Answer

Let the side of two squares be $a_1$ and $4_2$.
Then, perimeter of frist square $= 4a_1$
Given, perimeter of frist square $= 40$
According to the question,
$4a_1= 40 \Rightarrow a_1= 10$
Similarly, perimeter of second square $= 4a_2$
Given, perimeter of second square $= 96$
According to the question,
$ 4a_2= 96 \Rightarrow a_2= 24$
Let a be the side of another square.
Then, area of frist square $= a^2$
$\therefore$ Area of another square = Sum of the areas of frist and second squares
= Area of frist square + Area of second square
$=\text{a}^2_1+\text{a}^2_2$
$\Rightarrow\text{a}^2=(10)^2+(24)^2=100+576=676$
$\Rightarrow\text{a}=\sqrt{676}=26\text{m}$ [taking square root on both sides]
Perimeter of another square $= 4a = 4 \times 26 = 104\ m$.
So, the perimeter of another square is $104\ m$.

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Question 35 Marks

By what smallest number should $3600$ be multiplied so that the quotient is a perfect cube. Also find the cube root of the quotient.

Answer

Prime factors of $3600 = 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 \times 5$
Grouping the factors into triplets of equal factors, we get
$3600=\underline{2\times2\times2}\times2\times3\times3\times5\times5$
$\begin{array}{c|c}2 & 3600 \\ \hline2 & 1800\\\hline2&900\\\hline2&450\\\hline3&225\\\hline3&75\\\hline5&25\\\hline5&5\\\hline&1\end{array}$
We know that, if a number is to be perfect cube, then each of its prime factors must occur thrice.
We find that $2$ occurs once $3$ and $5$ occurs twice only.
Hence, the smallest number, by which the given number must be multiplied in order that the product is a perfect cube
$= 2 \times 2 \times 3 \times 5 = 60$
Also, product $= 3600 \times 60 = 216000$
Now, arranging into triplets of equal prime factors,
we have $216000=\underline{2\times2\times2}\times\underline{2\times2\times2}\times\underline{3\times3\times3}\times\underline{5\times5\times5}$
Taking one factor from each triplets,
we get $\sqrt[3]{216000}=2\times2\times3\times5=60$

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Question 45 Marks

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

Answer

Given number is 324.
Now, we subtract successive odd numbers starting from $1$ as follows:
Here, $324 - 1 = 323, $
$323 - 3 = 320 320 - 5 = 315, $
$315 - 7 = 308 308 - 9 = 299,$
$ 299 - 11 = 288 288 - 13 = 275, $
$275 - 15 = 260 260 - 17 = 243, $
$243 - 19 = 224 224 - 21 = 203,$
$ 203 - 23 = 180 180 - 25 = 155, $
$155 - 27 = 128 128 - 29 = 99,$
$ 99 - 31 = 68 68 - 33 = 35, $
$35 - 35 = 0$
We observe that the number $324$ reduced to zero $(0)$ after subracting $18$ odd numbers.
So, $34$ is a perfect square of $18$.
$\therefore\sqrt{324}=18$
Hence, the square root of $324$ is $18$.

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