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Ask QuestionPosted by Anjali Kumari 7 years, 8 months ago
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Posted by Niraj Kumar 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the ten's and unit's digits of the required number be x and y respectively.
Then, the number = (10x + y).
The number obtained on reversing the digits = (10y + x).
As per given condition
The sum of a two-digit number and the number obtained by reversing the order of its digits is 99.
{tex}\therefore{/tex} (10y + x) + (10x + y) = 99
{tex}\Rightarrow{/tex} 11(x + y) = 99
{tex}\Rightarrow{/tex} x + y = 9.
The digits differ by 3
So, (x - y) = ±3.
Thus, we have
x + y = 9 ........ (i)
x - y = 3 .......... (ii)
or x + y = 9 ......... (iii)
x - y = -3 ............ (iv)
From (i) and (ii), we get x = 6, y = 3.
From (iii) and (iv), we get x = 3, y = 6.
Hence, the required number is 63 or 36.
Posted by Ananya Mishra 7 years, 8 months ago
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Taufeeq Khan 7 years, 8 months ago
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Ananya Mishra 7 years, 8 months ago
Posted by Aditya Jha 7 years, 8 months ago
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Posted by Amit Chandra Jha 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
45 = 27 {tex} \times {/tex} 1 + 18
27 =18 {tex} \times {/tex} 1 + 9
18 = 9 {tex} \times {/tex} 2 + 0
So H.C.F. = 9
Now 9 = 27 – 18 {tex} \times {/tex} 1
{tex}\style{font-family:Arial}{\begin{array}{l}9=27-18\\=27-(45-27\times1)\times1\\=27-45+27\\=2\times27-(1)\times45=27x+45y\end{array}}{/tex}
⇒ x = 2, y = – 1.
Posted by Nistha Yadav 7 years, 8 months ago
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Sia ? 6 years, 6 months ago
{tex}\frac{2\sqrt{45\;\;}+3\sqrt{20}}{2\sqrt5}{/tex}
{tex}\text{=}\frac{\displaystyle2\sqrt{5\times9}+3\sqrt{5\times4}}{2\sqrt5}{/tex}
{tex}=\frac{6\sqrt5+6\sqrt5}{2\sqrt5}{/tex}
{tex}=\frac{\displaystyle12\sqrt5}{2\sqrt5}{/tex}
{tex}=6=\frac61{/tex}
Hence the given number can be written in the form of {tex}\frac pq{/tex} where p and q are integers.
So the given expression is a rational number.
Posted by Aish Ranglani 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
We have, α and β are the roots of the quadratic polynomial. f(x) =x2 - 5x + 4
Sum of zeros: {tex}\alpha+\beta=-\frac{b}{a}=-\frac{\text { coefficient of } x}{\text { coefficient of } x^{2}}{/tex}
product of zeros: {tex}\alpha \beta=\frac{c}{a}=\frac{\text { constant term }}{\text { coefficient of } x^{2}}{/tex}
We have a=1,b=-5 and c= 4.
Sum of the roots = α + β = 5
Product of the roots = αβ = 4
So,
{tex}\frac{1}{\alpha } + \frac{1}{\beta } - 2\alpha \beta = \frac{{\beta + \alpha }}{{\alpha \beta }} - 2\alpha \beta{/tex}
{tex}5/4-2\times4=5/4-8{/tex} ={tex}\left(5-32\right)/4{/tex}={tex}-27/4{/tex}
Hence,we get the result of {tex}\frac{{ 1 }}{\alpha} + \frac{{ 1 }}{\beta} - 2\alpha\beta{/tex} = {tex}-\frac{{ 27 }}{ 4}{/tex}
Posted by Vidhi Sagar 7 years, 8 months ago
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Shruti Mishra 7 years, 8 months ago
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Abhishek Reddy 7 years, 8 months ago
1Thank You