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Posted by Prerika Lamba 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
Check formulae in revision notes : https://mycbseguide.com/cbse-revision-notes.html
Posted by Dhanya Monu 7 years, 5 months ago
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Posted by Fathima Amal 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Let the man finishes the work in x days and that the boy finishes in y days.
One day's work of a man = {tex}\frac{1}{x}{/tex}
and one day's work of a boy ={tex}\frac{1}{y}{/tex}
Since, 4 men and 6 boys finish a piece of work in 5 days.
{tex}\therefore{/tex} One day's work of 4 men and 6 boys = {tex}\frac{1}{5}{/tex} of the work
{tex} \Rightarrow \frac{4}{x} + \frac{6}{y} = \frac{1}{5}{/tex}
Similarly, in second case,
One day's work of 3 men and 4 boys = {tex}\frac{1}{7}{/tex} part of the work
{tex}\therefore \frac{3}{x} + \frac{4}{y} = \frac{1}{7}{/tex}
Thus, we have the following equations
{tex}\frac{4}{x} + \frac{6}{y} = \frac{1}{5}{/tex} ......(i)
and
{tex}\frac{3}{x} + \frac{4}{y} = \frac{1}{7}{/tex} .....(ii)
Here , Eqs. (i) and (ii) are not in linear form , so we reduce them in linear form by putting {tex}\begin{array}{l}\frac1x\;=\;u\;and\;\frac1y\;=\;v\\\end{array}{/tex}
Now, Eq. (i) becomes {tex}4u + 6v = \frac{1}{5}{/tex} .....(iii)
and Eq(ii) becomes {tex}3u + 4v = \frac{1}{7}{/tex} .....(iv)
On multiplying Eq(iii) by 3 and Eq(iv) by 4 and then subtract Eq , we get
{tex}18v - 16v = \frac{3}{5} - \frac{4}{7}{/tex}
{tex}\Rightarrow 2v = \frac{{21 - 20}}{{35}} \Rightarrow 2v = \frac{1}{{35}} \Rightarrow v = \frac{1}{{70}}{/tex}
Put {tex}v=\frac{1}{70}{/tex} in Eq(iv), we get {tex}3u + \frac{4}{{70}} = \frac{1}{7}{/tex}
{tex} \Rightarrow 3u = \frac{1}{7} - \frac{4}{{70}}{/tex}
{tex} \Rightarrow 3u = \frac{6}{{70}}{/tex}
{tex} \Rightarrow u = \frac{1}{{35}}{/tex}
Thus {tex}u=\frac{1}{35}{/tex} and {tex}v=\frac{1}{70}{/tex}
{tex} \Rightarrow \frac{1}{x} = \frac{1}{{35}}{/tex}and {tex}\frac{1}{y} = \frac{1}{{70}}{/tex}
{tex} \Rightarrow x = 35{/tex} and {tex}y=70{/tex}
Hence, 1 man alone and 1 boy alone finishes the work in 35 and 70 days, respectively.
Posted by Pradhosan Pradhosan 7 years, 5 months ago
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Posted by Aryaman Pathania 7 years, 5 months ago
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Posted by Milan Jain 7 years, 5 months ago
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Alpna Suthar 7 years, 5 months ago
Shrishti Arora 7 years, 5 months ago
3x+4y=10 (i)
2x-2y=2 (ii)
Multiply eqn (i) by 2 and eqn (ii) by 3
We get,
6x+8y=20 - (iii)
6x-6y=6 - (iv)
Now solving eqn(iii) and (iv) we get 14y = 14 i.e y=1
Substituting y=1 in eqn (i) we get 3x=6 i. e. x=2
Posted by Yug Dangi 7 years, 5 months ago
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Shrishti Arora 7 years, 5 months ago
Cosec (theta)=1/sin (theta)
Sec (theta)=1/cos (theta)
Posted by Chinmoy Das 7 years, 5 months ago
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Posted by Prateek Pandey 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given polynomial is f(x) = x3 - 3x2 + x + 1
Let {tex} \alpha{/tex} = (a - b), {tex} \beta{/tex} = a and {tex} \gamma{/tex} = (a + b)
Now, {tex} \alpha + \beta + \gamma{/tex} = {tex} - \frac { ( - 3 ) } { 1 }{/tex}
⇒ (a - b) + a + ( a + b ) = 3
⇒ a - b + a + a+ b = 3
⇒ a + a + a = 3
⇒ 3a = 3
⇒ a = 3/3
⇒ a = 1
Also, {tex} \alpha \beta + \beta y + \gamma \alpha = \frac { 1 } { 1 }{/tex}
⇒ (a - b)a + a (a + b) + (a + b)(a - b) = 1
⇒ a2 - ab + a2 +ab + a2 - b2 = 1
⇒ 3a2 - b2 = 1 ( ∵ a = 1)
⇒ 3(1)2 - b2 = 1( ∵ a = 1)
⇒ 3 - b2 = 1
⇒ b2 = 2
⇒ b = {tex} \pm \sqrt{2}{/tex}
Hence, a = 1 and b = {tex} \pm \sqrt{2}{/tex}
Posted by Vikash Barnawal 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
we have, Dividend = divisor × quotient + remainder
{tex}\Rightarrow{/tex}{tex}{/tex}{tex}f(x)\;=\;g(x).\;q(x)\;+\;r(x){/tex}
{tex}\Rightarrow\ x^3\;–\;3x^2\;+\;x\;+\;2\;=\;\;g(x)\;\times\;(x\;–\;2)\;–\;2x\;+\;4{/tex}
{tex}\Rightarrow\ x^{3\;}–\;3x^2\;+\;x\;+\;2\;+\;2x\;–\;4\;=\;g(x)\;\times\;(x\;–\;2){/tex}
{tex}\Rightarrow{/tex} {tex}x^3\;–\;3x^2\;+3\;x\;\;\;–\;2\;=\;g(x)\;\times\;(x\;–\;2){/tex}
{tex} \Rightarrow\ g ( x ) = \frac { x ^ { 3 } - 3 x ^ { 2 } + 3 x - 2 } { ( x - 2 ) }{/tex}
{tex}\Rightarrow\ g ( x ) = \left( x ^ { 3 } - 3 x ^ { 2 } + 3 x - 2 \right) \div ( x - 2 ){/tex}
g(x) = x2 – x + 1
Posted by Prabhjot Kaur 7 years, 5 months ago
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Diksha Goyal 7 years, 5 months ago
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