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Sia ? 6 years, 6 months ago
Let us suppose that one man alone can finish the work in x days and one boy alone can finish it in y days. Then,
one man's one day's work {tex}= \frac { 1 } { x }{/tex}
One boy's one day's work {tex}= \frac { 1 } { y }{/tex}
{tex}\therefore{/tex} Eight men's one day's work = {tex}\frac { 8 } { x }{/tex}
{tex}12\ boy's{/tex} one day's work = {tex}\frac { 12 } { y }{/tex}
According to question it is given that {tex}8\ men{/tex} and {tex}12\ boys{/tex} can finish the work in {tex}10\ days{/tex}
{tex}10 \left( \frac { 8 } { x } + \frac { 12 } { y } \right) = 1 \Rightarrow \frac { 80 } { x } + \frac { 120 } { y } = 1{/tex} .................(i)
Again, {tex}6\ men{/tex} and {tex}8\ boys{/tex} can finish the work in {tex}14\ days{/tex}.
{tex}\therefore \quad 14 \left( \frac { 6 } { x } + \frac { 8 } { y } \right) = 1 \Rightarrow \frac { 84 } { x } + \frac { 112 } { y } = 1{/tex} ...........(ii)
Putting {tex}\frac { 1 } { x } = u{/tex} and {tex}\frac { 1 } { y } = v{/tex} in equations (i) and (ii), we get
{tex}80u + 120u - 1 = 0{/tex}
{tex}84u + 112v - 1 = 0{/tex}
By using cross-multiplication,
{tex}\Rightarrow \frac { u } { - 120 + 112 } = \frac { - v } { - 80 + 84 } = \frac { 1 } { 80 \times 112 - 120 \times 84 }{/tex}
{tex}\Rightarrow \quad \frac { u } { - 8 } = \frac { v } { - 4 } = \frac { 1 } { - 1120 }{/tex}
{tex}\Rightarrow \quad u = \frac { - 8 } { - 1120 } = \frac { 1 } { 140 } \text { and } v = \frac { - 4 } { - 1120 } = \frac { 1 } { 280 }{/tex}
{tex}u = \frac { 1 } { 140 } \Rightarrow \frac { 1 } { x } = \frac { 1 } { 140 } \Rightarrow x = 140{/tex}
{tex}v = \frac { 1 } { 280 } \Rightarrow \frac { 1 } { y } = \frac { 1 } { 280 } \Rightarrow y = 280{/tex}
One man alone can finish the work in {tex}140\ days{/tex} and one boy alone can finish the work in {tex}280\ days{/tex}.
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Sia ? 6 years, 6 months ago
According to the question, {tex}\triangle ABC {/tex} is an equilateral triangle.
In {tex}\triangle{/tex}ABD, using Pythagoras theorem,
{tex}\Rightarrow{/tex} AB2 = AD2 + BD2
{tex}\Rightarrow{/tex} BC2 = AD2 + BD2, (as AB = BC = CA)
{tex}\Rightarrow{/tex} (2 BD)2 = AD2 + BD2, (perpendicular is the median in an equilateral triangle)
{tex}\Rightarrow{/tex} 4BD2 - BD2 = AD2
{tex}\therefore{/tex} 3BD2 = AD2
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