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Ask QuestionPosted by Pawan Singh 7 years, 5 months ago
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Sia ? 6 years, 6 months ago
According to the question,A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot.
Let the speed of stream be x km/h
Then the speed of boat upstream = (18 - x) km/h
Speed of boat downstream = (18 + x) km/h
According to the question,
{tex}\frac { 24 } { 18 - x } - \frac { 24 } { 18 + x } = 1{/tex}
{tex}\frac { 24 ( 18 + x ) - 24 ( 18 - x ) } { (18 - x )(18+x)} = 1{/tex}
{tex}\frac { 24 ( 18 + x ) - 24 ( 18 - x ) } { 18 ^ { 2 } - x ^ { 2 } } = 1{/tex}
{tex}432 + 24 x - 432 + 24 x = 324 - x ^ { 2 }{/tex}
{tex}48 x = 324 - x ^ { 2 }{/tex}
{tex}x ^ { 2 } + 48 x - 324 = 0{/tex}
{tex}x ( x + 54 ) - 6 ( x + 54 ) = 0{/tex}
{tex}( x + 54 ) ( x - 6 ) = 0{/tex}
x + 54 = 0, x - 6 = 0
x = - 54, x = 6
Since speed cannot be negative
The speed of steam x = 6 km/h.
Posted by Vikash Vishwakarma 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
{tex}\vartriangle {/tex} ABC {tex} \sim {/tex} {tex}\vartriangle {/tex} DEF ...............Given
{tex}\therefore {/tex} {tex}{{ar(\Delta ABC)} \over {ar(\Delta DEF)}} = {\left( {{{BC} \over {EF}}} \right)^2}{/tex} ..........[The ratio of the area of two similar triangles is equal to the square of the ratio of their corresponding sides]
{tex}\Rightarrow {/tex} {tex}\frac{{64}}{{121}} = {\left( {\frac{{BC}}{{15.4}}} \right)^2} \Rightarrow {\left( {\frac{8}{{11}}} \right)^2} = {\left( {\frac{{BC}}{{15.4}}} \right)^2}{/tex}
{tex}\Rightarrow {/tex} {tex}\frac{8}{{11}} = \frac{{BC}}{{15.4}}{/tex} ...........Taking square root on both sides
{tex}\Rightarrow {/tex} BC={tex}\frac{{8 \times 15.4}}{{11}} \Rightarrow BC = 11.2\,cm{/tex}
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Sia ? 6 years, 6 months ago
If a and b are odd numbers then it should be in 2q+1 or 2q+3 form where q is a positive integer.
Let a = 2q + 3 , b = 2q + 1 and a > b
Now, {tex}\frac{a + b } {2} = \frac{ 2q + 3 + 2q + 1}{2}{/tex}
{tex}= \frac { 4 q + 4 } { 2 }{/tex}
= 2q + 2
{tex}\frac{a+b}{2}{/tex}=2(q+1) = an even number..........(1)
Now
{tex}\frac { a - b } { 2 } = \frac { ( 2 q + 3 ) - ( 2 q + 1 ) } { 2 }{/tex}
{tex}= \frac { 2 q + 3 - 2 q - 1 } { 2 }{/tex}
{tex}\frac{a-b}{2}= \frac { 2 } { 2 }{/tex} = 1 = an odd number..........(2)
Hence From (1) and (2) {tex}\frac{a+b}{2}{/tex} and {tex}\frac{a-b}{2}{/tex} are even and odd numbers respectively
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