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Sia ? 6 years, 6 months ago
Let the speed while going be x km/h
{tex}\therefore {/tex}Speed while returning = (x + 10) km/h
According to question
{tex}\frac { 150 } { x } - \frac { 150 } { x + 10 } = \frac { 5 } { 2 }{/tex}
or, {tex}x ^ { 2 } + 10 x - 600 = 0{/tex}
or, (x -f 30) (x - 20) = 0
or, x=20
{tex}\therefore {/tex} Speed while going =20 km/h
and speed while returning = 20 + 10 = 30 km/h
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Sia ? 6 years, 6 months ago

Let APB be the given line, and let a circle with centre O touch APB at P.
Then, {tex}\angle O P B{/tex}= 90°. Let there be another circle with centre O' which touches the line APB at P.
Then, {tex}\angle O ^ { \prime } P B{/tex}=90°.
This is possible only when O and O' lie on the same line O'OP. Hence,
the required locus is a line perpendicular to the given line at the point of contact.
Posted by Srushti Kunkolienkar 7 years, 2 months ago
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Sia ? 6 years, 6 months ago
Total outcomes when two dice are thrown = 36
Favourable outcomes = {(1, 6), (2, 3), (3, 2), (6, 1)} = 4
Probability(product of two numbers on the top of the dice is 6) = {tex}\frac{4}{36} = \frac{1}{9}{/tex}
Posted by Srushti Kunkolienkar 7 years, 2 months ago
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Kunal J 7 years, 1 month ago
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