Question

plz answer fast no wrong answer​

Answer 1

Given :

First Case,

• Shashi increases the speed of the tread mill by 1km/hr she taken 6 min less.

Second Case,

• Shashi reduces the speed of the tread mill by 1km/hr she taken 9 min more.

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To find :

• The distance that she has decided to walk.

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Formula to be used :

$\small\bf\:Distance=s \times t$

where,

• d denotes distance to be travelled
• t denotes the total time taken

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Unit Conversion :

As Speed is given in km/hr we need to convert the times from min to hr.

We know,

⠀⠀⠀⠀⠀⠀1 hr = 60 min

$\sf\:60~min~=~1~hr$

$\sf\:1~min~=~\frac{1}{60}~hr$

$\sf\:6~min~=~(\frac{1}{60} \times 6~)hr$

$\sf\:6~min~=~(\frac{1}{\cancel{60}} \times \cancel{6}~)hr$

$\sf\:6~min~=~(\frac{1}{10})hr$

$\sf\therefore\:6~min~=~0.1~hr$

Similarly,

$\sf\:9~min~=~(\frac{1}{60} \times 9~)hr$

$\sf\:9~min~=~(\frac{1}{\cancel{60}} \times \cancel{9}~)hr$

$\sf\:9~min~=~(\frac{3}{20})hr$

$\sf\therefore\:9~min~=~0.15~hr$

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Assumption :

Let,

⠀⠀Shashi walk x km/hr on First day.

⠀⠀Time taken by Shashi on First day by y hr.

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Solution :

Distance = speed × time

⠀⠀⠀⠀⠀⠀⠀⠀= xy km/hr

According to the question,

First case,

When Shashi increases the speed of the tread mill by 1km/hr she takes 0.1 hr less time.

$\sf\:Distance~=~xy$

$\sf\to\:(x+1) (y-0.1)~=~xy$

$\sf\to\:xy-0.1x+y-0.1~=~xy$

$\sf\to\:\cancel{xy}-0.1x+y-0.1~=~\cancel{xy}$

$\sf\to\:-0.1x+y-0.1~=~1$

$\sf\to\:-0.1x+y~=~0.1$ ---- $\large\red{\tt\pmb{1}}$

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Second case,

When Shashi reduces the speed of the tread mill by 1km/hr she takes 0.15 hr more time.

$\sf\:Distance~=~xy$

$\sf\to\:(x-1) (y+0.15)~=~xy$

$\sf\to\:xy+0.15x-y-0.15~=~xy$

$\sf\to\:\cancel{xy}+0.15x-y-0.15~=~\cancel{xy}$

$\sf\to\:0.15x-y-0.15~=~1$

$\sf\to\:0.15x-y~=~0.15$ ---- $\large\red{\tt\pmb{2}}$

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Adding equation 1 and 2,

$\sf\:-0.1x+\cancel{y}~=~0.1$

(+)$\sf\:0.15x-\cancel{y}~=~0.15$

_____________________

$\sf\:0.05x~=~0.25$

$\sf\to\:x~=~\cancel{\frac{0.25}{0.05}}$

$\small{\underline{\boxed{\mathrm{\to\:x~=~5}}}}$

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Substituting the value of x in 1,

$\sf\:-0.1x+y~=~0.1$

$\sf\to\:-0.1 (5)+y~=~0.1$

$\sf\to\:-0.5+y~=~0.1$

$\sf\to\:y~=~0.1+0.5$

$\small{\underline{\boxed{\mathrm{\to\:y~=~0.6}}}}$

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Therefore,

⠀⠀⠀⠀⠀⠀⠀Distance = speed × time

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀= xy km/hr

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ ⠀= (5 × 0.6) km/hr

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = $\small{\mathfrak\red{3~km/hr}}$

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