Question

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Here's a question :

Runners X, Y, Z compete in a 1000 m race. If X gives Y a 50 m start, they finish together; if X gives Z a 69 m start they finish together. What start does Y have to give Z so they finish together?

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Answer 1

$\huge\bold{Answer :-}$

= 20 metre

Refer to the attachment please :)

Answer 2

$\huge\star{\green{\underline{\mathfrak{Answer :-}}}}$

20 m

$\huge\star{\green{\underline{\mathfrak{Explanation :-}}}}$

Given :-

• X,T and Z are runners.
• X gives a start of 50 m to Y to end the race together.
• X gives start of 69 m to Z to end the race together.

To find :-

What start does Y have to give Z so they finish together?

Solution :-

We will use the formula :- Rate / Speed = Distance / Time

Case 1 :-

X runs 1000 m and Y runs 1000-50=950 m.

Applying the formula to find time.

We have equated both the equations because both of them equal to t.

$\therefore{\dfrac {1000}{X}=\dfrac {950}{Y}}$

$\leadsto {1000Y=950X}$

$\leadsto {Y=\dfrac {950}{1000} X}$

$\leadsto {Y= 0.950 X }$.....(1)

Case 2 :-

X runs 1000 m and Z runs 1000-69=931 m.

Applying the formula to find time.

We have equated both the equations because both of them equal to t.

$\therefore{\dfrac {1000}{X}=\dfrac {931}{Z}}$

$\leadsto {1000Z=931X}$

$\leadsto {Z =\dfrac {931}{1000} X}$

$\leadsto {Z =0.931 X }$......(2)

Now, Dividing equation (2) from the (1),

$\leadsto {\dfrac{Z}{Y}=\dfrac{0.931 \cancel {X}}{0.950 \cancel{X}}}$

$\leadsto {\dfrac{Z}{Y}=\dfrac{0.931}{0.950}}$

$\leadsto {\dfrac {Z}{Y}=0.98}$

$\leadsto {\dfrac {Z}{Y}=\dfrac {98}{100}}$

Multiplying 10 in denominator and numerator of the RHS to get the denominator as 1000.

$\leadsto {\dfrac {Z}{Y}=\dfrac {980}{1000}}$

Solving further,

$\leadsto {\dfrac {Z}{980}=\dfrac {Y}{1000}}$

So, it means Y runs 1000 m and Z runs 980 m.

Hence Y gives Z 100-980= 20 m start to finish the race together.

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