Question

Hola Brainlians. Here's a question to check your Logical Qúestion. No spam. Give your best answer.

(NTSE - 2015 , SAT - 2018)

Consider a circular track with a point A on it. There is another point B on that circular track. There are two persons P and Q. P starts moving in anticlockwise direction from A and Q starts moving in clockwise direction from A. Q starts from A 10 minutes later than P. P covers BA in anticlockwise direction in 24 minutes. Q covers BA in clockwise direction in 6 minutes. The speed of P is a and the speed of Q is b. Assume that speed of P and Q is uniform throughout. Calculate the ratio of speed of Q with respect to P.



You have to give diagram also for understanding.
Spammers stay away. No spam. Give best answer. Mods and stars answer.​

Answer 1

★ Concept :-

Here the concept of speed and quadratic equation has been used. We see that we are given certain values as well as relations. So firstly using those values, we can find different equations and relations. From these equations we can get another sets of values which on applying in main equation will give us a quadratic equation. This equation will give us the value of any one of the previously assumed variable. From this we shall find another values and finally we can get our ratio.

Let's do it !!

______________________________________

★ Solution :-

Given,

(AB = Distance covered from A to B)

(BA = Distance covered from B to A)

» A circular track

» Direction of movement of P from point A = anticlockwise direction

» Direction of movement of Q from point A = clockwise direction

» Starting point of P and Q = A

» Another point at circular track = B

» Q starts 10 minutes later than P from point A

» Time taken by P to cover BA in anticlockwise direction = 24 minutes

» Time taken by Q to cover BA in clockwise direction = 6 minutes

» Speed of both bodies is constant throughout.

Now let's start assuming our variables.

• Let the length of the track be x

• Unit of speed of both bodies = m/min (assumed, since in finding ratio they both will cancel each other)

• Time taken by P to cover AB in anticlockwise direction = t₁

• Time taken by Q to cover AB in clockwise direction = t₂

• According to qúestion, t₂ = t₁ - 10 (since Q starts 10 min later than P)

• Let the speed of P be p

• Let the speed of Q be q

• For the first relationship ::

We know the basic formula that,

→ Distance = Speed × Time

Since it is circular track, then distance covered by Q in clockwise direction as BA when added to distance covered by P in anticlockwise direction as BA will give us total length of circular track.

:: So, for Q, BA will be 6q (using the formula)

:: Also, for P, BA will be 24p (again using the formula)

Now on understanding the procedure, we get

>> 6q + 24p = x

• For body P ::

We already took the time taken by A to cover AB in anticlockwise direction as t₁ . Now if we add distance covered by P as AB with distance covered by P as BA in anticlockwise direction, then we can get the length of the circular track.

So,

>> pt₁ + 24p = x

On applying the value of x from first relationship, we get

>> pt₁ + 24p = 6q + 24p

On cancelling 24p we get,

>> pt₁ = 6q

From here, we will get

$\;\:\bf{\rightarrow\;\;\blue{\dfrac{p}{q}\;=\;\dfrac{6}{t_{1}}}}$

• For body Q ::

Similar like above process, if we add the distance covered by Q as AB with distance covered by Q as BA in clockwise direction, then the resultant will be equal to the length of the circular track.

So,

>> qt₂ + 6q = x

On applying the value of t₂ in terms of t₁, we get

>> q(t₁ - 10) + 6q = 6q + 24p

(From first relationship)

On cancelling 6q from both sides, we get

>> q(t₁ - 10) = 24p

This will give us,

$\;\;\tt{\rightarrow\;\;(t_{1}\:-\:10)\;=\;24\:\times\:\dfrac{p}{q}}$

On applying the value of p/q from above, we get

$\;\;\tt{\rightarrow\;\;(t_{1}\:-\:10)\;=\;24\:\times\:\dfrac{6}{t_{1}}}$

This will give,

$\;\;\tt{\rightarrow\;\;t_{1}(t_{1}\:-\:10)\;=\;24\:\times\:6}$

• Let t₁ be y

$\;\;\tt{\rightarrow\;\;y(y\:-\:10)\;=\;144}$

$\;\;\tt{\rightarrow\;\;y^{2}\:-\:10y\;=\;144}$

$\;\;\tt{\rightarrow\;\;y^{2}\:-\:10y\:-\:144\;=\;0}$

This gives us a quadratic equation. Then on splitting the middle term, we get

>> y² - 18y + 8y - 144 = 0

On taking common terms,

>> y(y - 18) + 8(y - 18) = 0

On grouping the common terms, we get

>> (y + 8)(y - 18) = 0

Here either (y + 8) = 0 or (y - 18) = 0. So,

>> y = -8 or y = 18

But since y = t₁ (time). And time cannot be negative so we have to neglect the negative value of y. So,

>> t₁ = 18 min

• For final ratio ::

We have to calculate the ratio of speed of Q with respect to the ratio of speed of P. This means we have to calculate the value of q/p .

We already have an equation that,

$\;\:\bf{\rightarrowtail\;\;\dfrac{p}{q}\;=\;\dfrac{6}{t_{1}}}$

On applying the value of t₁, we get

$\;\:\tt{\rightarrowtail\;\;\dfrac{p}{q}\;=\;\dfrac{6}{18}}$

On taking the reciprocal of L.H.S. and R.H.S. , we get

$\;\:\bf{\rightarrowtail\;\;\dfrac{q}{p}\;=\;\dfrac{18}{6}}$

$\;\:\underline{\boxed{\bf{\rightarrowtail\;\;\red{\dfrac{q}{p}\;=\;\dfrac{3}{1}\;=\;3}}}}$

$\;\:\underline{\boxed{\bf{\rightarrowtail\;\;\purple{q\;:\;p\;=\;3\;:\;1}}}}$

This is the required answer

(**min = minutes)