TG TSD question discussion
Hi TG sir,|
I was going through few of the problems in TSD workbook but didnt get the solutions provided for two problems,here are those problems-
1.Two cars A & B drive straight towards point P with speeds a & b respectively.At the start A,B & P form an equilateral triangle.After some time A& B move to new positions,B covers 80 miles & ABP becomes a right triangle.At moment when A arrives at P ,B is 120 miles from P .What is the distance between A& B in the beginning.
Please explain the solution in detail if possible.
2.KATE runs twice as fast as she walks.On her way to school,she walks for twice the length of time as she runs.In this way she takes 20 mins to get there.On her way home she runs for twice the length of time as she walks.How long does she take to get home.
The answer for above problem is 16/3 mins while I am getting ans as 16 mins .Kindly rectify my approach if I am wrong
Running speed of kate =R and running time = a
Walking speed of Kate=W and walking time =b
Now we have r/w=2/1
therefore time taken will be a/b=1/2 or a= b/2
from 1st sentence we get 2b + a = 20
substitute a=b/2 we get b = 8 mins and a = 4 mins
now according to second part we need to calculate 2a+b which comes out to be 2x4 + 8 =16 mins.
But the answer given is 16/3 in booklet kindly elaborate the approach.
Re: TG TSD question discussion
question 1. Initial distance is 240.|
let the equilateral triangle's side be d. so d/a=d-120/b
also as a reaches first so a>b hence right angle is formed in A's path thus after B travels 80 miles, A has traveled d-((d-80)cos60) i.e. hence second equation is 80/b=(d/2 + 40)/a.
equate a/b from both solve the quadratic in d.
question 2. this is quite straight forward. while going kate very astutely walks for twice the time as she runs (like everyone of us ;)) so eq. 1 is 80v/3=d where v is the walking speed and d is the dist. between her home and school. second eq. is 5x=d/v where x is the time she walks whilst coming back from school. hence x=16/3.