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19 Sep 2023, 10:45
Kudos
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
50% (03:13) correct
50%
(02:28)
wrong
based on 72
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Two cars P and Q start from two points A and B towards each other simultaneously. They meet for the first time 40 km from B. After meeting they exchange their speeds as well as directions and proceed to their respective starting points. On reaching their starting points, they turn back with the same speeds and meet at a point 20 km from A. What is the distance between A and B?
A. 100 km
B. 110 km
C. 120 km
D. 130 km
E. 140 km
A. 100 km
B. 110 km
C. 120 km
D. 130 km
E. 140 km
19 Sep 2023, 21:01
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Bunuel
As we are not concerned with P and Q but with the distance, let us think in terms of speed with which one way is covered.
Let the distance be d. Even after meeting, the speed in a particular direction remains the same.
(1) When P(speed p) and Q(speed q) meet for the first time.
…….(p)…………….(q)……
||….d-40……..|……..40……||
(a) 40 km with speed q and
(b) d-40 km with speed p.
Time travelled in each speed is same.
(2) When P(speed q) and Q(speed p) meet for the second time.
…….(q)…………….(p)……
||…...20…...|……....d-20……||
(a) \((d)+20\) km with speed q
(b) \((d)+(d-20)\) or \(2d-20\) km with speed p.
Time travelled in each speed is same again.
From the above two equating time travelled, we get => \(\frac{40}{d-40}=\frac{d+20}{2d-20}\)
\(80d-800=d^2-20d-800…..…d^2-100d=0……..d(d-100)=0\)
So either d is 0 or 100, but d cannot be 0.
Thus, d is 100.
A
21 Sep 2023, 09:14
Kudos
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The method that i have used as described in the attachment below :
Consider the distance between A and B as D.
Speed of P as x Kmph and Q as y Kmph.
Now keeping time as constant for the first meetup :
D-40/x = 40/y
From this we derive the ratio of the speeds :
D-40/ 40 = x/y.
Now moving to the next step, reversing the direction:
Distance that Q traveled post meetup = 40 + D - 20 Km
Distance that P traveled post meetup = D - 40 + 20 Km
So here again keeping the time as constant and reversing the speeds we can say :
D-20/40 = D+20/D-40
Solving the same we get D = 100.
Hence IMO A
*Apologies for the image, I figured that the arrow diagrams and calculation will be better explained like this.
If any moderator, recommends to alter or remove the attachment altogether will be happy to do so.
Consider the distance between A and B as D.
Speed of P as x Kmph and Q as y Kmph.
Now keeping time as constant for the first meetup :
D-40/x = 40/y
From this we derive the ratio of the speeds :
D-40/ 40 = x/y.
Now moving to the next step, reversing the direction:
Distance that Q traveled post meetup = 40 + D - 20 Km
Distance that P traveled post meetup = D - 40 + 20 Km
So here again keeping the time as constant and reversing the speeds we can say :
D-20/40 = D+20/D-40
Solving the same we get D = 100.
Hence IMO A
*Apologies for the image, I figured that the arrow diagrams and calculation will be better explained like this.
If any moderator, recommends to alter or remove the attachment altogether will be happy to do so.
Attachments
File comment: Basic Solution
TSD 1.png [ 337.95 KiB | Viewed 4654 times ]
