Two cars travel on a highway in the same direction. If car A travels

This problem is a "runaway." Faster car A starts at a "gap" distance of \(y\) miles ahead of B. Car A pulls away from B and until the \(y\)-mile gap between them doubles.

How far A must travel (to double \(y\)) is the key and the trap. \(y\) doubled is not the issue. 2y is not the answer.

The issue: Car A has to double \(y\) miles while B still moves during Leg 2.

One method: Choose a time traveled for Leg 1 to "assign" \(y\)

(1) Original gap of \(y\) miles?
• Assign a time based on relative speed
• SAME direction of travel? Subtract to get relative speed,
\(R=(Rate_{A}-Rate_{B})=(75-60)=15mph\)

• Let Leg 1 travel time = 2 hours
Car A traveled \((75*2) = 150\) mi
Car B traveled \((60*2) =120\) mi
Car A is \((150-120) = 30\) miles ahead of B
Starting gap distance \(y= 30\)

(2) Distance that A must travel to double \(y=30\)?
• start distance gap: 30 miles
• Desired distance gap: (30 * 2) = 60 miles
Car A travels (60-30) = 30 additional miles
• Car A does NOT cover 60 more miles. The 30-mile original gap never shrinks. It increases slowly, by 30 miles, to 60 miles

(3) Time for A to cover Leg 2 = 30 more miles? B still moves.
Use relative speed, R = 15 mph. D = 30
RT = D, so \(T=\frac{D}{R}\)
A's time: \(\frac{30mi}{15mph}=2\) hrs

(4) Distance that A traveled in Leg 2?
• (A's speed * A's time) = A's distance
• A's distance: \((75mph*2hrs)=150\) miles
• A traveled 150 miles to double the original gap

(5) \(150\) miles in terms of \(y=30\)? \(\frac{150}{30}=5\)
\(150=5y\)

Answer C

Given: Two cars travel on a highway in the same direction.

Asked: If car A travels at a rate of 75 mph and is \(y\) miles ahead of car B, which is traveling at 60 mph, in terms of \(y\), how many miles must car A travel to double the distance between itself and car B?

Relative speed = 75 - 60 = 15 mph
Relative Distance = y miles
TIme = y/15 hrs

Actual distance to be travelled = (y/15)*75 = 5y

IMO C

Car A's speed:75mph
Car B's speed:60mph

Ratio of speeds of Car A and Car B 5:4 = Ratio of their distances Distance A: Distance B = 5:4
Difference is 1. So when A travels 5 units B travels 4 unit.

Since current distance between them is y and we need to double it to 2y that means the gap between A and B has to increase by y more miles.

So when gap is 1 A trvels 5 unit distance and B travels 4 unit distance.
When gap is y, A travels 5y unit distance and B travels 4y unit distance.

Answer is 5y. Option C.

(75-60)*t= Y
t=Y/15
Distance travelled by A in time t=75*Y/15
=5Y
C:)

We know 3 facts.

A speed- 75 mph
B speed- 60 mph
Gap=Y

The trick is understanding the question is not asking us to find distance for distance traveled 2*Y but rather,
GAP+GAP=2*the gap. We already know the gap is "Y miles",we need to find distance A needs to travel to make another "Y miles" GAP.

So, the equation we can derive,

GAP=(Speed of A* Time)-(Speed of B*Time)

75T-60T=Y miles

T=Y/15 Miles
(Note: Time will be same for both the cars when they are "Y" miles apart)

Therefore,

Distance A Travels= (Time)*(Speed of A)

=75 mph * Y/15 = 5Y

This is a layman solution, you can solve this less than 2 mins as long as you get the trick.
Answer (C)

Simple answer:-

A is travelling at a higher speed than B, therefore, relative speed = \(R=(Rate_{A}-Rate_{B})=(75-60)=15mph\)

The time A will take to double the distance is \(\frac{y}{15}\) hours

So the distance covered by A in \(\frac{y}{15}\) hours is \(\,\,\,{\text{5y}}\,\,\,{\text{miles}}\) (Answer)

Distance = 75×y/15
=5y

Posted from my mobile device

I solved it in the way below
Ratio of speeds is
75:60= 5:4 which means if A travels 5m, B travels 4 miles.
Difference is distance (y)= 1 mile
Q asked is double the distance (2y)= 2 miles = (A=10miles & B=8miles)
Total travelled by A to achieve 2y = 10-5=5

Is my approach correct?

Answer- C

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