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Two coal carts, A and B, started simultaneously from opposite ends of
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10 Nov 2021, 01:28
10 Nov 2021, 01:28
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Question Stats:
81% (02:13) correct
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Two coal carts, A and B, started simultaneously from opposite ends of a 400-yard track. Cart A traveled at a constant rate of 40 feet per second; Cart B traveled at a constant rate of 56 feet per second. After how many seconds of travel did the two carts collide? (1 yard = 3 feet)
(A) 75
(B) 48
(C) 70/3
(D) 25/2
(E) 25/6
(A) 75
(B) 48
(C) 70/3
(D) 25/2
(E) 25/6
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Re: Two coal carts, A and B, started simultaneously from opposite ends of
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10 Nov 2021, 06:03
10 Nov 2021, 06:03
Carcass wrote:
Two coal carts, A and B, started simultaneously from opposite ends of a 400-yard track. Cart A traveled at a constant rate of 40 feet per second; Cart B traveled at a constant rate of 56 feet per second. After how many seconds of travel did the two carts collide? (1 yard = 3 feet)
(A) 75
(B) 48
(C) 70/3
(D) 25/2
(E) 25/6
(A) 75
(B) 48
(C) 70/3
(D) 25/2
(E) 25/6
Let's start with the word equation that features both travelers (i.e., Cart A and cart B).
Since the track is 400 yards long, we can write: (distance travelled by cart A) + (distance travelled by cart B) = 400 yards
Since the speeds are given in FEET per second, we should you convert 400 yards to 1200 feet
So our word equation becomes: (distance travelled by cart A) + (distance travelled by cart B) = 1200 feet
Let t = the travel time (in seconds) of cart A.
So, t = the travel time (in seconds) of cart B
Distance = (rate)(time)
Plug the relevant values into the word equation to get: 40t + 56t = 1200 feet
Simplify: 96t = 1200
Solve: t = 1200/96 = 600/48 = 100/8 = 25/2 (seconds)
Answer: D
Two coal carts, A and B, started simultaneously from opposite ends of
[#permalink]
10 Nov 2021, 22:07
10 Nov 2021, 22:07
Total length of yard, \(D = 400(3) = 1200 feet\)
Let \(x\) be the distance travelled by Car A from starting point to the point of collision with Car B
Hence, distance travelled by Car B will be \(1200 - x\)
Speed of Car A, \(S_A = 40 ft/s\)
Speed of Car B, \(S_B = 56 ft/s\)
Since time taken by both cars will be same,
\(t_A = t_B\)
\(\frac{x}{S_A} = \frac{1200 - x}{S_B}\)
\(\frac{x}{40} = \frac{1200 - x}{56}\)
\(56x = 1200*40 - 40x\)
\(96x = 1200*40\)
\(x = \frac{1200*40}{96}\)
Hence time taken will be = \(\frac{x}{40} = \frac{1200*40}{96*40} = \frac{1200}{96} = \frac{48(25)}{48(2)} = \frac{25}{2}\)
Hence, Answer is D
A general formula can be applied to such similar type of question,
Time \(t = \frac{D}{S_A + S_B}\)
Let \(x\) be the distance travelled by Car A from starting point to the point of collision with Car B
Hence, distance travelled by Car B will be \(1200 - x\)
Speed of Car A, \(S_A = 40 ft/s\)
Speed of Car B, \(S_B = 56 ft/s\)
Since time taken by both cars will be same,
\(t_A = t_B\)
\(\frac{x}{S_A} = \frac{1200 - x}{S_B}\)
\(\frac{x}{40} = \frac{1200 - x}{56}\)
\(56x = 1200*40 - 40x\)
\(96x = 1200*40\)
\(x = \frac{1200*40}{96}\)
Hence time taken will be = \(\frac{x}{40} = \frac{1200*40}{96*40} = \frac{1200}{96} = \frac{48(25)}{48(2)} = \frac{25}{2}\)
Hence, Answer is D
A general formula can be applied to such similar type of question,
Time \(t = \frac{D}{S_A + S_B}\)
