GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Apr 25Crafting an Impactful Resume: Balancing Detail with Brevity
10:00 AM EDT
-10:30 AM EDT
Showcase your achievements through impactful storytelling! Get ready for an LIVE session on April 25, Learn how to craft compelling narratives on your resume that resonate with admission committees.
Apr 24Get an Extra 30% Off Target Test Prep Plans
08:00 PM EDT
-11:59 PM EDT
Grab 30% off any Target Test Prep GRE plan during our Flash Sale and join Brian and many other students who have used Target Test Prep to score high on the GRE. Just enter the coupon code FLASH30 at checkout to save big.
Apr 26Get a Top GRE Score with Target Test Prep
12:00 PM EDT
-01:00 PM EDT
The Target Test Prep course represents a quantum leap forward in GRE preparation, a radical reinterpretation of the way that students should study. In fact, more and more Target Test Prep students are scoring 329 and above on the GRE.
Two coal carts, A and B, started simultaneously from opposite ends of
[#permalink]
10 Nov 2021, 01:28
10 Nov 2021, 01:28
Expert Reply
00:00
Question Stats:
81% (02:13) correct
18% (01:18) wrong based on 11 sessions
Hide Show timer Statistics
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
Retired Moderator
Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: Two coal carts, A and B, started simultaneously from opposite ends of
[#permalink]
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 \(t = \frac{D}{S_A + S_B}\)
