A portion of an automobile test track is divided into Segment A, Segme

Question

A portion of an automobile test track is divided into Segment A, Segment B, and Segment C, in that order. In a performance test on a car, the car traveled Segment A at a constant speed of 140 kilometers per hour (km/h). Immediately after this, the car rapidly slowed on Segment B and then traveled on Segment C at a constant speed of 70 km/h. The length of Segment C is 3 times the length of Segment A, and it took a total of 42 minutes for the car to travel both segments A and C. In the table, select the length of Segment A, in kilometers, and select the length of Segment C, in kilometers. Make only two selections, one in each column.

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ID: 100359

In the table, select the length of Segment A, in kilometers, and select the length of Segment C, in kilometers. Make only two selections, one in each column.

ID: 100359

Apt0810 · Accepted Answer

Let length of segment A=x
So length of segment C=3x

And total time taken is 42 min =42/60 hours= 0.7 hours

So we will use Total time= distance in A/Speed in A + distance in C/Speed in C

0.7= x/140 + 3x/70
0.7*140= x+6x
.7*140=7x
x=14

So length of segment A=14
Length of segment C=42

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Sajjad1994 · Answer

Official Explanation

We are asked to determine the lengths of two segments of an automobile test track. The passage states that a car traveled Segment A at a constant rate of 140 km/h and traveled Segment C at a constant rate of 70 km/h, which indicates that it took the car twice as much time to cover a given unit of distance in Segment C as it took to cover the same distance in Segment A. The passage further states that the length of Segment C is 3 times the length of Segment A. Because the car took twice the time to cover a given distance in Segment C as in Segment A, and because Segment C was 3 times as long as Segment A, it can be further determined that the car spent 6 times as much time traveling Segment C as it did traveling Segment A. The passage indicates that it took a total of 42 minutes for the car to travel both segments A and C, and 42 minutes is equal to 0.7 h (42/60). Therefore, given that the car spent 6 times as much of the total combined time on Segment C as it did on Segment A, it can be calculated that the car spent 0.1 h on Segment A and 0.6 h on Segment C. Because we know that the car was traveling at 140 km/h for 0.1 h on Segment A, we can determine that Segment A was 14 kilometers long.

The correct answer is 14.

As noted in the analysis above, the car spent 0.6 h traveling Segment C. Because we know that the car was traveling at 70 km/h for 0.6 h on Segment C, we can determine that Segment C was 42 kilometers long.

The correct answer is 42.

bM22 · Answer

Speed in Segment A = 140km, 
Speed in Segment B = unknown
Speed in Segment C = 70km.

Length of segment A = x,
Length of segment C = 3x.

total time taken to cover Segment A & B = 42/60 = 0.7 hrs.

The equation which will be formed is: x/140 + 3x/70 = 7x/140 = 7/10.

=> x = 14 km, 3x = 42 km

Answer : Segment A: 14 km, Segment B = 42 km.

BLTN · Answer

Let us employ weighted average.

3 segments
Distance Rate Time
A ......... 140 km/h
B
C......... 70 km/h

Length of C = 3A
Total Time for to pass A + C or 4A = 42 min = 42/60 = (7/10)h

Ratio of Segment A rate to C rate is  \frac{140}{70 }  = \frac{2}{1 } 
Ratio of Distance of Segment A to C =3A is  \frac{1}{3 } 
So, time = Distance/Ration
For segment A: 1/2
For segment C 3/1
Total 3.5 or 7/2

The proportion for segment A becomes  \frac{0.5}{3.5}  =  \frac{1}{7}  *  \frac{7}{10 } h =  \frac{1}{10} h Thus, 140km/h * 1/10 = 14km

The proportion for segment A becomes  \frac{3}{3.5}  =  \frac{6}{7} *  \frac{7}{10 } h =  \frac{6}{10} h Thus, 70km/h * 6/10 =42km

whatsarc · Answer

Can be solved using the options directly 
only pairs possible with the condition of x and 3x requirements
8,24
14,42
24,72

Test cases, 1st case calculated brings the time to about 33mins (given is 42 so eliminate) 
2nd case is 6mins and 36mins for Segment A and Segment C respectively.

So 14 and 42 are the correct options.

Sunny9681 · Answer

We can use the equation for this question which is very simple. But we can also form pairs and use trail and elimination method to get to the answer

Gemmie · Answer

Distance         Speed            Time
Segment A:           a              140 km/h
Segment B:
Segment C:          3a              70 km/h

42 min = \frac{42}{60} h = \frac{7}{10} h

\frac{a}{140} + \frac{3a}{70} = \frac{a}{140} + \frac{6a}{140} = \frac{7a}{140} = \frac{a}{20} = \frac{7}{10}

=> a = 14

=> 3a = 42

Segment A: 14
Segement C: 42

KarishmaB · Answer

Responding to a pm: How to do it using ratios? Speeds A : C = 2:1 Distance A : C = 1 : 3 Hence ratio of time taken A : C = 1/2 : 3 = 1 : 6 Of the total 42 mins then, 6 mins were spent on A and 36 mins on C. Distance of A = 140 * 6/60 = 14 km Distance of C = 70 * 36/60 = 42 km Select 14 and 42 Here is a discussion on Ratios: https://anaprep.com/arithmetic-ratios-t ... ll-starts/

A portion of an automobile test track is divided into Segment A, Segme

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Length of Segment A (kilometers)	Length of Segment C (kilometers)
		8
		14
		24
		42
		72
		126

Length of Segment A (kilometers)	Length of Segment C (kilometers)
		8
		14
		24
		42
		72
		126

GMAT Two-part Analysis (TPA) Questions

A portion of an automobile test track is divided into Segment A, Segme

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