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.

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/

LoneCase · Answer

could you perhaps show youre calculations ? thank you

FrozenDragon · Answer

I think basically the calculation here revolves around Distance, rate, and time.

Ex : for 14 and 42 case :

rate * time = Distance

140km/hr * time = 14 km

time = 0.1 hr

0.1 hr * 60 min / 1 hr =6 mins

and similarly,

70km/hr * time = 42 km

time = 0.6 hr

0.6 hr * 60 min/ 1 hr = 36 mins

so, 6 + 36 = 42 mins

I hope this helps

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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