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555-605 Level|   Math Related|               
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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.
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parkhydel
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.

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

Length of Segment A (kilometers): 14
Length of Segment C (kilometers): 42

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
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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.
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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
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                        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


 ­
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parkhydel
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.­

ID: 100359

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/
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