Sajjad1994 wrote:
Project IR Butler 2019-20 - Get one IR Question EverydayQuestion # 223, Date : 12-May-2020
This post is a part of Project IR Butler 2019-20.
Click here for Details 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