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Car A, B, and C participate in a car race. Car A beats car B by 45m,
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12 Mar 2022, 22:53
12 Mar 2022, 22:53
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Car A, B, and C participate in a car race. Car A beats car B by 45m, car B beats car C by 50m, and car A beats car C by 90m.
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Distance over which the race was conducted |
380m |
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Car A, B, and C participate in a car race. Car A beats car B by 45m,
[#permalink]
24 Mar 2022, 23:39
24 Mar 2022, 23:39
1
KarunMendiratta wrote:
Car A, B, and C participate in a car race. Car A beats car B by 45m, car B beats car C by 50m, and car A beats car C by 90m.
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Quantity A |
Quantity B |
Distance over which the race was conducted |
380m |
A. The quantity in Column A is greater.
B. The quantity in Column B is greater.
C. The two quantities are equal.
D. The relationship cannot be determined from the information given
Let the distance over which the race was conducted be \(D\)
When A finishes the race, B is 45m behind
i.e distance covered by B = (D - 45)m
When B finishes the race, C is 50m behind
i.e distance covered by C = (D - 50)m
When A finishes the race, C is 90m behind
i.e distance covered by C = (D - 90)m
We can say, (Ratio of speeds of A and B)(Ratio of speeds of B and C) = (Ratio of speeds of A and C)
\([\frac{D}{(D-45)}][\frac{D}{(D-50)}] = [\frac{D}{(D-90)}]\)
\([\frac{D}{(D-45)}] = [\frac{(D-50)}{(D-90)}]\)
Solving for D, we get D = 450
Col. A: 450
Col. B: 380
Col. A > Col. B
Hence, option A
Car A, B, and C participate in a car race. Car A beats car B by 45m,
[#permalink]
08 Apr 2022, 11:14
08 Apr 2022, 11:14
1
KarunMendiratta wrote:
Car A, B, and C participate in a car race. Car A beats car B by 45m, car B beats car C by 50m, and car A beats car C by 90m.
Quantity A |
Quantity B |
Distance over which the race was conducted |
380m |
Let the distance = d.
Car A beats car B by 45m.
In the time it takes A to travel d meters, B travels d-45 meters.
Thus:
\(\frac{A}{B }= \frac{d}{d-45}\)
Car A beats car C by 90m.
In the time it takes A to travel d meters, C travels d-90 meters.
Thus:
\(\frac{A}{C} = \frac{d}{d-90}\)
Combining the resulting ratios, we get:
\(\frac{B}{C} = \frac{B}{A} * \frac{A}{C} = \frac{d-45}{d} * \frac{d}{d-90} = \frac{d-45}{d-90}\)
Car B beats car C by 50m.
In the time it takes B to travel d meters, C travels d-50 meters.
Thus:
\(\frac{B}{C} = \frac{d}{d-50}\)
Since\( \frac{B}{C} = \frac{d-45}{d-90}\) and \(\frac{B}{C} = \frac{d}{d-50}\), we get:
\(\frac{d-45}{d-90} = \frac{d}{d-50}\)
Cross-multiplying, we get:
\(d^2 - 90d = d^2 - 95d + 2250\)
\(5d = 2250\)
\(d = 450\)
Quantity A is greater.
Show: ::
A
