Comprehension
These questions are based on the situation given below:
A road network (shown in the figure below) connects cities A, B, C and D. All road segments are straight lines. D is the midpoint on the road connecting A and C. Roads AB and BC are at right angles to each other with BC shorter than AB. The segment AB is 100 km long.
Ms. X and Mr. Y leave A at 8:00 am, take different routes to city C and reach at the same time. X takes the highway from A to B to C and travels at an average speed of 61.875 km per hour. Y takes the direct route AC and travels at 45 km per hour on segment AD. Y's speed on segment DC is 55 km per hour.
Question: 1
What is the average speed of Mr. Y for the journey in km per hour?
What is the average speed of Mr. Y for the journey in km per hour?
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In problems involving average speed, remember to calculate the total distance and the total time for the journey to determine the Correct Answer.
Updated On: Oct 29, 2026
- 47.5
- 49.5
- 50
- 52
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The Correct Option is B
Solution and Explanation
We are given that Mr. X and Mr. Y start at the same time and reach city C at the same time. Mr. X travels from A to B and then to C, while Mr. Y travels from A to D and then to C.
The total time taken by Mr. X is the time to travel from A to B and then from B to C. The time for Mr. X to travel from A to B is \( \frac{100}{61.875} = 1.615 \) hours.
To find Mr. Y's average speed, we use the total time taken by Mr. X and calculate the total distance traveled by Mr. Y.
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Question: 2
The total distance traveled by Y during the journey is approximately
The total distance traveled by Y during the journey is approximately
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Let BC = b. Equate the travel times of X and Y, and use Pythagoras in the right triangle ABC (right angle at B).
Updated On: Nov 4, 2026
- 130 km
- 150 km
- 105 km
- Cannot be determined
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The Correct Option is C
Solution and Explanation
Step 1: Set up:
Let \( BC = b \) km and \( AC = d \) km. The angle at B is a right angle, so \( d^2 = 100^2 + b^2 \).
Step 2: Time taken by X and Y:
X travels \( 100 + b \) km at 61.875 km/h, so \( t_X = \dfrac{100 + b}{61.875} \).
Y travels \( d/2 \) at 45 km/h and \( d/2 \) at 55 km/h: \( t_Y = \dfrac{d}{2}\left(\dfrac{1}{45} + \dfrac{1}{55}\right) = \dfrac{d}{2} \cdot \dfrac{2}{99} = \dfrac{d}{49.5} \).
Step 3: Equate the times:
\( \dfrac{100 + b}{61.875} = \dfrac{d}{49.5} \Rightarrow 100 + b = 1.25\,d \).
Step 4: Solve:
Square: \( (100 + b)^2 = \dfrac{25}{16}(10000 + b^2) \), which gives \( 9b^2 - 3200b + 90000 = 0 \). So \( b = \dfrac{3200 \pm 1000\sqrt{7}}{18} \), i.e. \( b \approx 30.79 \) or \( 324.8 \). Since \( BC \lt AB = 100 \), \( b \approx 30.79 \).
Then \( d = \dfrac{100 + 30.79}{1.25} \approx 104.6 \) km.
Step 5: Total distance of Y:
Y travels along AC, so the total distance is \( AC \approx 104.6 \approx 105 \) km.
Final Answer:
The correct option is (3): 105 km.
Let \( BC = b \) km and \( AC = d \) km. The angle at B is a right angle, so \( d^2 = 100^2 + b^2 \).
Step 2: Time taken by X and Y:
X travels \( 100 + b \) km at 61.875 km/h, so \( t_X = \dfrac{100 + b}{61.875} \).
Y travels \( d/2 \) at 45 km/h and \( d/2 \) at 55 km/h: \( t_Y = \dfrac{d}{2}\left(\dfrac{1}{45} + \dfrac{1}{55}\right) = \dfrac{d}{2} \cdot \dfrac{2}{99} = \dfrac{d}{49.5} \).
Step 3: Equate the times:
\( \dfrac{100 + b}{61.875} = \dfrac{d}{49.5} \Rightarrow 100 + b = 1.25\,d \).
Step 4: Solve:
Square: \( (100 + b)^2 = \dfrac{25}{16}(10000 + b^2) \), which gives \( 9b^2 - 3200b + 90000 = 0 \). So \( b = \dfrac{3200 \pm 1000\sqrt{7}}{18} \), i.e. \( b \approx 30.79 \) or \( 324.8 \). Since \( BC \lt AB = 100 \), \( b \approx 30.79 \).
Then \( d = \dfrac{100 + 30.79}{1.25} \approx 104.6 \) km.
Step 5: Total distance of Y:
Y travels along AC, so the total distance is \( AC \approx 104.6 \approx 105 \) km.
Final Answer:
The correct option is (3): 105 km.
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Question: 3
What is the length of the road segment BD?
What is the length of the road segment BD?
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D is the midpoint of the hypotenuse AC of the right triangle ABC (right angle at B). What is the length of the median to the hypotenuse?
Updated On: Nov 4, 2026
- 50 km
- Cannot be determined
- 55 km
- 52.5 km
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The Correct Option is D
Solution and Explanation
Step 1: Use the geometry:
Triangle ABC has a right angle at B, and D is the midpoint of the hypotenuse AC. The midpoint of the hypotenuse of a right triangle is the centre of the circumcircle, so \( DA = DB = DC \). Hence \( BD = \dfrac{AC}{2} \).
Step 2: Use the value of AC:
From the previous question (equal travel times of X and Y), \( AC \approx 104.6 \) km. So \( BD \approx 52.3 \) km.
Step 3: Choose the option:
The closest option to 52.3 km is 52.5 km.
Final Answer:
The correct option is (4): 52.5 km.
Triangle ABC has a right angle at B, and D is the midpoint of the hypotenuse AC. The midpoint of the hypotenuse of a right triangle is the centre of the circumcircle, so \( DA = DB = DC \). Hence \( BD = \dfrac{AC}{2} \).
Step 2: Use the value of AC:
From the previous question (equal travel times of X and Y), \( AC \approx 104.6 \) km. So \( BD \approx 52.3 \) km.
Step 3: Choose the option:
The closest option to 52.3 km is 52.5 km.
Final Answer:
The correct option is (4): 52.5 km.
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