Comprehension
DIRECTIONS for the following three questions: Answer the questions on the basis of the information given below.
A city has two perfectly circular and concentric ring roads, the outer ring road (OR) being twice as long as the inner ring road (IR). There are also four (straight line) chord roads from E1, the east end point of OR to N2, the north end point of IR; from N1, the north end point of OR to W2, the west end point of IR; from W1, the west end point of OR, to S2, the south end point of IR; and from S1 the south end point of OR to E2, the east end point of IR. Traffic moves at a constant speed of \(30\pi\) km/hr on the OR road, \(20\pi\) km/hr on the IR road, and \(15\sqrt{5}\) km/hr on all the chord roads.
Question: 1
The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is
The ratio of the sum of the lengths of all chord roads to the length of the outer ring road is
Show Hint
Call the inner radius half the outer radius and treat each chord as the slanting side of a right triangle formed with the centre. Find one chord, multiply by four, and divide by the outer circumference.
Updated On: Nov 7, 2026
- \(\sqrt{5} : 2\)
- \(\sqrt{5} : 2\pi\)
- \(\sqrt{5} : \pi\)
- None of the above.
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Read the set-up and fix a radius.
Both ring roads are circles with the same centre, call it O. A circle's length is \(2\pi\) times its radius, so if OR is twice as long as IR, then the radius of OR is twice the radius of IR.
Let the radius of OR be \(R\). Then the radius of IR is \(R/2\). This question asks only about lengths, so the speeds given in the passage are not needed here.
Step 2: Put the end points on axes.
Take O as the origin. The four end points of OR are E1 \(=(R,0)\), N1 \(=(0,R)\), W1 \(=(-R,0)\) and S1 \(=(0,-R)\).
The four end points of IR are E2 \(=(R/2,0)\), N2 \(=(0,R/2)\), W2 \(=(-R/2,0)\) and S2 \(=(0,-R/2)\).
Step 3: Find the length of the chord from E1 to N2.
E1 lies on the east axis and N2 lies on the north axis, so O, E1 and N2 form a right angled triangle with the right angle at O. The legs are \(OE_1 = R\) and \(ON_2 = R/2\).
By Pythagoras:
\[ E_1N_2 = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{\sqrt{5}}{2}R \]
The other three chords (N1 to W2, W1 to S2, S1 to E2) are each the hypotenuse of a triangle with legs \(R\) and \(R/2\), so each of them has the same length \(\frac{\sqrt{5}}{2}R\).
Step 4: Add the four chords and find the outer ring length.
Sum of all chord roads:
\[ 4 \times \frac{\sqrt{5}}{2}R = 2\sqrt{5}\,R \]
Length of the outer ring road:
\[ 2\pi R \]
Step 5: Form the ratio.
\[ \frac{2\sqrt{5}\,R}{2\pi R} = \frac{\sqrt{5}}{\pi} \]
The radius \(R\) cancels, so the ratio is \(\sqrt{5} : \pi\). This matches option 3.
Step 6: Why the other options fail.
Option 1, \(\sqrt{5} : 2\), has no \(\pi\) in it. It treats the outer ring as if its length were just \(2R\), which is the diameter and not the circumference.
Option 2, \(\sqrt{5} : 2\pi\), would need the outer ring to be \(4\pi R\) long, twice its real length.
Option 4 is wrong because option 3 is exactly the ratio we found.
Final Answer:
The chord roads add up to \(2\sqrt{5}R\) and the outer ring road is \(2\pi R\) long, so the ratio is \(\sqrt{5} : \pi\), which is option 3. \[ \boxed{3 \;(\sqrt{5} : \pi)} \]
Both ring roads are circles with the same centre, call it O. A circle's length is \(2\pi\) times its radius, so if OR is twice as long as IR, then the radius of OR is twice the radius of IR.
Let the radius of OR be \(R\). Then the radius of IR is \(R/2\). This question asks only about lengths, so the speeds given in the passage are not needed here.
Step 2: Put the end points on axes.
Take O as the origin. The four end points of OR are E1 \(=(R,0)\), N1 \(=(0,R)\), W1 \(=(-R,0)\) and S1 \(=(0,-R)\).
The four end points of IR are E2 \(=(R/2,0)\), N2 \(=(0,R/2)\), W2 \(=(-R/2,0)\) and S2 \(=(0,-R/2)\).
Step 3: Find the length of the chord from E1 to N2.
E1 lies on the east axis and N2 lies on the north axis, so O, E1 and N2 form a right angled triangle with the right angle at O. The legs are \(OE_1 = R\) and \(ON_2 = R/2\).
By Pythagoras:
\[ E_1N_2 = \sqrt{R^2 + \frac{R^2}{4}} = \sqrt{\frac{5R^2}{4}} = \frac{\sqrt{5}}{2}R \]
The other three chords (N1 to W2, W1 to S2, S1 to E2) are each the hypotenuse of a triangle with legs \(R\) and \(R/2\), so each of them has the same length \(\frac{\sqrt{5}}{2}R\).
Step 4: Add the four chords and find the outer ring length.
Sum of all chord roads:
\[ 4 \times \frac{\sqrt{5}}{2}R = 2\sqrt{5}\,R \]
Length of the outer ring road:
\[ 2\pi R \]
Step 5: Form the ratio.
\[ \frac{2\sqrt{5}\,R}{2\pi R} = \frac{\sqrt{5}}{\pi} \]
The radius \(R\) cancels, so the ratio is \(\sqrt{5} : \pi\). This matches option 3.
Step 6: Why the other options fail.
Option 1, \(\sqrt{5} : 2\), has no \(\pi\) in it. It treats the outer ring as if its length were just \(2R\), which is the diameter and not the circumference.
Option 2, \(\sqrt{5} : 2\pi\), would need the outer ring to be \(4\pi R\) long, twice its real length.
Option 4 is wrong because option 3 is exactly the ratio we found.
Final Answer:
The chord roads add up to \(2\sqrt{5}R\) and the outer ring road is \(2\pi R\) long, so the ratio is \(\sqrt{5} : \pi\), which is option 3. \[ \boxed{3 \;(\sqrt{5} : \pi)} \]
Was this answer helpful?
0
0
Question: 2
Amit wants to reach N2 from S1. It would take him 90 minutes if he goes on minor arc S1 - E1 on OR, and then on the chord road E1 - N2. What is the radius of the outer ring road in kms?
Amit wants to reach N2 from S1. It would take him 90 minutes if he goes on minor arc S1 - E1 on OR, and then on the chord road E1 - N2. What is the radius of the outer ring road in kms?
Show Hint
The minor arc from S1 to E1 is a quarter of the outer circle. Write the time for the arc and the time for the chord in terms of the radius, then equate their sum to 1.5 hours.
Updated On: Nov 7, 2026
- 60
- 40
- 30
- 20
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Set up the road map.
Let the outer radius be \(R\) km, so the inner radius is \(R/2\) (the outer ring is twice as long as the inner ring). With the centre as origin, S1 \(=(0,-R)\), E1 \(=(R,0)\) and N2 \(=(0,R/2)\).
Amit travels two parts: the arc from S1 to E1 on OR, then the chord from E1 to N2. Time is distance divided by speed for each part.
Step 2: Length and time on the arc S1 to E1.
S1 is the south end and E1 the east end of OR, so the radii to them are at right angles. The minor arc between them is one quarter of the circle:
\[ \text{arc} = \frac{1}{4}\times 2\pi R = \frac{\pi R}{2} \]
Speed on OR is \(30\pi\) km/hr, so
\[ t_1 = \frac{\pi R/2}{30\pi} = \frac{R}{60}\ \text{hr} \]
Step 3: Length and time on the chord E1 to N2.
The chord is the hypotenuse of a right triangle with legs \(R\) and \(R/2\):
\[ E_1N_2 = \sqrt{R^2 + \frac{R^2}{4}} = \frac{\sqrt{5}}{2}R \]
Speed on a chord road is \(15\sqrt{5}\) km/hr, so
\[ t_2 = \frac{\sqrt{5}R/2}{15\sqrt{5}} = \frac{R}{30}\ \text{hr} \]
Step 4: Use the 90 minutes.
90 minutes is 1.5 hours. Adding the two times:
\[ \frac{R}{60} + \frac{R}{30} = \frac{R + 2R}{60} = \frac{R}{20} \]
\[ \frac{R}{20} = 1.5 \Rightarrow R = 30 \]
Step 5: Check the other options.
The total time equals \(R/20\) hours, so the time grows straight with the radius.
With \(R = 60\) the trip takes 3 hours, with \(R = 40\) it takes 2 hours and with \(R = 20\) it takes 1 hour. None of these is 90 minutes, so only \(R = 30\) works.
Final Answer:
The outer ring road has radius 30 km, which is option 3. \[ \boxed{3 \;(30\ \text{km})} \]
Let the outer radius be \(R\) km, so the inner radius is \(R/2\) (the outer ring is twice as long as the inner ring). With the centre as origin, S1 \(=(0,-R)\), E1 \(=(R,0)\) and N2 \(=(0,R/2)\).
Amit travels two parts: the arc from S1 to E1 on OR, then the chord from E1 to N2. Time is distance divided by speed for each part.
Step 2: Length and time on the arc S1 to E1.
S1 is the south end and E1 the east end of OR, so the radii to them are at right angles. The minor arc between them is one quarter of the circle:
\[ \text{arc} = \frac{1}{4}\times 2\pi R = \frac{\pi R}{2} \]
Speed on OR is \(30\pi\) km/hr, so
\[ t_1 = \frac{\pi R/2}{30\pi} = \frac{R}{60}\ \text{hr} \]
Step 3: Length and time on the chord E1 to N2.
The chord is the hypotenuse of a right triangle with legs \(R\) and \(R/2\):
\[ E_1N_2 = \sqrt{R^2 + \frac{R^2}{4}} = \frac{\sqrt{5}}{2}R \]
Speed on a chord road is \(15\sqrt{5}\) km/hr, so
\[ t_2 = \frac{\sqrt{5}R/2}{15\sqrt{5}} = \frac{R}{30}\ \text{hr} \]
Step 4: Use the 90 minutes.
90 minutes is 1.5 hours. Adding the two times:
\[ \frac{R}{60} + \frac{R}{30} = \frac{R + 2R}{60} = \frac{R}{20} \]
\[ \frac{R}{20} = 1.5 \Rightarrow R = 30 \]
Step 5: Check the other options.
The total time equals \(R/20\) hours, so the time grows straight with the radius.
With \(R = 60\) the trip takes 3 hours, with \(R = 40\) it takes 2 hours and with \(R = 20\) it takes 1 hour. None of these is 90 minutes, so only \(R = 30\) works.
Final Answer:
The outer ring road has radius 30 km, which is option 3. \[ \boxed{3 \;(30\ \text{km})} \]
Was this answer helpful?
0
0
Question: 3
Amit wants to reach E2 from N1 using first the chord N1 - W2 and then the inner ring road. What will be his travel time in minutes on the basis of information given in the above question?
Amit wants to reach E2 from N1 using first the chord N1 - W2 and then the inner ring road. What will be his travel time in minutes on the basis of information given in the above question?
Show Hint
Use the radius found in the previous question. W2 and E2 sit at opposite ends of a diameter of the inner ring, so the ring part is a semicircle.
Updated On: Nov 7, 2026
- 60
- 45.
- 90
- 105
Show Solution
Verified By Collegedunia
The Correct Option is D
Solution and Explanation
Step 1: Recall the radius from the earlier information.
The question says to use the information of the previous question. That question gave the outer radius as \(R = 30\) km. So the inner radius is \(R/2 = 15\) km.
Amit's trip has two parts: the chord N1 to W2, then the inner ring road from W2 to E2.
Step 2: Time on the chord N1 to W2.
N1 \(=(0,30)\) and W2 \(=(-15,0)\). With the centre as origin, the chord is the hypotenuse of a right triangle with legs 30 and 15:
\[ \sqrt{30^2 + 15^2} = \sqrt{1125} = 15\sqrt{5}\ \text{km} \]
Speed on a chord road is \(15\sqrt{5}\) km/hr, so the time is
\[ \frac{15\sqrt{5}}{15\sqrt{5}} = 1\ \text{hr} = 60\ \text{minutes} \]
Step 3: Time on the inner ring from W2 to E2.
W2 is the west end and E2 is the east end of IR, so they are opposite ends of a diameter. The road between them is half of the inner circle:
\[ \frac{1}{2}\times 2\pi \times 15 = 15\pi\ \text{km} \]
Speed on IR is \(20\pi\) km/hr, so
\[ \frac{15\pi}{20\pi} = 0.75\ \text{hr} = 45\ \text{minutes} \]
Either semicircle has the same length, so it does not matter which way he goes round.
Step 4: Add the two times.
\[ 60 + 45 = 105\ \text{minutes} \]
Step 5: Why the other options fail.
60 is the chord time alone and 45 is the ring time alone, so each leaves out one part of the trip.
90 is the time quoted in the earlier question for a different route, so it is not this trip.
Only 105 is the sum of both parts.
Final Answer:
The total travel time is 105 minutes, which is option 4. \[ \boxed{4 \;(105\ \text{minutes})} \]
The question says to use the information of the previous question. That question gave the outer radius as \(R = 30\) km. So the inner radius is \(R/2 = 15\) km.
Amit's trip has two parts: the chord N1 to W2, then the inner ring road from W2 to E2.
Step 2: Time on the chord N1 to W2.
N1 \(=(0,30)\) and W2 \(=(-15,0)\). With the centre as origin, the chord is the hypotenuse of a right triangle with legs 30 and 15:
\[ \sqrt{30^2 + 15^2} = \sqrt{1125} = 15\sqrt{5}\ \text{km} \]
Speed on a chord road is \(15\sqrt{5}\) km/hr, so the time is
\[ \frac{15\sqrt{5}}{15\sqrt{5}} = 1\ \text{hr} = 60\ \text{minutes} \]
Step 3: Time on the inner ring from W2 to E2.
W2 is the west end and E2 is the east end of IR, so they are opposite ends of a diameter. The road between them is half of the inner circle:
\[ \frac{1}{2}\times 2\pi \times 15 = 15\pi\ \text{km} \]
Speed on IR is \(20\pi\) km/hr, so
\[ \frac{15\pi}{20\pi} = 0.75\ \text{hr} = 45\ \text{minutes} \]
Either semicircle has the same length, so it does not matter which way he goes round.
Step 4: Add the two times.
\[ 60 + 45 = 105\ \text{minutes} \]
Step 5: Why the other options fail.
60 is the chord time alone and 45 is the ring time alone, so each leaves out one part of the trip.
90 is the time quoted in the earlier question for a different route, so it is not this trip.
Only 105 is the sum of both parts.
Final Answer:
The total travel time is 105 minutes, which is option 4. \[ \boxed{4 \;(105\ \text{minutes})} \]
Was this answer helpful?
0
0
Top CAT Questions
- Compare and contrast between Portia and Jessica characters in Merchant of Venice.
- In a triangle ABC,AB=AC=8cm. A circle drawn with BC as diameter passes through A. Another circle drawn with center at A passes through B and C. Then the area,in sq.cm,of the overlapping region between the two circles is
- The simple interest accrued on an amount of Rs.25,000 at the end of three years is Rs.7,500. What would be the compound interest accrued on the same amount at the same rate in the same period?
- What is the sum of all roots of the equation \(|x+4|^2–10|x+4|=24\)?
- A solid metal sphere is melted and smaller spheres of equal radii are formed. 10% of the volume of the sphere is lost in the process. The smaller spheres have a radius which is \(\frac{1}{9}\text{th}\) the larger sphere. If 10 litres of paint were needed to paint the larger sphere,how many litres are needed to paint all the smaller spheres?
View More Questions