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Quantitative Aptitude Question on Time Speed and Distance

Two points A and B are 45 km apart. Anil started from A. Sunil started from B. They met each other after 1 hour 30 mins. and after meeting, they continued towards their destinations. Time taken by Anil to reach Point B was 1 hr 15 mins more than the time taken by Sunil to reach point A. Find speed of Anil

A

12

B

16

C

18

D

24

Answer

12

Explanation

Solution

Step 1: Define Variables
Let the speed of Anil be vAv_A km/h.
Let the speed of Sunil be vSv_S km/h.
They meet after 1 hour 30 minutes (1.51.5 hours). At this point, the total distance they covered is:
45 km.45 \,km.
Step 2: Relate the Distances Traveled
In the time before they met:
Distance covered by Anil=vA×1.5,\text{Distance covered by Anil} = v_A \times 1.5,
Distance covered by Sunil=vS×1.5.\text{Distance covered by Sunil} = v_S \times 1.5.
Since they meet after covering the total distance:
vA×1.5+vS×1.5=45.v_A \times 1.5 + v_S \times 1.5 = 45.
Simplify:
1.5(vA+vS)=45,1.5(v_A + v_S) = 45,
vA+vS=30.(Equation 1)v_A + v_S = 30. \quad \text{(Equation 1)}
Step 3: Relation Between Times After Meeting
After meeting, Anil takes 1 hour 15 minutes (1.251.25 hours) more than Sunil to reach their destinations:
Time taken by Anil after meeting=Time taken by Sunil after meeting+1.25.\text{Time taken by Anil after meeting} = \text{Time taken by Sunil after meeting} + 1.25.
The distances remaining after meeting are:
For Anil: Distance remaining=45−1.5vA,\text{For Anil: Distance remaining} = 45 - 1.5v_A,
For Sunil: Distance remaining=45−1.5vS.\text{For Sunil: Distance remaining} = 45 - 1.5v_S.
Using the formula time=distancespeed\text{time} = \frac{\text{distance}}{\text{speed}}, the equation becomes:
45−1.5vAvA=45−1.5vSvS+1.25.(Equation 2)\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5v_S}{v_S} + 1.25. \quad \text{(Equation 2)}
Step 4: Solve the Equations
From Equation 1:
vS=30−vA.v_S = 30 - v_A.
Substitute vS=30−vAv_S = 30 - v_A into Equation 2:
45−1.5vAvA=45−1.5(30−vA)30−vA+1.25.\frac{45 - 1.5v_A}{v_A} = \frac{45 - 1.5(30 - v_A)}{30 - v_A} + 1.25.
Simplify 1.5(30−vA)1.5(30 - v_A):
1.5(30−vA)=45−1.5vA.1.5(30 - v_A) = 45 - 1.5v_A.
Thus, the equation becomes:
45−1.5vAvA=1.5vA30−vA+1.25.\frac{45 - 1.5v_A}{v_A} = \frac{1.5v_A}{30 - v_A} + 1.25.
Multiply through by vA(30−vA)v_A(30 - v_A) to eliminate the denominators:
(45−1.5vA)(30−vA)=1.5vA2+1.25vA(30−vA).(45 - 1.5v_A)(30 - v_A) = 1.5v_A^2 + 1.25v_A(30 - v_A).
Expand both sides:
1350−45vA−1.5vA⋅30+1.5vA2=1.5vA2+37.5vA−1.25vA2.1350 - 45v_A - 1.5v_A \cdot 30 + 1.5v_A^2 = 1.5v_A^2 + 37.5v_A - 1.25v_A^2.
Simplify:
1350−90vA+1.5vA2=0.25vA2+37.5vA.1350 - 90v_A + 1.5v_A^2 = 0.25v_A^2 + 37.5v_A.
Combine terms:
1350+1.25vA2−127.5vA=0.1350 + 1.25v_A^2 - 127.5v_A = 0.
Divide through by 1.25:
vA2−102vA+1080=0.v_A^2 - 102v_A + 1080 = 0.
Solve the quadratic equation using the quadratic formula:
vA=−b±b2−4ac2a,v_A = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},
where a=1a = 1, b=−102b = -102, and c=1080c = 1080.
vA=−(−102)±(−102)2−4(1)(1080)2(1).v_A = \frac{-(-102) \pm \sqrt{(-102)^2 - 4(1)(1080)}}{2(1)}.
Simplify:
vA=102±10404−43202,v_A = \frac{102 \pm \sqrt{10404 - 4320}}{2},
vA=102±60842,v_A = \frac{102 \pm \sqrt{6084}}{2},
vA=102±782.v_A = \frac{102 \pm 78}{2}.
Two solutions:
vA=102+782=90,vA=102−782=12.v_A = \frac{102 + 78}{2} = 90, \quad v_A = \frac{102 - 78}{2} = 12.
Since vA=90 km/hv_A = 90 \, \text{km/h} is unrealistic for the scenario, we take:
vA=12 km/h.v_A = 12 \, \text{km/h}.
Final Answer
The speed of Anil is:
12 km/h.\boxed{12 \, \text{km/h}}.