The question is from the topic time and distance. Time taken while increasing and decreasing speed is given, we need to find out the distance. Time Speed and Distance is a favorite in CAT Exam, and appears more often than expected in the CAT Quantitative Aptitude section in the CAT Exam
Question 22 : Akash when going slower by 15 Km/hr, reaches late by 45 hours. If he goes faster by 10 Km/hr from his original speed, he reaches early by 20 hours than the original time. Find the distance he covers.
You may solve for D and S by writing the equations as:
\\frac{D}{s−15}\\) − \\frac{D}{s}\\) = 45
\\frac{D}{s}\\) - \\frac{D}{s+10}\\) = 20
You can solve these two equations to get D = 9750 Km which is a perfectly fine approach. However, if you are comfortable with alligation, you may save time on such TSD question as:
Both these equations can be solved quickly to get s = 65 Km/hr, t = 150 hr. Therefore, D = 9750 Km. I want to reiterate that using the first approach is perfectly fine. That's how I would do it. It is better to spend 20 seconds extra and get the right answer instead of getting it wrong with something you are not comfortable with.
Alternate Solution:
Let the normal speed be s
and the normal time be t
The distance D = st.
When speed decreases by 15 kmph, time increases by 45 hours. The distance will still be the same.
D = (s-15)(t+45)
st = (s-15)(t+45)
st = st +45s - 15t - 15*45
45s - 15t - 15*45 = 0
3s -t -45 = 0
3s = t + 45
t = 3s - 45
When speed increases by 10 kmph, time decreases by 20 hours. The distance will still be the same.
D = (s+10)(t-20)
st = (s+10)(t-20)
st = st -20s + 10t - 10*20
-20s + 10t - 10*20 = 0
-2s +t -20 = 0
t = 20 + 2s
So, t = 3s - 45 and t = 20 + 2s
3s - 45 = 20 + 2s
s = 65
So speed = s = 65 kmph
t = 20 + 2s
time = t = 20 + 2*65 = 150
Distance = st = 65 * 150 = 9750 km
The question is "Find the distance he covers."
Choice B is the correct answer.
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