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.

- 8750 Km
- 9750 Km
- 1000 Km
- 3750 Km

9750 km

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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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