Ramu went by car from Calcutta to Trivandrum via Madras, without any

thanks that helps with the answer

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


Ramu went by car from Calcutta to Trivandrum via Madras, without any stoppages. The average speed for the entire journey was 40 kmph. What was the average speed from Madras to Trivandrum?

(1) The distance from Madras to Trivandrum is 0.30 times the distance from Calcutta to Madras.
(2) The average speed from Madras to Trivandrum was twice that of the average speed from Calcutta to Madras.

In the original condition, you travel from Calcutta through Madra to Trivandrum. From Calcutta to Madras, speed is v1 and journey is d1. From Madras to Trivandrum, speed is v2 and journey is d2. So, there are 4 variables(v1,v2,d1,d2) and 1 equation(the average speed for the entire journey is 40kmph), which should match with the number of equations. So, you need 3 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), suppose d1=10, and then d2=3. From d2/t2=2(d1/t1) and 3/t2=2(10/t1), 20t2=3t1 is calculated. However, as the average speed for the entire journey is 40, (10+3)/(t1+t2)=40 -> 13=40(t1+t2)=40t1+40t2. When you substitute 20t2=3t1, 13=40t1+2(3t1)=46t1 is calculated. So, you can get values for t1 and t2. Therefore, you can get values in a unique way from v2=d2/t2=3/t2 as well, which is sufficient. So the answer is C.
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.


Ramu went by car from Calcutta to Trivandrum via Madras, without any stoppages. The average speed for the entire journey was 40 kmph. What was the average speed from Madras to Trivandrum?

(1) The distance from Madras to Trivandrum is 0.30 times the distance from Calcutta to Madras.
(2) The average speed from Madras to Trivandrum was twice that of the average speed from Calcutta to Madras.

In the original condition, you travel from Calcutta through Madra to Trivandrum. From Calcutta to Madras, speed is v1 and journey is d1. From Madras to Trivandrum, speed is v2 and journey is d2. So, there are 4 variables(v1,v2,d1,d2) and 1 equation(the average speed for the entire journey is 40kmph), which should match with the number of equations. So, you need 3 more equations. For 1) 1 equation, for 2) 1 equation, which is likely to make E the answer. In 1) & 2), suppose d1=10, and then d2=3. From d2/t2=2(d1/t1) and 3/t2=2(10/t1), 20t2=3t1 is calculated. However, as the average speed for the entire journey is 40, (10+3)/(t1+t2)=40 -> 13=40(t1+t2)=40t1+40t2. When you substitute 20t2=3t1, 13=40t1+2(3t1)=46t1 is calculated. So, you can get values for t1 and t2. Therefore, you can get values in a unique way from v2=d2/t2=3/t2 as well, which is sufficient. So the answer is C.


-> For cases where we need 3 more equations, such as original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 80% chance that E is the answer (especially about 90% of 2 by 2 questions where there are more than 3 variables), while C has 15% chance. These two are the majority. In case of common mistake type 3,4, the answer may be from A, B or D but there is only 5% chance. Since E is most likely to be the answer using 1) and 2) separately according to DS definition (It saves us time). Obviously there may be cases where the answer is A, B, C or D.

Try to minimise the number of variables.
We have two legs of the journey C to M and M to T. Average speed is 40 over the two legs.
We need avg speed from M to T.

(1) The distance from Madras to Trivandrum is 0.30 times the distance from Calcutta to Madras.
The relative distance is not enough because they could have equal speeds over the two legs or higher speed on one and lower on the other. You cannot find the speed on M to T.

(2) The average speed from Madras to Trivandrum was twice that of the average speed from Calcutta to Madras.
Now, you don't know the distance/time for which the higher speed was maintained. So again, this statement alone is not sufficient.

Using both, say speed from C to M was s. So speed from M to T must be 2s. Say, distance from C to M is 10d. Then distance from M to T is 3d.

\(40 = \frac{Total Distance}{Total Time} = \frac{10d + 3d}{\frac{10d}{s} + \frac{3d}{2s}}\)
Note that here d will get cancelled and you will easily be able to solve for s. Hence you will get the speed over M to T.

Answer (C)

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