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

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Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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Updated on: 11 Oct 2015, 04:08
4
00:00

Difficulty:

65% (hard)

Question Stats:

57% (01:56) correct 43% (02:08) wrong based on 128 sessions

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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.

Originally posted by excelingmat on 10 Oct 2015, 06:37.
Last edited by Bunuel on 11 Oct 2015, 04:08, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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10 Oct 2015, 14:21
1
excelingmat wrote:
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?

I. The distance from Madras to Trivandrum is 0.30 times the distance from Calcutta to Madras.
II. The average speed from Madras to Trivandrum was twice that of the average speed from Calcutta to Madras.

Let D1 = Distance from Calcutta to Madras, T1= Time taken to travel from Calcutta to Madras, R1= Speed at which car traveled during Calcutta to Madras
Let D2 = Distance from Madras to Trivandrum, T2 = Time taken to travel from Madras to Trivandrum, R2 = Speed at which car traveled during Madras to Trivandrum

Question stem is asking for, What is R2 = $$\frac{D2}{T2}$$ ?

The absolute differences between each individual rate with its average rate are in a ratio. The weights are the individual times taken for each segment of the trip.

Assuming R2 > R1, the allegation method yields:
$$(40-R1)*T1 = (R2-40) * T2$$ --> $$\frac{T1}{T2}=\frac{(R2-40)}{(40-R1)}$$ or $$\frac{(D1R2)}{(D2R1)}=\frac{(R2-40)}{(40-R1)}$$

(Statement 1): Substituting D2=.3D1 into $$\frac{(D1R2)}{(D2R1)}=\frac{(R2-40)}{(40-R1)}$$ yields $$\frac{(10*R2)}{(3*R1)}=\frac{(R2-40)}{(40-R1)}$$. An unique R2 can't be found from this equation. This statement alone is insufficient.

(Statement 2): Substituting R2= 2R1 into $$\frac{T1}{T2}=\frac{(R2-40)}{(40-R1)}$$ yields $$\frac{T1}{T2}=\frac{(2*R1-40)}{(40-R1)}$$ Since no information is given about ratio of T1 to T2, this statement alone is insufficient.

(Combined): After substituting R2= 2R1 into $$\frac{(10*R2)}{(3*R1)}=\frac{(R2-40)}{(40-R1)}$$, we now have $$\frac{20}{3}=\frac{(2*R1-40)}{(40-R1)}$$

R1=$$\frac{920}{26}$$ satisfies the equation, therefore R2= $$\frac{920}{13}$$

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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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17 Dec 2015, 11:20
2
Travel from A to B to C.

Average speed from B to C = ?

Statement 1 and Statement 2 give 1. The ratio of distance. 2. The ratio of speeds. Both are needed to find the asked.

Hence, C.
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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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19 Dec 2015, 10:07
thanks that helps with the answer
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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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21 Dec 2015, 20:44
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.
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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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21 Dec 2015, 21:34
3
excelingmat wrote:
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.

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.

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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any  [#permalink]

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21 Aug 2018, 17:29
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Re: Ramu went by car from Calcutta to Trivandrum via Madras, without any &nbs [#permalink] 21 Aug 2018, 17:29
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