# Data Interpretation Capsules - Shashank Prabhu - CAT 100 Percentiler - Set 2

• Question 1

Answer the questions on the basis of the information given below:
The annual sugarcane production (in million tonnes) in Meethagaon for the period 2000-2006 is shown in the bar graph given below.

What is the approximate average annual sugarcane production (in million tonnes) in Meethagaon for the period 2000-2005?
(a) 281.4 (b) 326.5 (c) 272.1 (d) 328.3

The sugarcane production in Meethagaon in the year 2007 increases by 15% over the year 2006. What is the approximate compounded annual growth rate of sugarcane production in Meethagaon over the period 2004-2007?
(a) 19% (b) 17% (c) 16% (d) 18%

Out of the following, which year has shown the highest percentage increase in sugarcane production in Meethagaon compared to the previous year?
(a) 2001 (b) 2004 (c) 2005 (d) 2006

Solution

(1) is very easy.. probably the shortest way is to take some base value and study the deviation around it. Take 280 for example. +15+17+7-47-43+1=-50 divided by 6 will be -8.33. So approximately 272.

(2) 2007 produce will be 337+33.7+16.8=387.5 approx.
CAGR over 3 years will be cube root of (387.5/237.09) which will be 1.178 or 17.8% which will be 18%. Heavy approximations plus calculation is required for this. Unless you have a calculator, doesnt make sense attempting this question.

(3) Compare 1.3/296, 1.25/234, 44/237 and 56/281. The doubt should be between c and d, but d is almost 20% while c is short of 20% by a fair bit. So, d is correct.

Question 2

Alok, Bharti, Chaman, Dinu, Ekant and Faisal are the only people available for selection in a team. The team should have at least two people subject to the following conditions.
• If Bharti is selected then Dinu should also be selected while Ekant should be rejected.
• If Alok is selected then exactly one from Bharti and Chaman should also be selected.
• If Chaman is selected then Ekant should also be selected while Faisal should be rejected.
• If the size of the team is less than 4 then Dinu and Faisal cannot be selected together.

(1) If Bharti and Faisal are not selected in the team then who should definitely be selected?
(a) Chaman
(b) Ekant
(c) Dinu
(d) Alok

(2) Which of the following pairs of two people cannot be selected along with any one else out of the remaining 4 people?
(a) Bharti and Dinu
(b) Chaman and Ekant
(c) Dinu and Ekant
(d) Ekant and Faisal

(3) If the team selected is of the maximum possible size, then who is/are definitely selected in the team?
(a) Alok
(b) Dinu
(c) Bharti
(d) Both Alok and Dinu

Solution

Teams with 2 members:
A cannot be in any such team as otherwise one of B and C must be selected and then subsequently one of D and E must be selected as well (hence exceeding 2 member size).
Possible teams are:
B, D
C, E
D, E
E, F

Teams with 3 members:
A, C, E
A, B, D
C, E, D

Teams with 4 members:
A, B, D, F
A, C, E, D

Note: No other team size is possible.

(1) b Possible teams without B and F are:
C, E
D, E
A, C, E
C, E, D
A, C, E, D

We can see that Ekant (E) should definitely be selected.

(2) d Ekant (E) and Faisal (F) (team number 4) appear together only once and their pair is not selected with any one else out of the remaining 4 people.

(3) d Team selected must be either
A, B, D, F
A, C, E, D
We can see that Alok (A) and Dinu (D) both are definitely selected in the team.

Question 3

Jack, Davy and Hector stole a few pieces of Aztec Gold from the flying Dutchman. As it was late in evening and as they were fairly intoxicated, they decided to divide the loot equally among themselves in the morning and went to sleep. During the night however, Hector tip toed to the chest, counted the pieces and took one. From the remaining pieces, he took a precise third and went back to sleep. After sometime Davy woke up. He counted the pieces, took one and then took exactly third of the remaining pieces and went to sleep. A little before sunrise Jack woke up, took one piece and like the other two pirates, took a precise third of the remainder. In the morning, all three sat together to divide the loot among themselves. Among them they found one piece, which was not real Aztec gold and so, they threw it away. From the remaining pieces they made an exact division. What could be the minimum number of pieces of Aztec gold (real or otherwise) that were initially stolen?

Solution:

Let total be 3x+1, after first person takes it, 2x left, after second person takes it 2(2x-1)/3 left, after third person takes it, 2/3[2(2x-1)/3-1] left ie (8x-10)/9 left. After one is removed, (8x-10)/9-1 left. This is in the form of 3k and so, 8x-19=27k. x=(27k+19)/8. As 19 gives remainder 3, 27k should give remainder 5 when divided by 8. The sequence of remainders are 3, 6, 1, 4, 7, 2, 5 (easier to do if you can notice the difference of 3 and treat it as a cycle from 1 to 8 and back to 1). So, k=7, x=26 and initial amount is 79 at least.

• Can somebody explain me solution / better solution for question no. 3?

• @Rohit-Rathore
Q3) Let say we had x coins in the beginning
After 1st loot : x - 1 - (x-1)/3 = 2/3(x-1) = (2x-2)/3
After 2nd loot : (2x - 2)/3 - 1 - [(2x - 2)/3 - 1]/3 = 2/3[(2x-2)/3 - 1] = (4x - 10)/9
After 3rd loot : (4x - 10)/9 - 1 - [((4x - 10)/9 - 1]/3 = 2/3 [(4x - 10)/9 - 1] = (8x - 38)/27
Now the final value (after throwing away the fake coin) = (8x - 38)/27 - 1 = (8x - 65)/27 is a multiple of 3.
So we can write, (8x - 65)/27 = 3k
x = (81k + 65)/8
x is an integer, 81K + 65 should be a multiple of 8, means (81K + 65) Mod 8 = 0
We know 81 Mod 8 = 1 and 65 mod 8 = 1. So to get remainder 0, we need to adjust K in a way that 81K mod 8 = 7.
Least value would be K = 7
For K = 7, x = (81 * 7 + 65)/8 = 79, which should (could) be our answer.
Let me know in case of any Logical/Calculation errors.

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