CAT 2018 Question Paper | CAT LR DI

CAT Questions

This is the easiest question of this entire CAT 2018 DILR set. You are required to find the true statement with the help of the given conditions. Try this beautiful 2018 CAT previous year paper question and crush your CAT Exam Preparation.

CAT DILR : CAT 2018 Question Paper Slot 1

Set 6 : Petrol Pumps

Fuel contamination levels at each of 20 petrol pumps P1, P2, …, P20 were recorded as either high, medium, or low.

1. Contamination levels at three pumps among P1 – P5 were recorded as high.
2. P6 was the only pump among P1 – P10 where the contamination level was recorded as low.
3. P7 and P8 were the only two consecutively numbered pumps where the same levels of contamination were recorded.
4. High contamination levels were not recorded at any of the pumps P16 – P20.
5. The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.

Question 3 : If the contamination level at P11 was recorded as low, then which of the following MUST be true?

1. The contamination level at P18 was recorded as low.
2. The contamination level at P15 was recorded as medium.
3. The contamination level at P14 was recorded as medium.
4. The contamination level at P12 was recorded as high.

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Video Explanation

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Explanatory Answer

Method of solving this CAT Question on CAT DILR

We know that three of P1 to P5 are recorded as High. This means the other two should be M. So, let us fill this data in.

P7 and P8 are both same, so both could be H or both M. Let us try both. We put P7 and P8 in, P9 and P10 get filled automatically.
Let us fill in the totals as well.

The number of pumps where high contamination levels were recorded was twice the number of pumps where low contamination levels were recorded.
So, of the number of low contaminations were x, then the number of high contaminations would be 2x.
H + L = 3x {3, 6, 9, 12, 15, or 18} or, M should be from {2, 5, 8, 11, 14, 17}
L + H is at least 5 so M cannot be 14 or 17. M cannot be 2 either. So, we can have M to be 5, 8 or 11.
So, we can have H, M, L to be {10, 5, 5} or {8, 8, 4} or {6, 11, 3}
There is no H in the P16 to P20 range, so total H cannot be 10. So, H, M, L being {10, 5, 5} can be ruled out.
So, we can have H, M, L to be {8, 8, 4} or {6, 11, 3}.
We cannot have more than 5 M’s in the range from P16 to P20.
So, we now know that {6, 11, 3} is also ruled out. So H, M, L has to be {8, 8, 4}.

We know that in the first 10 pipes we could have has H, M, L could have been {6, 3, 1} or {4, 5, 1}
So, H, M, L for the last 10 should be {2, 5, 3} or {4, 3, 3}. Since there is no H in the first 5 pipes, we cannot have 4 H in the P16 to P20 range.
So, we need to have only {6, 3, 1} in the first range.

The last 10 pipes should have H, M, L to be {2, 5, 3}.
High contamination levels were not recorded at any of the pumps P16 – P20. So, the last 5 should be either MLMLM or LMLML.
If the last 5 were LMLML, then we would not have one more L for P11 to P15. So, P11 to P15 should be MHMHM. Now, let us fine-tune this.

We know that P11 cannot be H otherwise P10 and P11 would be H.
So, we need to accommodate 2 H’s in the P12 to P15 range.
We could have them at P12 and P14, P12 and P15 and P13 and P15.

If P15 were not H, it has to be L. So, we can fill the first column completely.

P11 was L, then two possibilities get ruled out. P11 should have been L.
The remaining 2 should have been M’s. But even in this, one possibility has M in P13 and P14 – this is not possible.
So, we have only one combination.

Choice A , B and D are not true.
Choice C is true.

The question is "If the contamination level at P11 was recorded as low, then which of the following MUST be true?"

Hence, the answer is The contamination level at P14 was recorded as medium.

Choice C is the correct answer.