CAT 2018 Question Paper | CAT LR DI

CAT Previous Year Paper | CAT DILR | Set 5

CAT DI LR section has become increasingly tough beginning from 2015. DILR used to have distinct Data Interpretation sets and Logical reasoning puzzles. It used to be about computation and ability to read charts, graphs and tables for the Data Interpretation and Logical reasoning used to have Family tree, grid puzzles, arrangement, tournaments, cubes as some standard forms of puzzles.Since 2015 this pattern has been broken. With passing years, even the distinction between DI and LR has come down significantly. All you get in that one hour, are 8 high quality puzzles, with more than a few of them being significantly tough. Between CAT 2017 Question paper and CAT 2018 Question paper, you get to solve 32 actual CAT puzzles. This page intends to provide you just that. So, head on and crack those puzzles!

CAT DILR : CAT 2018 Question Paper Slot 2

Set 5 : Amusement Park Tickets

Each visitor to an amusement park needs to buy a ticket. Tickets can be Platinum, Gold, or Economy. Visitors are classified as Old, Middle-aged, or Young. The following facts are known about visitors and ticket sales on a particular day:
1. 140 tickets were sold.
2. The number of Middle-aged visitors was twice the number of Old visitors, while the number of Young visitors was twice the number of Middle-aged visitors.
3. Young visitors bought 38 of the 55 Economy tickets that were sold, and they bought half the total number of Platinum tickets that were sold.
4. The number of Gold tickets bought by Old visitors was equal to the number of Economy tickets bought by Old visitors.

Question 2 : If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number Middle-aged visitors buying Gold tickets was [TITA]

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Method of solving this CAT Question on CAT DILR

From second constraint,
OLD Visitors = x, MID = 2x, Young = 4x.
7x = 140 or, x = 20 , 2x = 40 , 4x = 80.
Let us fill this in. and then go to constraint 3 and 4.

From constraint 3, Total Platinum = 2y, then YNG platinum should be y.
From constraint 4, Let us fill OLD-GOLD and OLD-ECO has z.

OLD-GOLD > YNG-GOLD.
Z > 42 – y Or, y + z > 42
We need to find MID-GOLD. This is equal to 85 – 2y – z – (42 – y) = 43 – (y + z).
We know that y + z > 42. So, MID-GOLD has to be 0.

The question is "If the number of Old visitors buying Gold tickets was strictly greater than the number of Young visitors buying Gold tickets, then the number Middle-aged visitors buying Gold tickets was [TITA]"

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