Data Interpretation Capsules - Shashank Prabhu - CAT 100 Percentiler - Set 3

• Question 1

Fifty inabitants of the shire was surveyed to note down their posessions of Arkenstones, Elfstones and rings of power. Of then, 22 own an Arkenstone, 15 own an Elfstone and 14 own a ring of power. Nine of these inhabitants own exactly two items out of an Arkenstone, an Elfstone and a ring of power; and, one inhabitant owns all three. How many of the fifty inhabitants own none of three: Arkenstone, Elfstone or a ring of power?

Solution

Total number of objects will be given by a+2b+3c where a is the number of people who possess only one object, b is the number of people who possess exactly two objects and c is the number of people who possess exactly three objects. So, in this case, total number of objects is 22+15+14=51. Number of people who have exactly 2 objects is 9 and so, 2b=18. Similarly, number of people who have exactly 3 objects is 1 and so, 3c=3. So, a+18+3=51 and so, a=30. So, a+b+c=30+9+1=40 and so, 10 do not have any objects

Question 2

A, B, C, D, E are five students who took CAT 2015. Following are the sums of their overall scores, taken three at a time: 119, 121, 124, 125, 123, 126, 127, 128, 129 and 132. What is the highest and lowest score among the scores of A, B, C, D, E?

Solution

Let a > b > c > d > e
a + b + c = 132......(1)
c + d + e = 119......(2)
a + b + d = 129......(3)
b + d + e = 121......(4)
We know the values of all possible triplets. So, 5c3=10 cases in total. So, each element occurs 6 times.
6a + 6b + 6c + 6d + 6e = 1254
a + b + c + d + e = 209.....(5)
From (1), (2) and (5)
c = 42
From (2), (3) and (5)
d = 39... putting these values in (2)
e = 38
b = 44
a = 46

Question 3

An institute conducts 32 tests. The number of tests attempted by three students studying in the class is as follows:
Neil - 16
Nitin - 18
Mukesh - 20

The number of tests written by more than one student is at least:

a) 8
b) 11
c) 13
d) 16
e) 15

Solution

Let a be the number of students who have taken exactly 1 test, b be the number of students who have taken exactly 2 tests, c be the number of students who have taken exactly 3 tests. Total number tests will be equal to a + 2b + 3c = 16 + 18 + 20 = 54
Also, 32 tests have been conducted in total. So a+b+c=32
From the two equations
b + 2c = 22
To minimize overlap, we need to maximize c.. so c=11.

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