# Logical Reasoning Capsules - Edwin Jose, CAT DILR 100 Percentiler - Set 3

• DIRECTIONS : Answer the questions on the basis of the information given below.

A, B, C and D are four ladies who are friends of Elizabeth. On one Saturday the four of them visited Elizabeth at her weekend getaway.
I. The time of each visit was as follows: A at 8 O’clock, B at 9 O’clock, C at 10 O’clock and D at 11 O clock.
II. At least one woman visited Elizabeth between A and B.
III. At least one of C or D visited Elizabeth before A.
IV. C did not visit Elizabeth between B and D.

(1) Who visited Elizabeth first?
(a) A
(b) B
(c) C
(d) D

(2) Who visited Elizabeth last?
(a) A
(b) B
(c) C
(d) Insufficient data

Solutions:

We are given that A visited at 8 O’clock. Now from III we conclude that A visited at 8 p.m. Now from I we concluded that B has to visit at 9 a.m. otherwise nobody will be able to visit in between A & B. Now if D were to visit at 11 p.m. then condition IV will get violated hence we concluded that D visited at 11 a.m. and C visited at 10 p.m. From here all the questions are answered.
(1) b
(2) c

DIRECTIONS : Answer the questions on the basis of the information given below.

Each of the questions is followed by two statements. You have to decide whether the information provided in the statements is sufficient for answering the question.
Mark (a), If the question can be answered by using one of the statements alone, but cannot be answered by using the other statement alone.
Mark (b), If the question can be answered by using either statement alone.
Mark (c), If the question can be answered by using both the statements together, but cannot be answered using either statement alone.
Mark (d), If the question cannot be answered even by using both the statements together.

(1) Is one of X or Y a fraction?
If the question cannot be answered even by using both the statements together.
A. X^2Y^3 = 8
B. X is rational

(2) A is a prime number. Is B a prime number?
A. B = 9A + 7
B. AB is even

(3) Is 3x + 10y even?
A. x is even
B. y is odd

(4) Which amongst the three x, y and z (all real) is the greatest?
A. x:y:z::6:8:11
B. xyz–y^2 is positive

Solutions:

(1) C
As X^2Y^3 = 8, either X or Y or both are fractional. From statement B, we get that X is rational. Therefore, X^2 is either and integer or a fraction. Therefore, X^2 = 8 will either be an integer or γ^3 a fraction, So either X or Y is a fraction.
(2) A
Statement A alone is sufficient to say that B is not prime. However, statement B alone is not sufficient.
(3) C
From statement A we can conclude that 3x is even. As y is an integer (by virtue of it being odd) from statement B, we can conclude that 10y is even. The sum of two even numbers will be even.
(4) D
From statement A we cannot conclusively state which of the three is largest as they could all be positive or negative. From statement B we know that xyz is positive as xyz – y^2 (as y^2 will always be positive for real values of y) is positive. This is possible in two cases,
I. All of x, y and z are positive in this case z is the greatest
II. Y is negative and one of x and z is negative.
Hence we cannot conclude even if we combine the two statements.

DIRECTIONS : Answer the questions on the basis of the information given below.

Each question contains two statements in the question followed by four statements. Choose the alternative from the four statements which is logically related and follows the statements in the question.
(1) No M are P. Some M are S
(a) Some S are P .
(b) Some P are S
(c) No S are M
(d) Some S are not P

(2) All M are P. All S are M
(a) All P are S
(b) All M are S
(c) All S are P
(d) None of the above

(3) Some M are not P. All M are S
(a) No S are P
(b) All S are P
(c) Some S are not P
(d) No S are P

(4) No P are M. Some S are not M.
(a) No S are P
(b) All Pare S
(c) Some M could be S
(d) All S are P

(5) All Critical Thinking classes are full. John found a class that wasn’t full
(a)John might have found a Critical Thinking class
(b) The class John found was not a Critical Thinking class
(d) John was a dumb guy

Solutions:

(1) d Those S which are M are not P. Hence some S are not P.
(2) c
(3) c Those ‘M’ are not ‘P’ are ‘S’ because all ‘M’ are ‘S’
(4) c Important link here is ‘could’.
(5) b

1

1

4

1

1

2

1

1