LR - Operator Based Questions - Previous Year CAT/MBA Questions

You can practice all previous year CAT questions from the topic LR - Operator Based Questions. This will help you understand the type of questions asked in CAT. It would be best if you clear your concepts before you practice previous year CAT questions.

1. CAT 2000 LRDI | LR - Operator Based Questions

Choose 1; if the question can be answered by using one of the statements alone, but cannot be answered using the other statement alone.
Choose 2; if the question can be answered by using either statement alone.
Choose 3; if the question can be answered by using both statements together, but cannot be answered using either statement alone.
Choose 4; if the question cannot be answered even by using both statements together.

For any two real numbers

a ⊕ b = 1 if both a and b are positive or both a and b are negative.
= –1 if one of the two numbers a and b is positive and the other negative.

What is (2 ⊕ 0) ⊕ (–5 ⊕ –6)?

a ⊕ b is zero if a is zero.
a ⊕ b = b ⊕ a

A.
1
B.
2
C.
3
D.
4

Direction: Answer the question based on the following information.

The following operations are defined for real numbers.
a # b = a + b, if a and b both are positive else a # b = 1
a ∇ b = (a × b)^{a + b} if a × b is positive else a ∇ b = 1.

2. CAT 1998 LRDI | LR - Operator Based Questions

$\frac{(2 \neq 1)}{(1 \nabla 2)} =$

A.
1/8
B.
1
C.
3/8
D.
3

Answer: Option C

Explanation :

Since both 2 and 1 are positive, (2 # 1) = 2 + 1 = 3.
(1∇2)= (1× 2)¹⁺² = 23 = 8.
Thus, the given expression is equal to $\frac{3}{8} .$

3. CAT 1998 LRDI | LR - Operator Based Questions

$\frac{\{((1 # 1) # 2) - (10^{1.3} \nabla \log_{10} 0.1)\}}{(1 \nabla 2)} =$

A.
3/8
B.
$\frac{4 \times \log_{10} 0.1}{8}$
C.
$\frac{(4 + 10^{13})}{8}$
D.
None of these

Answer: Option A

Explanation :

Let us first simplify the numerator. Since 1 is positive,
(1 # 1) is 1 + 1 = 2 which again is positive. Then
(1 # 1) # 2 = 2 # 2 = 2 + 2 = 4
Now note that log₁₀ 0.1
= log₁₀ 10^–1 = –1
Then 10^1.3log₁₀ 0.1= 10^1.3× (–1) is negative.
So 10^1.3 ∇ log₁₀ 0.1 = 1
Hence, the numerator is equal to 4 –1 = 3
Since 1 × 2 = 2 is positive, (1∇2) = (1× 2)¹⁺² = 2³ = 8.
So the denominator = 8. Hence, the answer is $\frac{3}{8} .$

4. CAT 1998 LRDI | LR - Operator Based Questions

$(\frac{(X # - Y)}{(- X \nabla Y)}) = \frac{3}{8},$ then which of the following must be true?

A.
X = 2, Y = 1
B.
X > 0, Y < 0
C.
X, Y both positive
D.
X, Y both negative

5. CAT 1991 LRDI | LR - Operator Based Questions

If 8 + 12 = 2, 7 + 14 = 3 then 10 + 18 = ?

A.
10
B.
4
C.
6
D.
18

LR - Operator Based Questions - Previous Year CAT/MBA Questions

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