# 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.

**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

Answer: Option C

**Explanation** :

(2 ⊕ 0) ⊕ (–5 ⊕ –6)

= (2 ⊕ 0) ⊕ 1

From statement A

(0 ⊕ 2) = 0 but we don’t know the value of (2 ⊕ 0).

∴ Statement A alone is not sufficient to answer the question.

From statement B

(2 ⊕ 0) = (0 ⊕ 2)

∴ Statement B alone is not sufficient to answer the question.

After combining both the statements, we get,

(2 ⊕ 0) = (0 ⊕ 2) = 0

∴ (2 ⊕ 0) ⊕ (–5 ⊕ –6)

= (2 ⊕ 0) ⊕ 1

= 0 ⊕ 1

= 0

∴ Both the statements are required to answer the question.

Hence, option 3.

Workspace:

**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.

**CAT 1998 LRDI | LR - Operator Based Questions**

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

- 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 $\frac{3}{8}.$

Workspace:

**CAT 1998 LRDI | LR - Operator Based Questions**

$\frac{\left\{\left(\left(1\#1\right)\#2\right)-({10}^{1.3}\nabla {\mathrm{log}}_{10}0.1)\right\}}{(1\nabla 2)}=$

- A.
3/8

- B.
$\frac{4\times {\mathrm{log}}_{10}0.1}{8}$

- 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 log_{10} 0.1

= log_{10} 10^{–1} = –1

Then 10^{1.3 }log_{10} 0.1= 10^{1.3 }× (–1) is negative.

So 10^{1.3} ∇ log_{10} 0.1 = 1

Hence, the numerator is equal to 4 –1 = 3

Since 1 × 2 = 2 is positive, (1∇2) = (1× 2)^{1+2} = 2^{3} = 8.

So the denominator = 8. Hence, the answer is $\frac{3}{8}.$

Workspace:

**CAT 1998 LRDI | LR - Operator Based Questions**

$\left(\frac{\left(X\#-Y\right)}{\left(-X\nabla Y\right)}\right)=\frac{3}{8},$ 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

Answer: Option B

**Explanation** :

The best possible way to solve this is to check each of the given answer choices. In options (a), (c) and (d), either both X and Y are positive or both X and Y are negative.

Since we have (–Y) in the numerator of our expression and (–X) in the denominator, X and Y will never be both positive and neither will XY be positive.

Hence, both the numerator and the denominator of our expression will be 1 and the value will always be 1.

Hence, the only possible answer choice is (b).

Workspace:

**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

Answer: Option A

**Explanation** :

Here logic is : A + B = (A + B) – 18

Hence, 10 + 18 = {(10 + 18) – 18} = 10.

Hence, option 1.

Workspace:

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