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CAT 1998 — DILR Question 43

Operator Based QuestionsEasy
Passage / Data

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.

{((1#1}#2)-(101.3∇log100.1))(1∇2)=

Answer & solution

  • 3/8

  • B

    4×log100.18

  • C

    (4+1013)8

  • D

    None of these

Solution

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 101.3 log10 0.1= 101.3 × (–1) is negative.
So 101.3 ∇ 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 38.

CAT 1998 DILR Q43: 1 # 1 # 2 - ( 10 1 . 3 ∇ log 10 0 . 1 ) ( 1 ∇ 2 ) = — Solution | TheCATExam