Question 8447 – Aptitude
January 4, 2024GATE 1999
January 5, 2024Question 8448 – Aptitude
Consider the statement
“Not all that glitters is gold”
Predicate glitters(x) is true if x glitters and predicate gold(x) is true if x is gold. Which one of the following logical formulae represents the above statement?
Correct Answer: D
Question 31 Explanation:
Method 1:
Not all that glitters is gold.
Option A:
∀x:glitters(x) ⇒ ¬gold(x)
which means that every item (x), which glitters is not gold.
Option B:
∀x:gold(x) ⇒ glitters()
Every item (x) which is gold is a glitter.
(or)
Every golden item glitters.
Option C:
∃x: gold(x) ∧ ¬glitters(x)
There are some gold items which does not glitters.
Option D:
∃x:glitters(x) ∧ ¬gold(x)
There exists some glitters which are not gold.
(or)
Not all glitters are gold.
The answer is Option (D).
Not all that glitters is gold.
Option A:
∀x:glitters(x) ⇒ ¬gold(x)
which means that every item (x), which glitters is not gold.
Option B:
∀x:gold(x) ⇒ glitters()
Every item (x) which is gold is a glitter.
(or)
Every golden item glitters.
Option C:
∃x: gold(x) ∧ ¬glitters(x)
There are some gold items which does not glitters.
Option D:
∃x:glitters(x) ∧ ¬gold(x)
There exists some glitters which are not gold.
(or)
Not all glitters are gold.
The answer is Option (D).
Method 2:
⇒ (∼∀x) (∼ (glitters(x) ⇒ gold(x))
⇒ ∃x (∼ (∼glitters(x) ∨ gold(x))
⇒ ∃x (glitters ∧ gold(x))
∀x: glitters(x) ⇒ ¬gold(x)
∀x: gold(x) ⇒ glitters()
∃x: gold(x) ∧ ¬glitters(x)
∃x: glitters(x) ∧ ¬gold(x)