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CBSE Questions for Class 11 Engineering Maths Mathematical Reasoning Quiz 7 - MCQExams.com
CBSE
Class 11 Engineering Maths
Mathematical Reasoning
Quiz 7
The inverse of statement "
If a dog is barking, then it will not bite" is
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If a dog is not barking, then it will bite.
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If a dog is biting, then it will not bark.
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If a dog is not biting, then it will not bark.
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None of the above
Explanation
The inverse of statement without changing meaning of statement
statement becomes 'If a dog is not barking, then it will bite".
Which of the following statements is the inverse of
"Our pond floods whenever there is a thunderstorm."
?
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If there is a thunderstorm, then our pond floods.
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If we do not get a thunderstorm, then our pond does not flood.
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If our pond does not flood, then we did not get a thunderstorm.
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None of the above
Explanation
The inverse of statement is when the meaning of statement does not change.
Inverse of statement :- If we do not get a thunderstorm, then our pond does not flood.
Where the meaning of statement doesn't change.
Which of the following statements is the inverse
of
"If you do not understand geometry, then you do not know how to reason deductively." ?
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If you reason deductively, then you understand geometry.
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If you understand geometry, then you reason deductively.
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If the do not reason deductively, then you understand geometry.
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None of the above
Explanation
The inverse of statement is when the meaning of statement does not change.
Inverse of statement :- If you understand geometry, then you reason deductively.
Where the meaning of statement doesn't change.
If $$x+4=8$$, then $$x=4$$, Inverse of the statement is-
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If $$x+4=8$$, then $$x\neq4$$
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If $$x+4\neq8$$, then $$x\neq4$$
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If $$x+4\neq8$$, then $$x=4$$
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none of the above
Explanation
$$x+4\neq8$$, then $$x\neq4$$
this is the inverse obtained by negating hypothesis and conclusion condition
Which statement represents the inverse of the statement "If it is snowing, then Skeeter wears a sweater."?
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If Skeeter wears a sweater, then it is snowing.
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If Skeeter does not wear a sweater, then it is not snowing.
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If it is not snowing, then Skeeter does not wear a sweater.
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If it is not snowing, then Skeeter wears a sweater.
Explanation
To find the inverse we need to negate the hypothesis and conclusion, and hence on negating it is snowing we get it is not snowing and on negating skeeter wears a sweater we get skeeter does not wear a sweater.
$$C$$
Write negation of:
All natural numbers are integers and all integers are not natural numbers.
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All natural numbers are not integers and all integers are not natural numbers.
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All natural numbers are integers and all integers are natural numbers.
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Some natural numbers are integers and all integers are not natural numbers.
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All natural numbers are not integers and some integers are natural numbers.
Explanation
the negation of
All natural numbers are integers and all integers are not natural numbers. is All natural numbers are not integers and some integers are natural numbers. or Some natural numbers are not integers and some integers are natural numbers
more or less option d is correct.
$$D$$
If $$p, q, r$$ are simple proportions with truth values $$T, F, T$$, then the truth value of $$(\sim p\vee q)\wedge \sim r \Rightarrow p$$ is
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True
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False
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True, if $$r$$ is false
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True, if $$q$$ is true
Explanation
$$\sim p\vee q$$ means $$F\vee F = F, \sim r$$ means $$F$$
$$\therefore [(\sim p\vee q)\wedge \sim r)=F\wedge F = F$$ which imply $$ p = T$$ means $$F\Rightarrow T = T$$.
Hence, the truth value of the given statement is True.
$$\sim \left[ p\wedge (\sim q) \right] =$$
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$$\sim p \ \wedge \sim q$$
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$$\sim p \ \vee \sim q$$
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$$\sim p \ \wedge \ q$$
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$$\sim p\vee q$$
Explanation
$$\sim \left [p\wedge(\sim q)\right]=\sim p\vee\sim (\sim q)$$ (by demorgan's law)
$$=\sim p\vee q$$ ($$\because \sim(\sim p)=p$$)
Option $$D$$ is correct.
The inverse of the propositions $$(p \wedge \sim q) \rightarrow r$$ is____.
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$$(\sim r) \rightarrow (\sim p) \vee q$$
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$$(\sim p)\vee q \rightarrow (\sim p) $$
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$$r \rightarrow p \vee (\sim q)$$
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$$(\sim p) \wedge (\sim q) \rightarrow r$$
Explanation
In logic
, an
inverse
is a type of
c
onditional sentence
which is an immediate interference
made from another conditional sentence. Any conditional sentence has an inverse
. The inverse of
$$P\rightarrow Q\quad is\quad \neg P\rightarrow \neg Q$$
Hence, answer is
(
∼
r
) $$\rightarrow$$
(
∼
p
)
∨
q
$$p \leftrightarrow q$$ is equivalent to
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$$p \rightarrow q$$
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$$q \rightarrow p$$
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$$(p \rightarrow q) \vee (q \rightarrow p)$$
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$$(p \rightarrow q) \wedge (q \rightarrow p)$$
Explanation
Consider truth table of $$p\leftrightarrow q$$ $$($$ Let $$T$$ denotes true ans $$F$$ denotes false$$)$$
If $$p=T$$ and $$q=T$$ then $$p\leftrightarrow q = T$$
If $$p=T$$ and $$q=F$$ then $$p\leftrightarrow q = F$$
If $$p=F$$ and $$q=T$$ then $$p\leftrightarrow q = F$$
If $$p=F$$ and $$q=F$$ then $$p\leftrightarrow q = T$$
Now consider truth table of $$(p\rightarrow q)\wedge (q\rightarrow p)$$
If $$p=T$$ and $$q=T$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = T$$
If $$p=T$$ and $$q=F$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = F$$
If $$p=F$$ and $$q=T$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = F$$
If $$p=F$$ and $$q=F$$ then $$(p\rightarrow q)\wedge (q\rightarrow p) = T$$
Observe that both truth tables are equal , therefore the correct option is $$D$$
If $$p$$ is $$T$$ and $$q$$ is $$F$$, then which of the following have the truth value $$T$$?
$$(i)p\vee q$$
$$(ii)\sim p\vee q$$
$$(iii)p\vee (\sim q)$$
$$(iv)p\wedge (\sim q)$$
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$$(i),(ii),(iii)$$
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$$(i),(ii),(iv)$$
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$$(i),(iii),(iv)$$
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$$(ii),(iii),(iv)$$
Explanation
$$(i)p\vee q = T\vee F =T$$
$$(ii)\sim p\vee q=(\sim T)\vee F = F\vee F=F$$
$$(iii)p\vee (\sim q)=T\vee (\sim F)=T\vee T= T$$
$$(iv)p\wedge (\sim q)=T\wedge (\sim F)=T\wedge T=T$$
Hence option 'C' is correct
If $$p:3$$ is a prime number and $$q:$$ one plus one is three, then the compound statement "It is not that $$3$$ is a prime number or it is not that one plus one is three" is
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$$\sim p\vee q$$
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$$\sim (p\vee q)$$
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$$p\wedge \sim q$$
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$$\sim p\vee \sim q$$
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$$p\vee \sim q$$
Explanation
$$p:3$$ is a prime number
$$q:$$ one plus one is three
Compound statement: "It is not that $$3$$ is a prime number or it is not that one plus one is three" is
$$\sim p\vee \sim q$$
If $$p$$'s truth value is $$T$$ and $$q$$'s truth value is $$F$$, then which of the following have the truth value $$T$$?
(i) $$p\vee q$$
(ii) $$\sim p\vee q$$
(iii) $$p\vee (\sim q)$$
(iv) $$p\wedge (\sim q)$$
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(i), (ii), (iii)
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(i), (iii), (iv)
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(i), (ii), (iv)
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(ii), (iii), (iv)
Explanation
Given $$p's$$ truth value is $$T$$ and $$q's$$ truth value is $$F$$.
We have to determine which of the given have the truth value $$T$$.
$$(i) p\vee q$$
$$p$$
$$q$$
$$p \vee q$$
$$T$$
$$F$$
$$T$$
$$(ii) \sim p\vee q$$
$$p$$
$$\sim p$$
$$q$$
$$\sim p \vee q$$
$$T$$
$$F$$
$$F$$
$$F$$
$$(iii) p\vee (\sim q)$$
$$p$$
$$q$$
$$\sim q$$
$$p \vee (\sim q)$$
$$T$$
$$F$$
$$T$$
$$T$$
$$(iv) p\wedge (\sim q)$$
$$p$$
$$q$$
$$\sim q$$
$$p \wedge (\sim q)$$
$$T$$
$$F$$
$$T$$
$$T$$
From the above tables we see that, $$(i), (iii), (iv)$$ have truth value $$T$$.
Which of the following is not a logical statement?
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$$8$$ is less than $$6$$
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Every set is a finite set
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Kashmir is far from here
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The sun is a star
Explanation
A) $$8$$ is less than $$6$$ $$ \rightarrow $$ logical
B) Every set is a finite set
$$ \rightarrow $$ logical
C) Kashmir is far from here $$ \rightarrow $$ not a logical statement
D) The sun is a star $$ \rightarrow $$ logical
The truth values of $$p, q$$ and $$r$$ for which $$(p \wedge q) \vee (\sim r)$$ has truth value F are respectively
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F, T, F
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F, F, F
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T, T, T
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T, F, F
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F, F, T
Explanation
option
$$p$$
$$q$$
$$r$$
$$p\wedge q$$
$$\sim r$$
$$(p\wedge q)\vee (\sim r)$$
A
$$ F$$
$$T$$
$$ F$$
$$ F$$
$$T$$
$$T$$
B
$$ F$$
$$ F$$
$$ F$$
$$ F$$
$$T$$
$$T$$
C
$$T$$
$$T$$
$$T$$
$$T$$
$$ F$$
$$T$$
D
$$T$$
$$ F$$
$$ F$$
$$ F$$
$$T$$
$$T$$
E
$$ F$$
$$ F$$
$$T$$
$$ F$$
$$ F$$
$$ F$$
Option $$E$$ gives truth value $$F$$ .
So option $$E$$ is correct.
$$\left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right) $$ is a :
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A tautology
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A contradiction
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Both a tautology and a contradiction
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Neither a tautology nor a contradiction
Explanation
$$\left( p\wedge \sim q \right) \left( \sim p\wedge q \right) =\left( p\wedge \sim p \right) \wedge \left( \sim q\wedge q \right) $$
$$=f\wedge f=f$$
(By using associative laws and commutative laws)
Therefore, $$ \left( p\wedge \sim q \right) \wedge \left( \sim p\wedge q \right) $$ is a contradiction.
If p, q, r have truth values T, F, T respectively, then which of the following is $$\text{True}$$?
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$$(p\rightarrow q)\wedge r$$
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$$(p\rightarrow q)\wedge \sim r$$
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$$(p \wedge q)\wedge (p\vee r)$$
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$$q\rightarrow (p\wedge r)$$
Explanation
Since, p is true and q is false.
$$\therefore p\rightarrow q$$ has truth value F.
Statement r has truth value T.
$$\therefore (p\rightarrow q)\wedge r$$ has truth value F.
Also, $$(p\rightarrow q)\wedge \sim r$$ has truth value F.
Now, $$p\wedge q$$ has truth value F and $$p\vee r$$ has truth value T.
$$\therefore (p\wedge q)\wedge (p\vee q)$$ has truth value F.
As, $$p\wedge r$$ has truth value T.
Therefore, $$q\rightarrow (p\wedge r)$$ has truth value T.
$$\sim [(\sim p) \wedge q]$$ is logically equivalent to
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$$\sim (p \vee q)$$
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$$\sim [p \wedge (\sim q)]$$
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$$p \wedge (\sim q)$$
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$$p \vee (\sim q)$$
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$$(\sim p) \vee (\sim q)$$
Explanation
$$\sim [(\sim p)\wedge q]$$
By distributive property
$$ \Rightarrow \sim (\sim p)\vee (\sim q)\\ \Rightarrow p\vee (\sim q)$$
So option $$D$$ is correct.
The contrapositive statement of statement "If x is prime number, then x is odd" is
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If X is not is prime number, then x is not odd
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If X is not odd, then x is not a prime number
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If X is a prime number, then x is not odd
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If X is not a prime number, then x is odd
Explanation
If p, then q means $$p\rightarrow q$$
$$p\rightarrow q$$ is equivalent to its contrapositive
$$p\rightarrow q=\bar { p } +q=\sim p+q$$
Contrapositive: $$\sim q\rightarrow \sim p=\sim (\sim q)+\sim p$$
$$=q+\sim p$$
$$=\sim p+q$$
$$\therefore $$Contrapositive of "If x is prime,then x is odd" is
If x is not odd, then x not a prime number
Let
$$p:57$$ is an odd prime number
$$q:4$$ is a divisor of $$12$$
$$r:15$$ is the LCM of $$3$$ and $$5$$
be three simple logical statements. Which one of the following is true?
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$$p\vee (\sim q\wedge r)$$
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$$\sim p\vee (q\wedge r)$$
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$$(p\wedge q)\vee \sim r$$
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$$(p\vee (q\wedge r)$$
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$$(p\vee q)\wedge r$$
Explanation
$$p:57$$ is an odd number
$$q:4$$ is a divisor of $$12$$
$$r:15$$ is the LCM of $$3$$ and $$5$$
Then, $$p$$ is $$F$$
$$q$$ is $$T$$
$$r$$ is $$T$$ By the given options
$$\sim p\vee (q\wedge r)$$
$$=\sim F\vee (T\wedge T)=T\vee (T)=T$$
The converse of the contrapositive of the conditional $$p\rightarrow \sim q$$ is.
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$$p\rightarrow q$$
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$$\sim p \rightarrow \sim q$$
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$$\sim q \rightarrow p$$
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$$\sim p \rightarrow q$$
Explanation
The contrapositive of the statement $$p\longrightarrow \sim q\\ $$ is $$q\longrightarrow \sim p$$
The converse of the statement
$$q\longrightarrow \sim p$$
is $$\sim p \longrightarrow q$$
Given the true statement: If a quadrilateral is a square, then it is a rectangle. It follows that, of the converse and the inverse of this true statement.
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Only the converse is true
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Only the inverse is true
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Both are true
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Neither is true
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The inverse is true, but the converse is sometimes true
Explanation
The converse is: If a quadrilateral is a rectangle, then it is a square. The inverse is: If a quadrilateral is not a square, then it is not a rectangle. Both statements are false; or use Venn diagrams to picture the sets mentioned.
If $$p\vee q$$ is true and $$p\wedge q$$ is false, then which of the following is not true?
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$$p\vee q$$
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$$p\leftrightarrow q$$
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$$\sim p\vee \sim q$$
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$$q\vee \sim q$$
Given are three positive integers $$a, b,$$ and $$c$$. Their greatest common divisor is $$D$$; their least common multiple is $$M$$. Then, which two of the following statements are true?
$$(1)$$ The product $$MD$$ cannot be less than $$abc$$
$$(2)$$ The product $$MD$$ cannot be greater than $$abc$$
$$(3)$$ $$MD$$ equals $$abc$$ if and only if $$a, b, c$$ are each prime
$$(4)$$ $$MD$$ equals $$abc$$ if and only if $$a, b, c$$ are relatively prime in pairs
(This means : no two have a common factor greater than $$1$$.)
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$$1, 2$$
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$$1, 3$$
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$$1, 4$$
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$$2, 3$$
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$$2, 4$$
Explanation
Represent $$a, b, c$$ in terms of their prime factors. Then $$D$$ is the product of all the common prime factors, each factor taken as often as it appears the least number of times in $$a$$ or $$b$$ or $$c$$. $$M$$ is the product of all the non-common prime factors, each factor taken as often as it appears the greatest number of times in $$a$$ of $$b$$ or $$c$$.
Therefore, $$MD$$ may be less than $$abc$$, but it cannot exceed $$abc$$.
Obviously, $$MD$$ equals $$abc$$ when there are no common factors.
Given the following six statements:
(1) All women are good drivers
(2) Some women are good drivers
(3) No men are good drivers
(4) All men are bad drivers
(5) At least one man is a bad driver
(6) All men are good drivers.
The statement that negates statement (6) is:
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(1)
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(2)
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(3)
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(4)
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(5)
Explanation
Statement: All men are good drivers.
Negation of the above statement is:
Statement is false that all men are good drivers, that is, at least one man is bad driver.
Which one of the following statements is not true for the equation
$$i{ x }^{ 2 }-x+2i=0$$.
where $$i\equiv \sqrt { -1 } $$?
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The sum of the roots is $$2$$
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The discriminant is $$9$$
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The roots are imaginary
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The roots can be found by using the quadratic formula
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The roots can be found by factoring, using imaginary numbers
The moment of inertia of the plate about the $$x-$$axis is
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$$\dfrac{{M{L^2}}}{8}$$
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$$\dfrac{{M{L^2}}}{32}$$
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$$\dfrac{{M{L^2}}}{24}$$
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$$\dfrac{{M{L^2}}}{6}$$
Negation of $$p \rightarrow ( p \vee \sim q )$$ is
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$$\sim p \rightarrow ( \sim p \vee q )$$
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$$p \wedge ( \sim p \wedge q )$$
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$$\sim p \vee ( \sim p \vee \sim q )$$
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$$\sim p \leftarrow ( \sim p \rightarrow q )$$
The negation of $$q\vee \sim (p\wedge r)$$ is?
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$$\sim q\wedge \sim (p\vee r)$$
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$$\sim q\wedge (p\wedge r)$$
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$$\sim q\vee (p\wedge r)$$
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$$\sim r\vee(p\wedge r)$$
Explanation
$$q \vee \sim (p \wedge r)\,\,\,\,\,\,\,\,\,\,\left[ {\because \sim ( \sim p) = p} \right]$$
$$ \sim q \wedge \sim (p \wedge r)\,\,\,\,\,\left[ {\because \sim \left( {p \vee q} \right) = \sim p \wedge \sim q} \right]$$ [by De Morgans law]
Dual of $$( p \rightarrow q ) \rightarrow r$$ is _________________.
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$$p\vee (\sim q\wedge r)$$
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$$p\vee q\wedge r$$
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$$p\vee (\sim q\wedge \sim r)$$
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$$\sim p\vee (\sim q\wedge r)$$
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