Aptitude and Reasoning Questions

Q:

A number is selected at random from the numbers 1 to 30. What is the probability that it is divisible by either 3 or 7 ?

A) 1/30 B) 2/30
C) 1/25 D) 1/2
 
Answer & Explanation Answer: A) 1/30

Explanation:

Let A be event of selecting a number divisible by 3. B be the event of selecting a number divisible by 7.

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 A = { 3, 6, 9, 12, 15, 18, 21, 24, 27, 30 }, so n(A)=10
 B = { 7, 14, 21, 28 }, n(B)= 4 

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Since A and B are not mutually exclusive So :

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Q:

The probabilities that A and B will tell the truth are 2 / 3 and 4 / 5 respectively . What is the probability that they agree with each other ?

A) 3/5 B) 1/3
C) 2/5 D) 3/4
 
Answer & Explanation Answer: A) 3/5

Explanation:

Let A be the event of A will tell truth. B be the event of B tell truth

 

 P(A)=23,P(Ac)=1-P(A)=13

 

 

 

P(B)=45,P(Bc)=1-P(B)=15 

 

When both agree then they say true or they say false together, that is 

 

 AB or AcBc

 

 

 

Also these events will be mutually exclusive :

 

 P(AB)+P(AcBc)=P(A)P(B)P(C)=23*45+13*15=35

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Q:

The true discount on a bill due 9 months hence at 16% per annum is Rs. 189. The amount of the bill is

A) 1200 B) 1764
C) 1600 D) 1354
 
Answer & Explanation Answer: B) 1764

Explanation:

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Q:

The probabilities that drivers A, B and C will drive home safely after consuming liquor are 2 / 5, 3 / 7 and 3 / 4, respectively. What is the probability that they will drive home safely after consuming liquor ?

A) 3/70 B) 4/70
C) 9/70 D) 1/50
 
Answer & Explanation Answer: C) 9/70

Explanation:

Let A be the event of driver A drive safely after consuming liquor. 

 

Let B be the event of driver B drive safely after consuming liquor. 

 

Let C be the event of driver C drive safely after consuming liquor. 

 

P(A)=2/5,P(B)=3/7,P(C)=3/4

 

The events A, B and C are independent . Therefore,

 

PABC=P(A)P(B)P(C)=25*37*34=970

 

Therefore, The probability that all the drivers will drive home safely after consuming liquor is 9/70

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Q:

The odds favouring the event of a person hitting a target are 3 to 5. The odds against the event of another person hitting the target are 3 to 2. If each of them fire once at the target, find the probability that both of them hit the target.

A) 1/20 B) 4/20
C) 1/20 D) 3/20
 
Answer & Explanation Answer: D) 3/20

Explanation:

Let A be the event of first person hitting the target,

P(A) =33+5=38  (odd in favour)

Let B be the event of Second person hitting a target.

P(B)=23+2=25 (odd against)

Since both events are independent and both will hit the target so,

P(AB)= P(A)P(B)=38×25=320

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Q:

if logab+logba=loga+b,then

A) a + b = 1 B) a - b = 1
C) a = b D) ab=1
 
Answer & Explanation Answer: A) a + b = 1

Explanation:

 

if logab+logba=loga+b,then

loga+b=logab×ba=log 1

 

so, a+b=1

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Q:

A three-digit number is formed with the digits 1, 2, 3, 4, 5 at random. What is probability that number formed is divisible by 5 ?

A) 1/4 B) 1/5
C) 1/2 D) 3/4
 
Answer & Explanation Answer: B) 1/5

Explanation:

If a number ends with 5, 0 then the number will be divisible by 5
Here only 5 is present, end place will be fixed by 5 so, 

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Q:

The letters of the word CASTIGATION is arranged in different ways randomly. What is the chance that vowels occupy the even places ?

A) 0.04 B) 0.043
C) 0.047 D) 0.05
 
Answer & Explanation Answer: B) 0.043

Explanation:

 Vowels are A I A I O, 

 

 C    A   S    T   I    G   A    T   I    O   N

 

(O) (E) (O) (E) (O) (E) (O) (E) (O) (E) (O)

 

So there are 5 even places in which five vowels can be arranged and in rest of 6 places 6 constants can be arranged as follows :

 

n(E)=5P52!*2!(A,I are 2 times)*6P62!(T is 2 times)=21600

 

n(S)=11!2!*2!*2!(A, I are 2 times)=4989600

 

216004989600=0.043

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