Puzzle MCQ Quiz - Objective Question with Answer for Puzzle - Download Free PDF

Let's denote the floors from 1 (lowermost) to 6 (topmost).

O lives on floor numbered 4.

Only two people live between O and P.

Since O is on floor 4, P can be on floor 1 (two people between O and P: floors 3 and 2) or floor 7 (which doesn't exist).

So, P must be on floor 1.

Only M lives between O and N.

Y lives immediately below O.

O is on floor 4, so Y lives on floor 3.

Since O is on floor 4, M must be on floor 5, and N must be on floor 6.

The only remaining person is Z, and the only remaining floor is 2.

So, Z lives on floor 2.

Based on our deductions, Z lives on floor numbered 2.

Hence, the correct answer is "Option 4".

Given: A, B, C, D, E and F live on six different floors of the same building, numbered 1 to 6 from bottom to top.

Let's use the given conditions to deduce the floor arrangements:

1. The product of floors on which D and F live is 10.

Possible pairs for (D, F) or (F, D) that multiply to 10 are (2, 5) or (5, 2).

The sum of the floors on which B and D live is 11.

• Given D can be on floor 2 or 5:
• If D lives on floor 2, then B + 2 = 11 → B = 9 (case II is not possible as max floor is 6).
• If D lives on floor 5, then B + 5 = 11 → B = 6.

We can definitely say that D lives on floor 5 and B lives on floor 6.

2. C lives immediately above E.

This means C and E are on consecutive floors, with C on the higher floor.

The only person left is A, and the only floor left is 1. So, A lives on floor 1.

Final arrangement:

Thus, 1 person lives between A and E.

Hence, the correct answer is "Option 2".

Given: 1568 − 16 + 4 ÷ 5 × 22 = ?

Decoding the given information:

Now replacing the sign in the equation, we get:

⇒ 1568 ÷ 16 − 4 × 5 + 22

= 98 − 4 × 5 + 22

⇒ 98 − 20 + 22

= 78 + 22 = 100

Hence, the correct answer is "Option 2".

Let my current age = x years and my cousin’s age = y years.

Three-fifths of my current age is the same as five-sixths of that of one of my cousins’,

⇒ 3x/5 = 5y/6

⇒ 18x = 25y

My age ten years ago will be his age four years hence,

⇒ x – 10 = y + 4

⇒ y = x – 14,

⇒ 18x = 25(x – 14)

⇒ 18x = 25x – 350

⇒ 7x = 350

∴ x = 50 years

Let,

Price of one pen = x

Price of one notebook = y

Given:

1) Mohit bought 3 pens and 6 notebooks by paying an amount of Rs. 180

=> 3x + 6y = 180

=> 3x = 180 – 6y

=> x = (180 – 6y) ÷ 3

=> x = 60 – 2y --------------- (i)

2) Sudesh bought 5 pens and 2 notebooks by paying an amount of Rs. 116

=> 5x + 2y = 116

Putting the value of x from eq (i)

=> 5(60 – 2y) + 2y = 116

=> 300 – 10y + 2y = 116

=> 300 – 116 = 10y – 2y

=> 8y = 184

=> y = 23

Mohit spend on buying 6 notebooks

=> 6y = 6 × 23 = 138

Hence, Mohit spend “Rs. 138” on buying notebooks.

According to the question, after interchanging the given two signs and two numbers i.e :

• × and +
• Two numbers 3 and 9

So,

I. 7 + 3 – 8 ÷ 2 × 9

⇒ 7 + 3 - 4 × 9

⇒ 7 + 3 - 36

⇒ 10 - 36

⇒ -26

II. 4 + 3 – 9 × 8 ÷ 2

⇒ 4 + 3 - 9 × 4

⇒ 4 + 3 - 36

⇒ 7 - 36

⇒ -29

Let the age of father be F and that of son be S.

F + S = 50 (Given)

S = 50 – F     _____ (i)

Six years back father's age was 6 more than thrice his son's age.

According to problem:

(F – 6) = 3(S – 6) + 6     _____ (ii)

Substituting the value of equation (i) in (ii), we get:

F – 6 = 3(50 – F – 6) + 6

⇒ F – 6 = 3(44 – F) + 6

⇒ F – 6 = 132 – 3F + 6

⇒ F + 3F = 132 + 6 + 6

⇒ 4F = 144

⇒ F = 144/4

⇒ F = 36

So, father's age 6 years hence = (36 + 6) = 42

Hence, ‘42’ is the correct answer.

The logic followed here is:

1) 15 → 12 + 3 = 15 kg

2) 20 → 12 + 8 = 20 kg

3) 23 → 12 + 8 + 3 = 23 kg

4) 21 → It CANNOT be the total weight, in kg, of any combination of these boxes.

Hence, ‘21’ is the correct answer.

Let the number of bulls be ‘a’ and the number of hens be ‘b’.

So, the total numbers of heads are (a + b) and total numbers of legs are (4a + 2b).

According to question:

(4a + 2b) = 2(a + b) + 48

4a + 2b = 2a + 2b + 48

4a + 2b – 2a – 2b = 48

2a = 48

a = 24

So, the number of bulls is 24.

Hence, ‘24’ is the correct answer.

Given:

A side of a square-shaped park is 12 m.

• If a square-shaped garden with a side of 24 m is developed around the park, then the garden with the park will look like the picture below:

Formula:

Area of square = Side × Side

Calculation:

=> Total area of the park including the garden = Area of the outer square = 24 × 24

=> Area of square = 576 m2

Hence, the total area of the park including the garden is 576 m².

Let the present age of Anamika be A, Malini be M and that of Srinidhi be S.

The, according to the question:

1) Seven years from now, Anamika will be as old as Malini was 4 years ago. 

A + 7 = M - 4

⇒ M = A + 11

S = 2 years

And, 

2) Srinidhi was born 2 years ago. The average age of Anamika, Malini and Srinidhi 10 years from now will be 33 years.

A + M + S = 99 - 30

A + M + S = 69 

Now, by substituting the above values,

A + (A + 11) + 2 = 69 

2A + 13 = 69

A = 56 ÷ 2

A = 28 years

Hence, the present age of Anamika is 28 years is the correct answer.

For this question, We have to check the given equation as-

146 –  2 + 3 × 123 × 5 + 2

New equation after replacing symbols-

146 ÷ 2 × 3 – 123 - 5 × 2

According to the BODMAS rule-

⇒ 73 × 3 –  123 – 5 × 2

⇒ 219 –  123 –  10

⇒ 86

Hence, Option (4) is correct.

BODMAS Table: 

Given equation I: 55 + 42 – 36 × 20 ÷ 9 = 17

Now, if '20 and 36' are interchanged, then:

⇒ 55 + 42 – 20 × 36 ÷ 9 = 17

⇒ 55 + 42 – 20 × 4 = 17

⇒ 55 + 42 – 80 = 17

⇒ 97 – 80 = 17

= 17 = 17.

LHS = RHS.

Given equation II: 20 ÷ 2 × 36 + 81 – 41 = 400

Now, if '20 and 36' are interchanged, then:

⇒ 36 ÷ 2 × 20 + 81 – 41 = 400

⇒ 18 × 20 + 81 – 41 = 400

⇒ 360 + 81 – 41 = 400

⇒ 441 – 41 = 400

= 400 = 400

LHS = RHS.

Here, Both I and II are correct equation after interchanging the given symbol.

Hence, the correct answer is "Option 3".