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

Let the number be a.

According to question:

25² - a = 124a

⇒ 625 - a = 124a

⇒ 625 = 124a + a

⇒ 625 = 125a

⇒ 625/125 = a

⇒ 5 = a

Hence, '5' is the correct answer.

If we summarised the information:

→ Total heads = 200; it means total no of animals is also 200

M = monkeys

P = Pigeons

M + P = 200 → (1)

Monkeys have 4 legs and Pigeons have 2 legs;

It means; 4 M + 2 P = 580 → (2)

→ Equation (1) × 2

→ (M + P = 200) × 2

→ 2 M + 2 P = 400 → (3)

→ Equation (2) ─ (3)

→ 4 M + 2 P – 2M – 2P  = 580 – 400

→ 2 M = 180

→ M = 180 ÷ 2

→ M = 90 = no of Monkeys

By putting value in equation (1)

→ M + P = 200

→ 90 + P = 200

→ P = 200 ─ 90

→ P = 110 = no of Pigeons

Given:

A and B are extremities of the diameter of a circle. P and Q are any two points on the circle such that APBQ is a quadrilateral.

Formula used:

Angle in a semicircle is a right angle (i.e., ∠APB = ∠AQB = 90° if P and Q lie on the circle).

Calculation:

⇒ Since AB is the diameter, both ∠APB and ∠AQB subtend the semicircle.

⇒ ∠APB = ∠AQB = 90°

⇒ Hence, Conclusion I is true.

⇒ However, angles ∠PAQ and ∠QBP are not necessarily equal as they depend on the positions of P and Q on the circle.

⇒ So, Conclusion II is not necessarily true.

∴ Only Conclusion I follows.

Given:

The converse of a true statement must also be true.

Options Analysis:

1. If a triangle is an equilateral, then it is isosceles.

True statement.

Converse: "If a triangle is isosceles, then it is equilateral" is false.

2. If a number is divisible by 6, then it is divisible by 3.

True statement.

Converse: "If a number is divisible by 3, then it is divisible by 6" is false.

3. If the sum of all the digits of a number is divisible by 3, then the number is divisible by 3.

True statement.

Converse: "If a number is divisible by 3, then the sum of its digits is divisible by 3" is also true.

4. If a quadrilateral is a square, then its sides are equal.

True statement.

Converse: "If sides are equal, then it is a square" is false (it could be a rhombus).

∴ Correct answer is: If the sum of all the digits of a number is divisible by 3, then the number is divisible by 3.

Let there are "X" deer and "Y" peacocks.

Total heads are 80.

So, X + Y = 80 ( Equation 1)

Total legs are 200. Deer has 4 legs and peacock has 2.

So, 4X + 2Y = 200 ( Equation 2)

Multiplying Equation 1 by 4 and substracting with Equation 2, we get

Y = 60

Putting this value of Y in Equation 1, we get

X = 20

so, Total peacocks are Y = 60.

Hence, the correct answer is 60.

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

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

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.

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

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.

\({(A + 10 + M + 10 + S + 10) \over 3} = 33 \)

\(A + M + S + 30 = 33 \times 3\)

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.

Let total players = A

Players in Hotel X = 100/3% of A = A/3

Players in Hotel Y = A - A/3 = 2A/3

According to question:

A/3 + 20 = 50/100 × A

⇒ A/3 + 20 = A/2

⇒ A/2 - A/3 = 20

⇒ A/6 = 20

⇒ A = 20 × 6

⇒ A = 120

Now, Players in Hostel X = 120/3 = 40 

Players in Hostel Y = 120 - 40 = 80

Now, if 20 players from hotel X are shifted to hotel Y, then the number of players in hotel X becomes,

= 40 - 20 = 20

And the number of players in hotel X becomes what percent of the total number of players:

20/120 × 100

= 16.67%

Hence, ‘16.67%’ is the correct answer.

Let, the age of P and Q be measured in terms of ‘a’.

Six years ago the ratio of ages of P and Q was 6 : 5.

So, six years ago, P was 6a years old and Q was 5a years old.

Therefore, present age of P is (6a + 6) and present age of Q is (5a + 6).

Four years hence, it will be 11 : 10.

According to question:

[(6a + 6) + 4]/[(5a + 6) + 4] = 11/10

On solving, we get the value of a is 2.

So, the present age of P = (6a + 6) = (6 × 2 + 6) = 18 years.

Given fruit shop consists of two types of fruits i.e., Oranges and Mangoes

The number of oranges is three times the number of mangoes :

⇒ oranges = 3 × mangoes

⇒ oranges/mangoes = 3/1

Total ration of oranges and mangoes = 3 + 1 = 4

Total number of fruits = oranges + mangoes

Total number of fruits = 3 × mangoes + mangoes

Total number of fruits = 4 × mangoes

1) Consider total number of fruits = 44

⇒ 44 = 4 × mangoes

⇒ mangoes = 11

⇒ oranges = 3 × mangoes = 3 × 11 = 33

2) Consider total number of fruits = 42

⇒ 42 = 4 × mangoes

42 is not divisible by 4 so, 42 cannot the total number of fruits in the shop.

3) Consider total number of fruits = 48

⇒ 48 = 4 × mangoes

⇒ mangoes = 12

⇒ oranges = 3 × mangoes = 3 × 12 = 36

4) Consider total number of fruits = 40

⇒ 40 = 4 × mangoes

⇒ mangoes = 10

⇒ oranges = 3 × mangoes = 3 × 10 = 30

∴ Here, 'Total number of fruits in shop cannot be 42'.

Hence, the correct answer is "42".