Quant Based Puzzle MCQ Quiz - Objective Question with Answer for Quant Based Puzzle - Download Free PDF
Last updated on Jun 10, 2025
Latest Quant Based Puzzle MCQ Objective Questions
Quant Based Puzzle Question 1:
When a number is subtracted from the square of 25, it becomes 124 times of the number. Find the number.
Answer (Detailed Solution Below)
Quant Based Puzzle Question 1 Detailed Solution
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.
Quant Based Puzzle Question 2:
In a zoo, there are Monkeys and Pigeons. If heads are counted, there are 200 and if legs are counted, there are 580. How many Pigeons are there?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 2 Detailed Solution
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
Hence, 110 Pigeons were there.Quant Based Puzzle Question 3:
Directions: In the following question, a statement is given followed by two conclusions numbered I and II. You have to decide which of the conclusion(s) follow(s) the statement. Give your answer as
Statement : A and B are extremities of a diameter of a circle. P and Q are any two points on the circle such that APBQ is a quadrilateral.
Conclusions :
I) ∠APB = ∠AQB
II) ∠PAQ = ∠QBP
Answer (Detailed Solution Below)
Quant Based Puzzle Question 3 Detailed Solution
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.
Quant Based Puzzle Question 4:
The converse of the statement “If a number is even, then it is divisible by 2” is given by “If a number is divisible by 2, then it is even”. Here both the statement and its converse are true. With this information, choose the statement among the following four true statements whose converse is also true.
Answer (Detailed Solution Below)
Quant Based Puzzle Question 4 Detailed Solution
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.
Quant Based Puzzle Question 5:
There are deer and peacocks in a zoo. Their head and leg counts are 80 and 200, respectively. How many peacocks are there?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 5 Detailed Solution
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.
Top Quant Based Puzzle MCQ Objective Questions
Three-fifths of my current age is the same as five-sixths of that of one of my cousins’. My age ten years ago will be his age four years hence. My current age is ______ years.
Answer (Detailed Solution Below)
Quant Based Puzzle Question 6 Detailed Solution
Download Solution PDFLet 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 yearsMohit and Sudesh bought pens and notebooks from the same shop. Mohit bought 3 pens and 6 notebooks by paying an amount of Rs. 180. Sudesh bought 5 pens and 2 notebooks by paying an amount of Rs. 116. How much did Mohit spend on buying notebooks?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 7 Detailed Solution
Download Solution PDFLet,
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.Ages of father and son add up to 50. Six years back father's age was 6 more than thrice his son's age. What will be the father's age 6 years hence?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 8 Detailed Solution
Download Solution PDFLet 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 weights of three boxes are 3 kg, 8 kg and 12 kg. Which of the following CANNOT be the total weight, in kg, of any combination of these boxes?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 9 Detailed Solution
Download Solution PDFThe 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.In a group of bulls and hens, the number of legs is 48 more than twice the number of heads. The number of bulls is ________.
Answer (Detailed Solution Below)
Quant Based Puzzle Question 10 Detailed Solution
Download Solution PDFLet 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.
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, what will be the total area of the park including the garden?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 11 Detailed Solution
Download Solution PDFGiven:
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.Seven years from now, Anamika will be as old as Malini was 4 years ago. Srinidhi was born 2 years ago. The average age of Anamika, Malini and Srinidhi 10 years from now will be 33 years. What is the present age of Anamika?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 12 Detailed Solution
Download Solution PDFLet 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.
Out of the total number of players, 100/3% are in hotel X and the remaining are in hotel Y. If 20 players from hotel Y are shifted to hotel X, then the number of players in hotel X becomes 50% of the total number of players. If 20 players from hotel X are shifted to hotel Y, then the number of players in hotel X becomes what per cent of the total number of players?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 13 Detailed Solution
Download Solution PDFLet 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.
Six years ago the ratio of ages of P and Q was 6 : 5. Four years hence, it will be 11 : 10. What is the age of P now?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 14 Detailed Solution
Download Solution PDFLet, 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.
Hence, ‘18’ is the correct answer.A fruit shop consists of only two types of fruits namely oranges and mangoes. The number of oranges is three times the number of mangoes. Which of the following numbers CANNOT represent the total number of fruits in the shop?
Answer (Detailed Solution Below)
Quant Based Puzzle Question 15 Detailed Solution
Download Solution PDFGiven 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".