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All questions of Number System for Delhi Police Constable Exam

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Sum of three consecutive odd numbers & three consecutive even numbers together is 231. Difference between the smallest odd number and the smallest even number is 11. What is the sum of the largest even number and largest odd number?
  • a)
    71
  • b)
    91
  • c)
    101
  • d)
    81
  • e)
    Can not be determined
Correct answer is option 'D'. Can you explain this answer?

Kendrika answered
Let the three odd numbers be x, (x + 2), (x + 4) and
The three even numbers be (x + 11), (x + 13) and (x + 15)
Then,
⇔ x + (x + 2) + (x + 4) + (x + 11) + (x + 13) + (x + 15) = 231
⇔ 6x + 45 = 231
⇔ 6x = 186
⇔ x = 31
∴ Required sum :
= (x + 4) + (x + 15)
= 2x + 19
= 2 × 31 + 19
= 62 + 19
= 81

The sum of two even numbers is six more than twice of the smaller number. If the difference between these two numbers is 6, If the larger number lies between 15 to 25 Which is the smaller number?
  • a)
    16
  • b)
    6
  • c)
    24
  • d)
    12
  • e)
    Can not be determined
Correct answer is option 'D'. Can you explain this answer?

Kavya Saxena answered
If 12 is smaller number then larger number is 18
Sum = (12+18) = 30
Twice of the smaller number = 24.
The sum of two even numbers is six more than twice of the smaller number.
Therefore Number 12 satisfy both the conditions.

If three numbers are added in pairs, the sums equal to 11,16 and 23.Find the three numbers
  • a)
    8,6,16
  • b)
    9,2,14
  • c)
    9,5,14
  • d)
    7,2,8
Correct answer is option 'B'. Can you explain this answer?

Preeti Khanna answered
X+y = 11 - (1)
Y+z = 16 - (2)
Z+x = 23 -(3)
Add (1), (2), and (3)
2(x + y + z) = 50
X+y+z = 25
X = 25 – 16 = 9
y = 25 – 23 = 2
z = 25 – 11 = 14

Sum of eight consecutive odd numbers is 656. Average of four consecutive even numbers is 87. What is the sum of the largest even number and largest odd number?
  • a)
    171
  • b)
    191
  • c)
    101
  • d)
    181
  • e)
    179
Correct answer is option 'E'. Can you explain this answer?

Preeti Khanna answered
odd numbers — x-8, x-6, x-4, x-2, x, x+2, x+4, x+6
x-8 + x-6 + x-4 + x-2 + x + x+2 + x+4 + x+6 = 656
8x – 8 =656
x = 83
Even numbers — y-2, y, y+2, y+4
4y + 4 = 87 * 4
y = 86
sum of the largest even number and odd number = 89 + 90 = 179

If the numerator of a fraction is increased by 100% and the denominator is increased by 200%. The resultant fraction is 1 (1/15), What was the original fraction ?
  • a)
    1(4/3)
  • b)
    1(5/3)
  • c)
    1(3/5)
  • d)
    1(3/2)
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
Given:
- Numerator of a fraction increased by 100%
- Denominator of the same fraction increased by 200%
- Resultant fraction is 1 (1/15)

To find: Original fraction

Approach:
1. Let the original fraction be x/y
2. According to the given condition, (2x/y) / (3y) = 1(1/15)
3. Simplifying the above equation, we get 2x = 5y
4. Dividing both sides by y, we get 2x/y = 5
5. Therefore, the original fraction is 1(3/5)

Answer: Option C (1(3/5))

The difference between the digits of a two digit number is 5. Also the original number is 18 more than two times the number obtained by reversing its digits. Find the original number.
  • a)
    94
  • b)
    61
  • c)
    72
  • d)
    49
  • e)
    27
Correct answer is option 'C'. Can you explain this answer?

Ravi Singh answered
Let number is 10x+y
Then x-y = 5 or y-x = 5
Now given that, 10x+y = 2(10y+x) + 18 Solve, 8x – 19y = 18
Now solve: 8x – 19y = 18 and x-y = 5. In this y = 2, x = 7
And also solve; 8x – 19y = 18 and y-x = 5. In this y come to be negative which is not possible so discard this
So number is 10*7 + 2

A number gets reduced to its two-third when 24 is subtracted from it. Find oneeighth of the number?
  • a)
    7
  • b)
    8
  • c)
    9
  • d)
    10
  • e)
    None of these
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
Given: A number gets reduced to its two-third when 24 is subtracted from it.

Let the number be x.

According to the given condition, we can write:

x - 24 = (2/3)x

Multiplying both sides by 3, we get:

3x - 72 = 2x

Simplifying, we get:

x = 72

We need to find one-eighth of the number, which is:

(1/8) x 72 = 9

Therefore, the answer is option (c) 9.

When one-fourth of a number is added to 16, it becomes three-fourth of itself. Find the number?
  • a)
    24
  • b)
    32
  • c)
    36
  • d)
    48
  • e)
    None of these
Correct answer is option 'B'. Can you explain this answer?

Yash Patel answered
Solution:

Let the number be x.

According to the given condition:

- (1/4)x + 16 = (3/4)x

Solve for x:

- (1/4)x + 16 = (3/4)x
- 16 = (3/4)x - (1/4)x
- 16 = (2/4)x
- 16 = (1/2)x
- x = 32

Therefore, the number is 32.

Answer: B: 32

A number is divided by 2, 3, 4, 5 or 6, reminder in each case is one. But the number is exactly divisible by 7. The number lies between 250 and 350, the sum of digits of the number will be
  • a)
    4
  • b)
    7
  • c)
    6
  • d)
    10
  • e)
    Can not be determined
Correct answer is option 'A'. Can you explain this answer?

Preeti Khanna answered
To solve this problem, we need to find a number that satisfies the following conditions:
  1. When divided by 2, 3, 4, 5, or 6, the remainder is 1.
  2. The number is divisible by 7.
  3. The number lies between 250 and 350.
Let's start by finding the least common multiple (LCM) of 2, 3, 4, 5, and 6, which is the smallest number divisible by all of these numbers.
LCM(2, 3, 4, 5, 6) = 60
We need to find a number of the form 7k, where k is an integer, that leaves a remainder of 1 when divided by 60. The numbers in this sequence can be expressed as 60n + 1, where n is an integer.
Now, let's find the first few numbers of the form 60n + 1 that are divisible by 7 and lie between 250 and 350:
  • For n = 4: 60(4) + 1 = 241 (not divisible by 7)
  • For n = 5: 60(5) + 1 = 301 (divisible by 7)
So, the number we're looking for is 301.
Now, let's find the sum of its digits: 3 + 0 + 1 = 4
Therefore, the sum of the digits of the number is 4.

If 4 is added to the numerator of a fraction it becomes 1/3 and if 3 is added to the denominator it becomes 1/6 then find the numerator and denominator is
  • a)
    5/26
  • b)
    25/4
  • c)
    6/17
  • d)
    5/27
  • e)
    None of these
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Given information:

- Adding 4 to the numerator of a fraction gives 1/3.
- Adding 3 to the denominator of the same fraction gives 1/6.

Let's start by setting up equations to represent the given information:

- (numerator + 4) / denominator = 1/3
- numerator / (denominator + 3) = 1/6

We can simplify these equations by cross-multiplying and solving for the numerator and denominator separately.

- 3(numerator + 4) = denominator
- 6numerator = denominator + 3

Now we have two equations with two variables (numerator and denominator). We can use substitution or elimination to solve for these variables.

Using substitution:

- From the first equation, denominator = 3(numerator + 4)
- Substitute this into the second equation: 6numerator = 3(numerator + 4) + 3
- Simplifying: 3numerator = 15
- Solving for numerator: numerator = 5

Now we can substitute this value back into either of the original equations to find the denominator:

- From the first equation: 3(numerator + 4) = denominator
- Substituting numerator = 5: 3(5 + 4) = denominator
- Simplifying: denominator = 27

Therefore, the fraction is 5/27. The correct answer is option D.

a+b = c and abc = 30, then what is the value of a3 + b3 – c3 ?
  • a)
    60
  • b)
    90
  • c)
    -90
  • d)
    -60
Correct answer is option 'C'. Can you explain this answer?

Amal P answered
Given a+b =c 
also abc =  30, Now substitute c as a+b in ABC = 30

ab(a+b) = 30                                                                                                                         factorize 30, 30 can be (30 *1 ) or (15*2) or (6*5)

Iam taking 30 = 6 * 5  = ( 3 *2 ) (3 +2)  Compare with 
ab(a+b) = 30 


we can conclude a =3, b=2 and c= a+b = 2+3 =5

There fore 
 a^3 + b^3 – c^3 = 3^3 + 2^3 -5^3
                                                    
                                                               = 27 + 8 - 125 
      
                                                               = 35 - 125   = -90

When a number is multiplied by 13 and 13 is added to the product, the resultant is divisible by 5. Find the smallest product possible?
  • a)
    85
  • b)
    130
  • c)
    65
  • d)
    90
  • e)
    105
Correct answer is option 'C'. Can you explain this answer?

Alok Verma answered
13x + 13 which is divisible by 5, or 13(x+1) should be divisible by 5. The smallest value of x = 4 to be put here to make it divisible by 5. So the number is 13(4+1)

9/16 of a number is 51 greater than 50% of the number, then the number is
  • a)
    618
  • b)
    168
  • c)
    816
  • d)
    186
Correct answer is option 'C'. Can you explain this answer?

Aarav Sharma answered
Given:
9/16 of a number = 51 + 50% of the number

To find: The number

Solution:

Step 1: Let the number be 'x'

Step 2: According to the question,

9/16 x = 51 + 1/2 x

Step 3: Simplify the above equation

9/16 x - 1/2 x = 51

(9x - 8x)/16 = 51

x/16 = 51

x = 51 × 16

x = 816

Therefore, the number is 816.

Answer: Option (c) 816.

The number of prime factors of 510510
  • a)
    4
  • b)
    5
  • c)
    6
  • d)
    7
Correct answer is option 'D'. Can you explain this answer?

Aarav Sharma answered
Solution:

Prime factorization of 510510 can be written as:

510510 = 2 × 3 × 5 × 7 × 11 × 13

Therefore, the number of prime factors of 510510 is 6.

However, we can also count the prime factors in a different way:

- 2 is a prime factor of 510510
- 3 is a prime factor of 510510
- 5 is a prime factor of 510510
- 7 is a prime factor of 510510
- 11 is a prime factor of 510510
- 13 is a prime factor of 510510
- 17 is not a prime factor of 510510

Therefore, the number of prime factors of 510510 is 7 (counting each prime factor only once).

Hence, the correct option is (D) 7.

Chapter doubts & questions for Number System - Quantitative Aptitude for Competitive Exams 2025 is part of Delhi Police Constable exam preparation. The chapters have been prepared according to the Delhi Police Constable exam syllabus. The Chapter doubts & questions, notes, tests & MCQs are made for Delhi Police Constable 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests here.

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