RRB JE Number System Questions with Detailed Solutions

Get RRB JE Number System Questions to prepare for your upcoming exam. Solve the most expected questions and detailed solutions that can be asked in the exam.

Important RRB JE Number System Questions

Q1.

When 268 is added to 6K8, the result is 9P6. What is the least possible value of K and P for which 9P6 is divisible by 3?

  • A.

    K=4; P=6

  • B.

    K=6; P=9

  • C.

    K=0; P=3

  • D.

    K=3; P=0

    ✓ Correct

Answer & Solution

Correct option is D

Given:

268 + 6K8 = 9P6

9P6 is divisible by 3

Concept Used:

Addition of three-digit numbers.

Divisibility rule for 3: A number is divisible by 3 if the sum of its digits is divisible by 3.

Solution:

Perform the addition:

268

​6K8

9P6

Analyze the units column:

8 + 8 = 16. So, 6 is in the units place, and 1 is carried over.

Analyze the tens column:

6 + K + 1 (carry-over) = P

7 + K = P

Analyze the hundreds column:

2 + 6 = 8, but the result is 9, so there must have been a carry over of 1 from the tens column. Therefore K must be at least 3.

If K = 3, then 7+3 = 10, so P = 0, and a carry over of 1 occurs.

Apply the divisibility rule for 3 to 9P6:

9 + P + 6 = 15 + P

For 9P6 to be divisible by 3, 15 + P must be divisible by 3.

Find the least possible value of K:

From the addition, we know 7 + K = P.

If K = 3, then P = 10, but P is a single digit. Therefore we know there is a carry over from the tens column to the hundreds column.

Therefore, the addition in the tens column must result in a value of 10 or greater.

7+K = 10 + P.

K = 3+P.

Find the least possible value of P:

15 + P must be divisible by 3.

The smallest value of P that satisfies this is P = 0 (15 + 0 = 15, which is divisible by 3).

If P = 0, K = 3 + 0 = 3.

Verify the addition:

268

​638

906

906 is divisible by 3 because 9+0+6 = 15, which is divisible by 3.

The least possible value of K is 3, and the least possible value of P is 0.

Q2.

What is the average of all the natural numbers from 1 to 59?

  • A.

    29.5

  • B.

    31

  • C.

    30.5

  • D.

    30

    ✓ Correct

Answer & Solution

Correct option is D

Given:

The natural numbers from 1 to 59.

Concept Used:

The average of the first n natural numbers is given by the formula:

Average = Sum of numbersTotal numbers\frac{\text{Sum of numbers}}{\text{Total numbers}}​​

The sum of the first n natural numbers is given by the formula:

Sum = n(n+1)2\frac{n(n+1)}{2}​​

Solution:

Total numbers = 59

Sum of numbers from 1 to 59

= 59(59+1)2\frac{59(59 + 1)}{2}​​

= 59(60)2=1770\frac{59(60)}{2} = 1770​​

Average = 177059 \frac{1770}{59}​ = 30

Q3.

Which of the following ratio is smallest?

  • A.

    20:51

    ✓ Correct
  • B.

    22:47

  • C.

    21:49

  • D.

    28:39

Answer & Solution

Correct option is A

Solution:

Checking from the option:

For A (20:51):

​2051≈0.3922\frac{20}{51} \approx 0.3922​​

For B (22:47):

​2247≈0.4681\frac{22}{47} \approx 0.4681​​

For C (21:49):

​2149≈0.4286\frac{21}{49} \approx 0.4286​​

For D (28:39):

​2839≈0.7179\frac{28}{39} \approx 0.7179​​
The smallest ratio is A: 20:51

Q4.

Which of the following ratio is smallest?

  • A.

    17 : 39

  • B.

    23 : 48

  • C.

    10 : 36

    ✓ Correct
  • D.

    15 : 33

Answer & Solution

Correct option is C

Solution:

(A): 1739 \frac{17}{39}​ = 0.4359

(B): 2348\frac{23}{48}​ = 0.4792

(C): 1036\frac{10}{36}​ = 0.2778

(D): 1533\frac{15}{33}​ = 0.4545

The smallest value is 0.2778, which corresponds to the ratio 10:36.

Q5.

If 'M' is the smallest perfect square number, which is exactly divisible by 12, 15 and 18, then find the sum of the digits of the quotient obtained, when M is divided by 25.

  • A.

    8

  • B.

    9

    ✓ Correct
  • C.

    7

  • D.

    6

Answer & Solution

Correct option is B

Given:

​'M' is the smallest perfect square number, which is exactly divisible by 12, 15 and 18

Solution:

LCM of (12,15,18)=180\text{LCM of } (12, 15, 18) = 180 

180=2×2×3×3×5180 = 2\times2\times 3 \times 3 \times 5  (Multiply by one 5)

180×5=900 is the number180 \times 5 = 900 \text{ is the number} 

=90025=36=\frac{900}{25}= 36 

The sum of the digits of the quotient is 3+6=93 + 6 = 9​

​

Q6.

What is the average of all the natural numbers from 1 to 87?

  • A.

    45

  • B.

    44.5

  • C.

    43.5

  • D.

    44

    ✓ Correct

Answer & Solution

Correct option is D

Given: 

Numbers from 1 to 87

Formula used: 

Average=First number+Last number2Average = \frac{First \ number + Last \ number}{2} 

Solution: 

First term is 1 and last term is 87 

Average=1+872=882=44Average = \frac{1 + 87}{2} = \frac{88}{2} = 44 

Thus, the average of all natural numbers from 1 to 87 is 44

Q7.

A number when increased by 50 %, gives 2820. The number is:

  • A.

    3760

  • B.

    940

  • C.

    1880

    ✓ Correct
  • D.

    5640

Answer & Solution

Correct option is C

Given:

Final value = 2820

Increase percentage = 50%

Solution :

Let the number be x:

If the number is increased by 50% :

x + 0.5 x = 2820

1.5 x = 2820

x  = 28201.5\frac{2820}{1.5}​ = 1880

The number is 1880

Thus the correct answer is option (c)1880

Q8.
How many natural numbers up to 100 are divisible by both 2 and 3?
  • A.16
    ✓ Correct
  • B.13
  • C.17
  • D.14

Answer & Solution

Correct option is A

Given:
To find: the number of natural numbers up to 100 that are divisible by both 2 and 3.
Formula Used:
If a is the first term, d is the common difference, and l is the last term, then the number of terms n is: n = l−ad+1\frac{l - a}{d} + 1​​
Solution:
The first multiple of 6 is 6.
The largest multiple of 6 within 100 is 96.
This is an arithmetic sequence with the first term a = 6, the common difference d = 6, and the last term l = 96.
​n=96−66+1=906+1=15+1=16n = \frac{96 - 6}{6} + 1 = \frac{90}{6} + 1 = 15 + 1 = 16​​
Therefore, There are 16 natural numbers up to 100 that are divisible by both 2 and 3.
Q9.
Find the largest 4 digit number that is exactly divisible by 88.
  • A.9944
    ✓ Correct
  • B.8894
  • C.9844
  • D.9768

Answer & Solution

Correct option is A

Given:
We need to find the largest 4-digit number that is exactly divisible by 88.

Concept Used:
To find the largest 4-digit number divisible by a given number, we start from the largest 4-digit number (9999) and find the nearest number that is divisible by the given number.

Solution:

Start with the largest 4-digit number:

Largest 4-digit number = 9999

Find the remainder when 9999 is divided by 88:

Remainder = 9999÷88=559999 \div 88 = 55​​

Subtract the remainder from 9999 to get the largest 4-digit number divisible by 88:

Largest 4-digit number divisible by 88 = 9999 - 55 = 9944

The largest 4-digit number that is exactly divisible by 88 is 9944.

Q10.
Which of the following numbers is exactly divisible by 4?
  • A.10006
  • B.

    10012

    ✓ Correct
  • C.

    10002

  • D.

    10010

Answer & Solution

Correct option is B

​Concept Used:
A number is exactly divisible by 4 if the last two digits of the number form a number that is divisible by 4.
Solution:
For 10006: Last two digits are 06.
06 ÷\div 4 = remainder 2 (Not divisible by 4).
For 10012: Last two digits are 12.
12 ÷\div  4 = remainder 0 (Divisible by 4).
For 10002: Last two digits are 02.
02 ÷ 4 = remainder 2 (Not divisible by 4).
For 10010: Last two digits are 10.
10 ÷ 4 = remainder 2 (Not divisible by 4).
Thus, the only number divisible by 4 is: 10012
Q11.
Find the least multiple of 7 that leaves the remainder 4 when divided by 6, 9, 15 and 18.
  • A.364
    ✓ Correct
  • B.274
  • C.94
  • D.

    184

Answer & Solution

Correct option is A

Concept used
Set up an equation in the form N = LCM × k + 4, where N must also be a multiple of 7.
Solution:
LCM of 6, 9, 15, and 18 is 90
So, any number that satisfies the condition must be of the form:
N = 90k + 4
where k is an integer. (k = 1, 2, 3 ,4,….)
put k = 4
N = 90 × 4 + 4 = 364
Thus,  The least multiple of 7 that leaves a remainder of 4 when divided by 6, 9, 15, and 18 is 364.
Q12.

What is the remainder when 39393^{939}​ is divided by 8 ?

  • A.

    2

  • B.

    3

    ✓ Correct
  • C.

    5

  • D.

    0

Answer & Solution

Correct option is B

Given
Expression: 3939÷83^{939} ÷ 8​​
Formula Used
Remainder Theorem: If (ax+1)n(ax + 1)^n​ is divided by a, the remainder is always 1.
Solution
We need to express the base 3 in terms of a power that is close to a multiple of 8. We know that 323^2 ​= 9, and 9 divided by 8 leaves a remainder of 1.
​3939:3939=3938+1=3938×313939=(32)469×3 3^{939}: \\3^{939} = 3^{938 + 1} = 3^{938} × 3^1\\3^{939} = (3^2)^{469} × 3\\​​
Substitute 323^2​ with 9:
​=9469×3= 9^{469} × 3​​
When 9 is divided by 8, the remainder is 1. Therefore, for 9469÷8,9^{469} ÷ 8, ​the remainder is 1469=1.1^{469} = 1.​​
Now multiply this by the remaining part:
Remainder = 1 × 3 = 3
Final Answer
So the correct answer is (b)

Q13.

Express 5.973ˉ5.97\bar{3}​ as a vulgar fraction.

  • A.

    ​52192255 \frac{219}{225}​

    ✓ Correct
  • B.

    ​53192255 \frac{319}{225}​

  • C.

    ​52292255 \frac{229}{225}​

  • D.

    ​53092255 \frac{309}{225}​

Answer & Solution

Correct option is A

Given
Decimal number = 5.973ˉ5.97\bar{3}​​
Formula Used
Vulgar fraction for recurring decimal = Number - Non-recurring partNumber of 9s for recurring and 0s for non-recurring after decimal\frac{\text{Number - Non-recurring part}}{\text{Number of 9s for recurring and 0s for non-recurring after decimal}}​​
Solution
Let the fractional part be x = 0.97333...
Here, 97 does not repeat and 3 repeats.
Fraction value =  973−97900\frac{973 - 97}{900}​​
Fraction value = 876900 \frac{876}{900}​​
Simplify by dividing numerator and denominator by 4: 219225\frac{219}{225}​​
Total vulgar fraction =  52192255 \frac{219}{225}​​
Final Answer
So the correct answer is (a)

Q14.

The 7-digit number 57A001B is divisible by 9. What is the minimum value of (A + B)?

  • A.

    5

    ✓ Correct
  • B.

    8

  • C.

    3

  • D.

    2

Answer & Solution

Correct option is A

Given:
A 7-digit number 57A001B is divisible by 9.
Formula Used:
A number is divisible by 9 if the sum of its digits is a multiple of 9.
Solution:
Sum of the digits of 57A001B:
Sum = 5 + 7 + A + 0 + 0 + 1 + B = 13 + A + B
For the number to be divisible by 9, (13 + A + B) must be a multiple of 9.
Since A and B are digits (0 to 9), (A + B) \ge 0.
The smallest multiple of 9 greater than or equal to 13 is 18.
Equating the sum to 18 to find the minimum value of (A + B):
13 + A + B = 18
A + B = 18 - 13 = 5.
Therefore, the minimum value of (A + B) is 5.
Final Answer
So the correct answer is (a)

Q15.

Which of the following numbers is divisible by both 9 and 43?

  • A.

    10710

  • B.

    8901

    ✓ Correct
  • C.

    7230

  • D.

    7095

Answer & Solution

Correct option is B

Given
Options: 10710, 8901, 7230, 7095
Divisors: 9 and 43
Formula Used
A number is divisible by 9 if the sum of its digits is divisible by 9.
Solution
Check divisibility by 9 (Sum of digits):
10710: 1+0+7+1+0 = 9 (Divisible)
8901: 8+9+0+1 = 18 (Divisible)
7230: 7+2+3+0 = 12 (Not divisible)
7095: 7+0+9+5 = 21 (Not divisible)
Now check the remaining viable numbers for divisibility by 43:
10710 / 43 = 249.06 (Not divisible)
8901 / 43 = 207. Here, 43 × 200 = 8600. 8901 - 8600 = 301. 43 × 7 = 301. So 8901 = 43 × 207 (Divisible).
Final Answer
So the correct answer is (b)