RRB JE Number System Questions with Detailed Solutions

Given:

The natural numbers from 1 to 59.

Concept Used:

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

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

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

Sum = $$\frac{n(n+1)}{2}$$​​

Solution:

Total numbers = 59

Sum of numbers from 1 to 59

= $$\frac{59(59 + 1)}{2}$$​​

= $$\frac{59(60)}{2} = 1770$$​​

Average = $$\frac{1770}{59}$$​ = 30

Solution:

Checking from the option:

For A (20:51):

​$$\frac{20}{51} \approx 0.3922$$​​

For B (22:47):

​$$\frac{22}{47} \approx 0.4681$$​​

For C (21:49):

​$$\frac{21}{49} \approx 0.4286$$​​

For D (28:39):

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

Solution:

(A): $$\frac{17}{39}$$​ = 0.4359

(B): $$\frac{23}{48}$$​ = 0.4792

(C): $$\frac{10}{36}$$​ = 0.2778

(D): $$\frac{15}{33}$$​ = 0.4545

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

Given:

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

Solution:

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

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

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

$$=\frac{900}{25}= 36$$ 

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

​

Given: 

Numbers from 1 to 87

Formula used: 

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

Solution: 

First term is 1 and last term is 87 

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

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

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  = $$\frac{2820}{1.5}$$​ = 1880

The number is 1880

Thus the correct answer is option (c)1880

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 = $$9999 \div 88 = 55$$​​

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.

Given
Expression: $$3^{939} ÷ 8$$​​
Formula Used
Remainder Theorem: If $$(ax + 1)^n$$​ 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 $$3^2$$​= 9, and 9 divided by 8 leaves a remainder of 1.
​$$3^{939}: \\3^{939} = 3^{938 + 1} = 3^{938} × 3^1\\3^{939} = (3^2)^{469} × 3\\$$​​
Substitute $$3^2$$​ with 9:
​$$= 9^{469} × 3$$​​
When 9 is divided by 8, the remainder is 1. Therefore, for $$9^{469} ÷ 8,$$​the remainder is $$1^{469} = 1.$$​​
Now multiply this by the remaining part:
Remainder = 1 × 3 = 3
Final Answer
So the correct answer is (b)

Given
Decimal number = $$5.97\bar{3}$$​​
Formula Used
Vulgar fraction for recurring 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 =  $$\frac{973 - 97}{900}$$​​
Fraction value = $$\frac{876}{900}$$​​
Simplify by dividing numerator and denominator by 4: $$\frac{219}{225}$$​​
Total vulgar fraction =  $$5 \frac{219}{225}$$​​
Final Answer
So the correct answer 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)

Given
Sum of squares of two consecutive even natural numbers = 3700
Solution
Let the even numbers be x and x+2
​$$x^2 + (x+2)^2 = 3700 \\x^2 + x^2 + 4x + 4 = 3700\\2x^2 + 4x - 3696 = 0\\x^2 + 2x - 1848 = 0\\x^2 + 44x - 42x - 1848 = 0\\x(x+44) - 42(x+44) = 0\\(x-42)(x+44) = 0\\$$​​
x = 42 (since natural numbers must be positive)
The two consecutive even numbers are 42 and 44.
Sum of the numbers = 42 + 44 = 86
Final Answer
So the correct answer is (c)

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)