RRB JE Average Questions with Detailed Solutions

Given ​
The set: First 18 positive multiples of 8.
Formula Used
Average of an arithmetic progression = $$\frac{\text{First Term} + \text{Last Term}}{2}$$​​
Solution
The first positive multiple of 8 is $$8 \times 1 = 8.$$​​
The 18th positive multiple of 8 is$$8 \times 18 = 144.$$​​
Because the multiples form an arithmetic sequence, their average is the average of the first and last terms.
Average = $$\frac{8 + 144}{2} = \frac{152}{2} = 76.$$​​
Final Answer
So the correct answer is (a)

Given
First 18 positive multiples of 9.
Formula Used
Average of an arithmetic progression = (First Term + Last Term) / 2
Solution
First term = $$9 \times 1 = 9$$​​
18th term  $$= 9 \times 18 = 162$$​​
Average = $$\frac{9 + 162}{2}$$​​
Average = $$\frac{171}{2} = 85.5$$​​
Final Answer
So the correct answer is (c)

Given:
The average price of three items of furniture is Rs 16860.
The prices of the items are in the ratio 3:5:7.
Formula Used:

$$\text{Average Price} = \frac{\text{Total Price}}{\text{Number of Items}}$$​
Solution:
​$$\text{Average Price} = \frac{\text{Total Price}}{\text{Number of Items}}$$​
​$$16860 = \frac{\text{Total Price}}{3}$$​
$$\text{Total Price} = 16860 \times 3 = 50580$$​
Let the prices of the items be 3x,5x and 7x.
3x + 5x + 7x = 15x
15x = 50580
x =$$\frac{50580}{15} = 3372$$​
The cheapest item corresponds to the smallest ratio, which is 3x .
​$$\text{Price of Cheapest Item} = 3x = 3 \times 3372 = 10116$$​

Given:
Natural numbers from 1 to 94.
Formula Used:
Average =$$=\frac{\text{(Sum of numbers)}}{\text{(Total number)}}$$​​
Sum of first 'n' natural numbers = $$\frac{n(n+1)}{2}$$​​
Solution:
First, we need to find the sum of natural numbers from 1 to 94.
Using the formula, sum =$$\frac{ 94(94+1)}{2 }= \frac{94(95)}{2} = 4465.$$​​
The number of terms is 94.
Average = $$=\frac{\text{(Sum of numbers)}}{\text{(Total number)}}$$​= $$\frac{4465}{94}$$​​
Average = 47.5.
Alternate Method:
for a arithmetic progression the average is also equal to the average of the first and last term.
first term = 1
last term = 94
average = $$\frac{1+94}{2}$$​ = $$\frac{95}{2 }$$​= 47.5.
Therefore, the average of all natural numbers from 1 to 94 is 47.5.

Given:

Seven consecutive odd numbers.

Average of these numbers = 41.

Concept Used:

The average of consecutive odd (or even) numbers is the middle number.

Solution:

Since the average of seven consecutive odd numbers is 41, the middle (4th) number is 41.

The three odd numbers before 41 are 39, 37, and 35.

The three odd numbers after 41 are 43, 45, and 47.

Therefore the numbers are 35, 37, 39, 41, 43, 45, 47.

The smallest number is 35.

Alternate Method:

Formula Used:

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

Solution:

Let the smallest odd number be 'x'.

The seven consecutive odd numbers are: x, x+2, x+4, x+6, x+8, x+10, x+12.

The average is: $$\frac{x + (x+2) + (x+4) + (x+6) + (x+8) + (x+10) + (x+12)}{7}$$​ = 41.

​$$\frac{(7x + 42) }{ 7}$$​ = 41.

7x + 42 = 287.

7x = 245.

x = 35.

Therefore, the smallest number is 35.

Given:

Average price of three items = ₹16830

Price ratio = 3:5:7

Concept Used:

Average Price = $$\frac{\text{Total Price}}{\text{Number of Items}}$$​

Solution:

​$$16830 = \frac{3x + 5x + 7x}{3}$$​

16830 = $$\frac{15x}{3}$$​

16830 = 5x

x = $$\frac{16830}{5}$$​

x = 3366

Cheapest Item Price = 3x = 3×3366 = 10098

Given:

The average of 9 numbers  = 17.

After removing one number, the average of the remaining 8 numbers  = 15.

Concept Used:

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

Solution:

Sum of 9 numbers = 17 × 9 = 153

Sum of 8 numbers = 15 × 8 = 120

Removed number = Sum of  9 numbers − Sum of 8 numbers = 153 – 120 = 33

Given:

The average price of three items = Rs 16890

The prices are in the ratio 3:5:7

Concept Used:

Total Price = $$\text{Average Price} \times \text{Number of Items}$$​

Solution:

Three items be 3x, 5x, and 7x

Total Price = 3x + 5x + 7x = 15x

15x = 16890 × 3

15x = 50670

x = $$\frac{50670}{15}$$​ = 3378

Price of Cheapest Item= 3 × 3378 = 10134

​$$\boxed{10134}$$​​

Given:

The sum of 8 numbers is 336.

Formula Used:

Average = $$\frac{\text{Sum of the numbers}}{\text{Number of terms}}$$​​

Solution:

Average = $$\frac{336}{8}$$​​

Average = 42

The average of the 8 numbers is 42.

Given:

Initial average marks of some students = 75

After adding 5 students, the new average = 78

Marks of the 5 new students: 80, 85, 89, 81, 82

Formula Used:

The formula for average is:

Average = $$\frac{\text{Total Sum of Marks}}{\text{Total Number of Students}}$$​​

Solution:

Let the initial number of students be N.
Thus, the total marks before adding the 5 students:

Total Marks = 75N

Total marks of the 5 new students:

80 + 85 + 89 + 81 + 82 = 417

After adding 5 students, the total number of students = (N + 5)
New total marks = 75N + 417
New average = 78, so:

$$\frac{75N + 417}{N + 5}$$​ = 78

75N + 417 = 78(N + 5)

75N + 417 = 78N + 390

417 - 390 = 78N - 75N

27 = 3N

N = 9

Total number of students is 9.

Given:

The average price of three items = ₹16,455

Their prices are in the ratio 3:5:7

Formula Used:

Total Price = Average Price × Number of Items

Solution:

Let the prices be 3x, 5x, and 7x.
Then, 3x + 5x + 7x = Total Price

Total Price = 16,455 × 3 = 49,365

3x + 5x + 7x = 49,365

15x = 49,365

x = $$\frac{49,365}{15}$$​ = 3,291

The cheapest item corresponds to 3x:

3x = 3 × 3,291= 9,8733

The price of the cheapest item is ₹9,873.

Given:

1 to 97.

Concept Used:

Average = $$\frac{\text{Sum of all numbers}}{\text{Number of terms}}$$​

Solution:

Sum = $$\frac{n(n + 1)}{2}$$​

n = 97

Sum = $$\frac{97(97 + 1)}{2}$$​

= $$\frac{97 \times 98}{2}$$​

= $$\frac{9506}{2}$$​

= 4753

Total number = 97

Average = $$\frac{4753}{97}$$​ = 49

Given:

Average price of three items = ₹15765

Price ratio = 3:5:7

Concept Used:

Average Price = $$\frac{\text{Total Price}}{\text{Number of Items}}$$​

Solution:

15765 = $$\frac{3x + 5x + 7x}{3}$$​

3x + 5x + 7x = 15x

15765 = $$\frac{15x}{3}$$​

15x = 15765 × 3

x = $$\frac{47295}{15}$$​ = 3153

7x = 7 × 3153 = 22071

Given:

Average age of 40 boys = 18 years

One boy aged 26 years left the class.

Two new boys joined the class.

New average age remains 18 years.

Solution:

Total age = 40 $$\times$$​ 18 = 720 years

Total age after one boy left = 720 - 26 = 694 years

The new total number of boys is ( 40 - 1 + 2 = 41)

The new total age is ( 694 + x)

​$$\frac{694 + x}{41}$$​ = 18

694 + x = 18 $$\times 41 = 738 \implies$$​ x = 738 - 694 = 44 years

Average age = $$\frac{44}{2}$$​ = 22 years

Given:
The number of children in 10 families:
2, 1, 3, 4, 2, 3, 5, 1, 1, 3
Formula Used:
Mean (Average) Number of Children Per Family:
​$$\text{Mean} = \frac{\text{Sum of all observations}}{\text{Total number of observations}}$$​​
Solution:
Sum of all children = 2 + 1 + 3 + 4 + 2 + 3 + 5 + 1 + 1 + 3 = 25
Total number of families = 10
Mean = $$\frac{25}{10} = 2.5$$​​
Thus, the average number of children per family is 2.5.