: Find the average of these problems #IndiaNEWS #Education Today Hyderabad: This article is in continuation to the last article on arithmetic that will aid you in your preparation for government recruitment

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Find the average of these problems #IndiaNEWS #Education Today
Hyderabad: This article is in continuation to the last article on arithmetic that will aid you in your preparation for government recruitment exams. Here are some examples along with solutions on the Average topic.
Examples:
1. Find the average of first 15 whole numbers
a) 7 b) 8 c) 9 d) 10
Solution: 0,1,2,…. . 14
Here the gap is the same
(0 14 )/2 = 14/2 = 7
Note: If ‘0’ is one of the quantities of a given data, then that ‘0’ will also be included while calculating the average.
2. Find the average of 36, 28, 45, 63, 50
a) 40. 4 b) 42. 4 c) 44. 4 d) 46. 4
Solution:
Here the gap is not the same
Method 36 28 45 63 50/5 = 222/5
Average = 44. 4
Shortcut:
Assume that average = 50
36 50 = 14
28 50 = -22
45 50 = -5
63 50 = 13
50 50 = 0
-14 22 5 13 0 = -28
50 28/5 = 50 5. 6
= 44. 4
Similarly, you can take any number:
Let us say the average = 40
36 40 = -4
28 40 = -12
45 40 = 5
63 40 = 23
50 40 = 10
-4 12 5 23 10 = 22
40 22/5 = 40 4. 4
= 44. 4
3. The average of ‘n’ numbers x1, x2…. . xn is x- then the value of (xi x) is
a) 0 b) n c) x- d) n
4. What is the average of all the multiples of 6 from 20 to 80?
a) 48 b) 50 c) 51 d) 52
Solution:
24, 30, …. . 78
Here, the gap is the same
24 78/2 = 102/2 = 51
5. a, b, c, d, e, f, g are consecutive even numbers. j, k, l, m, n are consecutive odd numbers. Then, the average of all the numbers is
a) (a b m n)/4 b) (f c n g)/4
c) 3 ( a n/2) d) l d/n
Solution: a, b, c, d, e, f, g
Average = d
j, k, l, m, n
Average = l
Average of all numbers = (d l)/2
Note: Here the gap is same. So the average is the middle number.
6. The average of five consecutive odd numbers is 27. What is the product of the first and the last number?
a) 621 b) 667 c) 713 d)725
Solution: 23 25 27 29 31
23 x 31= 713
7. The average of x and 1/x is M then find the average of x3 and 1/x^3
a) 7M3 M/2 b) 8M3 6M/2
c) 8M3 M/2 d) 7M3 6M/2
Solution:
(x 1/x)/ 2 = M
x 1/x = 2M
(x 1/x)3 = (2M)3
x3 1/x x 3x 1/x (x 1/x) = 8M3
x3 1/x x 3(2M) = 8M3x3 1/x3 = 8M3 6M
Average of x3 and 1/x3 = 8M3 6M/2
Shortcut method:
Choose any two numbers
Let us say 1, 1/1
Average of 1, 1 is 1 ( = M)
Average of 1^3, 1/1^3 is 1
From given options
From option 1, (7(1)3 1)/2 = 6/2 = 3
From option 2, (8 (1)3 6 (1))/ 2 = 2/2 = 1
8. The average of 5 consecutive integers starting with ‘m’ is n. What is the average of 6 consecutive integers starting with (m 2)?
a) 2n 5/2 b) (n 2) c) (n 3) d) 2n 9/2
Solution:
m, m 1, m 2, m 3, m 4
Average = n
m 2 = n =>; m = n 2
m 2, m 3, m 4, m 5, m 6, m 7
average = 2m 9/2
(2(n-2) 9)/2 = 2n 4 9/2 = 2n 5/2
9. If a,b,c and d are four consecutive numbers. If the sum of a and c is 124.

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