Average

The average age of a group of 10 students is 14.
∴ the sum of the ages of all 10 of them = 10 × 14 = 140
When two students joins the group, the average increase by 1. New Average = 15
Now there are 12 students
∴ sum of all the ages of 12 students = ( 12 X 15 ) = 180
∴ the sum of the ages of two students who joined = 180 - 140 = 40
And the average age of these two students =20

⇒ Number of applicants that have been counted extra = 1857 − 1747 = 110
⇒ Hence, decrease in average = 110/222 = 0.495
⇒ Correct average = 1560 − 0.495 = 1559.505 = 1559.51
Hence, option D is correct.

Given, ratio of Boys to Girls = 5 : 4
Let's Boys in the class = 5x
and number of Girls in the class = 4x
So, Total number of students = No. of Boys + No. of Girls = 5x + 4x = 9x
Given, average weight of students in the whole class = 40 kg
W.K.T: Average Weight = Total weight/Total number of persons
⇒ Total weight of all students/Total no. of students = 40
⇒ Total weight of all students/(5x + 4x) = 40
⇒ Total weight of all students/9x = 40
⇒ Total weight of all students = 360x
Given, average weight of girls in the class is 10 kg less than the average weight of students in the whole class
⇒ Average weight of girls = 40 kg - 10 kg = 30 kg
⇒ Total weight of girls/Total no. of girls = 30
⇒ Total weight of girls/4x = 30
⇒ Total weight of girls = 120x
Now, Total weight of all students = Total weight of Boys + Total weight of Girls
⇒ 360x = Total weight of Boys + 120x
⇒ Total weight of Boys = 360x - 120x = 240x
Hence, Average weight of boys in the class = Total weight of Boys/Total no. of Boys
= 240x/5x
= 48 kg

We know that Average = sum of elements/number of elements
Given :
Average age of A, B, C = (A + B + C)/3 = 74
A + B + C = 3×74 = 222 years ⇒ eqn (1)
Average of A, B, C, D = (A + B + C + D)/4 = 68
A + B + C + D = 4×68 = 272 years ⇒ eqn (2)
Now subtracting eqn (1) from eqn (2), we get
D's age = 272 - 222 = 50 years.
Given, E's age = 4 + D's age
= 50 + 4 = 54 years.
So A is replaced by E
Average age of B, C, D, E = (B + C + D + E)/4 = 64
B + C + D + E = 64×4 = 256 years.
On substituting D's age and E's age in the above equation, we get
⇒ B + C + 50 + 54 = 256
⇒ B + C = 152
On substituting this in equation (1), we get
A's age = 222 - 152 = 70 years. (from 1)

Weight of first box = 200 kg
According to question,
Weight of 2nd box = 250 × 6/5 = 300 kg
Weight of 3rd box = 200 × 5/4 = 250 kg
Weight of 4th box = 350 kg
Weight of 5th box = 350 × 10/7 = 500 kg
Average of 3 heaviest box = (500 + 350 + 300)/3 = 383.33 kg
Average of 3 lightest box = (200 + 250 + 300)/3 = 250 kg
Required answer = 383.33 - 250
= 133.33 kg

Given:
Highest score - Lowest score = 172 runs.-----(1)
Total runs scored by player in 50 innings = 50×60
= 3000 runs.
Total runs scored by the player in 48 innings (after excluding two innings) = 48×58
= 2784 runs.
Sum of the scores of the excluded innings = 3000 - 2784
= 216 runs.
So,
Highest score + Lowest score = 216 runs.------(2)
Now adding eqn (1) and (2), we get
Highest score - Lowest score = 172 runs.
Highest score + Lowest score = 216 runs. -- (Here Lowest scores cancels each other on adding)
⇒ Highest score = (388/2) = 194.
Hence the highest score = 194 runs.

Sum of twelve number = 12 × 42 = 504
Sum of last five numbers = 5 × 40 = 200
Sum of first four numbers = 4 ×44 = 176
Sum of 5^th, 6^th and 7^th numbers
-⇒ 504 - (200 +176)
-⇒ 504 - 376
-⇒ 128
Let 5^th, 6^th and 7^th numbers are (x+6),×and (x+5). Then,
⇒ x+6+x+x+5 = 128
⇒ 3x = 128 -11
⇒ x = (117/3) = 39
Average of 5^th and 7^th number
⇒ ((x+6+x+x+5)/2)
⇒ = ((39+39+11)/2)
⇒ = (89/2) = 44.5
Hence the answer is : 44.5

Average of 34 numbers is = 0
Therefore sum of all 34 numbers = 0
It is quite possible that 33 of these numbers may be positive and if their sum is a then 34th number is (-a)

Total score = no of student × average mark
Total score of 50 students = (50 × 86) =4300
Total score of top 3 scorers = Total score of 50 students – score of without topper’s
Total score of top 3 scorers = 4300 – (47 × 85) = 305
To minimize the score of the top scorer,(from question)
we assume the other two top scorers score the maximum they can = 95 marks each.
Two top scorers score = (95+95) = 190
The top scorer scored = Total score of top 3 scorers - two top scorers score
So, the top scorer scored = 305 – 190 = 115 marks.
Hence, The the topper’s minimum score is 115 marks:

Let Santhosh's weight be y kg
According to Santhosh 54 < y < 63
According to Santhosh brother 50 < y < 60
According to Santhosh father y < 57
The value satisfying all the above conditions are 55, 56 and 57 Required average = 56 kg