Average

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QCS-6
Q 14

The average weight of 8 functional directors today = x ( we assume this information)

Therefore, the average weight of the same set of directors 3 years ago must have been --- x-3

The basic equation that should come to mind while solving this question is:

Initial weight of the 8 directors - weight of the old man who retired + the age of the new director = final total weight of the 8 directors

Initial weight of the 8 directors = 8(x)

Final weight of the 8 directors = 8 ( x-3)
Note that the average weight of the 8 director becomes the same as it was three years back

Putting the values above in our basic equation we get

8(x) - old + Young = 8( x-3)

=> Old - Young = 24

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QCS-6

Average Q15

Basic equation for this question is :

Total age of both the groups together = Total weight of the first group + Total age of the 2nd group

Total weight of the first group = n x 16 ( n is the number of people in the first group )

Total weight of the 2nd group = 15 x 20 = 300
Total weight of both the groups = 15.5 ( n+ 20)

Putting the above values in the equation above we have :

15 x 20 + 16 (n) = 15.5( n+20)

=> 300 + 16 n = 15.5 n + 310 or n = 20

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QCS6

Q16

average age of this family of four members ten years back = 20

Average age of the same four members of the family = 20 + 10 = 30

The basic equation, which governs the solving of the above question is:

Current total age of the family = Current total age of four members + Child 1 + Child 2 + Child -3

Current total age of the family = Number of people in the family now x Current average age

Current total age of the family = 7 ( 20)
Note that the question states that the average age of the family is the same as it was ten years ago.. so we must take the current average as x

Current total age of four members = 4 x current average age of the four members = 4 ( 20+10)

The ages of the three children must be y, y and y+2 ( Two children are identical twins and one child is 2 years older)

So putting the above values in our basic equation we get:
=> 7(20 ) = 4(20+10) + y + y + y +2

=> 20 = 3y + 2 or 18 = 3y => y = 6

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QCS-6

Q 19

Average of 16 innings = x
Total score after 16 innings = 16 x .................i
Total score in the next three innings= 82 .........ii

Thus the total score after 19 ( 16+3) innings = 16 x + 82
New average after 19 innings = x-2
Note that the average of the batsman reduces by 2 runs after the 19th inning

Thus the total score after 19 innings = 19 (x-2) ..............iii
logically, iii must be equal to the sum of i and ii

19 (x-2) = 16x + 82 => 3x = 120 => x = 40

average after 19 innings = 40 -2 = 38

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QCS-6
Average - Practice Questions

Q30

Average salary of officers, clerks, and bearers:1200, 450 and 180 respectively.
Average salary of the three classes = Rs 440
Raise in salary in the new scale of pay - officers, clerks, and bearers- 60, 30 and 20 Rs respectively
New monthly average salary= Rs 470
In the initial pay scale:
Initially,
total salary = salary of all officers + salary of all clerks + salary of all bearers
Assume the the number of officers, clerks and bearers to be a,c, and b
thus we have=> 440 (a+b+c) =1200(a) + 450(b) +180(c)
On simplifying the above equation we have : 440a + 440b + 440c = 1200a + 450c + 180b
=> 760a +260 c = 10b => 76 a + 10c = 260b => 76a + c = 26 b................(i)

Similarly, after the new payscale is launched -we can write a similar equation:
total salary = salary of all officers + salary of all clerks + salary of all bearers
Thus we have : 470 (a +b +c) = 1260 a + 480 c + 200 b
=> 790a + 10 c = 270 b => 79a + c = 27 b .............(ii)
From i and ii we have
76a + c = 26 b
79a + c = 27 b .
For such questions , we need to get a and c in terms of b. That is the number of officers and clerks in terms of the number of bearers.
Subtracting i from ii we get:
3a = b or the number of officers = a =b/3 ..........iii
Putting the value of iii in i we get
76 . (b/3) + c = 26b => c =2b/3
Thus the total number of employees = number of officers + number of clerks + number of bearers = b/3+ 2b/3+ b
=> Total employees = 2b
Number of officers = b/3
Percentage of officers = (Number of officers / total of employees ) x 100 = (b/3) /(2b) x 100 = 16.66 % or
Ans option (b)

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QCS 6 - Question 25

Average temp from Monday to Thursday = 58 and Average temperature from Tuesday to Friday = 66 degree

Monday + Tuesday + Wednesday + Thursday = 4 x 58

Tuesday + Wednesday + Thursday + Friday = 4 x 66

Subtracting equation 1 fro Equation 2

Friday - Monday = 4 x 66 - 4 x 58

Friday - Monday = 4 x 32

Friday : Monday = 11 :7

11X - 7X = 32 or X = 8

Temp on Friday = 11 x 8 = 88 degree

Temp on Monday = 7 x 8 = 56 degree

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QCS 6 Question 26

Average speed = Total Distance / Total time

Total Distance = 20 + 20 = 40 km

Average speed = 30 = 40 /(20/25 + 20/x) = > x = 37.5 kmph

Note : in the above equation => 20/25 is the time required to travel the first 20 km as time = distance / speed

and 20/x is the total time required to travel the last 20 km.. therefore the total time = (20/25 + 20/x)

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QCS-6- 27

Average speed = Total distance / Total time

Distance travelled at 60 kmph = 60 x 35/60 = 35 km

Distance travelled at 40 kmph = 40 x 35/60 = 70/3 km

Total distance = 70/3 km + 35 km = 175/3 km

Total time = 35 + 35 = 70 minutes which is equal to 70/60

Average speed = 175/3 divided by 70/60 = 175 x 60 /( 70 x 3) = 50 kmph

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QCS-6- 29

lET THE three sections be A, B and C

Average marks of section A = 40% AND the number of students in section A =30

Average marks of section B = 30% AND THE number of students in section B = 40

Average marks of section C= 50% AND the number of students in section C = 30

Assume that the total marks to be earned are 100 therefore average score of students in section A =40 AND THE total score of students in section A = 30 X 40 = 1200
Similarly, we can calculate the average marks in section B and C

oVERALL AVERAGE = Total marks of all the three sections / total number of students of all the three sections

= (40 x 30 + 40 x 30 + 50 x 30)/( 30 + 40 + 30) = 39