PRACTICE QUESTIONS ON AVERAGES

Total age of 50 students and 1 teacher = 25 x 51 = 1275
Total age of the students = 24 x 50 = 1200. { Average age reduces by 1 when teachers age is not included}
Therefore the age of the teacher is 1275 – 1200 = 75 years. Ans. (b)

Total age of A + B + C = 80 x 3 = 240 kgs.
Total age of A + B + C + D = 75 x 4 = 300 kgs.
Weight of D = 300 – 240 = 60 kgs.
Also the weight of E (2 kgs more than D) = 62 kgs.
A is replaced by E
E + B + C + D = 77 x 4 = 308
Difference between ages of E and A is 308 – 300 = 8 kgs
A’s weight = 62 – 8 = 54 kgs
Weight of A = 54 kgs. Ans. (a)

Total age of all the students is average age multiplied by total students.
⇒ 11 ¾ x 600 = 7050 — (1)
Let the total number of girls be A
Then, total number of boys = 600 – A
Total age of girls and boys = A x 11 and (600 – A) x 12 respectively
Therefore, ⇒ (11A) + (600 – A) x 12 = 7050
⇒ – A + 7200 + 7050
⇒ A = 7200 – 7050 = 150
Number of girls in school is 150. Ans. (b)

Total rainfall from Monday to Thursday
M, T, W, Th = 420.5 x 4 = 1,682 cms. — (1)
Total rainfall from Tuesday to Friday
T, W, Th, Fr = 440.5 x 4 = 1762 cms — (2)
Subtracting equation (1) from (2)
Difference between rainfall of Fr and M = 1,762 – 1682 = 80cms
20 : 21 can be written in numbers as 20x : 21x
Therefore 21x – 20x = 80 ⇒ x = 80
Rainfall on Monday = 20 x 80 = 1600
Rainfall on Friday = 21 x 80 = 1680
1600 cms and 1680 cms. Ans (d)

Let the average score till 14th inning be x.
Therefore total score till 14th inning = 14x
Total score after 15th inning = 14x + 58
Average score after 15th inning = 14x + 58/15 —(1)
Average after 15th inning = x + 3. — (as per the question) (2)
Equating (1) and (2)
⇒ 14x + 58/15 = x + 3 ⇒ 15x – 14x = 58 – 45 ⇒ x = 13
Average after 15th inning = 13 + 3 = 16. Ans (c)

Since, the average of 7 numbers = 8 their total = 8 x 7 = 56
One number is added then there are 8 numbers with the average 9.
Total of 8 numbers with average 9 = 8 x 9 = 72.
The added number is therefore:,
72 – 56 = 16. Ans. (c)

Let the number of males members be X and female members be Y.
Then,
Total age of male members = 34 x X
Total age of female members = 26 x Y
Total age of all the members = (X + Y) x 30
⇒ 34X + 26Y = (X + Y) x 30
⇒ 34X + 26Y = 30X + 30Y
⇒ 4X = 4Y ⇒ X = Y
Percentage of Males is 50%. Ans. (a)

Total of ages of jury of 5 = 40 x 5 = 200
A member resigns the total reduces = 200 – 35 = 165
When a new member joins, the total increases = 165 + 25 = 190
Average age of the new jury is
= 190/5
= 38. Answer (b).

Let the average of runs scored by Raju (R), Shyam (S), Hari (H) be x
Then their total = R + S + H = 3x → (1)
Average runs scored by Shyam (S), Hari (H) and Kishore (K) is x + 7
Then their total = S + H + K = (x + 7) x 3 → (2)
Subtracting (1) from (2)
R – K = 3x – 3x – 21
Difference between scores of Kishore and Raju = 21
Since Kishore’s score is 35, Raju’s score is 35 – 21 = 14
Raju’s score = 14. Ans. (a)

Total of ages of 30 boys = 30 x 12 = 360.
Total of ages of 32 boys = 32 x 12.25 = 392.
Total of ages of 2 new boys = 392 – 360 = 32 years
Average age of 2 new boys = 32/2 years
16 years. Ans (d)

Let the number be N.
∴ (0.1N + 0.25N + 0.5N + 0.75N)4 = 24
⇒ 1.6N/4 = 24
⇒ 1.6N = 96
⇒ N = 60

Let the total number of workers be W
Total salary of the workers = 8000W
Total salary of 7 technicians = 12000 x 7 = 84000
Total salary of others (W – 7) = 6000 x (W – 7)
∴ 8000W = 84000 + 6000W &ndash 42000
⇒ 2000W = 42000
⇒ W = 21

The total of 5 numbers = 27 x 5 = 135
The total of 4 numbers = 25 x 4 = 100
Therefore the excluded number = 135 – 100 = 35