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1. Find the average of first 67 numbers.
Sol:
average = \(\frac{{\frac{{n(n + 1)}}{2}}}{n}\) = \(\frac{{n + 1}}{2}\) = \(\frac{{67 + 1}}{2}\) = 34      Answer

2. The average of 35 numbers is 380. If each of the number is divided by 19. Find the new average.
Sol:
New Average = \(\frac{{old\;average}}{{19}}\) = \(\frac{{380}}{{19}}\) = 20     Answer

3. The average of 9 result is 60. If the average of first 5 result is 50 and the average of last 5 result is 70 then find the 5ᵗʰ result.
Sol:
Method(1):
5ᵗʰ result = 5 × 50 + 5 × 70 – 9 × 60 = 60      Answer

Method(2):555070500 × 5+10+1060Answer

4. The average of 50 numbers is 45. The average of 50 numbers and 3 new number is 51. Then find the average of three new numbers.
Sol:
Method(1):
Sum of three numbers = (50 + 3)×51 – 50×45
= 453
∴ average of three new numbers = \(\frac{{453}}{3}\) = 151      Answer

Method(2):
Increase in average = 51 – 45 = 6
A = 45
N = 50
 n = 3
∴ Average of new numbers = A + \(\left( {1 + \frac{N}{n}} \right)\).x
= 45 + \(\left( {1 + \frac{{50}}{3}} \right)\).6
151           Answer

Method(3):

5035345 ? 51-6×50- 300-6+100 i.e. ? value is100 above 51i.e. 151 Answer

5. The average salary of 30 workers in an office is Rs. 2400 per month. If the manager’s salary is added, the average becomes Rs. 2500 per month. Find the manager’s annual salary.
Sol:
Method(1):
Manager’s per month salary = (30 + 1)×2500 – 30×2400
= 5500
∴ annual salary of manager = 5500×12
66000 Rs.        Answer

Method(2):
N = 30
x = 2500 – 2400 = 100
A = 2400
n = 1
∴ per month salary of manager = A + \(\left( {1 + \frac{N}{n}} \right).x\)
= 2400 + \(\left( {1 + \frac{{30}}{1}} \right).100\)
 = 5500
∴ annual salary of manager = 5500 × 12
 = 66000 Rs.      Answer

Method(3):

301312400 ? 2500-100+300 more than2500= 3000 + 2500= 5500 per month Answer-100×30 = -3000∴ annual salary = 5500 × 12 = 6600 Rs. Answer

6. In a class there are 10 students at the age of 15 years. 15 students at the age of 10 years and 22 students of age 18 years. What is the average age of a student ?
Sol:
Method(1):
Average age of a student = \(\frac{{10 \times 15 + 15 \times 10 + 22 \times 18}}{{10 + 15 + 22}}\)
= \(\frac{{696}}{{47}}\)
= \(14\frac{{38}}{{47}}\) years/student       Answer

Method(2):

10152215 10 18+2×10= +20-3×15= -45+5×22= +11013 Answer+

7. A zoo has an average of 600 visitors on sunday and 390 visitors on other days. the average number of visitors per day in a month at 30 day beginning with sunday will be How many ?
Sol:
If the month starts with sunday then there will be 5 sunday in the month

525+200×5-10×25600 390+25400425 Answer

8. The average age of 44 students in a class is 21 years. If the teacher’s age is included then average becomes 21 year 6 months. Find the age of teacher.
Sol:
Method(1):
Age of teacher = (44 + 1)21.5 – 44×21
43.5 year        Answer

Method(2):
A = 21
N = 44
n = 1
x = 0.5
age of teacher = A + \(\left( {1 + \frac{N}{1}} \right).x\)
= 21 + \(\left( {1 + \frac{{44}}{1}} \right)0.5\)
= 43.5 years        Answer

Method(3):

44121 43.521.545Answer-0.5-0.5×44 = -22+22i.e. 22 above assumedaverage of 21.5

9. Find the average of following 12 numbers:
2, 10, 7, 9, 13, 12, 14, 18, 19, 5, 8, 3
Sol:
Method(1):

Average = \(\frac{{2 + 10 + 7 + 9 + 13 + 126 + 14 + 18 + 19 + 5 + 8 + 3}}{{12}}\)
10 Answer

Method(2):

2, 10, 7, 9, 13, 12, 14, 18, 19, 5, 8, 31210 -2 -5 -3 +1 0 +2 +6 +7 -7 -4 -9 -2 = - 210 Answer

10. Group A has 10 elements and their average is 5, Group B has 20 elements and their average is 8. If elements of these two groups are combined to form a third group C. Then find the average of Group C ?
Sol:
Method(1):

Average = \(\frac{{10 \times 5 + 20 \times 8}}{{10 + 20}}\) = 7    Answer

Method(2):
Using Alligation:

5 810 207Answer1 : 2×1×18 - 5 = 31211Method(3):10 205850 +3×20= 7 Answer