Practice
Problems on Averages – Set 6
1. On a journey across Chennai, a tourist bus went
at 10 km/h for 30% of the distance, at 40 km/h for 40% of distance and at 30
km/h for the remaining distance. The average speed for the whole journey was:
a) 10 kmph
b) 20 kmph
c) 5 kmph
d) 24 kmph
e) None of these
Solution:
Let the total distance be
x km.
We know, Distance = Speed
× Time ⇒ Time = Distance/Speed
Time required to cover
30% of the distance at 10 km/h = 0.3x/10 = 3x/100 hrs
Time required to cover
40% of the distance at 40 km/h = 0.4x/40 = x/100 hrs
Now, distance remaining =
x – 0.3x – 0.4x = 0.3x
∴ Time required to cover remaining 0.3x at 30 km/h = 0.3x/30)
= x/100 hrs
So, total time required
to complete x km distance = (3x/100) + (x/100) + (x/100) = 5x/100 hrs
∴ The average speed for the whole journey = x/(5x/100) = 20
km/h
2. Average age of a class of 25 students is 13
years. The average age of a group of 8 students is 12 years and the average age
of another group of 10 students is 11 years. What is the average age of the
rest of the students?
a) 10 years
b) 13 years
c) 15 years
d) 16 years
e) 17 years
Answer: E)
Solution:
We know that Average =
Sum of all quantities/total number of quantities
∴ Sum of ages of students = Average × number of students
The average age of 25
students is 13
∴ Sum of ages of 25 students = Average × number of students
⇒ Sum of ages of 25 students = 25 × 13 = 325 years
Average age of group of 8
students is 12.
⇒ Sum of ages of 8 students = 12 × 8 = 96 years
Average age of group of
10 students is 11.
⇒ Sum of ages of 10 students = 11 × 10 = 110 years
Let the number of
remaining students be x
∴ x + 8 + 10 = 25
⇒ x + 18 = 25
⇒ x = 7 students
∴ Sum of ages of 7 students = (Sum of ages of 25 students) – (Sum of ages of 8 students + Sum of
ages of 10 students)
⇒ Sum of age of 7 students = 325 – (96 + 110) = 325 – 206 =
119 years
∴ Average age of 7 students = Sum of ages of 7 students/7
⇒ Average age of 7 students = 119/7 = 17
∴ Average age of remaining students is 17 years
3. The average age of a family of five members is
24. If the present age of the youngest member is 8 years, what was the average
age of the family at the time, when the youngest member was about to born?
a) 20 years
b) 16 years
c) 12 years
d) 18 years
e) 22 years
Answer: A)
Solution:
Let present ages of five
family members be x1 , x2 , x3, x4 and
x5.
There average be (x1 +
x2 + x3 + x4 + x5)/5 = 24
⇒ x1 + x2 + x3+ x4 +x5=
24 × 5 = 120
Since, Age of
youngest member x5 is 8 years
x1 + x2 +
x3+ x4 + 8 = 120
⇒ x1 + x2 + x3+ x4 =
112. ----------(1)
Since youngest member has
taken birth 8 years ago, sum of ages of family members at that time was
(x1 - 8)
+ (x2 - 8) +(x3 - 8) +(x4 - 8)
= x1 + x2 +
x3+ x4 – 32
Putting equation one in
this equation, we get
8 years ago, sum of ages
of family members = 112 – 32 = 80
And average of their ages
at that time = 80/4 = 20 years
4. Nine men went to a hotel, 8 of them spent Rs. 3
each over their meals and the ninth spent Rs. 2 more than the average
expenditure of all the nine. The total money spent by all of them was
a) Rs.26.54
b) Rs.40.70
c) Rs.29.25
d) Rs.27.75
e) None of these
Answer: C)
Solution:
Let us assume the average
expenditure of all nine men = Rs. x
The amount spent by 8 men
= 8 × 3 = 24 (∵ it is given that 8 men spent Rs.3
each)
The amount spent by ninth
man = x + 2
∵It is stated that the amount spent by ninth man is 2 more
than average of all nine.
∴ Total expenditure of all nine men = 24 + (x + 2)
We know that, Average
expenditure = (Total expenditure)/(Total men)
Hence, total expenditure
= 9x
On equating total expenditures,
we get
⇒ 24 + x + 2 = 9x ⇒ 8x = 26 ⇒ x = 26/8 ⇒ x = Rs. 3.25
∴ Total expenditure of all nine men = 9 × 3.25 = Rs.29.25
Hence, the expenditure of
all nine men is Rs.29.25
5. The monthly salary of a person was Rs. 320 for
each of the first three years. He next got annual increment of Rs. 40 per month
for each of the following successive 12 years. His salary remained stationery
till retirement when he found that his average monthly salary during his
service period was Rs. 698. Find the period of his service.
a) 25 years
b) 15 years
c) 40 years
d) 50 years
e) 30 years
Answer: C)
Solution:
Total monthly salary for
1st 2 years = 320 + 320 = 640
The salary in next 13
years is in arithmetic progression with a common difference of 40. (Starting
from 3rd year)
nth term
of A.P = tn = a + (n – 1)d,
Where, a is 1st term
and d is the difference.
3rd year’s
salary = 1st term = a = 320/month
Salary after 12 years =
13th term = L = 320 + (13 – 1) × 40
= 800 /month
Sum of n terms of A.P =
n(a + L)/2
∴ Sum of 13 years of salary/month = 13 × (320 + 800)/2 = 7280
Along with the initial 2
years, Sum of 15 years of salary per month = 640 + 7280 = 7920
Let, he worked for x
years more until retirement on 800/month salary.
So, total salary/month
for (15 + x) years = 7920 + 800x
Average salary for (15 +
x) years = 698
∴ 698 = (7920 + 800x)/ (15 + x)
⇒ 698 × 15 + 698x =
7920 + 800x
⇒ 2550 =102x
⇒ x = 25
So, total years of
service = 15 + 25 = 40 years
6. In an exam, the average was found to be 50
marks. After deducting computational errors the marks of 100 candidates had to
be changed from 90 to 60 each, and the average came down to 45 marks. The total
number of candidates who took the exam was:
a) 300
b) 600
c) 200
d) 150
e) None of these
Answer: B)
Solution:
Let number of candidates
giving the exam = b
Then, total marks
obtained by b candidates = 50b.
Now,
⇒ Change in computational marks for 1 candidate = 30,
⇒ Change in computational marks for 100 candidates = 3000,
Then, Final marks of b
candidates = 45b.
Then according to the
condition,
⇒ 50b – 3000 = 45b
⇒ 5b = 3000
⇒ b = 600
Hence, 600 students took
the test.
7. In the month of April, Rahul’s average
expenditure for the first 10 days was Rs 800 per day. If his average
expenditure for the next 10 days decreases by Rs25 and after 10 days the
average expenditure again decrease by Rs.25 and he earns Rs 25, 000 per month.
Calculate his savings in the month of April.
a) Rs.1500
b) Rs.2000
c) Rs.1850
d) Rs.1950
e) Rs.1750
Answer: E)
Solution:
Average expenditure of
days 1 to 10 = Rs 800
Total expenditure for
days 1 to 10 = Rs 800 × 10 = 8000
Total expenditure for
days 11 to 20 = Rs (800 – 25) × 10 = Rs 775 × 10 = Rs 7750
Total expenditure for
days 21 to 30 = Rs (775 – 25) × 10 = Rs. 750 × 10 = Rs. 7500
Total expenditure= Rs
(7500 + 7750 + 8000) = Rs. 23250
Savings = Income –
Expenditure = Rs 25000 – Rs 23250 = Rs. 1750
8. The average age of 30 members in a group is 15.
One member aged 20 years left the class, but two new members come in his place
the difference of whose ages was 5 years. If the average age of all the boys
now in the class still remains 15 years, the age of the younger newcomer is:
a) 20 years
b) 15 years
c) 30 years
d) 10 years
e) 18 years
Answer: B)
Solution:
Now after 1 member of 20
year left and two members of age ‘X’ years and
‘x + 5’ year joined. The
number of members become 31.
∴ Required average age = Total age of 31 members/31
⇒ (450 - 20 + x + x + 5)/31=15 ⇒ 435 +
2x = 465 ⇒ 2x = 30
⇒ x = 15 years
9. The sum of five numbers is 290. The arithmetic
average of first two numbers is 48.5 and the arithmetic average of last two
numbers is 53.5. What is the third number?
a) 89
b) 84
c) 90
d) 87
e) 86
Answer: E)
Solution:
Let five numbers are a,
b, c, d, e
Average = sum of digits /
no. of digits
Given average of first
two numbers = 48.5
⇒ Sum of first 2 numbers = a + b = 2 × 48.5 = 97
Given Average of last two
numbers = 53.5
⇒ Sum of last 2 numbers = d + e = 53.5 × 2 = 107
Given, a + b + c +
d + e = 290
⇒ 97 + c + 107 = 290
⇒ c = 290 – 204
⇒ c = 86
10. The average of 8 observations was 25.5. It
was noticed later that two of those observations were wrongly taken. One
observation was 14 more than the original value and the other
observation was wrongly taken as 31 instead of 13. What
will be the correct average of those 8 observations?
a) 20.5
b) 21.5
c) 22.5
d) 20
e) 21
Answer: B)
Solution:
Actual total must be
lesser by 14 + (31 – 13) = 32
Therefore, actual average
must be lesser by 32/8 = 4
i.e., correct average = 25.5 – 4 = 21.5
