Problems on Average with steps : Kerala PSC Degree Level Questions
1. In the first 10 overs of a cricket game, the run rate
was only 3.2. What should be the run rate in the
remaining 40 overs to reach the target of 282 runs?
A. 6.25, B. 5.5, C. 7.4 ,D. 5
Answer : Option A
Explanation :
Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282
remaining runs to be scored = 282 - 32 = 250
remaining overs = 40
Run rate needed = 250/40=6.25
2. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855,
Rs. 7230 and Rs. 6562 for 5 consecutive months.
How much sale must he have in the sixth month so
that he gets an average sale of Rs.6500?
A. 4800 B. 4991 C. 5004 D. 5000
Answer : Option B
Explanation : Let the sale in the sixth month = x
Then( 6435+6927+6855+7230+6562+x)/6=6500
= > 6435 + 6927 + 6855+ 7230 + 6562 + x
= 6 × 6500 = 39000
= > 34009 + x = 39000
= > x = 39000 - 34009 = 4991
3. The average of 20 numbers is zero. Of them, How
many of them may be greater than zero, at the
most?
A. 1 B. 20 C. 0 D. 19
Answer : Option D
Explanation : Average of 20 numbers = 0
=>Sum of 20 numbers/20=0
=> Sum of 20 numbers = 0
Hence at the most, there can be 19 positive
numbers.
(Such that if the sum of these 19 positive
numbers is x, 20th number will be -x)
4. The captain of a cricket team of 11 members is 26
years old and the wicket keeper is 3 years older.
If the ages of these two are excluded, the
average age of the remaining players is one year
less than the average age of the whole team. Find
out the average age of the team.
A. 23 years B. 20 years C. 24 years D. 21 years
Answer : Option A
Explanation :Number of members in the team = 11
Let the average age of the team = x
=>(Sum of the ages of all the 11 members
of the team)/11=x
=> Sum of the ages of all the 11 members
of the team = 11x
Age of the captain = 26
Age of the wicket keeper = 26 + 3 = 29
Sum of the ages of 9 members of the team
excluding captain and wicket keeper
= 11x - 26 - 29 = 11x - 55
Average age of 9 members of the team
excluding captain and wicket keeper
=(11x”55)/9
Given that( 11x”55)/9=(x”1)
=> 11x - 55 = 9(x - 1)
=> 11x - 55 = 9x - 9
=> 2x = 46
=>x=46/2=23 years
5. The average monthly income of A and B is Rs. 5050.
The average monthly income of B and C is Rs. 6250
and the average monthly income of A and C is Rs.
5200. What is the monthly income of A?
A. 2000 B. 3000 C. 4000 D. 5000
Answer : Option C
Explanation :Let the monthly income of A = a
monthly income of B = b
monthly income of C = a
a +b=2×5050------------(Equation1)
b + c=2×6250------------(Equation2)
a +c=2×5200------------(Equation3)
(Equation 1) + (Equation 3) - (Equation 2)
=> ( a +b +a +c) “ (b +c)
=(2×5050)+(2×5200)”(2×6250)
=> 2a = 2(5050 + 5200 - 6250)
=> a = 4000
=> Monthly income of A = 4000
7. In Kiran’s opinion, his weight is greater than 65 kg
but less than 72 kg. His brother does not agree
with Kiran and he thinks that Kiran’s weight is
greater than 60 kg but less than 70 kg. His mother’s
view is that his weight cannot be greater than 68
kg. If all are them are correct in their estimation,
what is the average of different probable weights of
Kiran?
A. 70 kg B. 69 kg C. 61 kg D. 67 kg
Answer : Option D
Explanation : Let Kiran’s weight = x. Then
According to Kiran, 65 < x < 72 ——————
(equation 1)
According to brother, 60 < x < 70 ——————
(equation 2)
According to mother, xd”68 ——————(equation
3) Given that equation 1,equation 2 and equation
3 are correct. By combining these equations,
we can write as
65<xd”68
That is x = 66 or 67 or 68
average of different probable weights of Kiran =
(66+67+68)/3=67
8. The average weight of 16 boys in a class is 50.25
kg and that of the remaining 8 boys is 45.15 kg.
Find the average weights of all the boys in the
class.
A. 48.55 B. 42.25 C. 50 D. 51.25
Answer : Option A
Explanation :
Average weight of 16 boys = 50.25
Total Weight of 16 boys = 50.25×16
Average weight of remaining 8 boys = 45.15
Total Weight of remaining 8 boys = 45.15×8
Total weight of all boys in the class =
(50.25×16)+(45.15×8)
Total boys = 16+8=24
Average weight of all the boys =
(50.25×16)+(45.15×8)/24
= (50.25×2)+(45.15×1)/3
= (16.75×2) + 15.05
=33.5+15.05=48.55
9. A library has an average of 510 visitors on Sundays
and 240 on other days. What is the average number
of visitors per day in a month of 30 days beginning
with a Sunday?
A. 290 B. 304 C. 285 D. 270
Answer : Option C
Explanation :in a month of 30 days beginning with
a Sunday, there will be 4 complete weeks and
another two days which will be Sunday and Monday
Hence there will be 5 Sundays and 25 other days in
a month of 30 days beginning with
a Sunday
Average visitors on Sundays = 510
Total visitors of 5 Sundays = 510×5
Average visitors on other days = 240
Total visitors of other 25 days = 240×25
Total visitors = (510×5)+(240×25)
Total days= 30
Average number of visitors per day ={ (510×5)
+ (240×25)}/30
= { (51×5)+(24×25) }/3
= (17×5) + (8×25)
=85+200=285
10. A student’s mark was wrongly entered as 83
instead of 63. Due to that the average marks for
the class got increased by half 1/2. What is the
number of students in the class?
A. 45 B. 40 C. 35 D. 30
Answer : Option B
Explanation : Let the total number of students =
x
The average marks increased by 1/2 due to
anincrease of 83-63=20 marks.
But total increase in the marks = 1/2×x=x/2
Hence we can write asx/2=20•Ëx=20×2=40
Posts
