Quant Questions – Average Problems with Solution Set 1

Dear Readers, Important Average Problems for upcoming competitive Exams with Solutions. Average is an important topic from banking exam point of view. You could solve Average problems easily if you understand the basic concepts. Aspirants preparing for SBI Clerk, SBI PO, IBPS Clerk, IBPS PO, Insurance, RBI and other competitive examination can make of use these average problems. We have included some important average problems that are repeatedly asking in competitive exams. Practice lot of questions to become an expert in solving average problems and learn to use shortcuts.

Average Problems – Set 1

1).The average height of a group of 12 boys increases by 2 inches when a boy a height 52 inches is replace by another boy. What is the height of the new boy?
a)  54
b)  76
c)  72
d)  48
e)  None of these
2).The average age of members of a family in the year 2000 was 35 years. What is the average age of the family in year 2002?
a)  36
b)  34
c)  37
d)  Cannot be determined
e)  None of these
3).The average marks of Chintu in 4 subjects is 60. In the 5th subject he scored 75 marks. What marks should be scored in the 6th subject if his average marks in all the 6 subjects is 2 more than his average marks in first 5 subjects.
a)  70
b)  72
c)  81
d)  85
e)  None of these
4).Average marks obtained by Amitabh, Vinod& Shashi are 80 and the average marks obtained by Praveen and Smita are 65. If the average marks obtained by Amitabh, Vinod, Shashi, Praveen, Smita, Hema, Rekha and Jaya are 72.5, then find the average marks obtained by Hema, Rekha& Jaya.
a)  60
b)  70
c)  75
d)  73
e)  None of these
5).The average age of a family of 6 members, 3 years ago was 28 years. Meanwhile a child was born in the family and the average age of the family members is 27 years today. What would be the age of the child so born, 2 years from now?
a)  3 years
b)  5 years
c)  2 years
d)  1 years
e)  None of these
6).Find the average of first 50 natural number.
a)  50
b)  25
c)  25.5
d)  24
e)  None of these
7).Nitish travelled a distance of 50 km at speed of 10 kmph, next 60 km at speed of 15 kmph and remaining 40 km at speed of 20 kmph. Find the average speed of Nitish for the entire journey.
a)  16 kmph
b)  15 kmph
c)  13.6 kmph
d)  15.5 kmph
e)  None of these
8).The difference between the average of first 52 even numbers and first 49 odd numbers is ______.
a)  3
b)  2
c)  -3
d)  -2
e)  None of these
9).The average of 2 prime numbers lying between 20 and 40 is 30. The product of the 2 prime number is ________.
a)  851
b)  899
c)  951
d)  990
e)  Either (a) or (b)
Answers:                         
1). b) 2). c) 3). e) 4). b) 5). b) 6). c) 7). c) 8). e) 9). e)
Solutions:
1).Average height increases by 2 inches i.e. sum of height increases by 2×12=24 inches
:. Height of new boy = 52+24= 76 inches
Answer: b)
 
2).Since the age of all the members would increases by 2, hence the average age would also increase by 2 and hence average age in 2002 would be 35+2=37 years.
Answer: c)
 
3).Let Chintu’s score in 6th subject be “x” then [(60×4+75+x)/6]-2=(60×4+75)/5
(303+x)/6=315/5
à1515 + 5x = 1890
X=75
Alternate Method:
Average marks in five subject = 315/5 = 63, also sum total of marks in 6 subjects should be more by 6×2=12, hence marks in the 6th subject = 63 + 12 = 75 marks.
Answer: e)
 
4).Required average
= [(72.5×8)-(80×3)-(65×2)/3]
=(580-240-130)/3 =70
Answer: b)
 
5).Average age of 6 members today is 28 + 3 = 31 years
Let present age of the child be ‘x’ years.
So, (6×31+x)/7=27
186+x = 189,
x=3years, 2 years from now, the age of the child would be 5 years.
Answer: b)
 
6).Sum of ‘n’ natural no. = [n(n+1)]/2 and hence average = (n+1)/2
= (50+1)/2
=51/2=25.5
Answer: c)
 
7).Average speed = (Total distance/Total time)
Total distance= 50 + 60 + 40=150 km
Total time = (50/10)+(60/15)+(40/20)=5+4+2=11 hrs
Time=(Distance/Speed)
Average Speed = 150/11 = 13.66 kmph
Answer: c)
 
8).Average of first ‘n’ even no. = n+1, hence average of first 52 even no. = 52+1=53.
Also, average of first ‘n’ odd no. = n, hence average of first 49 odd numbers = 49
Required difference = 53-49=4
Answer: e)
 
9).Prime numbers lying between 20& 40 are: 23, 29, 31, 37
We have two sets of Prime numbers lying between 20& 40, whose average is 30, these sets are (23, 37) and (29, 31).
Product of the 2 prime numbers
= (23 × 37) or (29 × 31) = 851 or 899
Answer: e)
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