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Quiz Description :
Name : Average aptitude practice test
Subject : Aptitude
Topic : Average
Questions: 20 mcq
Time Allowed : 25 min
Important for : SBI, SSC, CTET, Bank Exam, GATE, CAT, State Level Competition, Class 10th,11th,12th etc…
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 Question 1 of 20
1. Question
1 pointsThe mean of 3,6,9,6 is…..
CorrectAverage = Sum of numbers/ total numbers
= 3+6+9+6/4 = 24/4 = 6
IncorrectAverage = Sum of numbers/ total numbers
= 3+6+9+6/4 = 24/4 = 6
UnattemptedAverage = Sum of numbers/ total numbers
= 3+6+9+6/4 = 24/4 = 6
 Question 2 of 20
2. Question
1 pointsWhat is the mean of first 5 multiple of 3?
CorrectExplanation Mean of first 5 multiple of 3 = 3+6+9+12+15/5
= 45/5 = 9
Short Method : 3 ( 1+2+3+4+5) /5
IncorrectExplanation Mean of first 5 multiple of 3 = 3+6+9+12+15/5
= 45/5 = 9
Short Method : 3 ( 1+2+3+4+5) /5
UnattemptedExplanation Mean of first 5 multiple of 3 = 3+6+9+12+15/5
= 45/5 = 9
Short Method : 3 ( 1+2+3+4+5) /5
 Question 3 of 20
3. Question
1 pointsFind the average of all prime numbers between 20 to 40.
CorrectExplanation There are four prime numbers between 20 to 40.
They are 23, 29, 31, and 37.
Required average= (23+29+31+37 / 4) =120/4 = 30
IncorrectExplanation There are four prime numbers between 20 to 40.
They are 23, 29, 31, and 37.
Required average= (23+29+31+37 / 4) =120/4 = 30
UnattemptedExplanation There are four prime numbers between 20 to 40.
They are 23, 29, 31, and 37.
Required average= (23+29+31+37 / 4) =120/4 = 30
 Question 4 of 20
4. Question
1 pointsFind the average of first 30 natural numbers.
CorrectExplanation Sum of first n natural numbers = n (n+1)/2
So, sum of first 30 natural numbers = 30(30+1)/2 = 30×31/2 = 465.
Required average= 465/ 30 = 15.5.
IncorrectExplanation Sum of first n natural numbers = n (n+1)/2
So, sum of first 30 natural numbers = 30(30+1)/2 = 30×31/2 = 465.
Required average= 465/ 30 = 15.5.
UnattemptedExplanation Sum of first n natural numbers = n (n+1)/2
So, sum of first 30 natural numbers = 30(30+1)/2 = 30×31/2 = 465.
Required average= 465/ 30 = 15.5.
 Question 5 of 20
5. Question
1 pointsThe average of 11 results is 60. If the average of first six results is 58 and that of the last six is 63, find the sixth result.
CorrectExplanation Sixth result = (58×6+63×660×11) = 66
IncorrectExplanation Sixth result = (58×6+63×660×11) = 66
UnattemptedExplanation Sixth result = (58×6+63×660×11) = 66
 Question 6 of 20
6. Question
1 pointsFind the average of first 25 multiple of 6.
CorrectRequired average = 6(1+2+3+……. +25) ÷25 = (6×25×26) ÷25×2 = 78.
IncorrectRequired average = 6(1+2+3+……. +25) ÷25 = (6×25×26) ÷25×2 = 78.
UnattemptedRequired average = 6(1+2+3+……. +25) ÷25 = (6×25×26) ÷25×2 = 78.
 Question 7 of 20
7. Question
1 pointsThe average of three consecutive even numbers is 30. Find the largest of these numbers.
CorrectExplanation – Let the numbers be x, (x+2), (x+4). Then,
X + (x+2) + (x+4) /3 =30
3x+6 = 3× 30, 3x =906 = 84
X = 84/3 = 28.
so X+4 = 32
IncorrectExplanation – Let the numbers be x, (x+2), (x+4). Then,
X + (x+2) + (x+4) /3 =30
3x+6 = 3× 30, 3x =906 = 84
X = 84/3 = 28.
so X+4 = 32
UnattemptedExplanation – Let the numbers be x, (x+2), (x+4). Then,
X + (x+2) + (x+4) /3 =30
3x+6 = 3× 30, 3x =906 = 84
X = 84/3 = 28.
so X+4 = 32
 Question 8 of 20
8. Question
1 pointsThe average weight of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew, who weight 53 kg, is replaced by a new man. Find the weight of the new man.
CorrectExplanation Total weight increased = (1.8 ×10) kg = 18 kg
Weight of the new man = (53 +18) kg = 71 kg.
IncorrectExplanation Total weight increased = (1.8 ×10) kg = 18 kg
Weight of the new man = (53 +18) kg = 71 kg.
UnattemptedExplanation Total weight increased = (1.8 ×10) kg = 18 kg
Weight of the new man = (53 +18) kg = 71 kg.
 Question 9 of 20
9. Question
1 pointsIf a, b, c, d,e,f and g are seven consecutive odd numbers, their average is:
CorrectExplanation clearly, b = (a+2), c= (a+4), d = (a+6), e = (a+8), f = (a+10), g= (a+12).
Average = a+(a+2)+ (a+4)+(a+6)+(a+8)+ (a+10)+(a+12)÷ 7 = 7a+42/7
= 7 (a+6)/7 = (a+6).
IncorrectExplanation clearly, b = (a+2), c= (a+4), d = (a+6), e = (a+8), f = (a+10), g= (a+12).
Average = a+(a+2)+ (a+4)+(a+6)+(a+8)+ (a+10)+(a+12)÷ 7 = 7a+42/7
= 7 (a+6)/7 = (a+6).
UnattemptedExplanation clearly, b = (a+2), c= (a+4), d = (a+6), e = (a+8), f = (a+10), g= (a+12).
Average = a+(a+2)+ (a+4)+(a+6)+(a+8)+ (a+10)+(a+12)÷ 7 = 7a+42/7
= 7 (a+6)/7 = (a+6).
 Question 10 of 20
10. Question
1 pointsThe average of a non zero number and its square is 4 times the number. The number is.
CorrectExplanation Let the no be x. then,
X+x²/2 = 4x, x²+x =8x
X²+x8x =0, x²7x = 0 » x(x7) =0
X7 = 0, x =7.
IncorrectExplanation Let the no be x. then,
X+x²/2 = 4x, x²+x =8x
X²+x8x =0, x²7x = 0 » x(x7) =0
X7 = 0, x =7.
UnattemptedExplanation Let the no be x. then,
X+x²/2 = 4x, x²+x =8x
X²+x8x =0, x²7x = 0 » x(x7) =0
X7 = 0, x =7.
 Question 11 of 20
11. Question
1 pointsA batsman makes a score of 77 runs in 17^{th} inning and thus increases his average by 3. Find his average after 17^{th} inning.
CorrectExplanation Let the average after 17^{th} inning= x
Then, average after 16^{th} inning (x3).
Hence, 16(x3)+ 77 = 17x
16x48+77 = 17x, 16x17x = 29
x = 29, x =29.
IncorrectExplanation Let the average after 17^{th} inning= x
Then, average after 16^{th} inning (x3).
Hence, 16(x3)+ 77 = 17x
16x48+77 = 17x, 16x17x = 29
x = 29, x =29.
UnattemptedExplanation Let the average after 17^{th} inning= x
Then, average after 16^{th} inning (x3).
Hence, 16(x3)+ 77 = 17x
16x48+77 = 17x, 16x17x = 29
x = 29, x =29.
 Question 12 of 20
12. Question
1 pointsA library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
CorrectExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
IncorrectExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
UnattemptedExplanation Since the month begins with Sunday, so there will be five Sunday in the month.
Required average = (510×5+240×25/30) = 8550÷ 30 = 285.
 Question 13 of 20
13. Question
1 pointsDistance between two stations A and B is 550 km. A train covers the journey from A to B at 70 km per hour and returns back to A with a uniform speed of 42 km per hour. Find the average speed of train during the whole journey.
CorrectExplanation Required average speed = (2xy/ x+y) km/hr. = 2×70×42/ (70+42)
= 5880/112 = 52.5 km/hr.
IncorrectExplanation Required average speed = (2xy/ x+y) km/hr. = 2×70×42/ (70+42)
= 5880/112 = 52.5 km/hr.
UnattemptedExplanation Required average speed = (2xy/ x+y) km/hr. = 2×70×42/ (70+42)
= 5880/112 = 52.5 km/hr.
 Question 14 of 20
14. Question
1 pointsThe average of ten numbers is 7. If each number is multiplied by 12, then the average of the new set of numbers is:
CorrectExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
IncorrectExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
UnattemptedExplanation Average of 10 numbers =7.
Sum of these 10 numbers = (10×7) =70.
Because, x₁+x₂+x₃+……..+x₁₀ = 70
» 12x₁+12x₂+12x₃+……..+12x₁₀ = 12×70 = 840
» 12x₁+12x₂+12x₃+……..+12x₁₀/10 = 84
Hence Average of new numbers is 84.
 Question 15 of 20
15. Question
1 pointsThe average of three numbers is 7. If the average of first two numbers is 10.What is the third number:
CorrectExplanation Average of three numbers is = 7.
Total of three numbers = 3×7 =21
Average of first number = 10,
Total of two numbers = 2× 10 = 20
Hence, Third number = 21 20 = 1
IncorrectExplanation Average of three numbers is = 7.
Total of three numbers = 3×7 =21
Average of first number = 10,
Total of two numbers = 2× 10 = 20
Hence, Third number = 21 20 = 1
UnattemptedExplanation Average of three numbers is = 7.
Total of three numbers = 3×7 =21
Average of first number = 10,
Total of two numbers = 2× 10 = 20
Hence, Third number = 21 20 = 1
 Question 16 of 20
16. Question
1 pointsA motorist travels at to a place 150 km away at an average speed of 50km/hr and returns at 30 km/hr. His average speed for the whole journey in km/hr is:
CorrectExplanation Average speed =2xy/(x+y) km/hr. = 2×50×30/50+30 =37.5 km/hr.
IncorrectExplanation Average speed =2xy/(x+y) km/hr. = 2×50×30/50+30 =37.5 km/hr.
UnattemptedExplanation Average speed =2xy/(x+y) km/hr. = 2×50×30/50+30 =37.5 km/hr.
 Question 17 of 20
17. Question
1 pointsThe average age of 30 students in a group is 10 years. When teacher’s age is included to it, the average increases by one. What is the teacher’s age in years?
CorrectExplanation Age of teacher = (31×1130×10) years = 341300 =41
Hence the age of teacher is 41 years.
IncorrectExplanation Age of teacher = (31×1130×10) years = 341300 =41
Hence the age of teacher is 41 years.
UnattemptedExplanation Age of teacher = (31×1130×10) years = 341300 =41
Hence the age of teacher is 41 years.
 Question 18 of 20
18. Question
1 pointsThe average of four numbers is 25.If one number is excluded, the average become 22. The excluded number is:
CorrectExplanation Excluded number = (25×4)(22×3) = 10066 = 34
IncorrectExplanation Excluded number = (25×4)(22×3) = 10066 = 34
UnattemptedExplanation Excluded number = (25×4)(22×3) = 10066 = 34
 Question 19 of 20
19. Question
1 pointsThree years ago, the average age of A and B was 18 years. With C joining them, the average age becomes 22 years. How old is C now?
CorrectExplanation Present age of (A+B) = (18×2+3×2) years = 42 years
Present age of (A+B+C) = (22×3) years =66 years
Hence, C’s age = (6642) = 24 years.
IncorrectExplanation Present age of (A+B) = (18×2+3×2) years = 42 years
Present age of (A+B+C) = (22×3) years =66 years
Hence, C’s age = (6642) = 24 years.
UnattemptedExplanation Present age of (A+B) = (18×2+3×2) years = 42 years
Present age of (A+B+C) = (22×3) years =66 years
Hence, C’s age = (6642) = 24 years.
 Question 20 of 20
20. Question
1 pointsA car travel 250 km in 6 hours and next 260 km also in 6 hours. What is average speed of car?
CorrectExplanation Total distance travel by car = (250+260) km. = 510 km
Total time = (6+6) hr. =12 hr.
Hence, Average speed of car = Total distance/ total time = 510/12 =42.5 km/hr.
IncorrectExplanation Total distance travel by car = (250+260) km. = 510 km
Total time = (6+6) hr. =12 hr.
Hence, Average speed of car = Total distance/ total time = 510/12 =42.5 km/hr.
UnattemptedExplanation Total distance travel by car = (250+260) km. = 510 km
Total time = (6+6) hr. =12 hr.
Hence, Average speed of car = Total distance/ total time = 510/12 =42.5 km/hr.