Practice Average - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Out of seven given numbers, the average of the first four numbers is 4 and that of the last four numbers is also 4. If the average of all the seven numbers is 3, fourth number is
(a)
(b)
(c)
(d)
Using Rule 1,
Average of two or more numbers/quantities is called the mean of these numbers, which is given by
$\text"Average(A)" = \text"Sumof observation / quantities"/\text"No of observation / quantities"$
∴ S = A × n
Fourth number
= (4×4+4×4 – 3×7)
= (16 +16-21) = 11
Q-2) The average of first three numbers is thrice the fourth number. If the average of all the four numbers is 5, then find the fourth number.
(a)
(b)
(c)
(d)
Let the numbers be a, b, c,
${\text"a + b + c"}/3$ = 3d
⇒ a + b + c = 9d
Again, ${\text"a + b + c + d"}/4$ = 5
⇒ a + b + c + d = 20
⇒ 9d + d = 20
⇒ 10d = 20 ⇒ d = 2
Q-3) The average marks obtained by 22 candidates in an examination are 45. The average marks of the first 10 candidates are 55 and those of the last eleven are 40. The number of marks obtained by the eleventh candidate is
(a)
(b)
(c)
(d)
Marks obtained by eleventh candidate
= 22 × 45 – (10 × 55 + 11 × 40)
= 990 – (550 + 440)
= 990 – 990 = 0
Q-4) A student finds the average of ten 2-digit numbers. While copying numbers, by mistake, he writes one number with its digits interchanged. As a result his answer is 1.8 less than the correct answer. The difference of the digits of the number, in which he made mistake, is
(a)
(b)
(c)
(d)
Difference in average = 1.8
∴ Difference between the number and the number formed by interchanging the digits
= 1.8 × 10 = 18
(Since 53 – 35 = 18)
∴ Number = 35
∴ Difference of digits = 5 – 3 = 2
Q-5) The average of 9 integers is found to be 11. But after the calculation, it was detected that, by mistake, the integer 23 was copied as 32, while calculating the average. After the due correction is made, the new average will be
(a)
(b)
(c)
(d)
Sum of 9 integers = 9 × 11 = 99.
New average = ${90+ 23- 32}/9$ = $90/9$ =10
Aliter : Using Rule 26,
The correct average will be = m +${\text"(a-b)"}/ \text"n"$.
Here, n = 9, m = 11
a = 23, b = 32
Correct mean = m + ${\text"(a - b)"}/\text"n"$
= 11 +${(23 -32)}/9$
= 11 +${(-9)}/9$
= 11 – 1 = 10
Q-6) A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was
(a)
(b)
(c)
(d)
Required number of wickets = x (let)
According to question,
12.4 × x + 26 = (x + 5) (12.4 – 0.4) = (x + 5) × 12
⇒ 12.4x + 26 = 12x + 60
⇒ 12.4x – 12x = 60 – 26
⇒ 0.4x = 34
⇒ x = $34/{0.4}$ = $340/4$ = 85
Q-7) The averages of runs scored by a cricket player in 11 innings is 63 and the average of his first six innings is 60 and the average of last six innings is 65. Find the runs scored by him in the sixth innings.
(a)
(b)
(c)
(d)
Runs scored by the cricketer in the 6th innings
= 6 × 60 + 6 × 65 – 11 × 63
= 360 + 390 – 693 = 57
Q-8) A batsman in his 12th innings makes a score of 63 runs and there by increases his average scores by 2. What is his average after the 12th innings?
(a)
(b)
(c)
(d)
Extra runs = 12 × 2 = 24
∴ Required average = 63 – 24 = 39
Q-9) The average of runs scored by a cricketer in his 99 innings is 99. How many runs will he have to score in his 100th innings so that his average of runs in 100 innings may be 100?
(a)
(b)
(c)
(d)
Number of runs scored in 100th innings
= 100 × 100 – 99 × 99
= 10000 – 9801 = 199
OR
Increase in average = 1 run
∴ Runs scored in 100th innings = 100 + 99 = 199
Q-10) A cricketer whose bowling average is 24.85, runs per wicket, takes 5 wickets for 52 runs and thereby decreases his average by 0.85. The number of wickets taken by him till the last match was :
(a)
(b)
(c)
(d)
Let the no. of wickets taken till the last match be n.
∴ Total runs at 24.85 runs per wicket = 24.85n
Total runs after the current match = 24.85n + 52
Total no. of wickets after the current match = n + 5
Bowling Average after the current match
⇒ ${24.8n+52}/{n+5}$ = 24.85 – 0.85
∴ ${24.8n+52}/{n+5}$ = 24
or 24.85n + 52 = 24n + 120
or 0.85n = 120 – 52
or n = $68/{0.85}$ = 80
Q-11) The average of 11 numbers is 63. If the average of first six numbers is 60 and the last six numbers is 65, then the 6th number is
(a)
(b)
(c)
(d)
Using Rule 1,
Average of two or more numbers/quantities is called the mean of these numbers, which is given by
$\text"Average(A)" = \text"Sum of observation / quantities"/\text"No of observation / quantities"$
∴ S = A × n
Sixth number
= 6 × 60 + 6 × 65 – 11 × 63
= 360 + 390 – 693 = 57
Q-12) The average of 20 numbers is 12. The average of the first 12 numbers is 11 and that of the next 7 numbers is 10. The last number is :
(a)
(b)
(c)
(d)
Using Rule 1,
Average of two or more numbers/quantities is called the mean of these numbers, which is given by
$\text"Average(A)" = \text"Sum of observation / quantities"/\text"No of observation / quantities"$
∴ S = A × n
Last number = Sum of 20 numbers
– sum of first 12 numbers – sum of next 7 numbers
= 20 × 12 – 11 × 12 – 7 × 10
= 240 – 132 – 70 = 38
Q-13) The mean of 11 numbers is 35. If the mean of first 6 numbers is 32 and that of the last 6 numbers is 37, find the sixth number.
(a)
(b)
(c)
(d)
Using Rule 1,
Average of two or more numbers/quantities is called the mean of these numbers, which is given by
$\text"Average(A)" = \text"Sum of observation / quantities"/\text"No of observation / quantities"$
∴ S = A × n
Sixth number
= 6 × 32 + 6 × 37 – 11 × 35
= 192 + 222 – 385 = 29
Q-14) The average of the first 100 positive integers is
(a)
(b)
(c)
(d)
Using Rule 3,
Average of 1,2,3, .....n =${n +1}/2$
Formula,
1 + 2 + 3 + ...... + n= ${n(n+1)}/2$
Average of these numbers = ${n+1}/2$
∴ Required average
${100+1}/2$ = 50.5
Q-15) The average of the squares of first ten natural numbers is
(a)
(b)
(c)
(d)
Using Rule 4,
Average of $1^2,2^2,3^2,4^2…x^2$ = ${(n+1)(2n+1)}/6$
${1^2+2^2+3^2+…+n^2}/n$ = ${(n+1)(2n+1)}/6$
∴ ${1^2+2^2+3^2+…+10^2}/10$ = ${(10+1)(2×10+1)}/6$
= ${11×21}/6$ = 38.5
Q-16) The average weight of 12 parcels is 1.8 kg. Addition of another new parcel reduces the average weight by 50 g. What is the weight of the new parcel ?
(a)
(b)
(c)
(d)
Total weight 12 parcels = 12 × 1.8 = 21.6 kg.
New average of 13 parcels = 1.8 – 0.05 = 1.75 kg.
Total weight of 13 parcels = 13 × 1.75 = 22.75 kg.
∴ Weight of new parcel = 22.75 – 21.6 = 1.15 kg.
Aliter : Using Rule 18,
( If in the group of N persons, a new person comes at the place of a person of 'T’ years, so that average age,
increases by 't’ years
Then, the age of the new person = T + N.t. )
Here, T = 1.8, N= (12 + 1) = 13, t = $50/1000$ = .05
Weight of new parcel = T – Nt
= 1.8 – 13 × 0.05
= 1.8 – 0.65 = 1.15 kg
Q-17) The average age of a husband and a wife was 27 years when they married 4 years ago. The average age of the husband, the wife and a new-born child is 21 years now. The present age of the child is
(a)
(b)
(c)
(d)
Sum of the present age of husband and wife
= 2 × 27 + 8 = 62 years
Sum of the present age of husband, wife and child
= 21 × 3 = 63 years
∴ present age of the child
= 63 – 62 = 1 year
Q-18) The average age of a husband and wife, who were married 4 years ago, was 25 years at the time of their marriage. The average age of the family consisting of husband, wife and a child, born during the interval is 20 years today. The age of the child is
(a)
(b)
(c)
(d)
Sum of the present age of husband and wife
= 2 × 25 + 2 × 4 = 58 years
Sum of the present age of husband, wife and child
= 3 × 20 = 60 years
&there; Child’s present age
= 60 – 58 = 2 years
Q-19) The average marks of 32 boys of section A of class X is 60 whereas the average marks of 40 boys of section B of class X is 33. The average marks for both the sections combined together is
(a)
(b)
(c)
(d)
Required average
${32×60+33×40}/72$
${1920+1320}/72$ = $3240/72$ = 45
Q-20) The average age of 12 players of a team is 25 years. If the captain’s age is included, the average age increases by 1 year. The age of the captain is :
(a)
(b)
(c)
(d)
Total age of 12 players = 12 × 25 = 300
Total age including captain = 13 × 26 = 338
∴ Age of the captain = 338 – 300 = 38 years
Aliter : Using Rule 23,
( If in a group, one member is replaced by a new member, then,
Age of new member = ( age of replaced member) ± xn
where, x = increase (+) or decrease (–) in average n = Number of members. )
Here, x = 1, n = 12
Age of the captain = Average + x (n + 1)
= 25 + 1 (12 + 1) = 38 years