AVERAGE SHORTCUT METHODS
1. Average = [Total of observations / No. of observations]
2. (i) When a person joins a group in case of increasing average Age
weight of new comer = [ (Previous Age + No. of persons) * Increase in Average ](ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age - No. of persons) * Decrease in Average ]
3. When a person leaves a group and another person joins the group in the place of person left, then
(i) In case of increasing average, Age (or) weight of new comer =
[ (Age of person left + No. of persons) * Increase in Average ]
(ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Age of person left - No. of persons) * Decrease in Average ]
4.When a person leaves the group but nobody joins this group, then
(i) In the case of increasing Average, Age (or) weight of man left =
[ (Previous Age - No. of present persons) * Increase in Average ]
(ii) In case of decreasing Average, Age (or) weight of new comer =
[ (Previous Age + No. of present persons) * Decrease in Average ]5. If a person travels a distance at a speed of x Km/hr returns to the
original place of y Km/hr then average speed is [ 2.x.y / ( x + y ) ]
6.If
half of the journey is travelled at speed of x km/hr and the next half
at a speed of x km/hr. Then average speed during the whole journey is [
2.x.y / (x + y) ]
7. If a
person travels 3 equal distances at a speed of x Km/hr, y Km/hr,z km/hr.
Then average speed during whole journey is [ 3.x.y / (x.y +y.x +z.x) ]
EXCERCISE
1. Find the average of all prime numbers between 30 and 50.
Sol. There are five prime numbers between 30 and 50.
They are 31, 37, 41, 43 and 47.
Required average = [(31 + 37 + 41 + 43 + 47) / 5] = 199/5 = 39.8.
2. Find the average of first 40 natural numbers.
Sol. Sum of first n natural numbers = 40 * 41 / 2 = 820.
Required average = 820 / 40 = 20.5
3. Find the average of first 20 multiples of 7.
Sol. Required average = 7(1+2+3 ... + 20) / 20 = [7 * 20 * 21 / 20 * 2] = [147 / 2] = 73.5.
4. The average weight
of 10 oarsmen in a boat is increased by 1.8 kg when one of the crew,
who weighs 53kg is replaced by a new man. find the weight of the new man
Sol. Total weight increased = (1.8 * 10)kg = 18kg.
Weight of the new man = (53 + 18)kg = 71kg.
5. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. find his average after 17th inning.
Sol. Let the average after 17th inning = x.
Then, average after 16th inning = (x - 3)
16(x - 3) + 87 = 17x or x = (87 - 48) = 39.
6.
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The average age
of the mother and her six children is 12 years which is reduced by 5
years if the age of the mother is excluded. How old is the mother?
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Sol: Therefore age of the mother
= ( 12 × 7 – 7 × 6)
= 42 years
7. The age of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is
Sol: Therefore excluded number
= (27 × 5) - ( 25 × 4)
= 135 – 100
= 35.
8. Three
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?
Sol: Present age of (A + B) = (18 × 2 + 3 × 2) years = 42 years.
Present age of (A + B + C) = (22 × 3) years = 66 years.
Therefore C’s age = (66 – 42) years
= 24 years.
9.
The average age of 36 students in a group is 14 years. When teacher’s
age is included to it, the average increases by one. What is the
teacher’s age in years?
SoL : Age of the teacher = ( 37 × 15 – 36 × 14 ) years = 51 years
10. The
average runs of a cricket player of 10 innings was 32. How many runs
must he makes in his next innings so as to increase his average of runs
by 4?
Sol: Average after 11 innings = 36.
Therefore required number of runs
= (36 × 11) – (32 × 10)
= 396 - 320
= 76.
11.
The average salary of all the workers in a workshop is Rs. 8000. The
average salary of 7 technicians is Rs. 12000 and the average salary of
the rest is Rs. 6000. The total number of workers in the workshop is
Sol: Let the total number of workers be x. Then,
8000x = (12000 × 7) + 6000 ( x – 7)
‹=› 2000x = 42000
‹=› x = 21.
12.
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Kamal
obtained 76, 65, 82, 67 and 85 marks(out of 100) in English,
Mathematics, Physics, Chemistry and Biology. What are his average marks?
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Sol: Average
= 76 + 65 + 82 + 67 + 85 / 5)
= (375 / 5) = 75.
13.
In the first 10 over’s of a cricket game, the run rate was only 3.2.
What should be the run rate in the remaining 40 0vers to reach the
target of 282 runs?
Sol: Required run rate = 282 – (3.2 × 10 / 40)
= 240 / 40
= 6.25.
14.
The average weight of a class of 24 students is 35 kg. If the weight of
the teacher be included, the average rises by 400 g. The weight of the
teacher is
Sol: Weight of the teacher
= (35.4 × 25 – 35 × 24)kg
= 45 kg.
15.
After replacing an old member by a new member, it was found that the
average age of five numbers of a club is the same as it was 3 years ago.
What is the difference between the ages of the replaced and the new
member?
Sol: Age decrease = (5 × 3) years = 15 years.
So, the required difference = 15 years
16.
The average weight of 8 person’s increases by 2.5 kg when a new person
comes in place of one of them weighing 65 kg. What might be the weight
of the new person?
Sol: Total weight increased
= (8 × 2.5) kg
= 20 kg.
Weight of new person
= (65 + 20) kg
= 85 kg.
17.
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?
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Sol: Runs scored in the first 10 overs = 10×3.2=32
Total runs = 282, remaining runs to be scored = 282 - 32 = 250
remaining over’s = 40, Run rate needed = 250 40 =6.25
18.
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?
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Sol: Let the sale in the sixth month = x
Then 6435+6927+6855+7230+6562+x 6 =6500
=> 6435 + 6927 + 6855+ 7230 + 6562 + x = 6 \times 6500 = 39000
=> 34009 + x = 39000, x = 39000 - 34009 = 4991
19. The average of 20 numbers is zero. Of them, How many of them may be greater than zero, at the most?
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Sol: 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)
20.
The average age of a class is 15.8 years. The average age of the boys
in the class is 16.4 yrs while that of the girls is 15.4 years. What is
the ratio of boys to girls in the class?
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Sol:
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Let the ratio is k: 1. Then, k*16.4 + 1 * 15.4= (k+l)*15.8
or(16.4 - 15.8)k=(15.8-15.4) or k=0.4/0.6= 2/3.
Required ratio = 2/3 : 1 = 2 : 3
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21. If a, b, c, d, e are five consecutive odd numbers, their average Is:
Sol :
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Clearly=a+2, c=a+4,d=a+6 and e=a+8.
Average (a + (a + 2) + (a + 4) + (a + 6) + (a + 8)) / 5 = (5a + 20) / 5 = (a + 4)
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22. The average of first five multiples of 3 is
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Sol:
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average = (1 + 2 + 3 + 4 + 5) * 3/5 = 15 * 3 / 5 = 9
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23.
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.
Sol:
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Required average
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=
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50.25 x 16 + 45.15 x 8
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16 + 8
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= 48.55
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A 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: |
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Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average
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= 285
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25. The
average weight of 8 person's increases by 2.5 kg when a new person comes
in place of one of them weighing 65 kg. What might be the weight of the
new person?
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Sol: Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
26. The average (arithmetic mean) of 3 numbers is 60. If two of the numbers are 50 and 60, what is the third number?
Sol : 3*60=180, which is the total number of points earned; the two numbers we do know are 50 and 60 which add up to 110
The third number is 180-110=70 since all three numbers must add up to 180
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27.
There are two sections A and B of a class, consisting of 36 and 44
students respectively. If the average weight of sections A is 40 kg and
that of section b is 35 kg. Find the average weight of the whole class?
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Sol: Total weight of(36+44)Students
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(36x40+44x35)
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= 2980 kg.
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Average weight of the whole class
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= (2980 / 80)
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=37.25.
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28.
A batsman makes a score of 87 runs in the 17th inning and thus
increases his averages by 3.Find his average after 17th inning?
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Sol: Let the average after 17th inning = x. Then, average after 16th inning = (x - 3)
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Average
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=16 (x-3)+87
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= 17x or x=(87-48)
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= 39.
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29.
A students was asked to find the arithmetic mean of the numbers 3, 11,
7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found the mean to be 12. What
should be the number in place of x?
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Sol: Clearly, we have (3+11+7+9+15+13+8+19+17+21+14+x/12)
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=12
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Number in place of x is
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137+x=144
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x= 144-137
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x= 7.
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30.David
obtained 76, 65, 82, 67 and 85 marks (out in 100) in English,
Mathematics, Physics, Chemistry and Biology.What are his average marks?
Average = (76+65+82+67+85/5) = 375/5 = 75
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31. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
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Average of 20 numbers
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= 0
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Sum of 20 numbers
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=(0 x 20) =0.
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It is quite possible that 19 of these numbers may be positive and if there sum id a, then 20th number is (-a).
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32. The average age of boys in a class is 16 years and that of the girls is 15 years. The average age for the whole class is
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Sol:
Clearly, to find the average, we ought to know the numbers of boys,
girls or students in the class, neither of which has been given. so the
data provided is inadequate.
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33.
The average age of 36 students in a group is 14 years. When teacher's
age is included to it, the average increases by one. What is the
teacher's age in years?
Sol: Age of the teacher is 37X15-34X14 years
=51 years
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34. The average of five numbers id 27. If one number is excluded,the average becomes 25. The excluded number is
Sol:
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= 27X5- 25X4
Excluded Number is = 135-100 = 35
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35.
The average score of a cricketer for ten matches is 38.9 runs. If the
average for the first six matches is 42. Then find the average for the
last four matches?
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Required average
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=(38.9 x 10)-(42 x 6)/ 4
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= 137 / 4.
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= 34.25
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36. A motorist travel to a place 150 km away at an average speed of 50 km/hr and returns ar 30 km/hr. His average speed for the whole journey in km/hr is
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Average Speed
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= (2xy/x +y )km/hr
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=(2x50 x30/ 50+30)
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= 37.5 km/hr.
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