AVERAGES EXAMPLE QUESTIONS PART 1
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. 6.5
C. 6.75
D. 7
Answer: Option A
Explanation:
Required run rate = (282 - (3.2 x 10))/40 = 250/40 = 6.25
2. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
A. 28 4/7 years
B. 31 5/7 years
C. 32 1/7 years
D. None of these
Answer: Option B
Explanation:
Required average = (67 x 2 + 35 x 2 + 6 x 3)/(2+2+3)
= (134 + 70 + 18)/7
= 222/7
= 31 5/7 years
3. 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. Rs. 4991
B. Rs. 5991
C. Rs. 6001
D. Rs. 6991
Answer: Option A
Explanation:
Total sale for 5 months = Rs. (6435 + 6927 + 6855 + 7230 + 6562) = Rs. 34009.
Required sale = Rs. [ (6500 x 6) - 34009 ]
= Rs. (39000 - 34009)
= Rs. 4991.
4. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
A. 0
B. 1
C. 10
D. 19
Answer: Option D
Explanation:
Average of 20 numbers = 0.
Sum of 20 numbers (0 x 20) = 0.
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a).
5. 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?
A. 76 kg
B. 76.5 kg
C. 85 kg
D. Data inadequate
E. None of these
Answer: Option C
Explanation:
Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
6. 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. What is the average age of the team?
A. 23 years
B. 24 years
C. 25 years
D. None of these
Answer: Option A
Explanation:
Let the average age of the whole team by x years.
11x - (26 + 29) = 9(x -1)
11x - 9x = 46
2x = 46
x = 23.
So, average age of the team is 23 years.
7. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
A. 3500
B. 4000
C. 4050
D. 5000
Answer: Option B
Explanation:
Let P, Q and R represent their respective monthly incomes. Then, we have:
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000.
P's monthly income = Rs. 4000.
8. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
A. 35 years
B. 40 years
C. 50 years
D. None of these
Answer: Option B
Explanation:
Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years = 90 years.
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years = 50 years.
Husband's present age = (90 - 50) years = 40 years.
9. A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
A. Rs. 7.98
B. Rs. 8
C. Rs. 8.50
D. Rs. 9
Answer: Option A
Explanation:
Total quantity of petrol
consumed in 3 years = (4000/7.50 + 4000/8 + 4000/8.50) litres
= 4000 (2/15 + 1/8 + 2/17) litres
= 76700/51 litres
Total amount spent = Rs. (3 x 4000) = Rs. 12000.
Average cost = Rs. (12000 x 51)/76700 = Rs. 6120/767 = Rs. 7.98
10. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun'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 Arun?
A. 67 kg.
B. 68 kg.
C. 69 kg.
D. Data inadequate
E. None of these
Answer: Option A
Explanation:
Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70.
According to Arun's mother, X <= 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average = (66 + 67 + 68)/3 = 201/3 = 67 kg.
11. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
A. 17 kg
B. 20 kg
C. 26 kg
D. 31 kg
Answer: Option D
Explanation:
Let A, B, C represent their respective weights. Then, we have:
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31.
B's weight = 31 kg.
12. 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. 47.55 kg
B. 48 kg
C. 48.55 kg
D. 49.25 kg
Answer: Option C
Explanation:
Required average = (50.25 x 16 + 45.15 x 8)/(16+8)
= (804 + 361.20)/24
= 1165.20/24
= 48.55
13. 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:
A. 250
B. 276
C. 280
D. 285
Answer: Option D
Explanation:
Since the month begins with a Sunday, to there will be five Sundays in the month.
Required average = (510 x 5 + 240 x 25)/30
= 8550/30
= 285
14. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
A. 53.33
B. 54.68
C. 55
D. None of these
Answer: Option B
Explanation:
Required average = (55 x 50 + 60 x 55 + 45 x 60)/(55+60+45)
= (2750 + 3300 + 2700)/160
= 8750/160
= 54.68
15. A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half (1/2). The number of pupils in the class is:
A. 10
B. 20
C. 40
D. 73
Answer: Option C
Explanation:
Let there be x pupils in the class.
Total increase in marks = x x 1/2 = x/2
x/2 = (83 - 63)
x/2 = 20
x= 40.
16. Ram's age was square of number last year and it will be cube of a number next year. How long must he wait before his age is again a cube of a number?
A. 10 year
B. 38 year
C. 39 year
D. 46 year
Solution:
Option(B) is correct
Ram's present age = 26.
He will be 43=6443=64
So, required time is,
=(64−26)=(64−26)
=38 years
17.The ages of Shivali and Tanisha are in the ratio of 11:7 respectively. After 8 years the ratio of their ages will be 15:11. What is the difference in years between their ages?
A. 4 year
B. 10 year
C. 6 year
D. 8 year
Solution:
Option(D) is correct
Let, the age of Shivali be, S=11x and age of Tanisha be, T=7x
Now, according to question after 8 years the ratio of their age become 15/11.
⇒11x+8/7x+8=15/11
⇒121x+88=105x+120
⇒x=2
Difference in ages =15×2−11×2
=8 years
18. In an objective examination of 90 questions, 5 marks are allotted for every correct answer and 2 marks are deducted for every wrong answer. After attempting all the 90 questions a students got a total of 387 marks. Find the number of questions that he attempted wrong.
A. 9
B. 10
C. 11
D. 12
Solution:
Option(A) is correct
Let the wrong questions be x
We get the equations:
(90−x)×5−x×2=387
⇒x=9
19. The average weight of a group of persons increased from 48 kg to 51 kg, when two persons weighing 78 kg and 93 kg join the group. Find the initial number of members in the group?
A. 21
B. 22
C. 23
D. 24
E. None of these
Answer & Explanation
Answer: Option C
Explanation:
Let the initial number of members in the group be n.
Initial total weight of all the members in the group = n(48)
From the data,
48n + 78 + 93 = 51(n + 2) => 51n - 48n = 69 => n = 23
Therefore there were 23 members in the group initially.
20. Sneh's age is 1/6th of her father's age. Sneh's father's age will be twice of Vimal's age after 10 years. If Vimal's eighth birthday was celebrated 2 years before, then what is Sneh's present age?
A. 7 year
B. 8 year
C. 6 year
D. None of these
Solution:
Option(D) is correct
Vimal's present age =8+2=10 year.
Father's age after 10 years =(10+10)×2=40years.
Father's present age = 30 years.
Sneh's present age
=16×30
=5 year
21.If 1 added to the age of the elder sister, then the ratio of the ages of two sisters becomes 0.5:1, but if 2 is subtracted from the age of the younger one, the ratio becomes 1:3, the age of the younger sister will be?
A. 7 year
B. 5 year
C. 8 year
D. 10 year
Solution:
Option(B) is correct
Suppose that age of elder sister be x year, and younger sister be y year. Then:
yx+1=0.51=12
⇒x−2y=−1 -------(i)
Again given:
y−2x=13
⇒x−3y=−6 -------- (ii)
From both the equations we get:
x=9 and y=5
So, the age of younger sister is 5 years
22. A club consists of members whose ages are in A.P. The common difference being 3 months. If the youngest member of the club is just 7 years old and the sum of the ages of all the members is 250, then number of members in the club are:
A. 18
B. 20
C. 25
D. 24
Solution:
Option(C) is correct
Let n be the number of members in the club
Then,
250=n/2[2×7+(n−1)×3/12]
n=25
23. Students of a class are made to stand in rows. If 4 students are extra in each row, there would be 2 rows less. If 4 students are less in each row, there would be 4 more rows. The number of students in the class is:
A. 90
B. 94
C. 92
D. 96
Solution:
Option(D) is correct
Let number of rows be x and number of students in each row be n.
Then, total number of students =x×n
Again,
(n+4)(x−2)=(x−4)(n−4)=xn
⇒n=12and x=8
Number of students =12×8=96
24. If Dennis is 1/3 rd the age of his father Keith now, and was 1/4 th the age of his father 5 year ago, then how old will his father Keith be 5 year from now?
A. 45 year
B. 40 year
C. 55 year
D. 50 year
Solution:
Option(D) is correct
Let the present age of Dennis and his father be x and y respectively.
and
(x−5)=(y−5)4
On solving both equation we get y=45 year.
Hence, required age =(y+5)= 50 year
25. Five years ago, Bina's age was three times that of the Arti. Ten years ago, Bina's age was half that of Chitra. If cc represents Chitra's current age, which of the following represent Arti's current age?
A. (c−10)3(c−10)3
B. c6+5c6+5
C. 3c−53c−5
D. 5c3−105c3−10
Solution:
Option(B) is correct
Let the Bina's current age =x
Arti's present age =y
Then, (x−5)=3(y−5)
⇒x−5=3y−15-------- (i)
Again,
x−10=1/2×(c−10)
x=(c−10)/2+10-------- (ii)
On putting value of x in equation (i),we get:
y=c/6+5
26.The total age of some 7 years old and some 5 years old children is 60 years. If I have to select a team from these children such that their total age is 48 years, In how many ways can it be done?
A. 3
B. 2
C. 1
D. 4
Solution:
Option(C) is correct
Let aa children of 7 years and bb children of 5 years be taken.
Then, 7a+5b=48
This is possible only when x=4 and b=4
Hence, only one combination is possible.
27.The digits of a three number are in AP. If the number is subtracted from the number formed by reversing its digits, the result is 396. What could be the original number?
A. 654
B. 135
C. 852
D. 753
Solution:
Option(B) is correct
The difference between 3-digit number and its reverse is 99 times the difference between its extreme (hundred and units) digits.
As the first difference is 396, the second is 4.
Further as the digits are in AP and the hundred's digits is less than the unit's digit, we have following possibilities:
135,246,357,468,579.….
28. In a family of husband, wife and a daughter, the sum of the husband’s age, twice the wife’s age, and thrice the daughter’s age is 85; while the sum of twice the husband’s age, four times the wife’s age, and six times the daughter’s age is 170. It is also given that the sum of five times the husband’s age, ten times the wife’s age and fifteen times the daughter’s age equals 450. The number of possible solutions, in terms of the ages of the husband, wife and the daughter, to this problem is:
A. 0
B. 1
C. 2
D. Infinitely many
Solution:
Option(A) is correct
Let the age of husband wife and daughter be denoted by h,wh,w and d respectively.
h+2w+3d=85-------- (i)
2h+4w+6d=170-------- (ii)
5h+10w+15d=450-------- (iii)
Multiplying the first equation by 5 we get
5h+10w+15d=425
but Eq (iii) gives 5h+10w+15d=450
So No solution possible.
29.Three persons Suresh, Devesh and Prashant were born on different days in the same year.
If the date and month of birth of Suresh, Devesh and Prashant are numerically equal (i.e. 1 Jan or 3 Mar or 5 May or 12 Dec etc), then what could be the minimum difference in the ages of youngest and oldest in days?
Note: Based on the comments, the question has been elaborated more.
A. 56
B. 60
C. 61
D. 62
Solution:
Option(C) is correct
As per the question date and month of birth of Suresh, Devesh and Prashant are numerically equal.
So, the possible dates could be, 1 Jan (1/1), or 2 Feb (2/2), or 3 Mar (3/3), or 4 Apr (4/4) and so on till 12 Dec (12/12).
To have the minimum differences in the ages of the oldest and youngest, one of them should be born in February since February is the smallest month.
Also, the other two should be born in either 'March and April' or 'January and March'.
Difference between the youngest, and the oldest is the number of days from
2nd Feb. to 4th Apr (when they are born on 2 Feb, 3 Mar, and 4 Apr).:
=(26+31+4)=61=(26+31+4)=61
or, 1st Jan to 3rd Mar (when they are born on 1Jan, 2 Feb, and 3 Mar).
=30+28+3
= 61 days.
30.The sum of the reciprocals of the ages of two colleagues is five times the difference of the reciprocals of their ages. If the ratio of the products of their ages to the sum of their ages is 14.4:1, the age (in years) of one of the colleagues must be between (both inclusive).
A. 20 and 23
B. 23 and 26
C. 26 and 30
D. 30 and 35
Solution:
Option(B) is correct
Suppose that age of the two colleagues be x year and y year
According to question:
1/x+1/y=5(1/x−1/y)
⇒y=3x2 --------(i)
Again,
Xy/x+y=14.4/1
5xy=72(x+y)
From equation (i):
5x×(3x/2)=72×(x+3x/2)
⇒x=24
i.e. Age of one of colleagues lies between 23 and 26 year.
31.The average of first five multiples of 3 is:
A. 8
B. 9
C. 10
D. 11
Solution:
Option(B) is correct
Basic Formula: 1,2,3...n
If nn is odd, the formula is (n+1//2) th term
The five multiples of 3 is 3,6,9,12,15
(n+/2)⇒(5+1/2)th term
⇒(62) th term = 3rd term
Here 3rd term is 9
32.There are two sections A and B of a class, consisting of 36 and 44 students’ respectively.
If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.
A. 30.00 kg
B. 35.00 kg
C. 37.25 kg
D. 42.50 kg
Solution:
Option(C) is correct
Total weight of (36+44=80) Students =(36×40+44×35) kg = 2980 kg
Therefore average weight of the whole class =(2980/80) kg
Therefore average weight = 37.25kg
33.
The distance between two stations A and B is 778 km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey.
A. 60.3 km/hr
B. 35.0 km/hr
C. 57.5 km/hr
D. 67.2 km/hr
Solution:
Option(D) is correct
Average Speed=(2xy/x+y) km/hr
=(2×84×56/84+56)
=(2×84×56/140)
=67.2 km/hr
34. The average of 50 numbers is 30. If two numbers, 35 and 40 are discarded, then the average of the remaining numbers is nearly:
A. 28.32
B. 29.68
C. 28.78
D. 29.27
Solution:
Option(B) is correct
Total sum of 48 numbers =(50×30)−(35+40)
=1500−75
=1425
Average =(1425/48)
= 29.68
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.
A. 33.25
B. 33.5
C. 34.25
D. 35
Solution:
Option(C) is correct
Total sum of last 4 matches
=(10×38.9)–(6×42)
=389–252=137
Average
=137/4
=34.25
36.
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.
A. 40
B. 39
C. 52
D. 55
Solution:
Option(B) is correct
Let the average after 17th innings = x
Then average after 16th innings = (x−3)
Therefore 16(x−3)+87=17x
Therefore x=39
37.
There were 35 students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.42 per day while the average expenditure per head diminished by Re 1. What was the original expenditure of the mess?
A. Rs. 450
B. Rs. 320
C. Rs. 550
D. Rs. 420
Solution:
Option(D) is correct
Let the original average expenditure be Rs.x then,
42(x−1)−35x=42
⇒7x=84
⇒x=12
Therefore original expenditure
=Rs.(35×12)
=Rs. 420
38.
Nine persons went to a hotel for taking their meals. Eight of them spent Rs.12 each on their meals and the ninth spent Rs.8 more than the average expenditure of all the nine. What was the total money spent by them.
A. Rs. 115
B. Rs. 116
C. Rs. 117
D. Rs. 118
Solution:
Option(C) is correct
Let the average expenditure of all the nine be Rs. x
Then, 12×8+(x+8)=9x
Therefore x=13
Total money spent = 9x=RS.(9×13) = Rs.117
39.
David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, mathematics, physics, chemistry and biology. What are his average marks?
A. 65
B. 69
C. 75
D. None of above
Solution:
Option(C) is correct
Average
=76+65+82+67+85/5
=375/5
=75
40.
The average of runs of a cricket player of 10 innings was 32. How many runes must be made in his next innings so as to increase his average of runs by 4?
A. 72
B. 74
C. 70
D. 76
Solution:
Option(D) is correct
Average after 11 innings = 36
Required number of runs = (36×11)−(32×10)
=396−320
= 76
41.
The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?
A. 83
B. 92
C. 90
D. 97
Solution:
Option(D) is correct
The total wages earned during the 15 days that the worker worked :
=15×90=Rs.1350.
The total wages earned during the first 7 days = 7×87 = Rs. 609.
The total wages earned during the last 7 days = 7×92 = Rs. 644.
Total wages earned during the 15 days = wages during first 7 days + wage on 8th day + wages during the last 7 days.
1350=609+ wage on 8th day +644
wage on 8th day = 1350−609−644 = Rs. 97
42.
The average temperature on Wednesday, Thursday and Friday was 25circ. The average temperature on Thursday, Friday and Saturday was 24circ. If the temperature on Saturday was 27circ, what was the temperature on Wednesday?
A. 24∘
B. 21∘
C. 27∘
D. 30∘
Solution:
Option(D) is correct
Total temperature on Wednesday, Thursday and Friday was 25×3=75∘
Total temperature on Thursday, Friday and Saturday was 24×3=72∘
Hence, difference between the temperature on Wednesday and Saturday= 3∘
If Saturday temperature =27∘, then Wednesday's temperature =27∘+3∘=30∘
43.
When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200 g. What is the average weight of the remaining 59 students?
A. 57
B. 56.8
C. 58.2
D. 52.2
Solution:
Option(A) is correct
Let the average weight of the 59 students be A.
Therefore, the total weight of the 59 of them will be 59 A.
The questions states that when the weight of this student who left is added, the total weight of the class
= 59A+45
When this student is also included, the average weight decreases by 0.2 kgs.
59A+4560=A−0.2
⇒59A+45=60A−12
⇒45+12=60A−59A
⇒A=57.
44.
The difference between two angles of a triangle is 24circ. The average of the same two angles is 54circ .Which one of the following is the value of the greatest angle of the triangle?
A. 45∘
B. 60∘
C. 66∘
D. 72∘
Solution:
Option(D) is correct
Let a and b be the two angles in the question, with a>b. We are given that the difference between the angles is 24∘.
⇒a–b=24.
Since the average of the two angles is 54∘, we have a+b/)2=54
Solving for b in the first equation yields b=a–24, and substituting this into the second equation yields
a+(a+24)/2=54
2a−24/2=54
2a−24=54×2
2a−24=108
2a=108+24
2a=132
a=66
Also,
b=a−24=66−24=42.
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is 180∘,
a+b+c=180.
Plugging the previous results into the equation yields 66+42+c=180.
Solving for c yields c=72
Hence, the greatest of the three angles a,b and c is c, which equals 72∘.
45.
Which one of the following numbers can be removed from the set S = {0, 2, 4, 5, 9} without changing the average of set S?
A. 0
B. 2
C. 4
D. 5
Solution:
Option(C) is correct
The average of the elements in the original set S is:
0+2+4+5+9/5=4
If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}.
The average of the elements is:
0+2+5+9/4=4.
46.
Average cost of 5 apples and 4 mangoes is Rs. 36. The average cost of 7 apples and 8 mangoes is Rs. 48. Find the total cost of 24 apples and 24 mangoes.
A. 1044
B. 2088
C. 720
D. 324
Solution:
Option(B) is correct
Average cost of 5 apples and 4 mangoes = Rs. 36
Total cost = 36×9=324
Average cost of 7 apples and 8 mangoes = 48
Total cost = 48×15=720
Total cost of 12 apples and 12 mangoes = 324+720=1044
Therefore, cost of 24 apples and 24 mangoes = 1044×2= 2088
47.
Average of ten positive numbers is x. If each number is increased by 10%, then x :
A. remains unchanged
B. may decrease
C. may increase
D. is increased by 10%
Solution:
Option(D) is correct
Let 10 numbers be x1,x2,x3…..x10
According to question average of these 10 numbers is 10.
(x1+x2+x3+……..+x10)/10=x
Now if each number is increased by 10% then new average, say y,
y=(1.1x1+1.1x2+1.1x3+………+1.1x10)/10
⇒y=1.1×((x1+x2+x3+……..+x10)/10)
⇒y=1.1x
⇒y is 10% increased.
48.
The average weight of three boys A,B and C is (54dfrac{1}{3}) kg, while the average weight of three boys B,D and E is 53 kg. What is the average weight of A,B,C,D and E?
A. 52.4 kg
B. 53.2 kg
C. 53.8 kg
D. Data inadequate
Solution:
Option(D) is correct
In this question, sum of numbers is provided, nut required sum (i.e. A+B+C+D+E) cannot be calculated by the given data.
Therefore the answer is Data inadequate.
49.
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:
A. 20
B. 21
C. 22
D. 23
Solution:
Option(B) is correct
Let the total number of workers in the workshop = x
Given Average of all workers
=8000⇒8000/x
Average of 7 members = 12000⇒7×12000=84000
Average of remaining workers = 6000⇒6000×(x–7)⇒6000x–42000
Find out the total number of workers
8000x=84000+6000x–42000
2000x=42000
x=21
Therefore the total number of workers in the workshop is = 21.
50.
40%40% of the employees in a factory are workers. All the remaining employees are executives. The annual income of each worker is Rs. 390. The annual income of each executive is Rs. 420. What is the average annual income of all the employees in the factory together?
A. 390
B. 405
C. 408
D. 415
Solution:
Option(C) is correct
Let e be the number of employees.
We are given that 40% of the employees are workers. Now, 40% of e is
40/100×e=0.4e
Hence, the number of workers is 2e/5
All the remaining employees are executives, so the number of executives equals
(The number of Employees)–(The number of Workers)
=e−2e/5
=3e/5
The annual income of each worker is Rs. 390. Hence, the total annual income of all the workers together is:
=2e/5×390
=156e
Also, the annual income of each executive is Rs. 420. Hence, the total income of all the executives together is
=3e/5×420
=252e
Hence, the total income of the employees is :
156e+252e=408e
The average income of all the employees together equals
The total income of all the employees/The number of employees
=408e/e
=408
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