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Applied Maths Chapter 3 Solutions
Quantitative Aptitude
EXERCISE- 3.1
Q.1 Find the mean of the first twelve natural numbers.
Ans. Natural numbers = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12
Mean = Sum of observations/Total no. of observations
= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12)/12
= 78/12
= 6.5
Q.2 Find the average of the first seven odd prime numbers.
Ans. Odd Prime numbers = 3, 5, 7, 11, 13, 17, 19
Mean = Sum of observations/Total no. of observations
= (3 + 5 + 7 + 11 + 13 + 17 + 19)/7
= 75/7
= 10 5/7
Q.3 Find the mean of the (positive) factors of 24.
Ans. Positive Factor of 24 = 1, 2, 3, 4, 6, 8, 12, 24
Mean = Sum of observations/Total no. of observations
= (1 + 2 + 3 + 4 + 6 + 8 + 12 + 24)/8
= 60/8
= 7.5
Q.4 5 people were asked about the time in a week they spend doing social work in their community. They replied 10, 7, 13, 20 and 15 hours, respectively. Find the average time in a week devoted by them to social work.
Ans.
Mean = Sum of observations/Total no. of observations
= (10 + 7 + 13 + 20 + 15)/5
= 65/5
= 13 hours
Q.5 The mean of 6 observations is 17.5. If five of them are 14, 9, 23, 25, and 10, find the sixth observation.
Ans.
The mean of 6 observations = 17.5
Total of 6 observations = 17.5 x 6
= 105.0 [By using Σxi = x̄n]
Sum of 5 observation = 14 + 9 + 23 + 25 + 10 = 81
6th observation = 105 – 81
= 24
Q.6 The average of 25 observations is 27. If one observation is included, the average remains at 27. Find the included observation.
Ans.
Average of 25 observations = 27
Sum of 25 observation = 27 x 25
= 675
If one more observation is included.
Therefore, the Sum of 26 observations = 27 x 26 = 702
Included observation = 26 observation – 25 observation
= 702 – 675
= 27
Q.7 The average of five numbers is 87. If one of the numbers is excluded, then the average gets decreased by 5. Find the excluded number.
Ans. Average of 5 numbers = 87
Sum of 5 numbers = 87 x 5
= 435
If one more number is excluded then the average becomes = 87 – 5
= 82
Sum of 4 numbers = 4 x 82
= 328
Therefore,
Excluded observation = Sum of 5 no.’s – Sum of 4 no.’s
= 435 – 328
= 107
Q.8 A nursery is closed on Sunday. The average plants sold in the remaining six days of a week is 156 plants and the average sale from Monday to Friday is 124 plants. Find the number of plants sold on Saturday.
Ans. The sum of plants sold from
(Monday – Saturday) = 6 x156 = 936
The sum of plants sold from
(Monday – Friday) = 5 x 124 = 620
Plants sold on Saturday = 936 – 620 = 316
Q.9 The average of five consecutive numbers is 125. Find the product at the number at the extreme positions.
Ans. Consecutive Numbers = x, x+1, x+2, x+3, x+4
Mean = Sum of observations/Total no. of observations
125 = (x + x+1 + x+2 + x+3 + x+4)/5
125 x 5 = 5x +10
5x = 615
x = 123
Extreme no. = 123, 127
Product = 123 x 127
= 15621
Q.10 The following are the weights (in kg) of 8 students of a class:
50, 44.5, 48.7, 45.1, 50.4, 43, 51, 49.3
(i) Find the average weight.
(ii) If a teacher, whose weight is 62, is also included then what will be the average weight?
Ans. (i) Average weight = Sum of observations/Total no. of observations
= (50 + 44.5 8.7 + 45.1 + 50.4 + 43 + 51 + 49.3)/8
= 382.0/8
= 47.75 kg
(ii) After adding teachers weight
Average = (62 + 382)/9
= 444/9 = 49.33 kg
Q.11 The average height of 10 students is 151.8 cm. Two more students of heights 157.6 cm and 154.4 cm join the group. What is the new average height?
Ans. The sum of the height of 10 students = 151.8 x 10
= 1518
After 2 more students,
Sum of height of 12 students = 1518 + 157.6 + 154.4 = 1830
Average = 1830/12
= 152.5 cm
Q.12 The mean marks of boys and girls in an examination are 70 and 73 respectively. If the mean marks of all the students in that examination are 71, find the ratio of number of boys to the number of girls.
Ans. Let, no. of boys be ‘x’
no. of girls be ‘y’
Total marks of boys = 70x
Total marks of girls = 73y
Total students = (x + y)
Total marks of all students = (x + y) 71 = 71x + 71y
ATQ,
Total marks of girls/boys = Total marks of all students
70x + 73y = 71x + 71y
73y – 71y = 71x – 70x
2y = x
y/x = 2/1
Therefore,
Ratio of boys and girls = 2: 1
Q.13 The average age of three students Vijay, Rahul and Rakhi is 15 years. If their ages are in the ratio 4: 5: 6 respectively, then find their ages.
Ans. The average age of 13 students = 15 years
Total age = 3 x 15 = 45 years
According to the given ratio,
4x + 5x + 6x = 45
15x = 45
x = 3
Therefore,
Vijay = 4 x 3 = 12 years
Rahul = 5 x 3 = 15 years
Rakhi = 6 x 3 = 18 years
Q.14 The average of 19 observations is 54. If the average of the first 10 observations is 56 and that of the last 10 observations is 53. Find the tenth observation.
Ans. Average of observation = 54
Sum of 19 observation = 54 x 19
= 1026
Sum of first 10 observation = 10 x 56 = 560
Sum of last 10 observation = 10 x 53 = 530
Therefore, 10th observation = 560 + 530 – 1026 = 64
Q.15 Average of 9 observations was found to be 35. Later on, it was detected that an observation 81 was misread as 18. Find the correct average.
Ans.
Average of 9 observations = 35
Sum of 9 observation = 9 x 35
= 315
Correct total = 315 – 18 + 81
= 315 + 63
= 378
Correct average = 378/9 = 42
Q.16 Under the MANREGA scheme 1000 new labourers are enrolled in Delhi. Earlier they were getting Rs.200 as daily wages, but now the authorities have increased the budget for them by 15 lakh per month:
(i) Calculate the present monthly budget of the ministry for 1000 labours.
(ii) Find the increase in daily income due to budget increase.
(iii) Find the new average monthly income per labour.
Ans. (i) Daily wages = Rs.200
Monthly wages = 200 x 30
= Rs.6,000
Total budget earlier = 6000 x 1000
= Rs.6000000
Increased budget = Rs.1500000
Present monthly budget =Rs.7500000
(ii) Increase in budget (monthly)
= 1500000/1000
= Rs.1500
Daily budget increased = Total budget/no. of labour
= 1500/30
= Rs.50
(iii) New Monthly Average = 7500000/1000
= Rs.7500
Q.17 The average annual PF contribution of a certain number of defence officers is Rs.28560 and that of other officers is Rs.22500. The number of defence officers is 22 times that of other officers. Then find the average savings of all the officers in total.
Ans. Let, other officers = x
Defence officers = 22x
Total of other officer’s PF = 22500x
Total of defence officer’s PF
= (28560)22x
Total = 22500x + 28560 x 22x
= x (22500 + 628320)
= x X 650820
Average of total officers = (x X 650820)/23x = 28296.52
Q.18 In a restaurant 35 visitors can have lunch at a time. If the number of visitors increases by 7, then the expense of the restaurant on food increases by Rs.42 while the average expenditure per head decreases by Rs.1. Find the original expenditure of the restaurant.
Ans. Let, the original expenditure be Rs.x
No. of visitors = 35
Average = x/35
If, no. of visitors = 35 + 7 = 42
Total expense = x + 42
Average = (x + 42)/42
Difference,
x/35 – (x + 42)/42 = 1
x/35 – 1 = (x + 42)/42
(x – 35)/35 = (x + 42)/42
(x – 35)/5 = (x + 42)/6
6x – 210 = 5x + 210 [By cross multiplying]
6x – 5x = 210 + 210
x = 420
Total expenditure = 420
Therefore, Average = 420/35 = 12
Q.19 The average of runs of Virat Kohli, the famous cricket player in the last 10 innings were 72. How many runs he must make in the 11th inning to increase the average by 3 runs?
Ans. Average innings = 72
Total runs of last 10 innings = 72 x 10 = 720
Required average = 72 + 3 = 75
(720 + x)/11 = 75
720 + x = 825
x = 825 – 720
x = 105
Q.20 Five years ago, the average age of a family of five was 32 years. The family includes Mrs and Mr Rai their son, daughter and daughter-in-law. Recently, the daughter Riya moved out of the house after marriage at the age of 30 years. Calculate the family’s present average age.
Ans.
Average age before 5 years = 32
Sum of age before 5 years = 5 x 32
= 160
Present sum of age of 5 person = 160 + 5 x 5 = 180
Now, total age of 4 person = 185 – 30 = 155
Therefore,
Average = 155/4
= 38.75 years
Q.21 An insect starts from point A and covers distance AB in 4 seconds. Then covers distance BC in 7 seconds, distance CD in 5 seconds and distance DA in 2 seconds. Calculate the average velocity and average speed.
Ans. Total Time = 4 + 7 + 5 + 2
= 18 sec
Average speed = Total Distance/Total Time
= 20/18
= 10/9 = 1.11 m/s
Average velocity = Total Displacement/Total Time
= 0/18
= 0 m/s
Q.22 Sara walks 7.2 km in one and a half hours and 3.5 km in 2 hours in the same direction. What is Sara’s average speed for the whole journey?
Ans.
d1 = 7.2 km, t1 = 1 1/2 = 0.5 3/2 h
d2 = 3.5 km, t2 = 2 h
Total distance = d1 + d2
= 7.2 + 3.5 = 10.7 km
Total time = t1 + t2
= 3/2 + 2 = 7/2 h
Average speed = Total distance/Total time
= 10.7/7/2
= (107 x 2)/(7 x 10)
= 107/35 = 3.0571
= 3.06 km/h
Q.23 In a public school, Maths grades for the year are calculated from assignments, tests and final exam.
Assignments count 30%, tests 20% and the final exam 50%. If Rohit scored 85 marks in assignments, 72 marks in tests and 61 marks in the final exam. What is Rohit’s overall grade in Maths ?
Ans.
Marks in assignment = 85 x 30/100
= 51/2 = 25.5
Marks in test = 72 x 20/100
= 14.4
Marks in final exam = 61 x 50/100
= 30.5
Therefore,
Total grade of Rahul = 25.5 + 14.4 + 30.5 = 70.4
Q.24 In a college, the overall grade in Physics is calculated with weights as follows:
| Assessment | Weightage |
| Labs and assignments in I semester | 30% |
| Labs and assignments in II Semester | 30% |
| Final exam | 40% |
Varan scored 66 marks for his labs and assignments in the first semester, and 82 marks for his labs and assignments in the second semester. How many marks should he score in the final exam to achieve 80% overall?
Ans.
Marks in Labs and assignment in I semester = 66 x 30/100 = 79.8
Marks in Labs and assignment in II semester
= 82 x 30/100 = 24.6
Let, Marks in final exam = x X 40/100 = 0.40x
Total of all marks = 80%
79.8 + 24.6 + 0.40x = 80
44.4 + 0.40x = 80
0.40x = 80 – 44.4
x = 35.6/0.4
= 89 Marks
FAQ’s related to Applied Maths Chapter 3 on Quantitative Aptitude:
Q.1 Why is Quantitative Aptitude important?
Ans. Quantitative aptitude skills are crucial for various competitive exams, entrance tests, and job interviews. They are also essential for everyday tasks such as budgeting, financial planning, and decision-making.
Q.2 What topics are covered in Applied Maths Chapter 3?
Ans. Applied Maths Chapter 3 on Quantitative Aptitude typically covers topics such as percentages, profit and loss, clock, time and distance, time and work, averages, ratio and proportion, etc.
Q.3 How will mastering Chapter 3 help you in future exams?
Ans. Mastering Chapter 3 will provide you with a strong foundation in quantitative aptitude, which is essential for scoring well in various competitive exams such as GRE, GMAT, SAT, CAT, banking exams, etc. These exams often contain sections dedicated to quantitative aptitude, and a good score in these sections can significantly enhance your overall performance.
These are few Frequently Asked Questions relating to Applied Maths Chapter 3
In Applied Maths chapter 3, you will explore fascinating topics that form the backbone of practical problem-solving techniques. Through clear explanations, illustrative examples, and step-by-step solutions, you’ll grasp complex concepts effortlessly. Whether you’re preparing for exams or simply eager to deepen your mathematical understanding, Applied Maths Chapter 3 promises an enriching learning experience that will set you on the path to success. Applied Maths Chapter 3, we delve deep into advanced mathematical concepts that are crucial for understanding.