213 + Interview Questions in QUANTITATIVE APTITUDE in AVERAGE Page 5 InterviewSolution

201.	What is the average of all the one digit, two digit and three digit natural numbers?1). 3002). 5003). 10004). 1250
Answer» As we know, all the one digit, two digit, and 3-digit natural numbers are from 1 – 999 ? SUM of n natural numbers = n(n + 1)/2 Sum of 999 natural numbers = (999 × 1000)/2 = 499500 ∴ REQUIRED AVERAGE = 499500/999 = 500

201.

What is the average of all the one digit, two digit and three digit natural numbers?1). 3002). 5003). 10004). 1250

Answer»

As we know, all the one digit, two digit, and 3-digit natural numbers are from 1 – 999

? SUM of n natural numbers = n(n + 1)/2

Sum of 999 natural numbers = (999 × 1000)/2 = 499500

∴ REQUIRED AVERAGE = 499500/999 = 500

Discussion

202.	1). 1112). 1313). 1744). 141
Answer» Total runs SCORED by PLAYER = 140 × 40 = 5600 Let the HIGHEST score be x His lowest score = x - 170 Total runs scored in the REMAINING 38 matches = 136 × 38 = 5168 Highest + lowest score = 5600 - 5168 x + (x – 170) = 432 x = 301 Lowest score = 301 – 170 = 131

202.

1). 1112). 1313). 1744). 141

Answer»

Total runs SCORED by PLAYER = 140 × 40 = 5600

Let the HIGHEST score be x

His lowest score = x - 170

Total runs scored in the REMAINING 38 matches = 136 × 38 = 5168

Highest + lowest score = 5600 - 5168

x + (x – 170) = 432

x = 301

Lowest score = 301 – 170 = 131

Discussion

203.	The mean of marks secured by 60 students in division A of class X is 66, 45 students of division B is 62 and that of 75 students of division C is 60. Find the mean of marks of the students of three divisions of Class X.1). 61.82). 62.53). 61.14). 63.9
Answer» TOTAL marks of division A = 60 × 66 = 3960 Total marks of division B = 45 × 62 = 2790 Total marks of division C = 75 × 60 = 4500 Total marks of all students = 3960 + 2790 + 4500 = 11250 Total number of students = 60 + 45 + 75 = 180 Average marks = 11250/180 = 62.5

203.

The mean of marks secured by 60 students in division A of class X is 66, 45 students of division B is 62 and that of 75 students of division C is 60. Find the mean of marks of the students of three divisions of Class X.1). 61.82). 62.53). 61.14). 63.9

Answer»

TOTAL marks of division A = 60 × 66 = 3960

Total marks of division B = 45 × 62 = 2790

Total marks of division C = 75 × 60 = 4500

Total marks of all students = 3960 + 2790 + 4500 = 11250

Total number of students = 60 + 45 + 75 = 180

Average marks = 11250/180 = 62.5

Discussion

204.	The average monthly salary of the workers in a workshop is Rs. 8,500. If the average monthly salary of 7 technicians is Rs. 10,000 and the average monthly salary of the rest is the workshop is 7800 then total number of workers1). 182). 203). 224). 24
Answer» LET TOTAL workers in a workshop be N. Given, Total monthly salary of workers in a workshop = 8500N Given, Total salary of 7 technicians = 7 × 10000 = 70000 Total REMAINING workers = N – 7 Total salary of remaining workers = 7800 × (N – 7) Then, ⇒ 8500N = 70000 + 7800 × (N – 7) ⇒ 8500N = 70000 + 7800N - 54600 ⇒ 700N = 15400 ⇒ N = 22 ∴ Total NUMBER of workers is 22.

204.

The average monthly salary of the workers in a workshop is Rs. 8,500. If the average monthly salary of 7 technicians is Rs. 10,000 and the average monthly salary of the rest is the workshop is 7800 then total number of workers1). 182). 203). 224). 24

Answer»

LET TOTAL workers in a workshop be N.

Given,

Total monthly salary of workers in a workshop = 8500N

Given,

Total salary of 7 technicians = 7 × 10000 = 70000

Total REMAINING workers = N – 7

Total salary of remaining workers = 7800 × (N – 7)

Then,

⇒ 8500N = 70000 + 7800 × (N – 7)

⇒ 8500N = 70000 + 7800N - 54600

⇒ 700N = 15400

⇒ N = 22

∴ Total NUMBER of workers is 22.

Discussion

205.	1). 432). 593). 674). Cannot be determined
Answer» Given, Average salary for chemical trade(in lacs per ANNUM) = Rs. 2.45 Average salary for electronics trade(in lacs per annum) = Rs. 3.56 Average salary for chemical trade and electronics trade(in lacs per annum) = Rs. 3.12 Let the number of students in Chemical be a and number of students in Electronics be b. $(\therefore\;\FRAC{{2.45\;\TIMES\;a\;+\;3.56\;\times\;b}}{{a\;+\;b}}\;=\;3.12)$ ⇒ 2.45a + 3.56b = 3.12a + 3.12b ⇒ 0.44b = 0.67a ⇒ b/a = 67/44 ⇒ b = 67a/44 For b to be an INTEGER the least value will be 67

205.

1). 432). 593). 674). Cannot be determined

Answer»

Given,

Average salary for chemical trade(in lacs per ANNUM) = Rs. 2.45

Average salary for electronics trade(in lacs per annum) = Rs. 3.56

Average salary for chemical trade and electronics trade(in lacs per annum) = Rs. 3.12

Let the number of students in Chemical be a and number of students in Electronics be b.

$(\therefore\;\FRAC{{2.45\;\TIMES\;a\;+\;3.56\;\times\;b}}{{a\;+\;b}}\;=\;3.12)$

⇒ 2.45a + 3.56b = 3.12a + 3.12b

⇒ 0.44b = 0.67a

⇒ b/a = 67/44

⇒ b = 67a/44

For b to be an INTEGER the least value will be 67

Discussion

206.	The average of first three out of four numbers is 18. The average of the last three numbers is 14. The sum of the first and last number is 16. The last number is1). 172). 133). 94). 2
Answer» Let the FOUR numbers be m, n, o, p. From the given data, Average of the FIRST three out of four numbers is 18 ⇒ (m + n + o) /3 = 18 ⇒ Sum of the first three numbers (m + n + o) = 54----(1) ALSO given that average of the LAST three numbers = 14 ⇒ (n + o + p)/3 = 14 ⇒ Sum of the last three numbers (n + o + p) = 42---- (2) Subtract equation (2) from equation (1), we get ⇒ m - p = 12----(3) Also given that sum of the first and last number = 16 ⇒ m + p = 16----(4) Subtract equation (4) from equation (3), we get ⇒ -2p = -4 ∴ p = 2 ∴ Last number p = 2

206.

The average of first three out of four numbers is 18. The average of the last three numbers is 14. The sum of the first and last number is 16. The last number is1). 172). 133). 94). 2

Answer»

Let the FOUR numbers be m, n, o, p.

From the given data,

Average of the FIRST three out of four numbers is 18

⇒ (m + n + o) /3 = 18

⇒ Sum of the first three numbers (m + n + o) = 54----(1)

ALSO given that average of the LAST three numbers = 14

⇒ (n + o + p)/3 = 14

⇒ Sum of the last three numbers (n + o + p) = 42---- (2)

Subtract equation (2) from equation (1), we get

⇒ m - p = 12----(3)

Also given that sum of the first and last number = 16

⇒ m + p = 16----(4)

Subtract equation (4) from equation (3), we get

⇒ -2p = -4

∴ p = 2

∴ Last number p = 2

Discussion

207.	The average of 100 items was found to be 30. If at the time of calculation, two terms were wrongly taken as 32 and 12 instead of 23 and 11, then the correct average is1). 29.82). 293). 29.94). 29.5
Answer» the SUM of 100 numbers = 30 × 100 = 3000 As 23 and 11 was added as 32 and 12 Therefore, actual sum = 3000 – 32 – 12 + 23 + 11 = 2990 correct mean = 2990/100 = 29.90

207.

The average of 100 items was found to be 30. If at the time of calculation, two terms were wrongly taken as 32 and 12 instead of 23 and 11, then the correct average is1). 29.82). 293). 29.94). 29.5

Answer»

the SUM of 100 numbers = 30 × 100 = 3000

As 23 and 11 was added as 32 and 12

Therefore, actual sum = 3000 – 32 – 12 + 23 + 11 = 2990

correct mean = 2990/100 = 29.90

Discussion

208.	1). 42). 53). 64). 7
Answer» Let the numbers be ‘a’, ‘B’, ‘c’ and ‘d’ Given, (a + b + c + d)/4 = 15 ⇒ a + b + c + d = 15 × 4 = 60----(1) Now, Average of a, b and c = 3 times d ⇒ (a + b + c)/3 = 3d ⇒ a + b + c = 9D----(2) Substituting EQUATION (2) in equation (1) ⇒ 9d + d = 60 ⇒ 10d = 60 ⇒ d = 60/10 = 6 ∴ Fourth number = d = 6

208.

1). 42). 53). 64). 7

Answer»

Let the numbers be ‘a’, ‘B’, ‘c’ and ‘d’

Given, (a + b + c + d)/4 = 15

⇒ a + b + c + d = 15 × 4 = 60----(1)

Now,

Average of a, b and c = 3 times d

⇒ (a + b + c)/3 = 3d

⇒ a + b + c = 9D----(2)

Substituting EQUATION (2) in equation (1)

⇒ 9d + d = 60

⇒ 10d = 60

⇒ d = 60/10 = 6

∴ Fourth number = d = 6

Discussion

209.	1). 30, 352). 30, 363). 35, 364). 36, 38
Answer» Given data - An average of 5 numbers = 40 Total summation of numbers = average × 5 Total summation of numbers = 40 × 5 Total summation of numbers = 200 When 2 numbers are being excluded the average of REMAINING 3 numbers becomes 45. Total of remaining 3 numbers = 45 × 3 Total of remaining 3 numbers = 135 From above we can see sum of excluded numbers = 200 - 135 = 65 Only option of numbers which add up to 65 is 30, 35 ∴ Excluded numbers are 30 & 35.

209.

1). 30, 352). 30, 363). 35, 364). 36, 38

Answer»

Given data -

An average of 5 numbers = 40

Total summation of numbers = average × 5

Total summation of numbers = 40 × 5

Total summation of numbers = 200

When 2 numbers are being excluded the average of REMAINING 3 numbers becomes 45.

Total of remaining 3 numbers = 45 × 3

Total of remaining 3 numbers = 135

From above we can see sum of excluded numbers = 200 - 135 = 65

Only option of numbers which add up to 65 is 30, 35

∴ Excluded numbers are 30 & 35.

Discussion

210.	In 7 days of a week starting Monday, on an average 300 boys attended a camp for the first five days and 350 boys on an average attended for the last three days. The average attendance in this week is 325. What was number of students present on Friday?1). 6602). 3003). 3254). 275
Answer» Average attendance in the week is 325 students per day. ∴ total number of students who attended throughout the week = 325 × 7 = 2275 Average attendance in first 5 DAYS (Mon-Fri) = 300 ∴ total attendance in first five days = 300 × 5 = 1500 Similarly, total attendance in last 3 days (Fri-Sun) = 3 × 350 = 1050 ∴ Attendance on Friday = 1500 + 1050 – 2275 = 275

210.

In 7 days of a week starting Monday, on an average 300 boys attended a camp for the first five days and 350 boys on an average attended for the last three days. The average attendance in this week is 325. What was number of students present on Friday?1). 6602). 3003). 3254). 275

Answer»

Average attendance in the week is 325 students per day.

∴ total number of students who attended throughout the week = 325 × 7 = 2275

Average attendance in first 5 DAYS (Mon-Fri) = 300

∴ total attendance in first five days = 300 × 5 = 1500

Similarly, total attendance in last 3 days (Fri-Sun) = 3 × 350 = 1050

∴ Attendance on Friday = 1500 + 1050 – 2275 = 275

Discussion

211.	In a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was 1). 30 years2). 29 years3). 25 years4). 24 years
Answer» Given, NUMBER of member = 5 AVERAGE present age = 33 years ∴ Sum of present ages of 5 member = (Number of members) × (Average present age) = 5 × 33 = 165 years Given, Age of youngest member = 9 years ∴ Sum of the ages of other 4 members just before the birth of youngest member = Sum of present ages of 5 member - Age of youngest member – 4 × 9 = 165 – 9 – 36 = 120 ∴ Average age of other 4 members just before the birth of youngest member = 120/4 = 30 years

211.

In a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was 1). 30 years2). 29 years3). 25 years4). 24 years

Answer»

Given,

NUMBER of member = 5

AVERAGE present age = 33 years

∴ Sum of present ages of 5 member

= (Number of members) × (Average present age)

= 5 × 33

= 165 years

Given, Age of youngest member = 9 years

∴ Sum of the ages of other 4 members just before the birth of youngest member

= Sum of present ages of 5 member - Age of youngest member – 4 × 9

= 165 – 9 – 36 = 120

∴ Average age of other 4 members just before the birth of youngest member

= 120/4 = 30 years

Discussion

212.	Find the average of cubes of 1, 3, 5, 7 and 9? 1). 2252). 12253). 2354). 245
Answer» ∴ REQUIRED AVERAGE = (1³ + 3³ + 5³ + 7³ + 9³)/5 = (1 + 27 + 125 + 343 + 729)/5 = 245

Discussion

213.	A goldsmith earns Rs.36000 in 12 days. In the first 2 days his average income was Rs.2600. What was his average income for the remaining days?1). Rs.3000 per day2). Rs.3080 per day3). Rs.3060 per day4). Rs.3050 per day
Answer» Given, TOTAL income of the goldsmith = Rs.36000 Given, average income in the FIRST TWO days = Rs.2600 ⇒ Total income in the first two days = 2600 × 2 = 5200 ⇒ Amount EARNED in the REMAINING days = 36000 – 5200 = 30800 ⇒ Average income for the remaining days = 30800/10 = Rs.3080

213.

A goldsmith earns Rs.36000 in 12 days. In the first 2 days his average income was Rs.2600. What was his average income for the remaining days?1). Rs.3000 per day2). Rs.3080 per day3). Rs.3060 per day4). Rs.3050 per day

Answer»

Given, TOTAL income of the goldsmith = Rs.36000

Given, average income in the FIRST TWO days = Rs.2600

⇒ Total income in the first two days = 2600 × 2 = 5200

⇒ Amount EARNED in the REMAINING days = 36000 – 5200 = 30800

⇒ Average income for the remaining days = 30800/10 = Rs.3080

Discussion

Explore topic-wise InterviewSolutions in .

What is the average of all the one digit, two digit and three digit natural numbers?1). 3002). 5003). 10004). 1250

1). 1112). 1313). 1744). 141

The mean of marks secured by 60 students in division A of class X is 66, 45 students of division B is 62 and that of 75 students of division C is 60. Find the mean of marks of the students of three divisions of Class X.1). 61.82). 62.53). 61.14). 63.9

The average monthly salary of the workers in a workshop is Rs. 8,500. If the average monthly salary of 7 technicians is Rs. 10,000 and the average monthly salary of the rest is the workshop is 7800 then total number of workers1). 182). 203). 224). 24

1). 432). 593). 674). Cannot be determined

The average of first three out of four numbers is 18. The average of the last three numbers is 14. The sum of the first and last number is 16. The last number is1). 172). 133). 94). 2

The average of 100 items was found to be 30. If at the time of calculation, two terms were wrongly taken as 32 and 12 instead of 23 and 11, then the correct average is1). 29.82). 293). 29.94). 29.5

1). 42). 53). 64). 7

1). 30, 352). 30, 363). 35, 364). 36, 38

In 7 days of a week starting Monday, on an average 300 boys attended a camp for the first five days and 350 boys on an average attended for the last three days. The average attendance in this week is 325. What was number of students present on Friday?1). 6602). 3003). 3254). 275

In a family of 5 members, the average age at present is 33 years. The youngest member is 9 years old. The average age of the family just before the birth of the youngest member was 1). 30 years2). 29 years3). 25 years4). 24 years

Find the average of cubes of 1, 3, 5, 7 and 9? 1). 2252). 12253). 2354). 245

A goldsmith earns Rs.36000 in 12 days. In the first 2 days his average income was Rs.2600. What was his average income for the remaining days?1). Rs.3000 per day2). Rs.3080 per day3). Rs.3060 per day4). Rs.3050 per day