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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
Of the 3 numbers whose average is 22, the first is 3/8th the sum of other two. What is the first number?1). 162). 203). 224). 18 |
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Answer» Let the THREE number are x, y and z GIVEN x = [3(y + z)] /8 ⇒ y + z = 8x/3 Now as given AVERAGE of three numbers is 22, ⇒ x + y + z = 66 ⇒ x + 8x/3 = 66 ⇒ 11x = 198 ⇒ x = 18 ∴ First number is 18. |
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| 102. |
A batsman makes 100 runs in the 25th match of his career. His average runs per match increases by 1.4. Find his average before the 25th match.1). 652). 553). 754). 45 |
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Answer» ⇒ Sum of score before 25th MATCH = 24 × t ⇒ Sum of score after 25th match = 24t + 100 ⇒ Average after 25th match = (24t + 100)/25 ATQ, ⇒ t + 1.4 = (24t + 100)/25 ⇒ 24t + 100 = 25t + 35 ⇒ t = 65 ∴ His average before 25th match is 65. |
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| 103. |
The average salary of Suresh from January to June is Rs. 7500, from July to September is Rs. 8000 and for last 3 months the average salary is Rs. 1000 more than July to sept. Find the average annual salary of Suresh.1). Rs. 88002). Rs. 90003). Rs. 85004). Rs. 8000 |
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Answer» Total salary from January to June = 7500 × 6 = 45000 Total salary from July to SEPTEMBER = 8000 × 3 = 24000 Average salary from OCT. to Dec. = Avg. salary from July to Sep. + 1000 ⇒ Total salary from Oct. to Dec. = (8000 + 1000) × 3 = 27000 ⇒ Annual average salary of SURESH = (45000 + 24000 + 27000)/12 ⇒ 96000/12 ∴ Annual average salary of Suresh = Rs. 8000 |
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| 104. |
1). 652). 663). 684). 70 |
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Answer» As PER the given data, Let the required MEAN SCORE be x, Then 20/100 × 60 + 30/100 × 40 + 50/100 × x = 58 ⇒ 1200 + 1200 + 50X = 5800 ⇒ 2400 + 50x = 5800 ⇒ 50x = 5800 - 2400 = 3400 ⇒ x = 3400/50 = 68 |
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| 105. |
A factory buys 8 machines. 3 Machine A, 4 Machine B and rest Machine C. Prices of the machines are Rs. 45000, Rs. 25000 and Rs. 35000 respectively. Calculate the average cost (in Rs) of these machines?1). 337502). 350003). 362504). 32250 |
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Answer» 3 Machine A, 4 Machine B and rest Machine C. So, the no. of machine C = 8 - 3 - 4 = 1 Prices of the machines are Rs. 45000, Rs. 25000 and Rs. 35000 respectively. So, the total price of all machines = Rs. (45000 × 3) + (25000 × 4) + (35000 × 1) = Rs. 135000 + 100000 + 35000 = Rs. 270000 ∴ The average price of machine = Rs. 270000/8 = Rs. 33750. |
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| 106. |
Five years ago, the average age of Raj and Shruti was 25 years. While Ram joining them now, the average becomes 29 years. What is the current age of Ram?1). 302). 313). 274). 33 |
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Answer» Five years ago, ⇒ SUM of Age of Raj and SHRUTI = 25 × 2 = 50[? Sum of quantity = AVERAGE × No. of quantity] ⇒ Sum of present age of Raj and Shruti = 50 + 10 = 60 ∴ Present age of Ram = 87 - 60 = 27 years |
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| 107. |
In a committee there are 15 members. If two members whose ages are 45 years and 55 years are replaced by the two new members, then average age of 15 members is increased by 2 years. What is the average age (in years) of the new members?1). 592). 613). 654). 68 |
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Answer» Let the initial sum x of ages of 15 members = 15A According to the question 15A - (45 + 55) + (2 new members) = 15(A + 2) ⇒ 15A - 100 + (2 new members) = 15A + 30 ⇒ Sum of ages of 2 new members = 130 ∴ Average age of the 2 new members = 130/2 = 65 years |
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| 108. |
The average height of three women is 164 cm. When another woman, the fourth one, joins the group, the average reduces by 4 cm. Now, a fifth woman with height 4 cm more than the fourth replaces one of the first three. The average height of these four is then 162 cm. What is the height of the replaced woman?1). 108 cm2). 112 cm3). 144 cm4). None of the above |
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Answer» Average height of 3 WOMEN = 164 ⇒ total height = 164 × 3 = 492 cm Now, as another woman joins the GROUP, average = total height of four woman/4 ⇒ Average = (height of 3 previous woman + height of forth one, h)/4 ⇒ 164 – 4 = (492 + h)/4 ⇒ 160 × 4 – 492 = h ⇒ h = 640 – 492 = 148 cm Now Fifth woman has height 4 cm more than the forth one ⇒ height of 5th woman = 148 + 4 = 152 cm 5th woman replaced one of the previous three women ⇒ New average of four people = (height of fifth woman + height of pervious 3 woman – height of replaced one, x + height of 4th woman)/4 Given, new average = 162 ⇒ 162 = (152 + 492 – x + 148)/4 ⇒ 648 = 792 – x ∴ x = 144 cm |
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| 109. |
1). 202). 243). 264). 28 |
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Answer» The MEAN of 14 observations = 11 ⇒ Sum of 14 observations = 11 × 14 = 154 Let the 15th observation be ‘X’. ⇒ (154 + x)/15 = 12 ⇒ 154 + x = 12 × 15 = 180 ⇒ x = 180 – 154 = 26 ∴ The 15th observation = 26 |
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| 110. |
The average revenues of 13 consecutive years of a company is Rs. 78 lakhs. If the average of first 7 years is Rs. 73 lakhs and that of last 7 years is Rs. 85 lakhs, what is the revenue for the 7th year?1). Rs. 94 lakhs2). Rs. 90 lakhs3). Rs. 88 lakhs4). Rs. 92 lakhs |
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Answer» The average revenues of 13 consecutive years of a COMPANY is Rs. 78 lakhs. So, the sum of revenues of 13 consecutive years of a company = Rs. 78 × 13 lakhs = Rs. 1014 lakh. The average of first 7 years is Rs. 73 lakhs and that of LAST 7 years is Rs. 85 lakhs. So, the total revenues of first 7 year = Rs. 73 × 7 lakhs = Rs. 511 lakhs. And, the total revenues of last 7 years = Rs. 85 × 7 lakhs = Rs. 595 lakhs. Then, the total revenues of last 6 years = Rs. 1014 – 511 lakhs = Rs. 503 lakhs. ∴ The REVENUE of the 7th year = Rs. 595 – 503 lakhs = Rs. 92 lakhs. |
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| 111. |
The average of 5 consecutive numbers is 18. If next 2 numbers are also included, then what will be the new average?1). 18.52). 193). 19.54). 20 |
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Answer» Let the 5 consecutive number be (x - 2), (x - 1), (x), (x + 1), (x + 2) The SUM of above consecutive number = 5x ⇒ Average of numbers = 18 ⇒ The sum of consecutive number/5 = 18 ⇒ 5x/5 = 18 ⇒ x = 18 If 2 more numbers are INCLUDED, we get The numbers are (x - 2), (x - 1), (x), (x + 1), (x + 2), (x + 3), (x + 4) ⇒ Sum = 7X + 7 Average = 7(x + 1)/7 = x + 1 = 18 + 1 = 19. |
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| 112. |
The mean of 22 observations is 10. Two more observations are included and the new mean becomes 11. The mean of two new observations is:1). 192). 203). 214). 22 |
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Answer» Total sum of 22 observations = 22 × 10 = 220 When two more observations are included and the new MEAN becomes 11, ⇒ Total sum of 24 observations = 24 × 11 = 264 ⇒ Difference of observations = 264 – 220 = 44 ∴ Mean of two new observations is = 44/2 = 22. |
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| 113. |
The average revenues of 9 consecutive years of a company is Rs 68 lakhs. If the average of first 5 years is Rs 63 lakhs and that of last 5 years is Rs 75 lakhs, find the revenue for the 5th year.1). Rs 80 lakhs2). Rs 76 lakhs3). Rs 78 lakhs4). Rs 74 lakhs |
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Answer» As per the given data Average revenues of 9 CONSECUTIVE years of a company = Rs.68 lakhs Total revenues of 9 consecutive years = 68 × 9 = Rs.612 lakhs Average REVENUE of the first 5 years = Rs.63 lakhs Revenue of the first 5 years = 63 × 5 = Rs.315 lakhs Average revenue of the last 5 years = Rs.75 lakhs Revenue of the last 5 years = 75 × 5 = Rs.375 lakhs ⇒ Revenue for the 5th year = (Revenue of the first 5 years + Revenue of the last 5 years) - total revenue for 9 consecutive years = (315 + 375) - 612 = 690 – 612 = Rs.78 lakhs |
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| 114. |
In a company 10 employees get a salary of Rs. 36,200 each and 15 employees get a salary of Rs. 33,550 each. What is the average salary per employee?1). 34,8752). 34,6103). 27,9004). 36, 410 |
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Answer» ⇒ TOTAL salary of 10 employees = 10 × 36200 = Rs. 362000 ⇒ Total salary of 15 employees = 15 × 33550 = Rs. 503250 ⇒ AVERAGE = (362000 + 503250)/(10 + 15) = 865250/25 = 34610 |
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| 115. |
1). 162). 183). 194). 17 |
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Answer» AVERAGE of numbers = Sum of numbers/ Total numbers Average of first 12 numbers is 15 Sum of FOUR numbers = 15 × 12 = 180 New number 41 is to be added Average of 13 numbers = (Sum of 12 numbers + 13th number)/13 Average = (180 + 41)/13 = 221/13 = 17 |
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| 116. |
1). 122). 153). 324). None of these |
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Answer» Sum of the age of all members = 23 × 4 = 92 Age of the YOUNGEST member = 5 Sum of the ages of all members at the time of birth of youngest member = 92 - (5 × 4) = 72 Average age at the time of birth of youngest member = 72/3 = 24 |
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| 117. |
1). 3.832). 4.153). 4.134). 4.70 |
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Answer» Sum of the GIVEN numbers = 2 + 2 + 2 + 4 + 6 + 7 = 23 Mean of the given values = 23/6 = 3.83 |
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| 118. |
1). 162). 83). 64). 12 |
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Answer» Given that mean of 10 observations is 17 ∴ Sum of 10 observations = 170 Also given that when ONE OBSERVATION is INCLUDED the new mean is 16 ∴ mean of 11 observations = 16 ∴ Sum of 11 observations = 11 × 16 = 176 11th observation = sum of 11 observations - sum of 10 observations = 176 - 170 = 6 |
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| 119. |
1). 6.92). 6.83). 6.74). 6.6 |
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Answer» TOTAL RUNS in FIRST 45 overs = 45 × 5.8 = 261 runs Total runs in 50 overs = 295 runs ∴ The required RUN rate in the remaining overs = (295 - 261)/5 = 6.8 runs per over |
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| 120. |
Average of 7 consecutive even numbers is 32. Which number is the largest of the 7 numbers?1). 362). 343). 384). 40 |
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Answer» Let the 7 consecutive even NUMBERS be x, (x + 2), (x + 4), (x + 6), (x + 8), (x + 10) and (x + 12). Given, Average of 7 consecutive even numbers = 32 ⇒ Sum of 7 consecutive even numbers = 32 × 7 = 224 ⇒ x + x + 2 + x + 4 + x + 6 + x + 8 + x + 10 + x + 12 = 224 ⇒ 7x + 42 = 224 ⇒ x = 182/7 = 26 ∴ The LARGEST of the 7 numbers = 26 + 12 = 38 |
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| 121. |
The average weight of a class is 36 kg. If a student with weight 15 kg joins the class, then new average weight becomes 33 kg. How many total number of students are there in the class?1). 52). 63). 84). 9 |
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Answer» LET the total number of student be X. Then, the total weight of STUDENTS = 36x If a student with weight 15 kg joins the class, then new average weight becomes 33 kg ⇒ 36x + 15 = 33(x + 1) ⇒ 36x - 33x = 33 - 15 = 18 ⇒ 3X = 18 ⇒ x = 6 Hence, the total number of students = 6 |
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| 122. |
The average attendance of a class for the first four days is 43 and that for the first five days is 41. The number of students present on the 5th day is1). 312). 323). 334). 34 |
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Answer» Total ATTENDANCE for first 4 days = 43 × 4 = 172 Total attendance for first 5 days = 41 × 5 = 205 Attendance for 5th day = 205 – 172 = 33 |
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| 123. |
Raj scored 67, 69, 78, and 88 in 4 of his subjects. What should be his score in the 5th subject so that his average equals 80?1). 882). 903). 964). 98 |
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Answer» LET the SCORE in 5th subject be X ⇒ Total score = 67 + 69 + 78 + 88 + x = 302 + x ⇒ Average = total marks/total subject ⇒ 80 = (302 + x)/5 ⇒ 400 = 302 + x ⇒ x = 98 ∴ score in 5th subject should be 98 |
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| 124. |
What will be the average of the first 7 positive odd numbers divisible by 5?1). 452). 353). 404). 32 |
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Answer» Average of the FIRST 7 positive odd numbers divisible by 5 First positive odd number which divisible by 5 = 5 Second positive odd number which divisible by 5 = 15 So on, SEVENTH positive odd number which divisible by 5 = 65 So, the series is in A.P (5, 15, 25, 35, 45, 55, 65) We know, Average of any AP = (first TERM + last term)/2 = (5 + 65)/2 = 70/2 = 35 |
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| 125. |
If one prime number is replaced by a number from the series of first five prime numbers then the average becomes 5.8. Find the replaced number.1). 92). 03). 64). 4 |
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Answer» Average = (sum of elements)/(number of elements) Given, The average of the first five prime NUMBERS is 5.6 - The first five prime number are 2, 3, 5, 7 and 11. Total sum of first five prime numbers = 2 + 3 + 5 + 7 + 11 = 28 New average after replacement = 5.8 Total sum of five numbers = 5.8 × 5 = 29 The DIFFERENCE between the sum of five numbers = 29 - 28 = 1 ∴ The replaced number is 1. |
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| 126. |
The average of 100 numbers is 45. The average of these 100 numbers and five other new numbers is 50. The average of the five new numbers will be1). 8002). 1503). 1764). 24 |
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Answer» GIVEN, average of 100 NUMBERS is 45. We know that the average of n numbers = (Total sum of n numbers)/n ∴ 45 = (Sum of 100 numbers)/100 Sum of 100 numbers = 4500 Also, the average of these 100 numbers and 5 new numbers is 50. $(\THEREFORE 50 = \frac{{Sum\;of\;100\;numbers + Sum\;of\;5\;new\;numbers}}{{100 + 5}})$ ⇒ 50 × 105 = 4500 + Sum of 5 new numbers ⇒ Sum of 5 new numbers = 5250 – 4500 = 750 THUS, the average of five new numbers = 750/5 = 150 |
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| 127. |
1). 52). 43). 64). 3 |
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Answer» AVERAGE of N numbers = 12 ∴ Sum of N numbers = 12N According to problem, ⇒ 12N - 6 = 14N - 14 ⇒ 2N = 8 ⇒ N = 4 |
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| 128. |
Find the average of first 21 multiples of 21?1). 2312). 2413). 214). 11 |
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Answer» First 21 multiples of 21 are - 21, 42, 63 …. 441 ⇒ AVERAGE of odd number of consecutive terms is always equal to the value of the middle term ⇒ If there are N number of consecutive terms, ⇒ Average = [(n + 1)/2]th term ⇒ Here n = 21 ∴ Average = [(21 + 1)/2]th term = 11th term ⇒ 11th term will be 11th multiple of 21 = 21 × 11 = 231 |
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| 129. |
1). Rs. 290.802). Rs. 285.603). Rs. 310.754). Rs. 225.96 |
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Answer» ⇒ Average = sum of elements / number of elements Given There are 2500 raincoats, 1500 schoolbags and 1000 water bottles were ordered in a month. Out of this, 60% of raincoats, 75% of schoolbags and 80% of water bottles were sold out. Number of sold raincoats = = (60/100) × 2500 = 1500 Number of sold schoolbags = = (75/100) × 1500 = 1125 Number of sold water bottles = = (80/100) × 1000 = 800 The price of a raincoat, a schoolbag and a water bottle is Rs. 300, Rs. 400 and Rs. 120. Total price of 300 raincoats = 300 × 1500 = Rs. 450000 Total price of 400 schoolbags = 400 × 1125 = Rs. 450000 Total price of 120 water bottles = 120 × 800 = Rs. 96000 Total price INCLUDING all three sold ITEMS = = 450000 + 450000 + 96000 = 996000 Total number of items = = 1500 + 1125 + 800 = 3425 Average income of that shop for that particular month. = (996000/3425) = 290.8 Average income of that shop for that particular month is Rs. 290.8 |
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| 130. |
Average age of 5 children of the family is 11 years. Average age of their parents and grandmother is 35 years. Find the average age of all the eight members of the family?1). 10 years2). 20 years3). 30 years4). 25 years |
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Answer» We know that, Average = Sum total of AGE of all the Members / Total no. of members ? According to Question ⇒ 11 = Sum total of Age of 5 children / 5 ⇒ Sum total of age of 5 children = 11 × 5 = 55 Similarly, according to Question ⇒ 35 = Sum total of Age of parents & GRANDMOTHER / 3 ⇒ Sum total of age of Parents & Grandmother = 35 × 3 = 105 ⇒ Total sum of ages of 8 members (5 Children, parents, grandmother) = 55 + 105 = 160 ⇒ Average = 160/8 = 20 |
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| 131. |
1). 151.52). 88.183). 78.34). 65.7 |
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Answer» ⇒ Number of ROOMS on first floor = 30 ⇒ Occupied ROOM on first floor = 60% of 30 = 18 ⇒ Number of rooms on second floor = 40 ⇒ Occupied room on second floor = 40% of 40 = 16 ⇒ Number of rooms on third floor = 40 ⇒ Occupied room on third floor = 75% of 40 = 30 ⇒ Total revenue = (18 × 200) + (16 × 100) + (30 × 150) = 9700 ∴ AVERAGE = 9700 / 110 = 88.18 |
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| 132. |
A survey was conducted over a group of 900 people asking about their usage of mobile, laptop and tablet. The average number of users that are using only one of them is 107, while 168 users are using all of the three items. If the total number of people using none of them is 5, how many users are using only two of the items?1). 3212). 4063). 5744). 620 |
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Answer» ⇒ Average no. of users using only ONE of them = 107 ⇒ Total no. of users using either mobile or laptop or tablet = 3 × 107 = 321 Now, ⇒ No. of users using all three of them = 168 ⇒ No. of people using none of them = 5 ⇒ Total no. of people in survey = 900 ∴ No. of people using only two of them = 900 – 321 – 168 – 5 = 900 – 494 = 406 |
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| 133. |
Average weight of a group of 11 students is 37 kg. When a student left the group, then average weight of the group becomes 35 kg. What is the weight (in kg) of the student who left?1). 652). 593). 574). 70 |
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Answer» ⇒ AVERAGE weight of 11 students = 37 ⇒ SUM of weights of 11 students = 37 × 11 = 407 Now, average weights of 10 students = 35 ⇒ Sum of weights of 10 students = 35 × 10 = 350 ∴ Weight of the student who LEFT = 407 - 350 = 57 KG |
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| 134. |
A New Year party was organised for which 150 people were invited, such that the average cost of food per person was Rs. 1840. But, due to the absence of some people at the party, the average cost of food per person increased by Rs. 160. How many of the invited people did not attend the party?1). 102). 123). 144). 16 |
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Answer» No. of people invited = 150 Average cost of food per person = Rs. 1840 ⇒ Total cost of food = 150 × 1840 = Rs. 276000 Let ‘x’ people attended the party ? The average cost of food increased by Rs. 160, ⇒ Average cost of food per person BECAME = 1840 + 160 = Rs. 2000 ⇒ 2000x = 276000 ⇒ x = 276000/2000 = 138 ∴ No. of invited people that did not ATTEND the party = 150 – 138 = 12 |
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| 135. |
1). 4 km2). 8 km3). 5 km4). 6 km |
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Answer» Let the DISTANCE covered in TWO more DAYS be x KM ⇒ (3 + 4 + 3.5 + 5 + 4.5 + x)/7 = 4 ⇒ 20 + x = 28 ⇒ x = 28 - 20 = 8 ∴ Distance covered in two more days = 8 km |
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| 136. |
1). 23 years2). 24 years3). 26 years4). 27.5 years |
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Answer» Given, Average AGE of 2 GRANDPARENTS = 62 years. Total age of grandparents = Average age × number of grandparents Total age of grandparents = 62 × 2 = 124 years & Average age of 2 PARENTS = 36 years. Total age of parents = average age × number of grandparents Total age of parents = 36 × 2 = 72 years & Average age of 4 kids = 62 years. Total age of kids = Average age × number of kids Total age of kids = 9 × 4 = 36 years Total age of all family members = 124 + 72 + 36 = 232 ⇒ Average age of family = Total of age/Number of people in family ∴ Average age of family = 232/8 = 29 years |
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| 137. |
1). 162). 143). 174). Data inadequate |
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Answer» Let US consider the number of boys to be x. Number of girls = 70% of x Therefore, x + 70% of x = 85 $(\begin{array}{l} \Rightarrow {\rm{}}x + \frac{{70 \times x}}{{100}} = 85\\ \Rightarrow {\rm{}}x + \frac{{7x}}{{10}} = 85 \END{array})$ ⇒ 10x + 7x = 850 ⇒ 17x = 850 ⇒ x = 850/17 ⇒ x = 50 Therefore, number of boys = 50 and the number of girls are (85 – 50) = 35 Number of boys playing only badminton = 50% of boys$( = \frac{{50}}{{100}} \times 50{\rm{}} = {\rm{}}25{\rm{}})$ ∴ Total number of boys playing Table TENNIS = 50 – 25 = 25 Total number of boys playing badminton = 60% of boys = $(\frac{{60}}{{100}} \times 50 = {\rm{}}30)$ ∴ Total number of boys playing both table tennis and badminton = 30 – 25 = 5 ∴ Total number of boys playing only Table Tennis = 25 – 5 = 20 Number of children playing only table tennis = 40% of all children $( = {\rm{}}\frac{{40}}{{100}} \times 85 = {\rm{}}34)$ ∴ Number of girls playing only Table tennis = 34 – 20 = 14 Total number of children playing badminton and tennis both = 12 ∴ total number of girls playing both badminton and table tennis both = 12 – 5 = 7 Therefore, number of girls playing only badminton = 35 – (14 + 7) = 14 |
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| 138. |
1). 35.332). 41.333). 27.54). 44.5 |
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Answer» ∴ Average of remaining numbers $( = \FRAC{{14\; \TIMES \;32\; - \;5\; \times \;26}}{{14\; - \;5}} = \frac{{448\; - \;130\;}}{9} = \frac{{318\;}}{9} = 35.33)$ |
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| 139. |
1). 30 years2). 26 years3). 35 years4). 25 years |
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Answer» Average age of 40 PERSONS including TEACHER = 16 YEARS 3 months $(\Rightarrow 16 + \frac{3}{{12}} = 16 + \frac{1}{4} = \frac{{65}}{4}{\rm{\;years}})$ We know, $(Average = \frac{{Sum\;of\;all\;observations}}{{Number\;of\;observations}})$ ⇒ Sum of all observations = Average × Number of observations Total age of 39 STUDENTS = 39 × 16 = 624 years Also, Total age of 40 persons = (65/4) × 40 = 650 years ∴ Age of the teacher = 650 years – 624 years = 26 years |
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| 140. |
A library has an average of 480 visitors on Sundays and 210 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is1). 2452). 2553). 2744). 280 |
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| 141. |
1). 122). 133). 154). 18 |
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Answer» As per given data - Average of first 9 numbers in series is 19 SUM of 9 numbers = average × 9 ⇒ Sum of 9 numbers = 19 × 9 = 171 & Average of 1st 4 numbers is 17 Sum of 1st 4 numbers = average × 4 Sum of 1st 4 numbers = 17 × 4 = 68 Also Average of LAST 4 numbers is 22 Sum of last 4 numbers = average × 4 Sum of last 4 numbers = 22 × 4 = 88 Sum of 9 numbers = sum of 1st 4 numbers + Middle number + sum of last 4 numbers ⇒ 171 = 68 + Middle number + 88 ∴ Middle number = 15 |
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| 142. |
1). 7272). 7283). 7294). 730 |
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Answer» As we KNOW, Sum of n natural NUMBERS = n(n + 1)/2 ⇒ AVERAGE of n natural numbers = (n + 1)/2 Given, Average of n natural numbers = 243 ⇒ (n + 1)/2 = 243 ⇒ n + 1 = 486 ⇒ n = 486 – 1 = 485 ⇒ 3n = 3(485) = 1455 ∴ Average of 3n natural numbers = (1455 + 1)/2 = 1456/2 = 728 |
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| 143. |
Out of 10 teachers of a school, one teacher retires and in his place, a new teacher of age 25 yr joins. As a result average age of teachers reduces by 3 yr. The age of the retired teacher is 1). 50 yr2). 55 yr3). 58 yr4). 60 yr |
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| 144. |
The average weight of P, Q and R is 47 kg. If the average weight of P and Q be 32.5 kg and that of Q and R be 48.5 kg, then what is the weight of Q (in kgs)?1). 252). 213). 294). 33 |
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Answer» <P>Let the weight of P, Q and R be p, q and r According to the problem statement ⇒ (p + q + r)/3 = 47 ⇒ p + q + r = 141----(i) Now GIVEN that average weight of P and Q is 32.5 kg ⇒ (p + q) /2 = 32.5 ⇒ p + q = 65 ----(ii) Now, (i) - (ii) Now given that average weight of Q and R is 48.5 kg ⇒ (q + r)/2 = 48.5 ⇒ q + r = 97 Using (iii) ⇒ q = 97 - 76 = 21 years ∴ the weight of q is 21 years |
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| 145. |
A shop of hardware store is closed on Monday. The average sales per day for remaining six days of a week are Rs. 13640 and the average sale of Tuesday to Thursday is Rs. 15152 while the average sales of Friday to Saturday is Rs. 8726. The sales on Sunday is:1). Rs. 179322). Rs. 189323). Rs. 198544). Rs. 20188 |
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Answer» Now as the FORMULA for AVERAGE can be given as AE = SE/NE Where AE = Average of entities SE = Sum of entities NE = Number of entities Now as the average sales per day for remaining six days of a week are RS. 13640 ∴ Sum of sales of six days = 13640 × 6 = 81840 Rs. Now as the average sale of Tuesday to Thursday is Rs. 15152 ∴ Sum of sales of Tuesday to Thursday = 15152 × 3 = 45456 Rs. Now as average sales of Friday to Saturday is Rs. 8726 ∴ Sum of sales of Friday to Saturday = 8726 × 2 = 17452 Rs. Now the sales of SUNDAY can be given as Total sales of six days - sales of Tuesday to Thursday - sales of Friday to Saturday = 81840 - 45456 - 17452 = Rs. 18932 |
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| 146. |
If there are four numbers such that average of A and B is 29, average of B and C is 34 and average of C and D is 50, then what will be the average of A and D?1). 342). 353). 444). 45 |
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Answer» A + B = 2 × 29 = 58----(1) B + C = 2 × 34 = 68----(2) C + D = 2 × 50 = 100----(3) Adding all these three EQUATIONS, we get A + 2B + 2C + D = 226 ⇒ A + D = 226 – 2 × 68 ⇒ A + D = 226 – 136 ⇒ A + D = 90 So, average of A + D = 90/2 = 45 |
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| 147. |
The average of a group of numbers is 64. When a number is added to the group, the average becomes 70. When the number is again added to the group, the average is again increased to 75. Find the number.1). 702). 903). 1104). 130 |
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Answer» Let there be ‘n’ numbers in the group and the number to be added be ‘x’ Sum of numbers in group = 64n When x is added to the group, sum of numbers in group = 70(n + 1) ⇒ 64n + x = 70(n + 1) ⇒ x = 70n + 70 – 64n ⇒ x = 6n + 70.... (1) Now, when x is again added, sum of numbers in group = 75(n + 2) ⇒ 70(n + 1) + x = 75(n + 2) ⇒ 70n + 70 + x = 75n + 150 ⇒ x = 5n + 80.... (2) EQUATING (1) and (2), ⇒ 6n + 70 = 5n + 80 ⇒ 6n – 5n = 80 – 70 ⇒ n = 10 Substituting in eq. (1), ⇒ x = 6(10) + 70 ∴ The number is 130 |
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| 148. |
The average of 17 results is 60. If the average of first 9 results is 57 and that of the last 9 results is 65, then what will be the value of 9th result?1). 392). 783). 1174). 156 |
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Answer» Average of all 17 results is = 60 Sum of all 17 results = 60 × 17 = 1020 Average of first 9 results = 57 Sum of first 9 results = 57 × 9 = 513 Average of LAST 9 results = 65 Sum of last 9 results = 65 × 9 = 585 Both SUMS I.e. of first 9 and last 9 digits include the 9 result So if we add both sums it will have the 9th result one extra time and we can subtract the actual sum from this ADDITION to GET the 9th result So sum of first and last 9 results = 513 + 585 = 1098 ∴ The 9th result will be = 1098 - 1020 = 78 |
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| 149. |
The average of 11 numbers is 8. If a number 12 is removed, then what will be the new average?1). 7.42). 7.63). 7.84). 8.2 |
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Answer» ∴ SUM of 11 numbers = 11 × 8 = 88 After a number 12 is removed Sum of remaining 10 numbers = 88 - 12 = 76 ∴ Average of remaining 10 numbers = 76/10 = 7.6 |
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| 150. |
The average weight of Jaffar, Guna and Ashik is 60 kg. If the average weight of Jaffar and Guna be 55 kg and that of Guna and Ashik be 50 kg, then the weight of Guna is: 1). 30 kg2). 20 kg3). 35 kg4). 40 kg |
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Answer» Let the WEIGHT of Jaffar, Guna and Ashik be a, b and c Average weight of Jaffar, Guna and Ashik = Sum of weights/3 ⇒ 60 = (a + b + c)/3 ⇒ a + b + c = 180----(1) Also, Average weight of Jaffar and Guna = sum of weights/2 ⇒ 55 = (a + b)/2 ⇒ a + b = 110 ⇒ a = 110 – b----(2) Also, Average weight of Guna and Ashik = Sum of weights/2 ⇒ 50 = (b + c)/2 ⇒ b + c = 100 ⇒ c = 100 – b----(3) Now, substituting equation (2) and (3) in equation (1), ⇒ (110 – b) + b + (100 – b) = 180 ⇒ 210 – b = 180 ⇒ b = 30 ∴ The weight of Guna is = 30 kg |
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