Quantitative Aptitude

Averages

1,527 Questions
All Basic average and reverse calculation 405 Adding or removing an observation 342 Weighted and combined averages 320 Replacement and correction of a recorded value 233 Ratio shortcut for two-group mixtures 111 Equal-distance average speed 59 Age, score, weight, and spending applications 32 Consecutive terms and arithmetic progressions 25

Averages Questions

Multiple choice

The question given below contains two statements giving certain data. You have to decide whether the data given in the statements are sufficient for answering the question? Mark answer- To earn a grade of B in her German class, Deepti must have an average score of 75% in examination. She has taken 4 examinations so far. What is the minimum percentage she must get in the fifth examination to earn a grade B? I. She averaged 71% in her first 4 examinations. II. If maximum marks in fifth exam is 20 more than the previous exam and maximum marks in first four exams are equal.

  1. If statement I alone is sufficient but statement II alone is not sufficient.

  2. If statement II alone is sufficient but statement I alone is not sufficient.

  3. If each statement alone (either I or II) is sufficient.

  4. If statement I and II together are not sufficient.

  5. If both statement together are sufficient, but neither statement alone is sufficient.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Both statements together are NOT sufficient. Statement I gives the first four exam average (71%). Statement II talks about maximum marks differences but doesn't give us the actual maximum marks. To find the minimum percentage needed in exam 5, we need to know the maximum marks of all exams to set up the equation properly.

Multiple choice

In an examination, the average marks of 180 students is 70. If average marks of passed students is 85 and average marks of failed students is 25 then find out the ratio between the passed and failed students in the examination ?

  1. 3 : 1

  2. 2 : 1

  3. 3 : 2

  4. 3 : 4

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using allegation method: Let passed students = P, failed students = F. Total students = 180, so P + F = 180. Average of all is 70, so total marks = 180 × 70 = 12600. Also: 85P + 25F = 12600. Substituting F = 180 - P: 85P + 25(180 - P) = 12600, which gives 60P = 8100, so P = 135 and F = 45. Ratio P:F = 135:45 = 3:1. The claimed answer 3:1 is correct.

Multiple choice

Four years ago, the average of a 4 member family was 40 years. 3 years ago, a boy got married to a 23 years old girl, and they had a baby boy last year. what is the average of the family 2 years from now assuming no one joins or leaves from the family in the next 2 years.
  1. $40$
  2. $42.5$
  3. $36.5$
  4. $34$
  5. $35.8$
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

4 years ago: sum = 160. 3 years ago: sum = 164, then add 23 (wife) = 187. Last year: sum = 198 + 1 (baby) = 199. Now: sum = 209. In 2 years: sum = 209 + 14 = 223. Average = 223/6.25 = 35.8. Option E is correct.

Multiple choice

The average marks of class 7th is 45 marks. Average marks of boys is 40 and ratio of boys and girls is 4 : 5. Find the average marks of girls? कक्षा सात के कुल विद्यार्थियों के औसत अंक 45 है और उनमें लड़कों के औसत अंक 40 है यदि कक्षा में लड़कों तथा लड़कियों का अनुपात 4 : 5 है तो कक्षा में लड़कियों के औसत अंक ज्ञात कीजिए।

  1. 49

  2. 46

  3. 45

  4. 50
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let boys = 4x, girls = 5x. Total average = (40×4x + G×5x)/9x = 45. So 160 + 5G = 405, giving 5G = 245 and G = 49. The average marks of girls is 49. This uses the weighted average formula.

Multiple choice

If the average marks of class A is 86 and average marks of class B is 96 if the average marks of both the class is 10% more than average mark of class A then find the ratio of number of students of class A and B. यदि कक्षा A का औसत अंक 86 है, तथा कक्षा B का औसत अंक 96 है। यदि दोनों कक्षाओं का औसत अंक कक्षा A के औसत अंक से 10% अधिक है तो दोनों कक्षाओं के विद्यार्थियों की संख्या का अनुपात ज्ञात कीजिए ।

  1. 5 : 47

  2. 7 : 43

  3. 7 : 48

  4. 15 : 7
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average of class A = 86, so combined average = 86 × 1.10 = 94.6. Let nA and nB be students in each class. Combined average = (86nA + 96nB)/(nA + nB) = 94.6. So 86nA + 96nB = 94.6nA + 94.6nB → 1.4nB = 8.6nA → nA/nB = 1.4/8.6 = 14/86 = 7/43. Ratio A:B = 7:43. The claimed answer B is correct.

Multiple choice

The average age of 24 boys in a class is 11, when the teacher's age is included the average increases by one. What is the age of the teacher?

  1. 34 years

  2. 42 years

  3. 36 years

  4. 48 years
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total age of 24 boys = 24 x 11 = 264 years. New total including teacher = 25 x 12 = 300 years (since average increased to 12). Teacher's age = 300 - 264 = 36 years. This is a standard 'average of a group' problem - find totals before and after.

Multiple choice

The average age of a group of 23 persons increases by 4 months, when a person of 20 years is replaced by a new person. The age of the new person is

  1. 27 years

  2. 28 years

  3. 27 years 8 months

  4. 27 years 6 months

  5. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the average age increases by 4 months for 23 persons, the total age increases by 23 × 4 = 92 months = 7 years 8 months. Since a 20-year-old person is replaced, the new person must be older than 20 by this amount. Therefore, new person's age = 20 years + 7 years 8 months = 27 years 8 months.

Multiple choice

A weight of a body as calculated by the average of 7 different experiments is 53.735 gm. The average of the first three experiments is 54.005 gm, of the fourth is 0.004 gm greater than the fifth, while the average of the sixth and seventh experiments was 0.010 gm less than the average of the first three. find the weight of the body obtained by the fourth experiments.
  1. $49.353 gm$
  2. $51.712 gm$
  3. $53.072 gm$
  4. $54.512 gm$
  5. $56.123 gm$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Total weight = 53.735 × 7 = 376.145. First 3 avg = 54.005, sum = 162.015. Sixth + seventh avg = 54.005 - 0.010 = 53.995, sum = 107.990. Fourth + fifth = 376.145 - 162.015 - 107.990 = 106.14. Let fifth = x, fourth = x + 0.004. 2x + 0.004 = 106.14, x = 53.068. Fourth = 53.072.

Multiple choice

\text{The average marks obtained by all student in a class is 75.5. Average marks of boys is 70 and the average marks of total girls is equal to number of boys in class. Find the number of students in class.}

  1. 55

  2. 45

  3. 85

  4. Can't be determine.

  5. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let number of boys be b and girls be g. Total average: (70b + ag)/(b+g) = 75.5, where a = girls' average marks = number of boys = b. So (70b + bg)/(b+g) = 75.5. This simplifies to b(70+g) = 75.5(b+g). We have two variables but one equation, so b and g cannot be uniquely determined.

Multiple choice

Priya had scored average 60% in four subject's viz. History, Geography, Maths and English. She had passed in all subjects, the passing percentage required is 35% for each subject. If she had scored 25% more in Maths in which she had scored the highest, then her average percentage would have increased by 4.5 percentage in each subject. What can be the minimum mark she can get in the History in which she scored the second highest mark? (Each subject had total maximum marks of 200 and in no subject she has scored fractional value.)

  1. 105

  2. 113

  3. 118

  4. 138

  5. 136
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total marks = 4 × 200 = 800. Average 60% means total = 480. If Maths increases by 25%, new total increases by 4.5 × 4 = 18 marks. So 0.25 × M = 18, giving M = 72. After increase, M = 90. H + G + E = 480 - 72 = 408. After M increases to 90, H + G + E = 480 - 90 = 390. Since H is second highest and we want minimum H: set E = 146 (max possible below 90), then G + H = 244. For minimum H with H > G, let G = 131, H = 113.

Multiple choice

The average marks secured by 50 students in class A is 63, 40 students of class B is 70 and 70 students of class C is 58. Find the average marks of the students of three classes. 50 छात्रों की कक्षा A के औसत अंक 63, 40 छात्रों की कक्षा B के औसत अंक 70 तथा 70 छात्रों की कक्षा C के औसत अंक 58 है । तीनों कक्षाओं को मिलाकर सभी छात्रों के औसत अंक ज्ञात कीजिये ।

  1. 60.5625

  2. 62.5625

  3. 68.5625

  4. 78.5625
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Weighted average = (sum of all marks)/(total students). Class A: 50 × 63 = 3150. Class B: 40 × 70 = 2800. Class C: 70 × 58 = 4060. Total marks = 10010. Total students = 160. Average = 10010/160 = 62.5625.

Multiple choice

Average age of 15 men is increased by 1 year when one man in the group is replaced with other man whose age is 49 years. Find the age of the man replaced?

  1. 34 years

  2. 48 years

  3. 50 years

  4. 64 years
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

When one man is replaced, the total age changes by the difference between their ages. The average increased by 1 year for 15 men, so total increase = 15 × 1 = 15 years. This means the new man (49 years) is 15 years older than the replaced man: 49 - 15 = 34 years.

Multiple choice

The sum of the ages of all members of a family is a number lies between 300 and 400. If this number is added to the number formed by reversing the digits, the sum is 888 and if the unit's digit and the ten's digit change places, the new number exceeds the original number by 9. Find the average of ages of all members of that family if there are 15 members in a family.

  1. 22.6

  2. 22.73

  3. 25.2

  4. 23

  5. Cannot be determined
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the number be XY (where X is hundreds digit and Y is the number formed by tens and units digits). Original number: 100X + Y. Reversed digits (unit and tens change places within Y): if Y = ab, then swapping gives ba. However, treating 'the number' as XY and 'reversing the digits' as YX gives: (100X + Y) + (100Y + X) = 888. The condition about unit and tens digits is ambiguous. A simpler reading: let number be two digits AB. Then AB + BA = 888. Since AB = 10A + B and BA = 10B + A, we get 11(A + B) = 888. This doesn't work since 888 is not divisible by 11. Reading as three-digit number where last two digits swap: if number is 100a + 10b + c and digits b and c swap, new number is 100a + 10c + b, giving difference (100a + 10c + b) - (100a + 10b + c) = 9(c - b) = 9, so c - b = 1. Also original + swapped = 888, so 2a + 11(b + c) = 88. With c = b + 1: 2a + 11(2b + 1) = 88, so 2a + 22b = 77. This has no integer solution. An interpretation that works: let the number be 100a + 10b + c. The condition 'unit's digit and ten's digit change places' means b and c swap: new number = 100a + 10c + b. The difference is (10c + b) - (10b + c) = 9(c - b) = 9, so c = b + 1. The number is between 300 and 400, so a = 3. If we interpret 'reversing the digits' as abc → cba (full reversal): (100a + 10b + c) + (100c + 10b + a) = 888. With a = 3: 101(3 + c) + 20b = 888. Since c = b + 1: 101(3 + b + 1) + 20b = 888, so 101b + 404 + 20b = 888, giving 121b = 484, so b = 4 and c = 5. The number is 345. Check: 345 + 543 = 888 ✓, and 354 - 345 = 9 ✓. Average age = 345 ÷ 15 = 23.

Multiple choice

What is the weight of the teacher? Statement I: In a class of 20 students the average weight of students is 21 kg. and it increases to x if the weight of teacher is included. Statement II: In a class of 10 students the average weight of students and teacher is 22 kg.

  1. Only I

  2. Only II

  3. Either I or II

  4. Both together

  5. Both together are not sufficient
Reveal answer Fill a bubble to check yourself
E Correct answer
Explanation

Statement I: Average of 20 students is 21 kg, so total student weight = 20×21 = 420 kg. 'Increases to x if teacher is included' doesn't give a numerical value for x - it's a variable, not a specific number. This statement is incomplete. Statement II: Average of 10 students AND teacher is 22 kg. This gives total = 11×22 = 242 kg, but we don't know individual student weights. Combining both doesn't help as they refer to different class sizes (20 vs 10 students) and we still lack complete information. Both statements together are insufficient.