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

Seven students played a game and their scores in ascending order were 30, 40, 50, 60, 70, 70 and 80. Which of the following statements is/are true? I. The average is greater than 60. II. The median is less than 60. III. The mode is greater than 50.

  1. II and III

  2. II only

  3. III only

  4. I and II
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Data: 30, 40, 50, 60, 70, 70, 80. Average = 400/7 = 57.14 (I is false). Median = 60 (II is false). Mode = 70, which is > 50 (III is true).

Multiple choice

A young man climbs a hill with a speed of 3 kilometers per hour and immediately returns from the top to the original place with a speed of 6 kilometers per hour. What is his average speed for the round trip?

  1. 3.75 Km/hr

  2. 4 Km/hr

  3. 4.5 Km/hr

  4. 4.75 Km/hr
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Average speed for a round trip with speeds v1 and v2 is 2 * v1 * v2 / (v1 + v2). Here v1 = 3 and v2 = 6. Average speed = 2 * 3 * 6 / (3 + 6) = 36 / 9 = 4 km/hr.

Multiple choice

The average age of a class is 16. If two students aged 17 and 15 leave the class, which of the following is true about the new average age?

  1. The average will not change as the average of the students leaving the class is also 16.

  2. The average will decrease because of the decrease in the total number of students.

  3. The average will increase because of the change in the number of students and the total age.

  4. Nothing can be said about the new average without knowing the exact number of students.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The average age of the class is 16. When two students with ages 17 and 15 leave, their combined age is 32, which is exactly 2 times the average of 16. Since the sum of the ages of the students leaving is equal to the original average multiplied by the number of students leaving, the total sum of the remaining ages divided by the new count remains 16.

Multiple choice

Directions: The following problem contains a question and two statements giving certain data. You have to decide whether the data given in the statements are sufficient for answering the question. During a 6-day local trade show, the least number of people registered in a single day was 80. Was the average (arithmetic mean) number of people registered per day for the 6 days greater than 90? (1) For the 4 days with the greatest number of people registered, the average (arithmetic mean) number registered per day was 100. (2) For the 3 days with the smallest number of people registered, the average (arithmetic mean) number registered per day was 85.

  1. Statement (1) alone is sufficient, but statement (2) alone is not.

  2. Statement (2) alone is sufficient, but statement (1) alone is not.

  3. Both (1) and (2) together are sufficient, but none of them alone is sufficient.

  4. Both of them independently are sufficient.

  5. Both the statements (1) and (2) together are not sufficient.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the 6 days be d1 <= d2 <= d3 <= d4 <= d5 <= d6. We know d1 >= 80. Statement 1: (d3+d4+d5+d6)/4 = 100, so d3+d4+d5+d6 = 400. Sum of all 6 days = d1+d2+d3+d4+d5+d6 = d1+d2+400. Since d1, d2 >= 80, sum >= 160+400 = 560. Average = 560/6 = 93.33 > 90. Sufficient.

Multiple choice

Directions: The question given below consists of a statement followed by Quantity I and Quantity II. Find the relationship between the quantities and choose the answer accordingly. The average marks obtained by Khalil in four tests were 73.5. When the marks of the fifth tests were added, the average increased by 2.9. Quantity I: If the marks obtained by Khalil in the first test were marked 76 but the original marks were 67, what would be the new average for five tests? Quantity II: If the marks obtained by Khalil in the each of the first three tests were increased by 2 and in each of the last two tests were decreased by 8, then what would be the new average for the five tests?

  1. Quantity I > Quantity II

  2. Quantity I < Quantity II

  3. Quantity I ≥ Quantity II

  4. Quantity I ≤ Quantity II

  5. Quantity I = Quantity II or relationship cannot be established
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Original sum of 4 tests = 73.5 * 4 = 294. Adding a 5th test increases the average by 2.9, so the new average is 76.4 and the sum is 382. The 5th test mark is 382 - 294 = 88. Quantity I: New sum = 382 - 67 + 76 = 391, new average = 391 / 5 = 78.2. Quantity II: Total change = (3 * 2) + (2 * -8) = 6 - 16 = -10. New average = 76.4 - (10 / 5) = 74.4. Since 78.2 > 74.4, Quantity I > Quantity II.

Multiple choice

The average marks of 6 students in a test are 64. All the students got different marks, one of the students obtained 70 marks and all other students scored 40 or above. The maximum possible difference between the second highest and the second lowest marks is

  1. 50

  2. 54

  3. 57

  4. 58
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Total marks = 6 * 64 = 384. One student = 70. Remaining 5 students sum = 314. To maximize the difference between 2nd highest (H2) and 2nd lowest (L2), we minimize L2 and maximize H2. Let L1=40, L2=40. Let H1=x, H2=y. Remaining marks = 384 - 70 - 40 - 40 = 234. To maximize y, we minimize x. Since x > y, let x = y + 1. 2y + 1 = 234. y = 116.5. This is impossible as marks are usually <= 100. Assuming max marks 100: H1=100, H2=94. Remaining = 384 - 70 - 100 - 94 = 120. L1=40, L2=40. Difference = 94 - 40 = 54.

Multiple choice

In a given village, there are only three sizes of families: families with 2 members, families with 4 members and families with 6 members. The proportion of families with 2, 4 and 6 members are roughly equal. A poll is conducted in this village wherein a person is chosen at random and asked about his/her family size. The average family size computed by sampling 1000 such persons from the village would be closest to

  1. 4

  2. 4.667

  3. 4.333

  4. 3.667
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When sampling people, you are more likely to pick a person from a larger family. The probability of picking a family of size n is proportional to n. The expected value is (2*2 + 4*4 + 6*6) / (2+4+6) = (4+16+36) / 12 = 56 / 12 = 4.667.

Multiple choice

A group of men and women gathered to compete in a marathon. Before the competition, each competitor was weighed and the average weight of the female competitors was found to be 120 lbs. What percentage of the competitors were women? (1) The average weight of the men was 150lb. (2) The average weight of the entire group was twice as close to the average weight of the men as it was to the average weight of the women.

  1. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient

  2. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient

  3. BOTH statements (1) and (2) TOGETHER are sufficient to answer

  4. EACH statement ALONE is sufficient to answer the question asked

  5. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice

Directions: Answer the questions independently of each other. 5 Scores in a classroom are broken into 5 different ranges, 51-60, 61-70, 71-80, 81-90 and 91-100. The number of students who have scored in each range is given below. 51 to 60 - 5 students, 61 to 70 - 10 students, 71 to 80 - 10 students, 81 to 90 - 4 students, 91 to 100 - 3 students. Furthermore, we know that the number of students who scored 80 or more is just one more than those who scored below 76. What is the maximum possible average overall of this class?

  1. 76.84

  2. 77.26

  3. 78.46

  4. 78.54
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let N1=5, N2=10, N3=10, N4=4, N5=3. Total students = 32. Let S be the number of students scoring 80 or more (N4+N5 = 7). Let X be the number of students scoring below 76. We are told S = X + 1, so X = 6. To maximize the average, we assume students in each range score at the upper bound. For 71-80, we have 10 students; 4 must be >= 76 to satisfy X=6 (since 5+10+1=16 students are below 71, but we need 6 below 76, so 6-5=1 student in the 71-75 range). Calculating the weighted sum with max values gives the result.

Multiple choice

Heights of the students of a class are being measured. The average height of boys is 130 cm and the average height of girls is 120 cm. If the average height of the whole class is 127 cm, what could be the possible strength of boys and girls in the class, respectively?

  1. 7 and 3

  2. 8 and 3

  3. 7 and 4

  4. 9 and 6

  5. 9 and 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the allegation method: (130-127) / (127-120) = 3 / 7. The ratio of boys to girls must be 3:7. Option A (7 and 3) is the inverse, but looking at the options, 7:3 is the only one that fits the ratio logic if the labels were swapped or if the question implies a specific order.

Multiple choice

There are 24 persons in a club. When three new members, whose average age is one-fourth of the average age of the initial group of 24 persons joined the club, the average age of all the members in the club decreased by 3 years. What is the average age (in years) of all the 27 members in the club?

  1. 9

  2. 33

  3. 36

  4. 42
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let A be the average of 24 persons. Total age = 24A. New members average = A/4. Total age of 3 new members = 3 * (A/4) = 0.75A. New total age = 24.75A. New average = 24.75A / 27 = A - 3. 24.75A = 27A - 81. 2.25A = 81. A = 36. New average = 36 - 3 = 33.