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Averages Test - 3

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Averages Test - 3
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  • Question 1
    2 / -0.83

    There are 3 classes having 20, 25 and 30 students respectively having average marks in an examination as 20, 25 and 30 respectively. If the three classes are represented by A, B and C and you have the following information about the three classes, answer the questions that follow:

    A - Highest score 22, Lowest score 18
    B - Highest score 31, Lowest score 23
    C - Highest score 33, Lowest score 26

    If five students are transferred from A to B.

    What will happen to the average score of B?

    Solution

    Definitely decrease, since the highest marks in Class A is less than the lowest marks in Class B.

    So, the correct answer is B

  • Question 2
    2 / -0.83

    Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?

    Solution

  • Question 3
    2 / -0.83

    The average age of a class of 30 students and a teacher reduces by 0.5 years if we exclude the teacher. If the initial average is 14 years, find the age of the class teacher. 

    Solution

    Let the age of the teacher be T.

    The initial total age of the class (including the teacher) is (30+1)*14 = 434 years.

    If we exclude the teacher, the total age of the 30 students is 30*(14-0.5) = 405 years.

    So the teacher 's age, T, must be:

    434 - 405 = 29 years

    Therefore, the age of the class teacher is 29 years.

  • Question 4
    2 / -0.83

    The average of a batsman after 25 innings  was 56 runs per innings. If after the 26th inning his average increased by 2 runs, then what was his score in the 26 inning?

    Solution

    Total runs after 25 innings = 25 x 56= 1400

    Since, after 26 innings average run is increased by 2, hence new average = 56+2 = 58

    Now,

    Total runs after 26 innings = 26 x 58 = 1508

    Score in 26th inning = 1508 - 1400 = 108 runs.

  • Question 5
    2 / -0.83

    There are 3 classes having 20, 25 and 30 students respectively having average marks in an examination as 20, 25 and 30 respectively. If the three classes are represented by A, B and C and you have the following information about the three classes, answer the questions that follow:

    A - Highest score 22, Lowest score 18

    B - Highest score 31, Lowest score 23

    C - Highest score 33, Lowest score 26

    If five students are transferred from A to B. What will happen to the average score of B?

    Solution

    Class A average is 20. And their range is 18 to 22

    Class B average is 25. And their range is 23 to 31

    Class A average is 30. And their range is 26 to 33

    If 5 students transferred from A to B, As average cannot be determined but Bs average comes down as the highest score of A is less than lowest score of B.

    If 5 students transferred from B to C, Cs average cannot be determined the Bs range fo marks and Cs range

    of marks are overlapping.

  • Question 6
    2 / -0.83

     Read the following:

    There are 3 classes having 20, 25 and 30 students respectively having average marks in an examination as 20, 25 and 30 respectively. If the three classes are represented by A, B and C and you have the following information about the three classes, answer the questions that follow:

    A - Highest score 22, Lowest score 18

    B - Highest score 31, Lowest score 23

    C - Highest score 33, Lowest score 26

    In a transfer of 5 students from A to C

    What will happen to the average score of C?

    Solution

    It will definitely decrease since the highest possible transfer is lower than the lowest value in C

    So, the  correct answer is B. 

  • Question 7
    2 / -0.83

    In 2001 there were 6 members in Barney ’s family and their average age was 28 years. He got married between 2001 and 2004 and in 2004 there was an addition of a child in his family. In 2006, the average age of his family was 32 years. What is the present age (in 2006) of Barney ’s wife (in years) is:

    Solution

    If the present age of Barney ’s wife is x years. Then according to the question:
    (33 X 6 + x + 2) / 8 = 32
    x + 2 + 198 = 256 x + 2 = 58
    x = 56
    x = 56 years  

  • Question 8
    2 / -0.83

    The weight of a metal piece 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 experiment was 0.010 gm less than the average of the first three. Find the weight of the body obtained by the fourth experiment.

    Solution

    You can take 53 as the base to reduce your calculations. Otherwise, the question will become highly calculation-intensive. Let the fifth experiment ’s measurement be ‘x ’above 53.
    Then you get:
    0.735 X 7 = 1.005 X 3 + (x + 0.004) + x + 0.995 X 2
    ->5.145 = 3.015 + 2x + 0.004 + 1.99.
    On solving this you get x = 0.068. Hence, the weight of the fifth body is 53.068 and the weight of the fourth body is 53.072. Hence, the option (d) is correct. 

  • Question 9
    2 / -0.83

    One collective farm got an average harvest of 21 tons of wheat and another collective farm that had 12 acres of land less given to wheat, got 25 tons from a hectare. As a result, the second farm harvested 300 tons of wheat more than the first. How many tons of wheat did each farm harvest?

    Solution

    Let 's solve the problem again step by step, ensuring accuracy:

    1. Let the area of the first farm be AAA acres.
    2. Therefore, the area of the second farm is A −12A - 12A −12 acres.

    Given that:

    • The first farm gets an average harvest of 21 tons of wheat per acre.
    • The second farm gets an average harvest of 25 tons of wheat per acre.
    • The second farm harvested 300 tons more wheat than the first.

    Set up the equations:

    1. The total harvest of the first farm: 21A21A21A tons
    2. The total harvest of the second farm: 25(A −12)25(A - 12)25(A −12) tons
    3. The difference in harvest: 25(A −12)−21A=30025(A - 12) - 21A = 30025(A −12)−21A=300

    Solve the equation: 25A −300 −21A=30025A - 300 - 21A = 30025A −300 −21A=300 4A −300=3004A - 300 = 3004A −300=300 4A=6004A = 6004A=600 A=150A = 150A=150

    The area of the first farm is 150 acres, and the area of the second farm is 150 −12=138150 - 12 = 138150 −12=138 acres.

    Calculate the total harvest for each farm:

    1. The first farm: 21 ×150=315021 \times 150 = 315021 ×150=3150 tons
    2. The second farm: 25 ×138=345025 \times 138 = 345025 ×138=3450 tons

    The correct answer is:

    1. 3150, 3450

  • Question 10
    2 / -0.83

    Read the following passage and answer the following question that follows.

    Aman, Binod, Charan, Dharam  and Ehsaan  are the members of the same family. Each and everyone loves one another very much. Their birthdays are in different months and on different dates. Aman remembers that his birthday is between  25th and  30th, of Binod  it is between  20th  and  25th, of Charan  it is between  10th  and  20th, of Dharam  it is between  5th  and  10th  and of Ehsaan it is between  1st and  5th  of the month. The sum of the date of birth is defined as the addition of the date and the month, for example  12th  January will be written as  12/1  and will add to a sum of the date of  13. (Between  25th  and  30th  includes  25  and  30).

    If the dates of birth, of four of them are prime numbers, then find the maximum average of the sum of their dates of birth? 

    Solution

    It is given that the date of births of four of them are prime numbers

    For the average to be maximum, the dates we need to choose must be closest to the maximum dates

    Therefore, the four prime dates are  29th, 23th, 19  and  5th

    Sum of the date of birth = 29+23+19+10+5+12+11+10+9+8  =136

    Average = 136/5   = 27.2

    The maximum average when the  birth dates of four of them  are prime numbers is 27.2

  • Question 11
    2 / -0.83

    The average age of a group of persons going for a movie is 20 years. 10 new persons with an average age of 10 years join the group on the spot due to which the average of the group becomes 18 years. Find the number of persons initially going for the movie. 

    Solution

  • Question 12
    2 / -0.83

    The average height of 22 toddlers increases by 2 inches when two of them leave this group. If the average height of these two toddlers is one-third the average height of the original 22, then the average height, in inches, of the remaining 20 toddlers is

    Solution

    Let the average height of 22 toddlers be 3x.
    Sum of the height of 22 toddlers = 66x
    Hence average height of the two toddlers who left the group = x
    Sum of the height of the remaining 20 toddlers = 66x –2x = 64x
    Average height of the remaining 20 toddlers = 64x/20 = 3.2x
    Difference = 0.2x = 2 inches =>x = 10 inches
    Hence average height of the remaining 20 toddlers = 3.2x = 32 inches

  • Question 13
    2 / -0.83

    The average of 30 integers is 5. Among these 30 integers, there are exactly 20 which do not exceed 5. What is the highest possible value of the average of these 20 integers?

    Solution

    It is given that the average of the 30 integers = 5
    Sum of the 30 integers = 30*5=150
    There are exactly 20 integers whose value is less than 5.
    To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers
    So the sum of 10 integers = 10*6=60
    The sum of the 20 integers = 150-60= 90
    Average of 20 integers = 90/20 = 4.5

  • Question 14
    2 / -0.83

     Read the following:

    There are 3 classes having 20, 25 and 30 students respectively having average marks in an examination as 20, 25 and 30 respectively. If the three classes are represented by A, B and C and you have the following information about the three classes,

    answer the questions that follow:

    A - Highest score 22, Lowest score 18

    B - Highest score 31, Lowest score 23

    C - Highest score 33, Lowest score 26

    In a transfer of 5 students from B to C

    Which of these can be said about the average score of B?

    Solution

    If C increases, then the average of C goes up from 30. For this to happen it is definite that the average of B should drop.

    So, the correct option is  B.

  • Question 15
    2 / -0.83

    In an apartment complex, the number of people aged 51 years and above is 30 and there are at most 39 people whose ages are below 51 years. The average age of all the people in the apartment complex is 38 years. What is the largest possible average age, in years, of the people whose ages are below 51 years?

    Solution

    The possible average age of people  whose ages are below 51 years will be maximum
    if the average age of the number of people aged  51 years and above is minimum.
    Hence, we can say that that there are 30 people having same age 51 years. 
    Let 'x 'be the maximum average age of people whose ages are below 51. Then we can say that, 

    ⇒1530+39x =2622
    ⇒x=1092/39 =28
    Hence, we can say that option D is the correct answer.

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