AVERAGE

Q1. The mean of 29 test scores is 89. What is the sum of these test scores?  

  Ans.  Total Tests = 29, Mean = 89.

            Mean =  Sum of n quantities/ Total Number

            Sum of quantities      = Mean × Total number

                                      = 89 × 29 = 2581 

            The sum of these test scores is 2581

    Q2.  The mean of some numbers is 64. The sum of the numbers is 1472. How many numbers are there?   

  Ans.  Mean = 64, Sum of numbers = 1472 

            Mean = Sum of n quantities/ Total Number

            Total Number             = Sum  of n quantities/ Total Number

                                 =  1472/64= 23

    Q3.  Arun’s test scores are 95, 82, 75, and 88. What score must she get on the fifth test in order to achieve an average of 84 on all five tests?   

  Ans.  We are given four of the five test scores. The sum of these 4 test scores is 340. Let x represent the fifth test score, then the sum of all five test scores is 340 + x. 

            Now,    Mean = 84, Sum of numbers = 340 + x,

            Total number = 5

            5 * 84 = 340 + x 

            420   = 340 + x 

            420 – 340   = x 

            x                = 80

            so the fifth test score should be 80 to make the average of 84.

    Q4.  Find the mean of 1, 2, 3 ……. 100.

  Ans.  This can be done in two ways, one way is to go by the formula, through which:

            Mean = (1 = 2 + 3 + … + 100)/50 = 5050/100 = 50.5

            But here adding 1 to 100 will take lot of time.

            The other ways is using the formula for sum of first n natural numbers = n(n + 1)/2 to find the sum of 1 to 100. We have n = 100,

            Sum = (100 * 101)/2 = 101 × 50 = 5050

            Mean = 5050/100 = 50.5

    Q5.  Hari traveled for 2 hours at a rate of 70 kilometers per hour and for 5 hours at a rate of 60 kilometers per hour. What was his average speed for the 7-hour period?

  Ans.  Here the Average speed is given by

Total Distance/Total Time

            Total distance = 2 × 70 + 5 × 60 = 140 + 300 = 440

            Total time = 2 + 5 = 7 hrs

            Average speed =  440/4 = 62.85

    Q6.  The average weight of a group of 30 students increases by 1 kg when the weight of their teacher was added. The average weight of the students is 31kgs including the weight of the teacher, what is the weight of the teacher?

  Ans.  Average weight including the Teacher = 31

            As the new average is 1kg more than the old average, old average = 31 – 1 = 30 kgs.

            Weight of the 30 students = 30 × 30 = 900 kgs

            Weight of the 30 students and their teacher

            = 31 × 31 = 961 kgs

            Therefore, the weight of the Teacher = 961 – 900

                                                                 = 61 kgs.

    Q7.  The average marks scored by 50 students in the class were 43, the marks of one student were altered from 86 to 75; the marks of another student were altered from 15 to 33. What is the new average?

  Ans.  Total marks = 50 × 43 = 2150

            Change I = 86 – 75 = –11

            Change II = 33 – 15 = 18

            Net Change = 18 – 11 = 7

            New total = 2150 + 7 = 2157

            Hence average = 2157/50 = 43.14

    Q8.  In a certain game, each of the 5 players received a score between 0 and 100 inclusive. If their average score was 80, what is the greatest possible number of the 5 players who could have received a score of 50?

  Ans.  Average = 80

            Now let’s go by trial and error, lets look at following table:

            Marks                                          Average

            1     2      3        4       5        

            50   50  50      50      50                     50

            50   50  50      50     100                    60

            50   50  50     100   100                    70

            50   50 100   100   100                    80

            50                                                      10

            In first scenario, all students get 50, so average is 50, in second four students get 50, even if fifth get 100, average is 60. Now similarly in the third scenario after three students getting 50 and others getting 100, still the average is 70. Only in the fourth scenario can average be 80 in case the where 2 students getting 50 and rest maximum. Maximum students who can get 50 marks are 2.

    Q9.  The average of the test scores of a class of X students is 70, and the average of the test scores of a class of Y students is 92. When the scores of both classes are combined, the average is 86. What is the value of (X/Y)?

  Ans.  Average of I class

            =>            70 = Total Marks 1/X

            Total marks I  = 70X

            Average of II class

            =>            92 = Total Marks 11/Y

            Total marks II = 92Y

            Average of I and II

            =>            86 = (Total Marks 1 + Total Marks 11)/X + Y

            From first two equations,

                             86 = (70X + 92Y)/X + Y

               86X + 86Y = 70X + 92Y

                          16X = 6Y

                             x/y = 3/18

Q10. Ram finds the average of 10 positive integers. Each integer contains two digits. By mistake he interchanges the digits of one number say YX for XY. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits X and Y?

  Ans.  Let the original number be XY = 10Y + X

                              So YX = 10X + Y

            Total number = 10, difference in average = 1.8,

            difference in number = 10 * 1.8 = 18

            Therefore 10Y + X      = 10X + Y + 18

                         9X – 9Y = 18,

                              X – Y = 2

Q11. There are 10 compartments in passenger train, which carries on an average 20 passengers per compartment. If atleast 12 passengers were sitting in each compartment and no any compartment has equal number of passengers then maximum how many passengers can be accommodated in any compartment?

Ans11.           As the average is 20 passengers, total number of passengers = 10 × 20 = 200

            Minimum number in a compartment = 12

            And since number of passengers are not same, therefore if in one compartment there are 12, in the rest it must be 13, 14, 15, 16, 17, 18, 19, 20, adding all (including 12) = 144, so the passengers in 10th compartment can be 200 – 144 = 56

Q12.   There is twice the number of two wheelers as there are three wheelers and the numbers of 4 wheelers are equal to the number of two wheelers. What is the average number of wheel per vehicle?

Ans12.           Let the number of three wheelers be X, therefore number of two wheelers will be 2X, and number of 4 wheelers be 2X, Number of wheels are then 3X, 4X and 8X, total number of wheels are 15X, and the average wheels per vehicle is 15X/5X = 3.

Q13.   Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders, the average expenses 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? (CAT 1999)

Ans13.                       Let X be the fixed cost and Y be the variable cost, then as per question:

            X + 25Y = 17500(25 × 700), X + 50Y = 30000(600 × 50), Solving X = 5000 and Y = 500. Average expenses for 100 boarders = (5000+500x100)/100 = 550

Q14.   A shipping clark has five boxes of different but unknown weights each weighing less than 100 kg. The clerk weighs the boxes and pairs. The weights obtained are 110, 112, 113, 114, 115,116, 177, 118, 120 and 121 kgs. What is the weight of the heaviest box?(CAT 2001)

Ans14.           Now the clerk weighed 10 pairs, which are made from 5 boxes (5C2). Now consider the maximum weights to be 61 kg and 60 kg to make 121 kg, which cannot be true to make 120 kg(which will need two 60 kgs). Consider 121 kg is made from 64 kg and 57 kg, it is also untrue as to make 120 you a 63 kg box, which makes the 63, 64 kg pair to be 127 kg. Now, Consider 1212 kg is made out of 62 kg and 59 kg, then you need a 58 kg for 120, 56 kg for 118, and 54 kg for 116. Now these weights make all the given 10 combinations, which concludes highest weight is 62.

Q15.   Three math’s classes: X, Y and Z take an algebra test, the average score of class X is 83, the average score of class Y is 76 and the average score of class Z is 85. The average score of class X and Y is 79 and the average score of class Y and Z is 81. What is the average score of classes X, Y and Z? (CAT 2001)

Ans15.           Let Class X has x students and Y has y students and X have z students. Therefore total score of class X = 83x, Y = 76y and Z = 85z. Average of X and Y is 70, therefore

(83x + 76y)/(x + y) = 79, solving 4x = 3y    ...(1)

            Average of Y and Z is 81

            Therefore (76y + 85z)/(y + z) = 81, solving

            4z = 5y            ...(2)

            Overall average of three classes = (83x + 76y + 85z)/(x + y + 5)

                        Using (1) and (2) in this, average is 81.5

Q16. You can collect rubies and emeralds as many as you can. Each ruby is of rupees four crore and each emerald is of Rs. 5 crore. Each ruby weighs 0.3 kg, and each emerald weighs 0.4 kgs. Your bag can carry at the most 12 kgs. What you should collect to get the maximum wealth?(CAT 1998)

Ans16.           1 kg ruby is worth 13.33 crore  4/0.3 1 kg emerald is worth 12.5 crore 5/0.4.

            The rubies need to be maximized, therefore 40 rubies (making 12 kg)

Q17.   Mayak, Mirza, Rohit and jaspal bought a motorbike for $60,000, mayak paid 1/2 of the sum of the amount paid by the other boys. Mirza paid 1/3rd of the sum of the amounts paid by the other boys and Rohit

Ans17.           Since mayak paid ½ of other three, suppose others paid X, therefore

            X + ½X = 60000, solving X = 40000

            Now Mirza paid 1/3 of other three, suppose others paid X, therefore X + 1/3X = 60000, solving X = 15000

            Similarly Rohit paid 12000, therefore Jaspal paid 60000 – (20000 + 15000 + 12000) = $13000

Q18. A piece of string is 40 cm long. It is cut into 3 pieces. The longest piece is 3 times as long as the middle sized and shortest piece is 23 cms shorter than the longest piece. Find the length of the shortest piece in cms.(CAT 2002)

Ans18.           Let the largest piece is x, then middle sized is x/3, and shortest is x – 23, therefore x + x/3 + (x – 23) = 4, solving x = 27, and the shortest piece is 4 cms

Q19.   Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight, what is the weight of the dry grapes available from 20 kgs of fresh grapes.(CAT 2001)

Ans19.           Solid part is fresh grapes is .10, therefore solid part of 20 kg = 2 kg and in dry grapes is 0.80, 0.8 x weight of dry grapes = 2

                        Weight of dry grapes = 2/0.8  = 2.5 kg

Q20. Using only 2, 5, 10, 25 and 50 paise coins, what will be the minimum number of coins required to pay exactly 78 paise, 69 paise and rupee 1.01 to three different persons? (CAT 2003)

Ans20.           Here 1.01 can be made by 50 + 25 + 10 +10 + 2 + 2 + 2, total 7 coins 69 can be made by 50 + 10 + 5 + 2 + 2, total 5 coins 78 can be made by 50 + 10 + 10 + 2 + 2 + 2 + 2, total 7 coins. Total 19 coins.

Q21. A milkman mixes 20 liters of water with 80 liters of milk after selling 1/4th of this mixture, he adds water to replenish the quantity that he has sold. What is the current proportion of water to milk?(CAT 2004).

Ans21.           Initial mixture is 80 milk + 20 water = 100 liters

            Mixture sold = 1/4^th  = 25 liters (which is 20 liter milk and 5 liter water)

            Water added = 25 liters

            Total water now is 25 + (20 – 5) = 40 liters

            Total milk is 60 liters

            Water:milk = 40:60 = 2:3