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AVERAGES

PROF: RAVINDRA P. MANGRULKARAPTITUDE TRAINERPIBM PUNE

QUANTITATIVE APTITUDE

AVERAGES - INTRODUCTIONAN AVERAGE, OR MORE ACCURATELY AN ARITHMETIC MEAN IS IN SIMPLE TERMS, THE SUM OF N DIFFERENT DATA DIVIDED BY N.

FOR EXAMPLE : IF A BATSMAN SCORES 35, 45 AND 37 RUNS IN 1ST, 2ND, AND 3RD INNINGS RESPECTIVELY, THEN HIS AVERAGE RUNS IN 3 INNINGS IS EQUAL TO (35 + 45 + 37) / 3 = 39 RUNS.

FORMULAAVERAGE = TOTAL OF DATA / NUMBER OF DATA

TOTAL = AVERAGE * NUMBER OF DATA

WORKED OUT PROBLEMSPROBLEM 1 : THE AVERAGE OF A BATSMAN AFTER 25 INNINGS WAS 56 RUNS PER INNING. IF AFTER THE 26TH INNING HIS AVERAGE INCREASED BY 2 RUNS, THEN WHAT WAS HIS SCORE IN THE 26TH INNING?

SOLUTION : RUNS IN 26TH INNING = RUNS TOTAL AFTER 26 INNINGS RUNS TOTAL AFTER 25 INNINGS. = 26 * 58 25 * 56 TRY TO UNDERSTAND FOLLOWING METHOD FOR CALCULATION = 26*(56+2) 25*56------------------------(HOW??!) = 26*2 + (26*56-25*56)------------------(HOW??!) = 52 + 56----------------------------------------(HOW??!) = 108 .

SHORT CUT METHODAVERAGE IN FIRST 25 INNINGS565656--56 -25 TIMES (LAST)

AVERAGE AFTER 26 INNINGS58 (56 +2)58 (56 +2)58 (56 +2)--58 (56 +2) -25 TIMES58 (56 +2) -26 TIMES(LAST)

RUNS IN 26 TH INNING = DIFFERENCE IN ABOVE 2 COLUMNS IT WILL BE EQUAL TO 25 * 2 + 58 = 108

SHORTCUT FORMULA IS = NEW NUMBER= OLD NUMBER * DIFFERENCE + NEW AVERAGE

DIFFERENCE5

WORKED OUT PROBLEMSPROBLEM 2 : 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 A CLASSTEACHER?

USING SHORTCUT FORMULA NEW NUMBER = OLD NUMBER * DIFFERENCE + NEW AVERAGEANSWER : = 30 * 0.5 + 14 = 29

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DETAIL SOLUTION :

AVERAGE INCLUDING A TEACHER (31 NUMBERS)14 (13.5 + 0.5)14 (13.5 + 0.5)14 (13.5 + 0.5)--14 (13.5 + 0.5) 30 TIMES14 (13.5 + 0.5) 31 TIMES(LAST)

AVERAGE EXCLUDING A TEACHER (30 NUMBERS)13.513.513.5--13.5 30 TIMES (LAST)

TEACHERS AGE WILL BE EQUAL TO DIFFERENCE BETWEEN 2 COLUMNS = 14 + 30*0.5 = 14 +15 = 29

PROBLEM 3 : THE AVERAGE WEIGHT OF 4 MEN IS INCREASED BY 3 KG WHEN 1 OF THEM WHO WEIGHS 120 KG IS REPLACED BY ANOTHER MAN WHAT IS THE WEIGHT OF A NEW MAN?

SOLUTION :

USING SHORTCUT FORMULA WEIGHT OF NEW PERSON = NUMBER OF PERSONS * DIFFERENCE IN AVERAGE + WEIGHT OF REMOVED PERSON

= 4 * 3+ 120 = 132

PROBLEM 4:

THE AVERAGE OF MARKS OBTAINED BY 120 CANDIDATES IN A CERTAIN EXAMINATION IS 35. IF THE AVERAGE MARKS OF PASSED CANDIDATES IS 39 AND THAT OF THE FAILED CANDIDATES IS 15. WHAT IS THE NUMBER OF CANDIDATES WHO PASSED THE EXAMINATION?SOLUTION : LET THE NO. OF PASSED CANDIDATES BE X.THEN TOTAL MARKS = 120 * 35 = 39X + (120 X)15 4200 = 39X + 1800 15X 4200 1800 = 24X 24X = 2400 X = 100

SOLUTION9

SHORT CUT FORMULA

NUMBER OF PASSED CANDIDATES = TOTAL CANDIDATES*(TOTAL AVERAGE FAILED AVERAGE) PASSED AVERAGE FAILED AVERAGE

NUMBER OF FAILED CANDIDATES = TOTAL CANDIDATES*(PASSED AVERAGE TOTAL AVERAGE) PASSED AVERAGE FAILED AVERAGE

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USING SHORTCUT FORMULA

NUMBER OF PASSED CANDIDATES = TOTAL CANDIDATES*(TOTAL AVERAGE FAILED AVERAGE) PASSED AVERAGE FAILED AVERAGE

= {120 ( 35 15 )} (39 15) = (120 * 20) 24 = 100

2. NUMBER OF FAILED CANDIDATES = TOTAL CANDIDATES*(PASSED AVERAGE TOTAL AVERAGE) PASSED AVERAGE FAILED AVERAGE

= {120 ( 39 35 )} (39 15) = (120 * 4) 24 = 20

SHORT CUT TO FIND AVERAGE OF SERIES

PROBLEM 5:FIND THE AVERAGE OF FIRST 43 MULTIPLES OF 18?

SOLUTION: = (18*1+18*2+18*3+18*418*43) / 45= 18*( 1+ 2 +3 +4 +.+43) / 43= 18 * AVERAGE OF 1 43 NUMBERS.

43 CAN BE DIVIDED IN TO 2 GROUPS OF 21 NUMBERS WITH A NUMBER IN THE MIDDLE. (1-21) 22 (23-43) AS 22 IS IN THE MIDDLE . SO IT IS THE AVERAGE OF THE NUMBERS FROM 1 TO 43.

ANSWER WILL BE 18 * 22 = 396.

PROBLEM12

SHORT CUT TO FIND AVERAGE OF SERIES

PROBLEM 6 :FIND THE AVERAGE OF NUMBERS FROM 21 TO 60?

SOLUTION: THERE ARE 40 NUMBERS IN THE SERIES. NOW 40 IS THE EVEN NUMBER.

40 CAN BE DIVIDED IN TO 2 GROUPS OF 20S WITH A NUMBER IN THE MIDDLE. 21 - 60 CAN BE DIVIDED IN TO 2 GROUPS OF 20S. ( 21 40 ) ( 41 60 ) .SO THE AVERAGE WILL BE AVERAGE OF MIDDLE 2 NUMBERS.MIDDLE 2 NUMBERS ARE 40 AND 41.HENCE AVERAGE = (40 + 41) / 2 = 81 / 2 = 40.5

SHORT CUT TO FIND AVERAGE OF SERIES PROBLEM 7 :FIND THE AVERAGE OF NUMBERS FROM 71 TO 97?

SOLUTION: THERE ARE 27 NUMBERS IN THE SERIES. NOW 27 IS THE ODD NUMBER.

71 - 97 CAN BE DIVIDED IN TO 2 GROUPS OF 13 NUMBERS WITH A NUMBER IN THE MIDDLE.

27 NUMBERS (71 97) CAN BE DIVIDED IN TO 2 GROUPS OF 13S WITH A NUMBER IN THE MIDDLE. ( 71 83 ) 84 ( 85 97 ) .

AS 84 IS IN THE MIDDLE . SO IT IS THE AVERAGE OF THE NUMBERS FROM 71 TO 97.ANSWER WILL BE = 84.

PROBLEM 8 :

THE AVERAGE OF 11 RESULTS IS 50. IF THE AVERAGE OF FIRST 6 RESULTS IS 49 AND THAT OF THE LAST 6 IS 52. FIND 6TH RESULT.

SOLUTION:

THE TOTAL OF 11 RESULTS IS 11 * 50 = 550. (6TH RESULT IS CONSIDERED ONCE.)

THE TOTAL OF FIRST 6 RESULTS IS 6 * 49 = 294. (INCLUDES 6TH RESULT)THE TOTAL OF LAST 6 RESULTS IS 6 * 52 = 312. (INCLUDES 6TH RESULT)

THE TOTAL OF ABOVE 12 RESULTS IS 294 + 312 = 606.6TH RESULT IS COMMON TO BOTH SO IT IS CONSIDERED TWICE.

SO, 6TH RESULT = 12 REULTS 11 RESULTS = 606 550 = 56

AVERAGE RELATED TO SPEEDIF A PERSON TRAVELS A DISTANCE AT A SPEED OF X KM / HR AND THE SAME DISTSNCE AT A SPEED OF Y KM / HR THEN AVERAGE SPEED IS 2XY / (X + Y).

IF A PERSON TRAVELS 3 EQUAL DISTANCES AT A SPEED OF X KM / HR, Y KM / HR AND Z KM / HR THEN AVERAGE SPEED IS 3XYZ / (XY + YZ+XZ).

IF A16

PROBLEM 9:

A TRAIN TRAVELS FROM A TO B AT A RATE OF 20 KM/HR AND FROM B TO A AT THE RATE OF 30 KM/HR. WHAT IS THE AVERAGE RATE FOR WHOLE JOURNEY?

SOLUTION: USING FORMULA : AVERAGE SPEED = 2XY / (X + Y).

= 2*20*30 / (20 + 30) = 1200 / 50 = 24 KM/HR

PROBLEM17

PROBLEM 10:

A PERSON TRAVELS 3 EQUAL PARTS OF DISTANCE WITH SPEEDS OF 40, 30 AND 15 KM/HR RESPECTIVELY. FIND THE AVERAGE SPEED DURING JOURNEY?

SOLUTION: USING FORMULA : AVERAGE SPEED = 3XYZ / (XY + YZ+XZ).

= 3*40*30*15 / (40*30 + 30*15 + 15*40) = 120 * 450 / 2250 = 24 KM/HR

PROBLEM 11:

WITH AN AVERAGE SPEED OF 40 KM/HR, A TRAIN REACHES ITS DESTINATION IN TIME. IF IT GOES WITH AN AVERAGE SPEED OF 35 KM/HR, IT IS LATE BY 15 MINUTES. THE LENGTH OF THE WHOLE JOURNEY IS?

SOLUTION: SPEED = DISTANCE / TIMELET THE DISTANCE BE X.TIME T1 WITH AVERAGE SPEED OF 40 KM/HR = X / 40 HR.TIME T2 WITH AVERAGE SPEED OF 35 KM/HR = X / 35 HR.DIFFERENCE IN TIME T1 AND T2 GIVEN IS 15 MINUTE = 15/60 HR = HR. SO (X / 35 ) (X / 40) = (8X-7X) / 280 = X / 280 = THE LENGTH OF THE WHOLE JOURNEY X = 280 / 4 = 70

PROBLEM 12:

1/3RD OF A WHOLE JOURNEY IS COVERED AT THE RATE OF 25 KM/HR. 1/4TH AT THE RATE OF 30 KM/HR AND THE REST AT 50 KM/HR. FIND THE AVERAGE SPEED FOR THE WHOLE JOURNEY?

SOLUTION: LET THE DISTANCE BE X.TIME T1 FOR 1/3RD JOURNEY (X/3) WITH AVERAGE SPEED OF 25 KM/HR = X / (3*25) HR.TIME T2 FOR 1/4TH JOURNEY (X/4) WITH AVERAGE SPEED OF 30 KM/HR = X / (4*30) HR.TIME T3 FOR REST JOURNEY (5X/12) WITH AVERAGE SPEED OF 50 KM/HR = 5X / (12*50) HR.TOTAL TIME = T1 + T2 + T3 = X / (3*25) + X / (4*30) + 5X / (12*50) = (X / 75) + (X / 120) + (X / 120) = (8 X + 5 X + 5 X) / 600 = 18 X / 600 = 3 X / 100.

AVERAGE SPEED = DISTANCE / TIME = X / (3 X / 100) = 100 X / 3 X = 100 / 3 = 33.33 KM/HR

THANK YOU

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