2. AVERAGES - INTRODUCTION
AN AVERAGE, OR MORE ACCURATELYAN
ARITHMETIC MEAN IS IN SIMPLE TERMS, THE SUM
OF N DIFFERENT DATADIVIDED BYN.
FOR EXAMPLE : IF A BATSMAN SCORES 35, 45AND
37 RUNS IN 1ST, 2ND, AND 3RD INNINGS
RESPECTIVELY, THEN HIS AVERAGE RUNS IN 3
INNINGS IS EQUALTO
(35 + 45 + 37) / 3 = 39 RUNS.
4. WORKED OUT PROBLEMS
PROBLEM 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 TOTALAFTER26
INNINGS – RUNS TOTALAFTER 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 .
5. SHORT CUT METHOD
AVERAGE IN FIRST 25 INNINGS AVERAGE AFTER 26 INNINGS
56 58 (56 +2)
56 58 (56 +2)
56 58 (56 +2)
- -
- -
56 -25 TIMES (LAST) 58 (56 +2) -25 TIMES
58 (56 +2) -26 TIMES(LAST)
RUNS IN 26 TH INNING = DIFFERENCE IN ABOVE 2 COLUMNS
IT WILL BE EQUALTO
25 * 2 + 58 = 108
SHORTCUT FORMULA IS = NEW NUMBER
= OLD NUMBER * DIFFERENCE + NEWAVERAGE
6. WORKED OUT PROBLEMS
PROBLEM 2 :
THE AVERAGE AGE OF A CLASS OF 30 STUDENTS ANDA
TEACHER REDUCES BY 0.5 YEARS IF WE EXCLUDE THE
TEACHER. IF THE INITIALAVERAGE IS 14 YEARS, FIND
THE AGE OF ACLASSTEACHER?
USING SHORTCUT FORMULA
NEW NUMBER
= OLD NUMBER * DIFFERENCE + NEWAVERAGE
ANSWER :
= 30 * 0.5 + 14 = 29
7. 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 TIMES
•14 (13.5 + 0.5)– 31TIMES(LAST)
AVERAGE EXCLUDINGA
TEACHER (30 NUMBERS)
•13.5
•13.5
•13.5
•-
•-
•13.5 – 30 TIMES (LAST)
•TEACHER’S AGE WILL BE EQUALTO DIFFERENCE
BETWEEN 2 COLUMNS = 14 + 30*0.5
= 14 +15 = 29
8. PROBLEM 3 :
THE AVERAGE WEIGHT OF 4 MEN IS INCREASED BY 3 KG WHEN1
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 NEWPERSON
= NUMBER OF PERSONS * DIFFERENCE IN AVERAGE +WEIGHT
OF REMOVED PERSON
= 4 * 3+ 120 = 132
9. 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
10. SHORT CUT FORMULA
NUMBER OF PASSED CANDIDATES =
TOTALCANDIDATES*(TOTALAVERAGE – FAILEDAVERAGE)
÷
PASSED AVERAGE – FAILEDAVERAGE
NUMBER OF FAILED CANDIDATES =
TOTALCANDIDATES*(PASSEDAVERAGE – TOTALAVERAGE)
÷
PASSED AVERAGE – FAILEDAVERAGE
11. USING SHORTCUT FORMULA
1. NUMBER OF PASSED CANDIDATES =
TOTALCANDIDATES*(TOTALAVERAGE – FAILEDAVERAGE)
÷
PASSED AVERAGE – FAILEDAVERAGE
= {120 ( 35 – 15 )} ÷ (39 – 15)
= (120 * 20) ÷ 24
= 100
2. NUMBER OF FAILED CANDIDATES =
TOTALCANDIDATES*(PASSEDAVERAGE – TOTALAVERAGE)
÷
PASSED AVERAGE – FAILEDAVERAGE
= {120 ( 39 – 35 )} ÷ (39 – 15)
= (120 * 4) ÷ 24
= 20
12. 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*4……………18*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 WITHA
NUMBER IN THE MIDDLE.
(1-21) 22 (23-43)
AS 22 IS IN THE MIDDLE . SO IT IS THE AVERAGE OFTHE
NUMBERS FROM 1 TO 43.
ANSWER WILL BE 18 * 22 = 396.
13. 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 ANUMBER 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 2NUMBERS.
MIDDLE 2 NUMBERS ARE 40 AND 41.
HENCEAVERAGE = (40 + 41) / 2
= 81 / 2
= 40.5
14. 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 WITHA
NUMBER IN THE MIDDLE.
27 NUMBERS (71 – 97) CAN BE DIVIDED IN TO 2 GROUPS OF 13S WITHA
NUMBER IN THE MIDDLE.
( 71 – 83 ) 84 ( 85 – 97 ) .
AS 84 IS IN THE MIDDLE . SO IT IS THE AVERAGE OF THENUMBERS
FROM 71 TO 97.
ANSWER WILL BE = 84.
15. PROBLEM 8 :
THE AVERAGE OF 11 RESULTS IS 50. IF THE AVERAGE OF FIRST6
RESULTS IS 49 AND THAT OF THE LAST 6 IS 52. FIND 6TH RESULT.
SOLUTION:
THE TOTALOF 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 TOTALOFABOVE 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
16. AVERAGE RELATED TO SPEED
1. IFAPERSON TRAVELS ADISTANCEATASPEED
OF X KM / HR AND THE SAME DISTSNCE ATA
SPEED OF Y KM / HR THEN AVERAGE SPEED IS
2XY / (X + Y).
2. 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).
17. PROBLEM 9:
A TRAIN TRAVELS FROM A TO B ATA RATE OF 20
KM/HR AND FROM B TO AATTHE 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
18. 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
19. PROBLEM 11:
WITH AN AVERAGE SPEED OF 40 KM/HR, A TRAIN REACHES ITS
DESTINATION IN TIME. IF IT GOES WITH AN AVERAGE SPEEDOF 35
KM/HR, IT IS LATE BY 15 MINUTES. THE LENGTH OF THE WHOLE
JOURNEY IS?
SOLUTION:
SPEED = DISTANCE / TIME
LET 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 = ¼ HR.
SO (X / 35 ) – (X / 40) = ¼
(8X-7X) / 280 = ¼
X / 280 = ¼
THE LENGTH OF THE WHOLE JOURNEY X = 280 / 4 =70
20. PROBLEM 12:
1/3RD OF A WHOLE JOURNEY IS COVERED ATTHE RATE OF 25 KM/HR. 1/4TH ATTHE
RATE OF 30 KM/HR AND THE REST AT50 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 50KM/HR
= 5X / (12*50) HR.
TOTALTIME = T1 + T2 + T3 = X / (3*25)
= (X / 75)
+ X / (4*30) +
+ (X / 120) +
5X / (12*50)
(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
