Replacement
Content
QA – Replacement
ADDITION OR REMOVAL OF ITEMS AND CHANGE IN AVERAGE
Since, Average =
So, the change in number of items will change the average. The number of items may change if:
1) When an item is added to the existing group of items
Say the average of N items = A
Now, ‘n’ New items are added and the average increases or decreases by x, then
Average of New items added = A + ( 1+ )x
When only one item is added, i.e. n = 1, then Value of the New item added = A + ( 1+ N)x
2) When items are removed from the existing group
Say the average of N items = A
Now, ‘n’ items are removed and the average increases or decreases by x, then
Average of items removed = A + ( 1- )x
When only one item is removed, i.e. n = 1, then Value of the item removed = A + ( 1- N)x
REPLACEMENT OF SOME OF THE ITEMS
Sometimes, when a number of items of a group are removed and these are replaced with
equal number of different items, then the average of the group changes, increases or
decreases) by x.
Let there are N items in the group, then
Sum of New items added – Sum of removed items = Nx
-)ve, when average decreases
+)ve, when average increases
Examples
1) By scoring a 100 in his 20th innings, Tendulkar increased his average by 2. What was
his average for the first 19 innings?
a) 120
b) 138
c) 140
d) None of the above
Ans:
If he would have scored 100 runs, his average ould have remained 100. However his
average increases by 2. Therefore in a way in his 20th innings he has scored 2 extra runs
fro all his 20 innings. So he ahs scored 2 x20=40 extra runs. So runs cored in 20 th innings=
100+40=140
2) A batsman scored 100 runs in his 20th innings .After this innings his new average became
62 (considering that he is out in every match).Then what is his old average?
(a) 58
(b) 60
(c) 54
(d) 55
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QA – Replacement
Ans:Let his average for the first 19 innings be x. Hence total runs scored by him in the first 19
innings will be 19x. Since he scored a 100 in his 20th innings, total runs scored by him in 20
innings will be (19x + 100). Hence the average of 20 innings will be (19x + 100) / 20, and this
is equal to (x + 2). If we solve this equation, we will get x = 60.
3) The average score of a batsman after 25 innings is 42 runs per innings. If after the
26th innings, his average runs increased by 2 runs, then what is his score in the 26th
inning?
(a) 92
(b) 93
(c) 94
(d) 96
Ans:Runs in 26th inning = Total runs after 26th innings – Total runs after 25th innings
= 26 44 – 25 42 = 94
Alternatively, this question can be done by the above given central value meaning of
average. Since the average increases by 2 runs per innings, we can assume that 2 runs have
been added to his score in each of the first 25 innings. Now, the total runs added in these
innings have been contributed by the runs scored in the 26th inning, which must be equal to
25 2 = 50 runs.
And after contributing 50 runs, his score in the 26th inning is 48 runs.
Hence, runs scored in the 26th inning = new average + old innings change in average
= 44 + 25 2 = 94.
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ABS Classes
2
ADDITION OR REMOVAL OF ITEMS AND CHANGE IN AVERAGE
Since, Average =
So, the change in number of items will change the average. The number of items may change if:
1) When an item is added to the existing group of items
Say the average of N items = A
Now, ‘n’ New items are added and the average increases or decreases by x, then
Average of New items added = A + ( 1+ )x
When only one item is added, i.e. n = 1, then Value of the New item added = A + ( 1+ N)x
2) When items are removed from the existing group
Say the average of N items = A
Now, ‘n’ items are removed and the average increases or decreases by x, then
Average of items removed = A + ( 1- )x
When only one item is removed, i.e. n = 1, then Value of the item removed = A + ( 1- N)x
REPLACEMENT OF SOME OF THE ITEMS
Sometimes, when a number of items of a group are removed and these are replaced with
equal number of different items, then the average of the group changes, increases or
decreases) by x.
Let there are N items in the group, then
Sum of New items added – Sum of removed items = Nx
-)ve, when average decreases
+)ve, when average increases
Examples
1) By scoring a 100 in his 20th innings, Tendulkar increased his average by 2. What was
his average for the first 19 innings?
a) 120
b) 138
c) 140
d) None of the above
Ans:
If he would have scored 100 runs, his average ould have remained 100. However his
average increases by 2. Therefore in a way in his 20th innings he has scored 2 extra runs
fro all his 20 innings. So he ahs scored 2 x20=40 extra runs. So runs cored in 20 th innings=
100+40=140
2) A batsman scored 100 runs in his 20th innings .After this innings his new average became
62 (considering that he is out in every match).Then what is his old average?
(a) 58
(b) 60
(c) 54
(d) 55
Proprietary and Confidential
ABS Classes
1
QA – Replacement
Ans:Let his average for the first 19 innings be x. Hence total runs scored by him in the first 19
innings will be 19x. Since he scored a 100 in his 20th innings, total runs scored by him in 20
innings will be (19x + 100). Hence the average of 20 innings will be (19x + 100) / 20, and this
is equal to (x + 2). If we solve this equation, we will get x = 60.
3) The average score of a batsman after 25 innings is 42 runs per innings. If after the
26th innings, his average runs increased by 2 runs, then what is his score in the 26th
inning?
(a) 92
(b) 93
(c) 94
(d) 96
Ans:Runs in 26th inning = Total runs after 26th innings – Total runs after 25th innings
= 26 44 – 25 42 = 94
Alternatively, this question can be done by the above given central value meaning of
average. Since the average increases by 2 runs per innings, we can assume that 2 runs have
been added to his score in each of the first 25 innings. Now, the total runs added in these
innings have been contributed by the runs scored in the 26th inning, which must be equal to
25 2 = 50 runs.
And after contributing 50 runs, his score in the 26th inning is 48 runs.
Hence, runs scored in the 26th inning = new average + old innings change in average
= 44 + 25 2 = 94.
Proprietary and Confidential
ABS Classes
2
