Average
- The mean of 20 items is 47. Later it is found that the item 62 is wrongly written as 26. Find the correct mean.
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Later it is found that the item 62 is wrongly written as 26 ,
∴ Difference = 62 – 26 = 36∴ Required average = 47 + 36 20
Required average = 47 + 1.8 = 48.8
Second method to solve this question with the help of given formula :
Here, n = 20 , m = 47 , a = 62 , b = 26Correct Average = m + (a - b) n
Correct Option: A
Later it is found that the item 62 is wrongly written as 26 ,
∴ Difference = 62 – 26 = 36∴ Required average = 47 + 36 20
Required average = 47 + 1.8 = 48.8
Second method to solve this question with the help of given formula :
Here, n = 20 , m = 47 , a = 62 , b = 26Correct Average = m + (a - b) n Correct Average = 47 + (62 - 26) 20 Correct Average = 47 + 36 20
Correct Average = 47 + 1.8 = 48.8
- A tabulator while calculating the average marks of 100 students of an examination, by mistake enters 68, instead of 86 and obtained the average as 58; the actual average marks of those students is
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On the basis of given details in question , we have
Difference = 86 – 68 = 18∴ Actual average = 58 + 18 100
Actual average = 58.18
Second method to solve this question with the help of given formula :
Here, n = 100 , m = 58 , a = 86 , b = 68Correct Average = m + (a - b) n
Correct Option: A
On the basis of given details in question , we have
Difference = 86 – 68 = 18∴ Actual average = 58 + 18 100
Actual average = 58.18
Second method to solve this question with the help of given formula :
Here, n = 100 , m = 58 , a = 86 , b = 68Correct Average = m + (a - b) n Correct Average = 58 + (86 - 68) 100
Correct Average = 58 + 18 100
Correct Average = 58 + 0.18 = 58.18
- Mean of 10 numbers is 30. Later on it was observed that numbers 15, 23 are wrongly taken as 51, 32. The correct mean is
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Later on it was observed that numbers 15, 23 are wrongly taken as 51, 32 .
∴ Difference = 15 + 23 – 51 – 32 = –45∴ Correct average = 30 - 45 = 25.5 10
Second method to solve this question with the help of given formula :
Here, n = 10, m = 30 , a = 15, b = 23 , p = 51 , q = 32Correct Average = m + (a + b - p - q) n Correct Average = 30 + (15 + 23 - 51 - 32) 10
Correct Option: A
Later on it was observed that numbers 15, 23 are wrongly taken as 51, 32 .
∴ Difference = 15 + 23 – 51 – 32 = –45∴ Correct average = 30 - 45 = 25.5 10
Second method to solve this question with the help of given formula :
Here, n = 10, m = 30 , a = 15, b = 23 , p = 51 , q = 32Correct Average = m + (a + b - p - q) n Correct Average = 30 + (15 + 23 - 51 - 32) 10 Correct Average = 30 + 
38 - 83 
10 Correct Average = 30 - 45 = 25.5 10
Correct Average = 30 – 4.5 = 25.5
- The mean of 50 numbers is 30. Later it was discovered that two entries were wrongly entered as 82 and 13 instead of 28 and 31. Find the correct mean.
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Required average = 30 + (28 + 31 - 82 - 13) 50 Required average = 30 + - 36 50
Required average = 30 – 0.72 = 29.28
Second method to solve this question with the help of given formula :
Here, n = 50, m = 30 , a = 28, b = 31 , p = 82, q = 13Correct Average = m + (a + b - p - q) n
Correct Option: C
Required average = 30 + (28 + 31 - 82 - 13) 50 Required average = 30 + - 36 50
Required average = 30 – 0.72 = 29.28
Second method to solve this question with the help of given formula :
Here, n = 50, m = 30 , a = 28, b = 31 , p = 82, q = 13Correct Average = m + (a + b - p - q) n Correct Average = 30 + (28 + 31 - 82 - 13) 50 Correct Average = 30 + - 59 - 95 50 Correct Average = 30 - 36 50
Correct Average = 30 – 0.72 = 29.28
- Of the three numbers, the first number is twice of the second and the second is thrice of the third number. If the average of these 3 numbers is 20, then the sum of the largest and smallest numbers is
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Let the third number be p.
∴ Second number = 3p
and first number = 6p
∴ 6p + 3p + p = 3 × 20
⇒ 10p = 60
⇒ p = 6
∴ Required sum = 6p + p = 7p = 7 × 6 = 42
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 3, X = 20Largest Number = 3ab X 1 + b + ab Largest Number = 3 × 2 × 3 × 20 1 + 3 + 2 × 3
Correct Option: B
Let the third number be p.
∴ Second number = 3p
and first number = 6p
∴ 6p + 3p + p = 3 × 20
⇒ 10p = 60
⇒ p = 6
∴ Required sum = 6p + p = 7p = 7 × 6 = 42
Second method to solve this question with the help of given formulas :
Here, a = 2, b = 3, X = 20Largest Number = 3ab X 1 + b + ab Largest Number = 3 × 2 × 3 × 20 1 + 3 + 2 × 3 Largest Number = 18 × 20 = 36 10 Smallest Number = 3 X 1 + b + ab Smallest Number = 3 × 20 1 + 3 + 2 × 3 Smallest Number = 3 × 20 = 6 10
Sum = 36 + 6 = 42
