Average
- Average weight of 25 students of a class is 50 kg. If the weight of the class teacher is included, the average is increased by 1 kg. The weight of the teacher is
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Average weight of 25 students = 50 kg
If the weight of the class teacher is included, the average is increased by 1 kg .
Total increase in weight = 26 × 1
∴ Weight of teacher = 50 + 26 × 1 = 76 kg
Second method to solve this question with the help of given formula :Here, N = 50, T = 45, t = 1 = 0.1 10
Weight of teacher = Average + × (n + 1)
Correct Option: A
Average weight of 25 students = 50 kg
If the weight of the class teacher is included, the average is increased by 1 kg .
Total increase in weight = 26 × 1
∴ Weight of teacher = 50 + 26 × 1 = 76 kg
Second method to solve this question with the help of given formula :Here, N = 50, T = 45, t = 1 = 0.1 10
Weight of teacher = Average + × (n + 1)
Weight of teacher = 50 + 1 (25 + 1)
Weight of teacher = 50 + 26 = 76 kg
- The batting average for 40 innings of a cricket player is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, the average of the remaining 38 innings is 48 runs. The highest score of the player is
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Let the highest score of cricketer be n runs.
∴ His lowest score = (n – 172) runs
According to the question, 38 × 48 + n + n – 172 = 40 × 50
⇒ 1824 – 172 + 2n = 2000
⇒ 1652 + 2n = 2000Correct Option: D
Let the highest score of cricketer be n runs.
∴ His lowest score = (n – 172) runs
According to the question, 38 × 48 + n + n – 172 = 40 × 50
⇒ 1824 – 172 + 2n = 2000
⇒ 1652 + 2n = 2000
⇒ 2n = 2000 – 1652 = 348
∴ n = 348 ÷ 2 = 174 runs
- The average run of a player is 32 out of 10 innings. How many runs must he make in the next innings so as to increase his average by 6 ?
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Let runs scored in the next innings = n
According to the question,
⇒ 10 × 32 + n = 11 × 38
⇒ 320 + n = 418
⇒ n = 418 – 320 = 98
Second method to solve this question with the help of given formula :
Here, T = 32, N= (10 + 1) = 11, t = 6Correct Option: D
Let runs scored in the next innings = n
According to the question,
⇒ 10 × 32 + n = 11 × 38
⇒ 320 + n = 418
⇒ n = 418 – 320 = 98
Second method to solve this question with the help of given formula :
Here, T = 32, N= (10 + 1) = 11, t = 6
Required Runs = T + Nt
Hence , Required Runs = 32 + 11 × 6 = 32 + 66 = 98
- A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was
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Let required number of wickets = n
According to question,
⇒ 12.4 × n + 26 = (n + 5) (12.4 – 0.4)
⇒ 12.4 × n + 26 = (n + 5) × 12
⇒ 12.4n + 26 = 12n + 60
⇒ 12.4n – 12n = 60 – 26
⇒ 0.4n = 34Correct Option: D
Let required number of wickets = n
According to question,
⇒ 12.4 × n + 26 = (n + 5) (12.4 – 0.4)
⇒ 12.4 × n + 26 = (n + 5) × 12
⇒ 12.4n + 26 = 12n + 60
⇒ 12.4n – 12n = 60 – 26
⇒ 0.4n = 34
⇒ n = 34 ÷ 0.4∴ n = 340 = 85 4
- A cricket player after playing 10 tests scored 100 runs in the 11th test. As a result, the average of his runs is increased by 5. The present average of runs is
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Here , A cricket player after playing 10 tests scored in the 11th match = 100 runs
If the average in 10 tests be n , then
According to question ,
⇒ n × 10 + 100 = (n + 5) × 11
⇒ 11n – 10n = 100 – 55Correct Option: C
Here , A cricket player after playing 10 tests scored in the 11th match = 100 runs
If the average in 10 tests be n , then
According to question ,
⇒ n × 10 + 100 = (n + 5) × 11
⇒ 11n – 10n = 100 – 55
⇒ n = 45
∴ Required average = 50
