Average ( SET 4 : Q16-Q20 )
Solution: 2
Let the sum of the nos. in the first set be N, and that of the second set be M.
⇒ N/x = y, and M/y = x
⇒ N = xy, and M = xy.
⇒ M + N = 2xy
Since there are x + y numbers, we have
Average = (M+N)/(x+y) = 2xy/(x+y)Solution: 3
Average marks secured by 55 boys = 67
Total marks secured by 55 boys = 67 × 55 = 3685
An item 85 is misread as 30
⇒ New average = (3685 - 30 + 85)/55 = 3740/55 = 68
∴ The Correct mean of marks = 68
Smart trick
New average = Old average + (Error/Number of boys) = 67 + (85 - 30)/55 = 67 + (55/55) = 67 + 1 = 68
Solution: 4
Let average score of first 11 innings be x runs.
Then, total runs made by the batsman in 11 innings = 11x
According to the question, after 12th inning, average score = x + 1
⇒ 79 + 11x = 12x + 12
⇒ x = 79 – 12 = 67
Thus, average score after 12th inning = 67 + 1 = 68 runs.Solution: 3
Let the second number be x
∵ first number is twice the second, the first number = 2x
Also, since first number is half of the third, third number = 4x
Now, the average of the 3 numbers is 63.
We know that, average = Sum of all quantities/Number of quantities
∴ (2x + x +4x)/3 = 63
⇒ 7x =189
⇒ x = 27
Thus, the numbers are 54, 27 and 108
The difference of first and third number = 108 – 54 = 54Solution: 4
Let the old average be x runs
We know that,
Average = Sum of observations/No. of observations
∴ Sum of runs scored in 13 innings = Average × No. of innings
⇒ Sum of runs = x × 13 = 13x
Runs scored in the 14th inning = 150
New average = x + 3
∴ Sum of runs = (x + 3) × 14 = (14x + 42)
Now, 13x + 150 = 14x + 42
∴ x = 150 – 42 = 108
⇒ New average = x + 3 = 108 + 3 = 111 runs
No comments:
Post a Comment