Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.
Average speed = (2xy) /(x + y) km/hr
= (2 * 50 * 30) / (50 + 30) km/hr.
37.5 km/hr.
Let the initial number of persons be x. Then,
16x + 20 * 15 = 15.5 (x + 20) <=> 0.5x = 10 <=> x = 20.
Let the avg age of 7 boys be 'p' years
Let the age of the girl be 'q' years
From given data,
The age of 7 boys = 7p years
Now the new average = (p + 1) when 22 yrs is replaced by q
Now the age of all 7 will become = 7(p + 1) yrs
Hence, 7p - 22 + q = 7(p + 1) yrs
7p - 22 + q = 7p + 7
q = 22 + 7 = 29
Therefore, the age of girl = q = 29 years.
Let the total distance to be covered is 84 kms.
Time taken to cover the distance without stoppage = 84/42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 84/28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.
Given,
mean of 100 observations is 40
=> Total of 100 observations = 40 x 100 = 4000
84 is misread as 48
=> Difference = 84 - 48 = 36
=> Now, new total of 100 observations = 4000 + 36 = 4036
Correct Mean = 4036/100 = 40.36
Let the speeds of the car, train and bus be 5x, 9x and 4x km/hr respectively.
Average speed = 5x + 9x + 4x/3 = 18x /3 = 6x km/hr.
Also, 6x = 72 => x = 12 km/hr
Therefore, the average speed of the car and train together is =
= 7x = 7 x 12 = 84 km/hr.
Let the three numbers be x, y, z.
From the gien data,
x = 2y ....(1)
x = z/2 => z = 2(2y) = 4y .....(From 1) ...........(2)
Given average of three numbers = 56
Then,
Now,
x = 2y => x = 2 x 24 = 48
z = 4y = 4 x 24 = 96
Now, the highest number is z = 96 & smallest number is y = 24
Hence, required sum of highest number and smallest number
= z + y
= 96 + 24
= 120.
Manager's monthly salary Rs. (1600 * 21 - 1500 * 20) = Rs. 3600.
Let avg ages of 5 members at present is G
And age of new member is m and an older is n
Required = (5G + m - n)/5 = G - 3
m - n = 15 years
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