Average Questions

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 third subject marks be 'x'
=> Second subject marks = 2x
=> Third subject marks = 4x
Given avg = 224
x + 2x + 4x = 224 x 3
=> 7x = 224 x 3
=> x = 96
Hence, Second subject marks = 2x = 2 x 96 = 192.

Let the ratio of initial quantity of oils be 'x' => 4x, 5x & 8x.
Let k be the quantity of third variety of oil in the final mixture.
Let the ratio of initial quantity of oils be 'y'
From given details,
4x + 5 = 5y ..... (1)
5x + 10 = 7y .....(2)
8x + k = 9y ......(3)

By solving (1) & (2), we get
x = 5 & y = 5

From (3) => k = 5

Therefore, quantity of third variety of oil was 9y = 9(5) = 45kg.

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,

$$\begin{gathered} x + y + z3 = 56 \\ 2y + y + 4y3 = 56 (From 1 & 2) \\ 7y = 56 x 3 \\ y = 24 \end{gathered}$$

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.

Total money decided to contribute = 750 x 12 = 9000

Let 'b' boys dropped

The rest paid 150/- more

=> (12 - b) x 900 = 9000

=> b = 2

Hence, the number of boys who dropped out is 2.

Average speed =  km/hr

                      =   km/hr

     D _  C  _ A          --------(1)

     D > B > C           --------(2)

from (1) and (2)

     D > B > C > A   ---------(3)

Again      E _ B _ A

But B > A, from (3)

So          E > D > B > C > A      [ Since B is the average of E and A so it is eqidistant from both E and A]