If ?4096 =64 ,then the value of ?40.96 + ?0.4096 + ?0.004096 + ?0.00004096 is?

The amount of petrol left after 4 operations =  200 × 1 - 40 200 4 =81.92 

Hence the amount of kerosene = 200 - 81.92 = 118.08 liters

Let the initial amount of honey in the jar was K, then 

  512 = K 1 - 1 5 4     ? 20 % = 20 100 = 1 5

        or      

512 = K 4 5 4

Therefore, K = 1250

Hence initially the honey in the jar= 1.25 kg

Rate of rice of quantity 280 kg = Rs. 15.60/kg

Rate of rice of quantity 120 kg = Rs. 14.40/kg

He want to earn a profit of Rs. 10.45/kg

Rate of Mix to sell to get profit of 10.45 =  

$$\begin{gathered} 280 x 15 . 60 + 120 x 14 . 40 280 + 120 + 10 . 45 \\ 4368 + 1728 400 + 10 . 45 \\ = > 15 . 24 + 10 . 45 \\ = 25 . 69 \end{gathered}$$

L . C . M of 2, 4, 6, 8, 10 and 12 is 120
So the bells will toll together after 120 second i,e 2 minutes. In 30 min. they will together in (30/2) + 1 times, i.e 16 times.

Milk in 1-litre mixture of A = 4/7 litre. 

Milk in 1-litre mixture of B = 2/5 litre. 

Milk in 1-litre mixture of C = 1/2 litre.

By rule of alligation we have required ratio X:Y   

   X                  :                 Y  

  4/7                                2/5  

           \                      /

              (Mean ratio)
                   (1/2) 

           /                      \

 (1/2 ? 2/5)     :       (4/7 ? 1/2)  

    1/10                      1/1 4 

So Required ratio = X : Y = 1/10 : 1/14 = 7:5

Customer ratio of Milk and Water is given by

Milk          ::        Water

  6.4                         0

          8 1 + 3 8 = 64 11

  64 11                       64 10 - 64 11

=> Milk : Water = 110 : 11 = 10 : 1

Therefore, the proportionate of Water to Milk for Customer is 1 : 10

Initial quantity of copper = 80 100 x 50 = 40 g 

And that of Bronze = 50 - 40 = 10 g

Let 'p' gm of copper is added to the mixture

=> 50 + p x 90 100 = 40 + p

=> 45 + 0.9p = 40 + p

=> p = 50 g

Hence, 50 gms of copper is added to the mixture, so that the copper is increased to 90%.

General Formula:

Final or reduced concentration = initial concentration x  1 - a m o u n t b e i n g r e p l a c e d i n e a c h o p e r a t i o n t o t a l a m o u n t n

where n is the number of times the same operation is being repeated. The "amount being replaced" could be pure or mixture as per the case. similarly ,"total amount" could also be either pure or mixture. Here amount being replaced denotes the quantity which is to be withdrawn in each time.

Therefore,     50 × 1 - 5 50 3

= 36.45 L

W i n e ( l e f t ) W a t e r ( a d d e d ) = 343 169

It means    W i n e ( l e f t ) W i n e ( i n i t i a l a m o u n t ) = 343 512   ? 343 + 169 = 512

Thus ,       

343 x = 512 x 1 - 15 K 3

  ? 343 512 = 7 8 3 = 1 - 15 k 3

=> K = 120

Thus the initial amount of wine was 120 liters.

Let M litres milk be added

=> 60 + M 40 = 3 1

=> 60 + M = 120

=> M = 60 lit.