A milkman has 20 liters of milk. If he mixes 5 liters of water, which is freely available, in 20 liters of pure milk. If the cost of pure milk is Rs. 18 per litre, then the profit of the milkman, when he sells all the mixture at cost price, the profit %?

Correct Answer: 50 gm

Explanation:

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%.

Correct Answer: 7:5

Explanation:

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

Correct Answer: 2:1

Correct Answer: 193 : 122

Explanation:

Given the three mixtures ratio as (1:2),(2:3),(3:4)

(1+2),(2+3),(3+4)

Total content = 3,5,7

 

Given equal quantities of the three mixtures are mixed, then LCM of 3, 5, 7 = 105

105/3 = 35 , 105/5 = 21 , 105/7 = 15

 

Now, the individual equal quantity ratios are (35x1, 35x2), (21x2, 21x3), (15x3, 15x4)

 i.e (35,70), (42,63), (45,60)

 

So overall mixture ratio of milk and water is

35+42+45 : 70+63+60

122:193

 

But in the question asked the ratio of water to milk = 193 : 122

Correct Answer: 27 %

Explanation:

Let the percentage of benzene = X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135  
X = 27

So, required percentage of benzene = 27 %

Correct Answer: 11:9

Explanation:

From the given data,

 

The part of honey in the first mixture = 1/4

 

The part of honey in the second mixture = 3/4

 

Let the part of honey in the third mixture = x

 

Then,

 

1/4             3/4

 

           x

 

(3/4)-x     x-(1/4) 

 

Given from mixtures 1 & 2 the ratio of mixture taken out is 2 : 3

 

=>  3 4 - x x - 1 4 = 2 3

 

=> Solving we get the part of honey in the third mixture as 11/20

 

=> the remaining part of the mixture is water = 9/20

 

Hence, the ratio of the mixture of honey and water in the third mixture is 11 : 9 .

Correct Answer: 2:1

Explanation:

Given rate of wheat at cheap = Rs. 2.90/kg

Rate of wheat at cost = Rs. 3.20/kg

Mixture rate = Rs. 3/kg

Ratio of mixture = 

          2.90                              3.20

                              3

(3.20 - 3 = 0.20)            (3 - 2.90 = 0.10)

 

0.20 : 0.10 = 2:1

 

Hence, wheat at Rs. 3.20/kg be mixed with wheat at Rs. 2.90/kg in the ratio of 2:1, so that the mixture be worth Rs. 3/kg.

Correct Answer: 118.08L

Explanation:

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

Correct Answer: 3:2

Explanation:

Let he mixes the oils in the ratio = x : y

Then, the cost price of the oils = 60x + 65y

Given selling price = Rs. 68.20

=> Selling price = 68.20(x+y)

Given profit = 10% = SP - CP

=> 10/100 (60x + 65y) = 68.20(x+y)-(60x + 65y)

=> 6x + 6.5y = 8.20x + 3.20y

=>2.2x = 3.3y

=> x : y = 3 : 2

Correct Answer: 36.45L

Explanation:

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