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.
Initial quantity of copper = = 40 g
And that of Bronze = 50 - 40 = 10 g
Let 'p' gm of copper is added to the mixture
=> = 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%.
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
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
Let the percentage of benzene = X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135
X = 27
So, required percentage of benzene = 27 %
Concentration of glucose are in the ratio =
Quantity of glucose taken from A = 1 liter out of 2
Quantity of glucose taken from B = 3/5 x 3 = 1.5 lit
Quantity of glucose taken from C = 0.8 lit
So, total quantity of glucose taken from A,B and C = 3.6 lit
So, quantity of alcohol = (2 + 3 + 1) - 3.6 = 2.4 lit
Ratio of glucose to alcohol = 3.6/2.4 = 3:2
As water costs free, water sold at cost price of milk gives the profit.
Required profit % = 5/20 x 100 = 5 x 5 = 25%.
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
=>
=> 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 .
The amount of petrol left after 4 operations = 81.92
Hence the amount of kerosene = 200 - 81.92 = 118.08 liters
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
General Formula:
Final or reduced concentration = initial concentration x
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,
= 36.45 L
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