For these type of problems,
Quantity of Diesel remained =
Here p = 12 , q = 120
=> = 120 x 0.9 x 0.9 x 0.9 = 87.48 lit.
As given equal amounts of alloys are melted, let it be 1 kg.
Required ratio of gold and silver =
Hence, ratio of gold and silver in the resulting alloy = 105/103.
Let M litres milk be added
=>
=> 60 + M = 120
=> M = 60 lit.
He has gain = 15 - 12 = 3,
Gain% = (3/12) x 100 = (100/4) = 25.
He has 25% gain.
Given, Manideep purchases 30kg of barley at the rate of 11.50/kg nad 20kg at the rate of 14.25/kg.
Total cost of the mixture of barley = (30 x 11.50) + (20 x 14.25)
=> Total cost of the mixture = Rs. 630
Total kgs of the mixture = 30 + 20 = 50kg
Cost of mixture/kg = 630/50 = 12.6/kg
To make 30% of profit
=> Selling price for manideep = 12.6 + 30% x 12.6
=> Selling price for manideep = 12.6 + 3.78 = 16.38/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 =
% of milk in first bottle = 64%
% of milk in second bottle = 100 - 26 = 74%
Now, ATQ
64% 74%
68%
6 4
Hence, by using allegation method,
Required ratio = 3 : 2
The amount of petrol left after 4 operations
=
=
=
= 81.92 litres
Hence the amount of kerosene = 200 - 81.92 = 118. 08 litres
Here total parts of milk and water in the solution is 6+2 = 8 parts.
1part = 640/8 = 80
old mixture contains 6parts of milk and 2 parts of water(6:2).
To get new mixture containing half milk and half water, add 4parts of water to the old mixture then 6:(2+4) to make the ratio same.
i.e, 4 x 80 = 320 ml.
Number of liters of water in 150 liters of the mixture = 20% of 150 = 20/100 x 150 = 30 liters.
P liters of water added to the mixture to make water 25% of the new mixture.
Total amount of water becomes (30 + P) and total volume of mixture is (150 + P).
(30 + P) = 25/100 x (150 + P)
120 + 4P = 150 + P => P = 10 liters.
Copyright ©CuriousTab. All rights reserved.