Directions: Refer the following data to answer the questions given below.
Let x be the quantity of alcohol from 1st vessel, hence (1 – x) quantity is mixed from 2nd vessel to get the resulting mixture. Also let the quantity of each vessel be 10 litres.
Vessel 1 Vessel 2 Vessel 3
Alcohal 8 5 7
Water 2 5 3
8x + 5(1 – x) = 7 -> 8x + 5 – 5x = 7 -> 3x = 2 -> x = 2/3
2x + 5(1 – x) = 3 -> 2x + 5 – 5x = 3 -> 3x = 2 -> x = 2/3
1 – x = 1/3
Therefore the ratio = 2/3 : 1/3 = 2:1
Directions: Refer the following data to answer the questions given below.
Solution :
As the mixture is in the ratio 5:1, the quantity of alcohol is 5/6 and that of water is 1/6.
As the seller earns a profit of Rs.25, the cost of alcohol is Rs.75 for 5/6th quantity of alcohol.
Therefore the cost of pure alcohol = 75/(5/6) = 75*6/5 = Rs.90
Directions: Refer the following data to answer the questions given below.
Solution :As vessel A contains 150 litres of mixture of milk and water in the ratio 3:1, the quantity of milk is 3*150/4 litres, and that of water is 150/4 litres.
Similarly, as vessel B contains 200 litres of mixture of milk and water in the ratio 5:2, the quantity of milk is 5*200/7, and that of water is 2*200/7 litres.
After mixing the total quantity is 350 litres.
The quantity of milk is {(3*150/4) + (5*200/7)} = 7150/28
The quantity of water is {(150/4) + (2*200/7)} = 2650/28
The ratio of the resulting mixture = 7150/2650 = 143 : 53
Directions: Refer the following data to answer the questions given below.
Solution :
Let the cost of 1 kg rice be Re.1. ...(A)
Good quality in 1 kg mixture in 1st bag = 1 - 40% = 1-(2/5) = 3/5 kg ...(1)
Cost price of 1 kg mixture in 1st bag = (3/5) x 1 = Re.3/5 (Based on our above equation and assumption A)
Good quality in 1 kg mixture in 2nd bag = 1 - 60% = 1-(3/5) = 2/5 kg ...(2)
Cost price of 1 kg mixture in 2nd bag = (2/5) x 1 = Re.2/5 (Based on our above equation and assumption A)
It is given that the final mixture has worst quality to good quality of rice in ration 3:4.
Good quality in 1 kg final mixture = 4/(3+4) x 1 = 4/7kg ...(3)
Then its mean price = Re.4/7 (Based on our above equation and assumption A)
From 1,2 and 3 we get m = 4/7, d = 3/5 and c = 2/5. (Refer formula given in introduction before question 1.)
d - m = (3/5)-(4/7) = 1/35
m - c = (4/7)-(2/5) = 6/35
Applying formula, cheaper quantity:dearer quantity = d-m / m-c = 1/35 : 6/35 = 1:6
Final mixture should contain 63Kg in total as per question.
so, quantity of mixture taken from 2nd bag = 1/(1+6) x 63 = 9 kg
and the quantity of mixture taken from 1st bag = 6/(1+6) x 63 = 54 kg.
hence the answer is 54kg,9kg
Directions: Refer the following data to answer the questions given below.
Solution:
S.P of 1 litre of mixture = Rs.10.80 and Gain 20%.
Therefore, C.P of 1 litre of mixture = Rs.100/120 x 10.80 = Rs.9 ...(1)
C.P of 1 litre milk of 1st kind = d = Rs.12
C.p of 1 litre milk of 2nd kind = c = Rs.8
Mean Price = m = Rs.9 (from equation 1)
Then d - m = 3 , m - c = 1
Applying above values to our allegations formula (refer introduction before question)
Ratio of quantities of 1st and 2nd kind = 1:3 .
Let X litres of milk of 1st be mixed with 60litres of 2nd kind.
Then, 1 : 3 = X : 60
1/3 = X/60
X = 60/3 = 20.