Let us assume P liters of Diesel added to the mixture so that Diesel will be 25% in the new mixture.
According to Question,
Quantity of Diesel in 150 liters of the mixture = 20% of 150 = 150 x 20/100 = 30 liters.
After adding P liters of Diesel, The total quantity of Diesel becomes (30 + P) and total volume of mixture will be (150 + P).
Again According to Question,
After adding the P liters of Diesel in the mixture, the Diesel quantity becomes 25% of the new mixture.
Quantity of Diesel in new mixture = 25% of Total mixture.
? (30 + P) = (150 + P) x 25 %
? 30 + P = (150 + P) x 25/100
? 30 + P = (150 + P) x 1/4
? 120 + 4P = 150 + P
? 4P - P = 150 - 120
? 3P = 30
? P = 10 liters.
First of all we write the fraction of milk present in three mixtures. In A : 7 /12
In B : 17 /24
In combination of A and B : 5 /8
We now apply allegation rule on these fractions from figure.
So, Ratio of A: B = 2 : 1
2nd Method
Let us assume P mixture taken from first vessel and Q mixture taken from second vessel to form a new mixture.
Part of Milk in P mixture from first vessel = 7P/12
Part of Milk in Q mixture from Second vessel = 17Q/24
Part of Water in P mixture from first vessel = 5P/12
Part of Milk in Q mixture from Second vessel = 7Q/24
According to question,
After mixing the P and Q, we will get mixture.
Milk in New Mixture / Water in New Mixture = 5/3
{(7P/12) + (17Q/24 ) } /{ (5P/12) + (7Q/24) } = 5/3
{(14P + 17Q)/24 } /{ (10P + (7Q)/24 } = 5/3
{(14P + 17Q) } /{ (10P + (7Q)} = 5/3
(14P + 17Q) x 3 = 5 x (10P + (7Q)
42P + 51Q = 50P + 35Q
51Q - 35Q = 50P - 42P
8P = 16Q
P = 2Q
P/Q = 2
P : Q = 2 : 1
Average speed = (100 / 10) = 10 km/hr.
Ratio of time taken at 7 km/hr to 12 km/hr = 2 : 3
Time taken at 7 km/hr = 2 (2 + 3) × 10 = 4 hrs.
Distance covered at 7 km/hr = 7 × 4 = 28 km.
Distance covered at 12 km/hr = 100 ? 28 = 72 km.
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