Let the percentage of benzene =X
(30 - X)/(X- 25) = 6/4 = 3/2
=> 5X = 135 or
X = 27 so,
required percentage of benzene = 27 %
%age of liquid Base in the original mixture =1/5 x 100 = 20% In the final mixture % of the liquid Base = 3/5 x 100 = 60% Now using the rule of allegation Hence reduced quantity of the first mixture and the quantity of mixture B which is to be added are the same. Total mixture = 10 + 10 = 20 liters and quantity of liquid A = 20/5 x 4 = 16 lts
ln a mixture of 1,000 ml sodium carbonate : benzene solution = 3 : 1.
Hence, sodium carbonate: 750 ml benzene solution 250 ml
A 250 ml of 3 : 2 solution contains 150 ml sodium carbonate solution and l00ml benzene solution.
Total sodium carbonate solution = 900 ml, total benzene solution = 350 ml
After using 250 ml to make strong organic solution , sodium carbonate solution used
= 250/1250 x 900 = 180 ml
pure sodium carbonate solution left = 900 -180 =720ml
If X be the required number, then
(45 x X)1/2 = 3 x (5 x 22)1/2
45X = 9 x 110
X = 22
The first alloy is prepared by mixing metals in proportions 7 : 2,
Then the weight of the first alloy will be = 7x + 2x
Assume ’x’ = 1Kg
The weight of the first alloy will be 9kg
The second alloy is prepared by mixing metals in proportions 7 : 11,
The weight of the second alloy will be 18kg
If equal quantities of the two alloys are melted to form a third alloy,
Then 18Kg of first alloy should be used, then its ratio will become 14 : 4,
Now, the proportion of argentum and brass in third alloy = 7 + 14 : 11 + 4 = 21 : 15 = 7 : 5
Let X = K/(Y2 - 1).
Now 24 = K/(100 - 1)
Or K = 24 x 99.
X = K/(Y2 – 1)
When Y = 5, X = 24 × 99/(25 - 1 ) = 99
Let total number of employees be X.
Then, 8000 x X = 7 x 12000 + (X - 7) x 6000
X = 26.
Thus, the total number of employees in the organization is 26.
Let the total distance to be covered is 48 kms.
Time taken to cover the distance without stoppage = 48/42 hrs = 2 hrs
Time taken to cover the distance with stoppage = 48/28 = 3 hrs.
Thus, he takes 60 minutes to cover the same distance with stoppage.
Therefore, in 1 hour he stops for 20 minutes.
Total marks of 10 papers = 80 x 10 = 800
Total marks of 8 papers = 81 x 8 = 648
Total marks of two papers = (800 - 648) = 152
If highest total is 92, then the lowest total is
(152 - 92) = 60.
Let the number of students in classes X, Y and Z be A, B and C respectively.
Then, total score of X= 83A, total score of Y = 76B, total score of Z = 85C.
Also given that,
(83A + 76B) / (A + B) = 79
=>4A = 3B.