Answer: 3 : 2

Explanation:
Quantity of Metal Gold=19, Copper = 9;
Resultant = 15
Proportion 6 : 4
Gold : Copper = 6 : 4 = 3 : 2

Answer: 30 kg and 70 kg

Explanation:
If the cost price of each kind of sugar be Rs. 100 then the selling price of one kind is Rs .107 (7% profit),Selling price of the other kind is Rs . 117 (17% profit).
Selling price of the mixture = Rs.110
SP of Each R1=117,R2=107
Mixture = 110
Proportion 3:7
Let the quantity of each in 100 kg be 3x & 7x
3x + 7x = 100=>x=10.The quantity of each in 100 kg is 30kg and 70 kg

Answer: 69 kg

Explanation:
R1=Sugar costing Rs 2.85 per kg
R2= Sugar costing Rs 6.10 per kg
RM=Average cost of mixture
We know that the mixture is being sold at a profit of 20%
ie, cost price of mixture=(100/120)*4.80=Rs 4

RM = Rs 4 = 400 paise

R1=285 paise
R2=610 paise

N1=210
N2=115

N1/N2 = 210/115 = 42/23

If quantity of sugar at 285 paise per kg is 42 kg,
Then sugar at 610 paise per kg = 23 kg

∴ if sugar at 285 paise per kg is 126 kg,
Then sugar at 610 paise per kg=69 kg

Answer: 21

Explanation:

Answer: 143 : 53

Explanation:
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

Answer: 5.25

Explanation:
Let the amount of milk be x.
36x = 28(x + 1.5)
36x = 28x + 42
8x = 42
x = 5.25 litres

Answer: 72.9%

Explanation:
We know [(a – b) / a]n = c.
Where a = 500 ml, b = 50ml, n = 3
Proportion of the milk = [(500 – 50) / 500]3
= (450/500)3
= (9/10)3
= 0.729
Percentage of milk in the vessel = 100c = 72.9%

Answer: Rs 0.25/gm

Explanation:
Let the weight of the gold be x gm
Weight of other metal = (18 - x) gm
Let the price of the other metal be Rs y / gm
7.20x + y(18 – x) = 84
7.20(18 – x) + yx =60.10
Solving these tow equations simultaneously , x =10,y = 0.25
Quality of the other metal = 18 – 10=8 gm
Price of other metal = Rs 0.25/gm

Answer: 30 liters

Explanation:
The mixture contains (10/100)70 = 7 liters of water
It contains (70 - 7) = 63 liters of wine
Let x liters of water be added
(7 + x)/(70 + x)=37/100
X = 30 liters

Answer: 16kg

Explanation:
Price of the cheaper item = 4
Price of the costlier item = 10
Mean price = 5

Quantity of Costlier/ Quantity of Cheaper = 1/5
= Quantity of costlier = Quantity of the cheaper / 5
= Quantity of costlier = 80 / 5 = 16 kg.

Answer: 40 m

Explanation:
Let length = 3x m and breadth = 2x m
Then, (3x + 5) * 2x = 2600 => 6x2 + 10x – 2600 = 0
3x2 + 5x – 1300 = 0 (3x + 65) (x – 20) = 0 , x = 20
Breadth = 2x = 40 m