problem for word problem experts

Question

                                      comparison

 

A and B contain milk and water in the ratio 7:3 and 3:2             Two vessels contain mixture of milk and water in
respectively find the ratio in which the quantity be taken           ratio 5:2 and 3:1 respectively the ratio of milk and
from the vessels so that the ratio of milk and water in water      in new solution. If two mixtures are mixed
new mixture is 2:1 equal amount

   m=milk and w=water                                                             let there is x milk and water
(7/10)m + (3/5)m milk : (3/10)w + (2/5)w water                          5x/7+3x/1=41/28

(7/10)m + (6/10)m    2                                                                  2x/7+x/4=15/28
-----------------       = ---
(3/10)w  + (4/10)w    1                                                                  milk 41 and water 15

13m/10 / 7w/10=2/1

13m/7w=2/1

 

 

it goes wrong 

i did what second problem did i did gather milk with milk and water with water so what is my mistake

 

 

• My doubt is in these two example we add milk to milk and water to water. is this rule that we only can add same quantity if it is then why first problem get wrong
• Why we cant add one vessel milk and water to other vessel milk and water

 

 

Answer 1

Here is one way to think about the second problem:

 

Let X be the quantity of mixture in the 5:2 ratio, and Y be the quantity of the second mixture in 3:1 ratio...thus we add

 

X(5/7Milk + 2/7Water) + Y(3/4Milk + 1/4Water) together.

 

The second problem says we have two vessels (assuming they are the same size, or X=Y),

 

we can combine the Milk and Water quantities to see that:

 

Milk = (5/7+3/4)X = 41/28

Water = (2/7+1/4)X = 15/28

 

So the ratios of milk to water in the new combined mixture is (41/28):(15/28) or they are in a 41:15 ratio.

 

 

 

Using this setup for the first problem will help us to understand what is going on!

 

Let X be the quantity of the 7:3 mixture (A), and Y be the quantity of the 3:2 mixture (B)

 

Thus our new mixture has:

 

X(7/10Milk+3/10Water)+Y(3/5Milk+2/5Water) 

 

grouping Milk and Water together we get:

 Milk = 7x/10+3y/5

Water = 3x/10+2y/5

 

And this must be in a 2:1 ratio, so 

Milk/Water = 2, or Milk=2*Water...

 

Thus 7x/10+3y/5 = 2(3x/10+2y/5)...distributing and grouping x's and y's yields:

 

x/10=y/5, or x=2y...

 

So we should add twice as much of the mixture A as we do for the mixture B.

 

Let's check how this would work in the context of the first problem:

 

Suppose we add: 2(7/10Milk + 3/10Water) +(3/5Milk+2/5Water) together...we get:

 

20/10Milk+10/10Water...this the Milk:Water ratio of the new mixture is 2:1, as desired!

 

 

 

In mixture problems it is assumed that the liquids combine through stirring, thus labeling quantities of each mixture with variables (like I did with x and y) allows us to combine quantities as necessary and enforce ratios as indicated in the problem.

 

Hope this solution method helps you to organize the information and setup systems of equations that can be solved!