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Bunuel
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from given info
2 * (3x+5y) and 3*(4x+7y)
we get
milk = 18 and water = 31
ratio ; 18:31



Bunuel
Two solutions having milk and water in the ratios 3 : 5 and 4 : 7 are mixed in the ratio 2 : 3. What is the ratio of milk and water in the resultant mixture?

A. 80 : 139
B. 81 : 139
C. 82 : 139
D. 83 : 139
E. 84 : 139
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Answer is choice B.
the amount in 1 lit for sol1 : 3/8 and 5/8
the amount in 1 lit sol2 : 4/11 and 7/11

so the final solution will have ratio : 6/8 + 12/11 : 10/8 + 21/11
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Hello lacktutor,
I tried same way to solve this problem but my answer is not matching with the that is given.so can you please help me?
Q: two alloys A and B are composed of two basic elements. The ratios of the compositions of the two basic elements in the two alloys are 5:3 and 1:2. A new alloy X is formed by mixing the two alloys A and B in the ratio 4:3. What is the ratio the composition of the two basic elements in alloy X? (Answer; 1:1) lacktutor
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first, [m:w] = [ 3 : 5 ] ; sum=8
second, [m:w]=[ 4 : 7 ] ; sum =11
mixture [m:w]=[ 2 : 3 ]

lcm of first and second is (8,11)= 88
Now,
back to first ,
[m:w]= [3 : 5]*[88/8]
=[3 : 5]*11
=[33:55]

back to second,
[m:w]=[4:7]*[88/11]
= [32:56]

now, multiplying first and second corresponding to mixture,we get,

[m:w]=[33:55]*2
=[66:110]

& [m:w]=[32:56]*3
=[96:168]

Now ,combine this two milk and water, we get,
[M:W]=[66+96:110+168]
=[162:278]
=[81:139]

so, Ans:B.

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We can use the weighted Average formula A1*w1+ A2*w2/ w1+w2
We will use only milk quantity
In first solution milk quantity is 3/8 and in second solution milk quantity is 4/11 and the weights as 2 and 3
Use the above formula and substitute the values

3/8*2 + 4/11* 3 / 2+3
=88/220 which is milk to total

As the question requires milk : water we get 81:(220-81)
=81:139

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from solution 1 and 2 new amount of milk taken
3/8 (2/5) + 4/11((3/5) = new milk

new milk = 81/220

from solution 1 and 2 new amount of water taken

5/8(2/5) + 7/11 (3/5) = new water

new water= 139/220

New ratio = 81/139
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Bunuel
Two solutions having milk and water in the ratios 3 : 5 and 4 : 7 are mixed in the ratio 2 : 3. What is the ratio of milk and water in the resultant mixture?

A. 80 : 139
B. 81 : 139
C. 82 : 139
D. 83 : 139
E. 84 : 139


I dont know why but my brain literally hurts while trying to solve Mixture problems...i am totally hopeless at this stage
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thakurrani81
Bunuel
Two solutions having milk and water in the ratios 3 : 5 and 4 : 7 are mixed in the ratio 2 : 3. What is the ratio of milk and water in the resultant mixture?

A. 80 : 139
B. 81 : 139
C. 82 : 139
D. 83 : 139
E. 84 : 139


I dont know why but my brain literally hurts while trying to solve Mixture problems...i am totally hopeless at this stage

Hope the links below help:

18. Mixture Problems



For more check Ultimate GMAT Quantitative Megathread

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Use alligation ,
let the amount of milk in the final soultion be = x

(3/8 - x ) / 3 = (x - 4/11)/2

33 - 88x = 132x - 48
220x = 81
x= 81/220

Total amount = 220
MIlk = 81
water = 220 - 81 =139

Refer to the alliagation diagram attached­
Attachments

alligation_29th.png
alligation_29th.png [ 11.25 KiB | Viewed 2730 times ]

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sayan640
Use alligation ,
let the amount of milk in the final soultion be = x

(3/8 - x ) / 3 = (x - 4/11)/2

33 - 88x = 132x - 48
220x = 81
x= 81/220

Total amount = 220
MIlk = 81
water = 220 - 81 =139

Refer to the alliagation diagram attached­
­Since we are considering milk, should;t we take 2:5 as ratio??
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