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How many liters of a 40% iodine solution need to be mixed with 35 lite

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How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 02:31
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 29 Mar 2017, 06:36
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Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


One approach is to sketch the solutions with their components separated.

Let's start with the 35 liters of 20% iodine solution:
Image
20% of 35 liters = 7 liters, so the original solution contains 7 liters of iodine.


To this solution, we'll add x liters of the solution that's 40% iodine.
Image
40% of x liters = 0.4x liters. So, we're adding 0.4x liters of iodine.


When we combine the two solutions, it's easy to see how the resulting mixture looks, since we need only add the volumes of iodine in each solution.
Image


As you can see, our resulting mixture has a TOTAL volume of 35 + x liters.
We can also see that the resulting mixture contains 7 + 0.4x liters of iodine.
Since the resulting mixture must by 35% (aka 35/100) iodine, we can write the following equation:
(7 + 0.4x)/(35 + x liters) = 35/100
First simplify 35/100 to get: (7 + 0.4x)/(35 + x liters) = 7/20
Cross multiply to get: 20(7 + 0.4x) = 7(35 + x liters)
Expand: 140 + 8x = 245 + 7x
Solve: x = 105

Answer: D

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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 02:51
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Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


These Qs are best done with weighted Average method....

Two solutions - x of 40% and 35 of 20% results in a solution of 35% solution..

SO ratio = \(\frac{QTY- of- 40}{Qty- of- 20%} =\frac{35-20}{40-35} = \frac{15}{5}.....\)
so \(\frac{x}{35} = \frac{15}{5} = 3..................x = 35*3 = 105\)

D
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 02:44
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Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


Solution 1:
Assume the iodine solution to be mixed = x lts.
Iodine = 0.4x lts, Water = 0.6x lts.

Solution 2: 35 liters of a 20% iodine solution
Iodine = 7 lts, Water = 28 lts.

Total iodine = 0.4x + 7
Total water = 0.6x + 28

The resultant is a 35% idoine solution.
Hence (0.4x + 7) / (x + 35) = 35/100
40x + 700 = 35x + 1225
5x = 525
x = 105 lts

Correct Option: D
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 03:33
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 09:32
1
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


60% water solution mixed with 80% water solution to get 65% water solution.

Solution will be mixed in the ratio::-

x/35= 80-65/65-60
x/35= 15/5
x= 3*35= 105

D is the answer
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 09 Jun 2016, 09:44
D it will be

weighted avg concept :-



20 --------------------35-------------------40

qyantity of 20%/quantity of 40% = 40- 35/35-20

35/x = 5/15
x = 105
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 10 Jun 2016, 06:07
Divyadisha wrote:
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


60% water solution mixed with 80% water solution to get 65% water solution.

Solution will be mixed in the ratio::-

x/35= 80-65/65-60
x/35= 15/5
x= 3*35= 105

D is the answer



Hi, Divyadisha ,, may i knw how did u get that formula
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 10 Jun 2016, 06:08
chetan2u wrote:
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


These Qs are best done with weighted Average method....

Two solutions - x of 40% and 35 of 20% results in a solution of 35% solution..

SO ratio = \(\frac{QTY- of- 40}{Qty- of- 20%} =\frac{35-20}{40-35} = \frac{15}{5}.....\)
so \(\frac{x}{35} = \frac{15}{5} = 3..................x = 35*3 = 105\)

D


hi chetan2u can u plz explain tat formula,
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 10 Jun 2016, 09:34
If x is the number of liter of the 40% iodine solution, we have :
0.4 *x + 0.2 * 35 = 0.35 ( 35 + x)
0.4 *x + 0.2 * 35 = 0.35 * 35 + 0.35 * x
0.05 * x = 35 (0.35 - 0.2)
0.05 * x = 35 * 0.15
x = 35 * 3 = 105
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 10 Jun 2016, 20:17
let x=liters of 40% iodine solution needed
.4x+(.2)(35)=.35(x+35)
x=105 liters
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 11 Jun 2016, 05:35
Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


Attachment:
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Capture.PNG [ 4.45 KiB | Viewed 4906 times ]


35% ----> 7 + 0.4x
100%---->\(\frac{(7 + 0.4x)}{35} *100\) = \(\frac{(7 + 0.4x)}{7}*20\)


Or, \(\frac{(140 + 8x)}{7}\) = 35 + x

Or, 140 + 8x = 245 + 7x

Or, x = 245 - 140

Or, x = 105

Hence answer will be (D) 105
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 29 Mar 2017, 06:49
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Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


Another approach is to use Weighted Averages:

Weighted average of groups combined = (group A proportion)(group A average) + (group B proportion)(group B average) + (group C proportion)(group C average) + ...

In this case, we have:
35% iodine = (proportion of 20% solution)(20% iodine ) + (proportion of 40% solution)(40% iodine)

To determine the proportions, let's let x = the volume of 40% solution needed
We already know we have 35 liters of 20% solution
This means the TOTAL volume = 35 + x
So, the proportion of 20% solution 35/(35 + x)
And the proportion of 40% solution x/(35 + x)

Plug these values into formula to get: 35 = [35/(35 + x)](20) + [x/(35 + x)](40)
Multiply both sides by (35 + x) to get: 35(35 + x) = (35)(20) + (x)(40)
Expand: 1225 + 35x = 700 + 40x
Subtract 700 from both sides: 525 + 35x = 40x
Subtract 35x from both sides: 525 = 5x
Solve: x = 105

Answer: D

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How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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New post 15 Sep 2018, 05:30
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Bunuel wrote:
How many liters of a 40% iodine solution need to be mixed with 35 liters of a 20% iodine solution to create a 35% iodine solution?

A. 35
B. 49
C. 100
D. 105
E. 140


Dear Moderator,
Please change tag from " work/rate" to "mixture Problems ". Thank you.
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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite  [#permalink]

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Re: How many liters of a 40% iodine solution need to be mixed with 35 lite   [#permalink] 15 Sep 2018, 05:31
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