How much pure alcohol should be added to 400ml of a 15% solu

Question

How much pure alcohol should be added to 400ml of a 15% solution to make the strength of solution 32%?

A. 100 ml
B. 60 ml
C. 120 ml
D. 130 ml
E. 150 ml

jeckll · Accepted Answer

Here's how I did it:

(400)(.15) = 60  mL of Alcohol

.32 = \frac{60+x}{400+x}

(.32)(400+x)=60+x

128+.32x=60+x

68=.68x

x=100

kp1811 · Answer

400ml has 15% alcohol i.e 60ml . This means we have 340ml of water or the liquid with which alcohol is mixed 
Let V be the volume of alcohol added to make the strength 32%
then V/340+V = 32/100. solving this we get V as 160ml

so we need to add 160 -60 = 100ml of alcohol

A-100ml

raghavs · Answer

we can also go by answer choices Tke 100 ml for eg 400 (old)+100 (new concentr)ml 500*32/100 = 160 ml (60ml is de old concentration +100 ml (newly added) Let me know if im wrong :roll:

Cheers, Raghav

kp1811 · Answer

yes we can go by answer choices as well....but in GMAT we will see questions having values either in ascending or descending order unlike the options in the question here. Plugging in/POE is always suggested to be done taking the middle option first and then proceed. 
I feel taking answer choices will take more time to solve this question.

xcusemeplz2009 · Answer

let xml is added 
15% of 400=60 ml
after addition
(60+x)/(400+x)=32%
solving x=100 ml

generis · Answer

Let x = the amount of alcohol to be added
x = 100% alcohol = 1.00 (decimal percent)

Weighted average, slight variation. 
For each part, A and B, to be mixed:

1) multiply its alcohol concentration (weight) by its volume

2) sum the separate results, which =

3) (Desired alcohol concentration of resultant solution) * (Volume of resultant solution, usually A + B)

.15(400) + 1(x) = .32(400 + x)

60 + x = 128 + .32x

.68x = 68

x =   \frac{68}{.68}= 100

Answer A

Henry S. Hudson Jr. · Answer

Method 1:
Equation in terms of alcohol. 
400(.15)+x(1.0)=(400+x)(.32)

60+x=128+.32x
.68x=68
x=100

Method 2: Equation in terms of water

.85(400)+x(0)=(.68)(400+x)

solving for x will give you the same answer.

Method 3- alligation

15____________32_____________100

68 17

4 1

the ratio of the 15 percent solution to pure alcohol = 4:1
Since there are 400 ml of the 15% solution, the multiplier is 100. 100(1)=100

Henry S. Hudson Jr. · Answer

Method 1:

Equation in terms of alcohol.

400(.15)+x(1.0)=(400+x)(.32)

60+x=128+.32x

.68x=68

x=100

Method 2: Equation in terms of water

.85(400)+x(0)=(.68)(400+x)

solving for x will give you the same answer.

Method 3- alligation

15____________32_____________100

68__________________________________ 17

4 _________________________________ 1

the ratio of the 15 percent solution to pure alcohol = 4:1
Since there are 400 ml of the 15% solution, the multiplier is 100. 100(1)=100

Method 4-Weighted average (in terms of alcohol)

/ =32/100

Method 5-Weighted average (in terms of water)

/ =68/100

gracie · Answer

let a=alcohol to be added
.15*400+a=.32(400+a)
a=100 ml
A

NinetyFour · Answer

I felt that using allegation was the easiest way to solve this question.

We need the solutions to be mixed in a 4:1 ratio, thus if we have 400ml of the first mixture, we will need  100ml  of pure alcohol to make a mixture with ratio 4:1.

bumpbot · Answer

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How much pure alcohol should be added to 400ml of a 15% solu

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How much pure alcohol should be added to 400ml of a 15% solu

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