A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg

liquid 1 = a, 1l = 1kg
liquid 2 = b, 1l=0.8kg
mixture 0.48kg=0.5l, 0.96kg=1l

Ans (C)

Ans: C

1lit mixture weighs= 960
1000 800

960
160 40

160:40=4:1

so, (4/5)*100=80

liquid 1 = 1000 gm and liquid 2 = 800 gms
half will be 500 and 400 gms
allegation method
we get
----500 ==== 400
---- ---480----
-----80-------20
ratio ; 4--------1
% liquid 1 to total ; 4/5 * 100 ; 80
IMO C:

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

A. 70
B. 75
C. 80
D. 85
E. 90

It is a weighted mean question.

Better to solve it through alligation method . Very quick and easy .

480 gms of mixture amounts to 1/2 lt of mixture .
480 * 2 gms = 960 gms of mixture amounts to 1 lt of mixture .

Applying the " Cross method "

Liq 1 : Liq 2 = 4:1
so, Liq 1 = 80% by weight

A chemist mixes two liquids 1 and 2. One liter of liquid 1 weighs 1 kg and one liter of liquid 2 weighs 800 gm. If half liter of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

--> The mixture is 960 grams in total.
L2------------------Mixture-----------L1
800------------------960------------1000
*-------160-----------*------40------*

--> \(L2 : L1= \frac{40}{160} = 1 : 4\)
--> The percentage of liquid 1 in the mixture is \(\frac{4}{(1+4)}= \frac{4}{5} =0.8\) or \(80\)%

Answer (C)

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is

A. 70
B. 75
C. 80
D. 85
E. 90

The question is basically testing weighted average concept with two liquids of different concentration.
Conc. of Liquid 1 = \(\frac{1kg}{1Litre} = \frac{1000g}{1000ml}\) = 1 g/ml
Conc. of Liquid 2 = \(\frac{800g}{1Litre} = \frac{800g}{1000ml}\) = 0.8 g/ml

Final concentration(weighted average) of solution = 0.480 * 2 = 0.960 g/ml (in 1 litre of final solution)
Now let volume of liquid 1 mixed = a
volume of liquid 2 mixed = b
So,
Percentage of liquid 1 = \(\frac{a }{ a + b} * 100\) ?

As per weighted average concept = \(\frac{1*a + 0.8*b}{a + b} = 0.96\)
\(\frac{a }{ b} = 4\)

Hence \(\frac{a}{a + b}*100 = \frac{4}{5}*100\) = 80%

Answer C.

A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is


So we get two liquids with a different density (one is heavier than the other), that now we have mixed and we are asked to determine the proportion of one of the liquids on that mixture.

As we know the density of each of the liquids and the average density of the mixture, the solution comes from a simple function.

Being the liters of liquid 1 "x" and of liquid 2 "y":

1000x+800y=480 & x+y=.5

So x= .5-y then 1000(.5-y) + 800y=480; so -200y + 500=480; so y= 20/200 = 1/10

It is important to notice the units we are using, liters for the volume, grams for the weight.
So the 1/10 means that there will be 100ml of Liquid 2 in the mixture, then the 400ml left would be of Liquid 1.

Important also to remember that we are asked the proportion of Liquid 1 on the mixture, so we have that 400/500=80%.

Then C is the right answer.
The equation solution is long, so would be much faster and convenient to just plug the solutions % with the equation (double checking and being careful with the measures).


A. 70
B. 75
C. 80
D. 85
E. 90

500 ml of Liquid 1 contains 500 gms and 500 ml of Liquid 2 contains 400 gms

— Liquid 1 ——— Liquid 2 ——
— —500 ————— 400 ———
— ———— 480 ———————
— (480-400) —— (500-480)——
—— 80 —————— 20 ———

So, percentage of Liquid 1 = 80/(80+20)*100
= 80

Option C

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Half an litre weighs 480 kg means 1l will weigh \(480*2 = 960\)
now use allegation method\( liquid2:liquid 1 = \frac{1000-960}{ 960-800 }= \frac{1}{4}\)
thus percentage of liquid 1 =\( \frac{4}{5} *100 = 80\)
hence C

let x and y=volume of liquids 1 and 2 respectively in mixture
x(1/2)+y(2/5)=(x+y)(12/25)→
x/y=4/1
4/(4+1)=liquid 1 80% volume of mixture
C

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