Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
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Mixtures and Alligation Questions

Multiple choice chemistry separation of substances classification of mixtures mixtures: examples and properties types of solutions

$3$ litre of mixture of propane $(C {3}H _{8})$ $ butane $(C{4}H_{10})$ on complete combustion gives $10$ litre $CO_{2}$. Find the composition of mixture.

  1. $C _{3}H _{8}2L$ and $C _{4}H _{10}\ 1L$
  2. $C _{3}H _{8}3L$ and $C _{4}H _{10}\ 0L$
  3. $C _{3}H _{8}\ 1.5L$ and $C _{4}H _{10}\ 1.5L$
  4. $C _{3}H _{8}\ 0L$ and $C _{4}H _{10}\ 3L$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let x be the volume of propane and y be the volume of butane. We have x + y = 3. The combustion reactions are C3H8 + 5O2 -> 3CO2 + 4H2O and C4H10 + 6.5O2 -> 4CO2 + 5H2O. Thus, 3x + 4y = 10. Solving these equations gives x = 2 and y = 1.

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

A pot contains $81$ litres of pure milk. $\displaystyle \frac{1}{3} $ of the milk is replaced by the same amount of water. Again $\dfrac{1}{3} $ of the mixture is replaced by that amount of water. The ratio of milk and water in the new mixture is:

  1. $1:2$
  2. $1:1$
  3. $2:1$
  4. $4:5$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Initially, Milk = 81 litres and waterr = 0 litre Afte 1st operation,
Milk = $ \displaystyle[81 - \frac{1}{3} \times 81] litres = (81 - 27) litres = 54$ litres
Water =  $\displaystyle [ 0+ \frac{1}{3} \times 54] litres $ = $(54-18)$ litres $= 36$ litres
Water= $\displaystyle [27 - \frac{1}{3} \times 27] litres + [\frac{1}{3} \times 54 + \frac{1}{3} \times 27 ] litres$ = $(27-9)$ litres + $(18+9)$ litres $= 45 $litres 
$\therefore $ Required ratio of milk and water in the new mixture $= 36: 45 = 4:5 $

Multiple choice maths fundamental concept of ratio and proportion division problem dividing a quantity in a given ratio problems on ratios

In a mixture $60$ litres, the ratio of milk and water $2:1$. If this ratio is to be $1:2$, then the quantity of water to be further added is

  1. $30$
  2. $40$
  3. $50$
  4. $60$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Quantity of milk $=60\times \dfrac{2}{3}=40$ litres
Quantity of water $=60-40=20$ litres
As per question we need to add water to get quantity $2:1$
$\Longrightarrow \dfrac{40}{20+x}=\dfrac{1}{2}$
$\Rightarrow\;20+x=80$
$\Rightarrow\;x=60$ litres

Multiple choice zoology economic biology dairy farm management cattle farming production of food from animals

How many litres of water should be added to $45$ litres of $90\%$ concentrated milk to reduce the concentration to $75\%$?

  1. $8$ litres
  2. $9$ litres
  3. $10$ litres
  4. $12$ litres
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total milk by volume=90% of 45=40.5 litre of milk.

In the diluted case, let the total volume be 'x' litre. So, 75% of x=40.5 l.
So, x=54 litre, thus volume of added water = (54-45)l= 9 litre.

So the correct option is '9 litres'.

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

Brine has a density of 1.2 g/cc. 40 cc of it is mixed with 30 cc of water. The density of the resulting solution will be

  1. $2.11$ g/cc
  2. $1.11$ g/cc
  3. $12.2$ g/cc
  4. $20.4$ g/cc
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Density of Brine$=\rho _{b}=1.2g/cc$

Volume of Brine$=v _{b}=40cc$
Mass of Brine$=m _{b}=\rho _{b}\times v _{b}=1.2\times 40=48g$
Density of Water$=\rho _{w}=1g/cc$
Volume of Water$=v _{w}=30cc$

Mass of Water$=m _{w}=\rho _{w}\times v _{w}=1\times 30=30g$
Density of mixture$=\dfrac{\text{Mass of mixture}}{\text{ Volume of mixture}}=\dfrac{m _{b}+m _{w}}{v _{b}+v _{w}}=\dfrac{48+30}{40+30}=\dfrac{78}{70}=1.11g/cc$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

When two liquid of same volume but different densities $\rho _{1}$ and $\rho _{2}$ are mixed together, then the density of the mixture is

  1. $\dfrac {p _{1}+p _{2}}{2}$
  2. $p _{1}+p _{2}$
  3. $\dfrac {2p _{1}p _{2}}{p _{1}+p _{2}}$
  4. $2p _{1}+2p _{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$Density \,\,of \,\,mixture = \dfrac{Total\,\, mass }{ Total \,\,volume}$

Let $1^{st}$ liquid have mass $M _1$, density $p _1$ and Volume $V$

and
$2^{st}$ liquid have mass $M _2$, density $p _2$ and Volume $V$
So,Density of mixture  $= \dfrac{M _1 +M _2}{2V}$$=$$\dfrac {p _1V+p _2V}{2V}$

                                 $\rho _m=\dfrac{p _1+p _2}{2}$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

The densities of three liquids are D, 2D and 3D. What will be the density of the resulting mixture if equal volumes of the three liquids are mixed? 

  1. 6D

  2. 1.4D

  3. 2D

  4. 3D
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

let $V$ be the volume of each liquid then the total volume of the mixture becomes $3V$. 

$\rho = \dfrac{total\ mass}{total\ volume}$

Therefore, the mass of the liquids can be written as:
$m _1=D\times V=DV$
$m _1=2D\times V=2DV$
$m _1=3D\times V=3DV$

the total mass of the liquids is
$M= DV+2DV+3DV=6DV$

Therefore, the density of the mixture is:
$\rho=\dfrac{6DV}{3V}$

$\rho = 2D$

Multiple choice chemistry substances in the surroundings - their states and properties measurement of density properties of substances fundamental and derived units

$60  cc$ of a liquid of relative density $1.4$ are mixed with $40  cc$ of another liquid of relative density $0.8$. The density of the mixture is

  1. $1.16 {g}/{cc}$
  2. $2.26 {g}/{cc}$
  3. $11.6 {g}/{cc}$
  4. $116 {g}/{cc}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
given : - ${ S } _{ 1 }=1.4\quad \quad \quad \quad { V } _{ 1 }=60cc$
               ${ S } _{ 2 }=0.89/cc\quad \quad { V } _{ 2 }=40cc$


The density of the mixture is given by:
$S _{mixture }= \dfrac { { S } _{ 1 }{ V } _{ 1 }+{ S } _{ 2 }{ V } _{ 2 } }{ { V } _{ 1 }+{ V } _{ 2 } } $

$S _{mixture} = \dfrac { 1.4\times 60+0.8\times 40 }{ 100 } $

$S _{mixture} = 1.16 g/cc$

Multiple choice maths calculations and mental strategies 1 equations from statements forming equations from statements writing mathematical statements

The ratio of the volumes of water and glycerine in 240cc of mixture is 1:3 .The quantity of water (in cc) that should be added to the mixture so the volumes of water and glycerine 2:3 is: 

  1. $55$
  2. $60$
  3. $62.5$
  4. $64$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
The ratio of the volume of water and Glycerine in $240$cc of mixture is $1:3$. 

The quantity of water(in cc) that should be added to the mixture so that the new ratio of the volume of water and glycerine becomes $2:3$

Let the ratio of the volume of water and glycerine be $x$ and $3x$.

$x+3x=240$

$\Rightarrow 4x=240$

$\Rightarrow x=60$

So the volume of water is $60cc$

Volume of glycerine is $60cc$

Volume of glycerine is $(3\times 60)cc=180$cc

Let x litre of water is added

$\dfrac{60+x}{180}=\dfrac{2}{3}$

$\Rightarrow 3(60+x)=360$

$\Rightarrow 180+3x=360$

$\Rightarrow 3x=360-180$

$\Rightarrow 3x=180$

$\Rightarrow x=60$.
Multiple choice maths ways to multiply and divide mental multiplication multiplication methods multiplication of numbers

Choose the correct answer from the alternatives given.
In a mixture of $35$ litres, the ratio of milk and water is $4: 1$. How many litres of water must be added to make the ratio $2 : 3$?

  1. $28$
  2. $40$
  3. $35$
  4. $70$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Initial quantity of milk and water is Milk  $=\dfrac{4}{5} \times $ $35 = 28 $litres
Water $=\dfrac{1}{5} \times $ $35 =7 $litres.
Let $x$ liters of water is added, then, 
$\dfrac{28}{7 + x} = \dfrac{2}{3}$
$x = 35$
Hence , 35 liters of water is added.

Multiple choice maths ratio, proportion and unitary method converting to ratios finding ratios other quantities

In what ratio must a grocer mix two varieties of pulses costing Rs.$15$ and Rs.$20$ per kg respectively so as to get a mixture worth Rs.$16.50$ kg?

  1. $3 : 7$
  2. $5 : 7$
  3. $7 : 3$
  4. $7 : 5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Consider the amount of pulse of price $Rs15$ $=x$


And the amount of pulse of price $Rs20$ $=y$

Then the total amount of mixture $=15x+20y$

But the price per $kg$ of mixture $=Rs16.50$

So, total price of $x+y kg$ $=16.50(x+y)$

Now according to the equation 

$16.50(x+y)=15x+20y$

$16.50x+16.50y=15x+20y$

$1.50x=3.50y$

$\frac { x }{ y } =\frac { 3.50 }{ 1.50 } \ =\frac { 0.7 }{ 0.3 } =\frac { 7 }{ 3 } $

Hence, required Ratio is $7:3$

So, the Option $C$ is the correct answer.

Multiple choice

An alloy contains zinc, copper and Iron in the ratio 4 : 3 : 2. If 20 kg of copper is added to 90 kg of this alloy, what will be the ratio of zinc and copper in the alloy?

  1. 1 : 1

  2. 1 : 2

  3. 4 : 5

  4. 5 : 4

  5. 3 : 5
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Original alloy: 90 kg total, ratio 4:3:2. Sum of parts = 4+3+2 = 9. Zinc = (4/9)*90 = 40 kg. Copper = (3/9)*90 = 30 kg. Iron = (2/9)*90 = 20 kg. Add 20 kg copper: New Copper = 30 + 20 = 50 kg. Zinc remains 40 kg. Ratio of Zinc to Copper = 40:50 = 4:5.