Quantitative Aptitude

Mixtures and Alligation

1,816 Questions
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Mixtures and Alligation Questions

Multiple choice

A container originally contains 10 litres of pure spirit. From this container, 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container?

  1. 7.58 litres

  2. 7.84 litres

  3. 7 litres

  4. 7.29 litres
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The amount of spirit remaining after n replacements is given by the formula: Final = Initial * (1 - volume_replaced/total_volume)^n. Here, 10 * (1 - 1/10)^3 = 10 * (0.9)^3 = 10 * 0.729 = 7.29 litres.

Multiple choice physics heat - measurement sources of heat introduction to heat measuring temperature

Three liquids A, B and C are at temperature of $60^{o}C, 55^{o}C$ respectively $4g$ of A mixed with $3g$ of C gives $65^{o} C$ and $2g$ of A mixed with $3g$ of B gives $57^{o}C$. The temperature of the mixture when equal masses of $B$ and $C$ are mixed is 

  1. $52.1^{o}C$
  2. $55^{o}C$
  3. $52.5^{o}C$
  4. $53^{o}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the principle of calorimetry (heat lost = heat gained), we can determine the relative specific heats of the liquids. Solving the equations for the mixtures allows us to find the final equilibrium temperature when B and C are mixed.

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

A goldsmith desires to test the purity of a gold ornament suspected to the mixed with copper. The ornament weights $0.25\ kg$ in air and is observe to displace $0.015$ litre of water when immersed in it. Densities of gold and copper with respect to water are, respectively, $19.3$ and $8.9$. The approximate percentage of copper in the ornament is

  1. $5\%$
  2. $10\%$
  3. $15\%$
  4. $25\%$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let volume of gold in ornament $V _1$ and that of copper $=V _2$ and density of gold $=\rho g$ and that of copper$=\rho _c$

$\Rightarrow \rho g V _1+\rho _cV-2=0.25\rightarrow (1)$
Volume of ornament $=$Volume of water displaced
$\Rightarrow V _1+V-2=0.15\times10^{-3}ms\Rightarrow V _2=(0.015\times10^{-3}-V _1)$
According to question-
$\cfrac{\rho _g}{\rho _w}=19.3$ and $\cfrac{\rho _c}{\rho _w}=8.9\ \rho _g=19.3\times10^3 kg/m^3$
$\rho _c=8.9\times10^3 kg/m^3$
Putting these value in equation $(1)$
$(19.3\times 10^{ 3 })V _{ 1 }+(1.9\times 10^{ 3 })(0.0015\times 10^{ -3 }-V _{ 1 })0.25\ [(19.3\times 10^{ 3 })-(1.9\times 10^{ 3 })]V _{ 1 }+(8.9\times 0.015)=0.25\ 10.4\times 10^{ 3 }V _{ 1 }=0.12\ V _{ 1 }=\cfrac { 0.12 }{ 10.4\times 10^{ 3 } } =0.0115\times 10^{ -3 }=1.15\times 10^{ -5 }m^{ 3 }$
$V _2=(0.015\times10^3-0.0115\times10^{-3})=0.0035\times10^{-3}\ \% \quad of\quad copper=(\cfrac{V _2}{V _1+V _2})\times100=[\cfrac{0.0035\times10^{-3}}{(0.0115+0.0035)\times10^{-3}}]\times100$
$\approx 23.34\%\ \approx 25\%$

Multiple choice physics upthrust in fluids, archimedes' principle and floatation density of a fluid density of fluid density and relative density

A liquid mixture of volume $V$ has two liquids as its ingredients with densities $\alpha  \; and\; \beta $. If the density of the mixture is $\sigma $, then the mass of the first liquid in the mixture is :

  1. $\dfrac{\alpha V[\sigma \beta +1]}{\beta [\alpha +\sigma ]}$
  2. $\dfrac{\alpha V[\sigma -\beta ]}{ [\sigma +\beta]}$
  3. $\dfrac{\alpha V[\beta-\sigma ]}{ \beta-\alpha }$
  4. $\dfrac{\alpha V[1-\sigma\alpha ]}{ \beta[\alpha-\sigma ] }$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Let mass of liquid with density $\alpha =M _1$
mass of liquid with density $\beta =M _2$
Total volume$=V$
Net density of mixture$=\sigma$
Total mass$=M _1+M _2$
$\Rightarrow V\sigma =M _1+M _2$
$\Rightarrow M _2=V\sigma -M _1$ ......$(1)$

$\left[\because \dfrac{Total \, Mass}{v}=\sigma\right]$

$T=\dfrac{Total \,mass}{Total \, volume}=\dfrac{M _1+M _2}{\dfrac{M _1}{\alpha}+\dfrac{M _2}{\beta}}$ .......$(2)$
sub $(1)$ in $(2)$

$\Rightarrow \sigma =\dfrac{M _1+(v\sigma -M _1)}{\dfrac{M _1}{\alpha}+\left(\dfrac{v\sigma -M _1}{\beta}\right)}$

$\Rightarrow M _1=\dfrac{\alpha V(\beta -\sigma)}{\beta -\alpha}$.
Multiple choice physics heat - measurement thermometers application of various thermometric scales different types of thermometers

One litre of water at $30^{o}C$ is mixed with one litre of water at $50^{o}C$. The temperature of the mixture will be

  1. 80C

  2. more than 50C but less than 80C

  3. 20C

  4. between 30C and 50C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Let the first volume of water be V1. Let the second volume of water be V2. Let the initial temperature of the first volume of water be θ1. Let the initial temperature of the second volume of water be θ2.

Let the energy required to heat a unit volume of water by unit temperature by C.

So, if the final temperature is θfinal, we have

heat gained by the first volume of water

$=V _1×C×(θ _{final}−θ _1)$

and

heat lost by the second volume of water

$=V _2×C×(θ _2−θ _{final}).$

By the law of conservation of energy,

heat lost = heat gained

∴$V _1×C×(θ _{final}−θ _1)=V _2×C×(θ _2−θ _{final})$

∴$V _1θ _{final}−V _1θ _1=V _2θ _2−V _2θ _{final}$

∴$θ _{final}=V _1θ _1+V _2θ _2V _1+V _2$

Substituting the given values,

$θ _{final}=\dfrac{30+50}{2}=40^∘C$.

The answer is option D
Multiple choice physics isothermal and adiabatic processes work done by an ideal gas in isothermal expansion thermodynamic processes heat and thermodynamics

Identical cylinders contain helium at 2.5 atm and agron at 1 atm respectively. If the are filled  in one of the cylinder,the pressure would be.

  1. 3.6 atm

  2. 1.75 atm

  3. 1.5 atm

  4. 1.0 atm
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given that,

Pressure ${{p} _{1}}=2.5\,atm$

Pressure ${{P} _{2}}=1\,atm$

Volume ${{V} _{1}}={{V} _{2}}=V$

Both the cylinders are similar, volume of both the gases is equal

Let the volume of gases be V

The amount of pressure P in one of the cylinder will be equal to the total pressure at equilibrium

Now,

  $ P=\dfrac{{{P} _{1}}{{V} _{1}}+{{P} _{2}}{{V} _{2}}}{{{V} _{1}}+{{V} _{2}}} $

 $ P=\dfrac{2.5\times V+1\times V}{2V} $

 $ P=1.75\ atm $

Hence, the pressure is $1.75\ atm$

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

$100\ ml$ of an aqueous solution contains $6.0\times {10}^{21}$ solute molecules. The solution is diluted to $1$ lit. The number of solute molecules present in $10\ ml$ of the dilute solution is:

  1. $6.0\times {10}^{20}$
  2. $6.0\times {10}^{19}$
  3. $6.0\times {10}^{18}$
  4. $6.0\times {10}^{17}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
100 ml solution diluted to 1 liters (1000) ml contains $6.0 \times 10^{21}$ solute molecular

No of molecules present in 10 ml

$ = \dfrac{10 \times 6.0 \times 10 ^{21}}{1000} = 6 \times 10 ^{19} $

Multiple choice chemistry mix and separate different types of solutions various mixtures introduction to solutions

How much water, in liters, must be added to 0.5 L of 6 M HCl to make it 2 M?

  1. 0.33

  2. 0.5

  3. 1

  4. 1.5

  5. 2
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Molarity =$\dfrac{Mass \,of\, the\, solute}{(Molar \,mass\, of\, the\, solute)\times {(Vol. of\, soln.\, in\, liters)}}$

Mass of solute will remain same before and after mixing water.
so 
       $M _1V _1$$=$$M _2V _2$
or 
       $V _2$$=$$6\times0.5/2$$=$$1.5L$
this is final total volume so water added $=$$1.5-0.5$$=$$1L$ 
Multiple choice chemistry acids and alkalis neutralisations in everyday life acids and bases in daily life neutralisation reaction

A 26 ml of $ N-Na _{2}CO _{3} $ solution is neutralized by the solutions of acids A and B in different experiments. The volumes of the acids A and B required were $10 ml$ and $40 ml$, respectively. How many volumes of A and B are to be mixed in order to prepare 1 litre of normal acid solution? 

  1. $179.4, 820.6$
  2. $820.6, 179.4$
  3. $500, 500$
  4. $474.3, 525.7$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2} (A)$
$ N _{1}V _{1} (Na _{2}CO _{3}) = N _{2}V _{2}(B)$
$ N _{1} = 1 , V _{1} = 26ml, V _{2} = 10ml, V _{3} = 40ml$
Normality of A
$ N _{2} = \dfrac{N _{1}V _{1}}{V _{2}} = \dfrac{1 \times 26}{10} = 2.6 N $
Normality of B
$ N _{3}= \dfrac{N _{1}V _{1}}{V _{3}} = \dfrac{1 \times 26}{40} = 0.65 N $
if we mix A & B then
$ N _{1}V _{1} + N _{2} V _{2} = N _{3} (V _{1}+V _{2})$
$ 2.6 V _{1}+0.65V _{2} = 1 \times 1000 $
$ 2.6V _{1}+ V _{2} + 0.65 V _{2} = 1000 ...(1)$
$ V _{1}+V _{2} =1000 ...(2)$
multiply $eq^{n}$ (2) by 0.65 
$ 0.65 V _{1}+0.65V _{2} = 650 ...(3)$
$ V _{2} = 820.6 ml$
$ V _{1} = 1000 - 820.6 = 179.4 ml $
option "a" correct.

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

A jar contained a mixture of two liquids A and B in the ratio 7 : 2. When 18 litres of mixture was taken out and 18 litres of liquid B was poured into the jar. This ratio became 2 : 3. The quantity of liquid A contained in the jar initially was: 

  1. $\frac { 450 } { 17 }$ litres
  2. $\frac { 490 } { 19 }$ litres
  3. $\frac { 490 } { 17 }$ litres
  4. $\frac { 450 } { 19 }$ litres
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let initial quantities be 7x and 2x. After removing 18 litres of mixture, the remaining quantities are proportional, and adding 18 litres of pure liquid B changes the ratio to 2:3. Solving the algebraic equation yields 490/17 litres for liquid A.

Multiple choice maths mixture types of ratios ratios in proportion mathematical logic

Choose the correct answer form the alternatives given.
The ratio of spirit and water in a mixture is 1: 3. If the volume of the solution is increased by 25% by adding spirit only. What is the resultant ratio of spirit and water? 

  1. $2:3$
  2. $1:4$
  3. $1: 2$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the volume of spirit and water be x and 3x Then, total volume = 4x. Resultant  volume of solution = 1.25 $\times$ 4x = 5x
Therefore, increase in volume = $5x - 4x = x$
So, the new ratio of spirit of water $2x : 3x = 2:3$
It is to be noted that increase in volume is due to addition of spirit only.

Multiple choice maths calculating and mental strategies 3 finding percentage of a number how many in all? problems on percentage

15 litres of a mixture contain 20% milk and the rest water. If 3 litres of water be mixed in it, the percentage of milk in the new mixture will be ..............

  1. 17%
  2. 16$\frac{2}{3}$%
  3. 18$\frac{1}{2}$%

  4. 15%
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

15 litres of a mixture contain 20% milk
So, milk=15*20/100=3 litres
water=15-3=12 litres
3 litres of water is mixed, so 
 new 18 litres of mixture has 15 litres of water and 3 litres of milk
So, percentage of milk= quantity of milk*100/total mixture
$M$ %$=3*100/18$
$=100/6=16\frac { 2 }{ 3 }$ %
Answer (B) 16$\frac{2}{3}$%


Multiple choice maths percent and percentage use of percentages converting between fractions or decimals and percentages percentage of a quantity

Choose the correct answer from the alternatives given.
Prabhu purchased 30 kg of rice at the rate of Rs. 17.50 per kg and another 30 kg rice at a certain rate. He mixed the variety of two rice and sold the entire quantity at the rate of Rs.18 per kg and made 20% overall profit. At what price per kg did he purchase the lot of another 30 kg rice?

  1. Rs. 14.50

  2. Rs. 12.50

  3. Rs. 15.50

  4. Rs. 13.50
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let the price at which the other $30$ kg was bought be Rs. x.
CP of mixture = $\frac{100}{120}\times 18 = 15$
$\frac{2.5}{15 x} = \frac{30}{30}$
$15x=2.5$
Hence, x= Rs. $12.5$

Multiple choice chemistry introduction to analytical chemistry different types of solutions various mixtures introduction to solutions

What volume would you dilute 0.2 L of a 15 M solution to obtain a 3 M solution?

  1. 1L

  2. 225L

  3. 10L

  4. 0.4L

  5. 0.1L
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that :

$\displaystyle M _1V _1 = M _2V _2 $ 

Given :

$\displaystyle  M _1 = 15 $  M
$\displaystyle V _1 = 0.2 $ L
$\displaystyle  M _2 =3$  M
$\displaystyle  V _2 =$  ????

$\displaystyle  15 \times 0.2 = 3 V _2$
Thus, $\displaystyle  V _2 = 1$  L

Hence, the correct option is A.
Multiple choice business maths functions and graphs some functions and their graphs -i graphs of the form y=ax^2+bx+c introduction to sets

A large mixing tank currently contains $200$ gallons of water into which $10$ pounds of sugar have been mixed. A tap will open pouring $20$ gallons per minute of water into the tank at the same time sugar is poured into the tank at a rate of $2$ pound per minute. Find the concentration (pounds per gallon) of sugar in the tank after $14$ minutes. Then 

  1. the concentration is greater than at the beginning?

  2. the concentration lesser than at the beginning?

  3. the concentration equal to the concentration at the beginning?

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $t$ be the number of minutes since the tap opened. 

Since the water increases at $20$ gallons per minute, and the sugar increases at $2$ pound per minute, these are constant rates of change. 

This tells us the amount of water in the tank is changing linearly, as is the amount of sugar in the tank. 

We can write an equation independently for each:

$\text{Water}: \: W(t)=200+20t$ in gallons

$\text{Sugar}: \: S(t)=10+2t$ in pounds​​

The concentration $C$ will be the ratio of pounds of sugar to gallons of water.

$C(t)=\dfrac{10+2t}{200+20t}$​​

The concentration after $14$ minutes is given by evaluating $C(t)$ at $t=14$

$\therefore C(14)=\dfrac{10+28}{200+280}=\dfrac{19}{240}$

This means the concentration is $19$ pounds of sugar to $240$ gallons of water.

At the beginning, the concentration is $C(0)=\dfrac{10+0}{200+0}=\dfrac{1}{20}$.

Since $\dfrac{19}{240}\approx 0.08 > \dfrac{1}{20}=0.05$, the concentration is greater after $14$ minutes than at the beginning.