Quantitative Aptitude
Mixtures and Alligation
1,816 Questions
Mixtures and Alligation Questions
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$\displaystyle\frac{7}{3}$
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$\displaystyle\frac{5}{4}$
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$\displaystyle\frac{19}{13}$
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$\displaystyle\frac{15}{19}$
C
Correct answer
Explanation
For a mixture, the adiabatic index gamma_mix = (n1*Cv1 + n2*Cv2) / (n1*Cp1 + n2*Cp2) is not the standard formula. Use Cv_mix = (n1*Cv1 + n2*Cv2) / (n1+n2) and Cp_mix = (n1*Cp1 + n2*Cp2) / (n1+n2). For diatomic, Cv=5R/2, Cp=7R/2. For monatomic, Cv=3R/2, Cp=5R/2. Cv_mix = (2*2.5R + 1*1.5R)/3 = 6.5R/3. Cp_mix = (2*3.5R + 1*2.5R)/3 = 9.5R/3. Ratio = 9.5/6.5 = 19/13.
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$X_m=X_p=X_v$
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$X_p=\frac {1}{X_v}$
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$X_m=\dfrac {1}{X_P}=\dfrac {1}{X_V}$
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$X_V=\dfrac {1}{X_p}=\dfrac {1}{X_m}$
A
Correct answer
Explanation
For an ideal gas mixture, the mole fraction of a component is equal to its volume fraction and its pressure fraction (Dalton's Law of Partial Pressures). Therefore, Xm = Xp = Xv.
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$3.3$ atm
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$3.1$ atm
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$0.43$ atm
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$\cfrac{1.1}{3}$ atm
B
Correct answer
Explanation
Total pressure = P_water_vapour + P_oxygen. Aqueous tension (P_water_vapour) is constant at 0.1 atm. P_oxygen = 1.1 - 0.1 = 1.0 atm. If volume is 1/3, P_oxygen becomes 1.0 * 3 = 3.0 atm. Total pressure = 3.0 + 0.1 = 3.1 atm.
A
Correct answer
Explanation
Total pressure = 200 + 150 + 320 + 130 = 800 mm. Volume fraction of hydrogen = (Partial pressure of H) / (Total pressure) = 200 / 800 = 0.25.
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$34.5$
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$25.0$
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$23.0$
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$20.0$
A
Correct answer
Explanation
The reaction is N2O4 -> 2NO2. If NO2 is 50% of the volume, the mole fraction of NO2 is 0.5 and N2O4 is 0.5. The average molar mass is (0.5 * 46) + (0.5 * 92) = 23 + 46 = 69. Vapour density is molar mass / 2, which is 69 / 2 = 34.5.
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$2$ litres and $14$ litres
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$8$ litres and $35$ litres
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$7$ litres and $25$ litres
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$6$ litres and $15$ litres
D
Correct answer
Explanation
Let x litres of 90% and (21 - x) litres of 97% be mixed. Then 0.90x + 0.97(21 - x) = 0.95*21, giving x = 6 litres of 90% solution and 15 litres of 97% solution.
C
Correct answer
Explanation
Using Boyle's Law (P1V1 = P2V2) for each gas in the new volume (1L): Oxygen: 1 atm * 1 L = P_O2 * 1 L => P_O2 = 1 atm. Nitrogen: 0.5 atm * 2 L = P_N2 * 1 L => P_N2 = 1 atm. Total pressure = 1 + 1 = 2 atm.
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$50$ $mm$
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$400$ $mm$
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$60$ $mm$
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$1000$ $mm$
C
Correct answer
Explanation
According to Dalton's Law of partial pressures, the pressure exerted by a gas in a mixture is proportional to its volume percentage. 60% of 100 mm is 60 mm.
D
Correct answer
Explanation
Glass 1: 1/2 milk, 1/2 water. Glass 2: 3/4 milk, 1/4 water. Total milk = 1/2 + 3/4 = 5/4. Total water = 1/2 + 1/4 = 3/4. Ratio of milk to water = (5/4) / (3/4) = 5/3.
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$70$ litre.
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$75$ litre.
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$67$ litre.
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$98$ litre.
B
Correct answer
Explanation
Initial solute = 25% of 300 = 75. Let x be water added. New concentration = 75 / (300 + x) = 20/100 = 1/5. 375 = 300 + x, so x = 75.
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$220\ g$
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$330\ g$
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$230\ g$
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$150\ g$
A
Correct answer
Explanation
Total weight = 520 + 580 = 1100 g. Divided into 5 packets = 1100 / 5 = 220 g per packet.
C
Correct answer
Explanation
Initial wine = 30, water = 10. Let x be the water added. (30 / (10 + x)) = 5/2. Cross-multiplying gives 60 = 50 + 5x, so 5x = 10, x = 2.
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5$\mathrm { Ltr }$
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7.5$\mathrm { Ltr }$
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10$\mathrm { Ltr }$
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12$\mathrm { Ltr }$
C
Correct answer
Explanation
Initial: A=4x, W=3x. Added 5L water: A=4x, W=3x+5. New ratio: 4x/(3x+5) = 4/5. 20x = 12x + 20. 8x = 20. x = 2.5. Alcohol = 4 * 2.5 = 10 L.
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$\cfrac { 1 }{ 4 } $
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$\cfrac { 81 }{ 256 } $
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$\cfrac { 27 }{ 64 } $
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$\cfrac { 37 }{ 64 } $
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$\cfrac { 175 }{ 256 } $
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$53$ kgs
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$80$ kgs
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$36$ kgs
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$44$ kgs
D
Correct answer
Explanation
Alloy X (60 kg) has lead and tin in ratio 3:2, so tin = (2/5)*60 = 24 kg. Alloy Y (100 kg) has tin and copper in ratio 1:4, so tin = (1/5)*100 = 20 kg. Total tin = 24 + 20 = 44 kg.