Q1): To prepare 36 L of a 20% solution, in what quantities should 12% and 30% solutions be mixed? (Find litres of 30% solution)
A) 14 L
B) 16 L
C) 18 L
D) 20 L
Answer: B) 16 L
Explanation:
- Step 1: Use alligation ratio (Lower : Higher) = (Higher − Mean) : (Mean − Lower)
- Step 2: Lower = 12, Mean = 20, Higher = 30
- Step 3: Ratio = (30 − 20) : (20 − 12) = 10 : 8 = 5 : 4
- Step 4: Total parts = 5 + 4 = 9, total quantity = 36 L
- Final: 30% part = (4/9)×36 = 16 L, so option B
Q2): 60 kg rice is made by mixing ₹45/kg and ₹60/kg rice such that the mixture costs ₹54/kg. How much ₹60/kg rice is used?
A) 24 kg
B) 30 kg
C) 36 kg
D) 40 kg
Answer: C) 36 kg
Explanation:
- Step 1: Apply alligation on prices
- Step 2: Cheaper = 45, Mean = 54, Dearer = 60
- Step 3: Ratio (Cheaper : Dearer) = (60 − 54) : (54 − 45) = 6 : 9 = 2 : 3
- Step 4: Dearer fraction = 3/(2+3) = 3/5
- Final: ₹60 rice = (3/5)×60 = 36 kg, so option C
Q3): A 20 L milk-water mixture is in the ratio 3:2. If 5 L of mixture is removed and replaced by water, what is the new percentage of milk?
A) 40%
B) 45%
C) 50%
D) 55%
Answer: B) 45%
Explanation:
- Step 1: Milk = (3/5)×20 = 12 L, Water = 8 L
- Step 2: In 5 L removed, milk removed = (12/20)×5 = 3 L
- Step 3: Remaining milk = 12 − 3 = 9 L; remaining total = 15 L
- Step 4: Add 5 L water ⇒ total = 20 L, milk stays 9 L
- Final: Milk% = (9/20)×100 = 45%, so option B
Q4): A 25 L solution contains 40% acid. How many litres should be removed and replaced with water so that acid becomes 32%?
A) 3 L
B) 4 L
C) 5 L
D) 6 L
Answer: C) 5 L
Explanation:
- Step 1: Initial acid = 40% of 25 = 10 L
- Step 2: Remove x L of 40% solution ⇒ acid removed = 0.4x
- Step 3: New acid = 10 − 0.4x (after adding x L water)
- Step 4: New acid must be 32% of 25 = 8 L ⇒ 10 − 0.4x = 8
- Final: 0.4x = 2 ⇒ x = 5 L, so option C
Q5): 8 L of 25% alcohol solution is mixed with 12 L of 10% alcohol solution. What is the alcohol percentage in the mixture?
A) 14%
B) 15%
C) 16%
D) 18%
Answer: C) 16%
Explanation:
- Step 1: Alcohol in 8 L = 0.25×8 = 2.0 L
- Step 2: Alcohol in 12 L = 0.10×12 = 1.2 L
- Step 3: Total alcohol = 2.0 + 1.2 = 3.2 L; total liquid = 20 L
- Step 4: Percentage = (3.2/20)×100
- Final: 16%, so option C
Q6): Wheat at ₹30/kg is mixed with wheat at ₹45/kg to get 60 kg mixture worth ₹36/kg. How much ₹45/kg wheat is used?
A) 20 kg
B) 22 kg
C) 24 kg
D) 26 kg
Answer: C) 24 kg
Explanation:
- Step 1: Apply alligation on prices
- Step 2: Cheaper = 30, Mean = 36, Dearer = 45
- Step 3: Ratio (Cheaper : Dearer) = (45 − 36) : (36 − 30) = 9 : 6 = 3 : 2
- Step 4: Dearer fraction = 2/(3+2) = 2/5
- Final: ₹45 wheat = (2/5)×60 = 24 kg, so option C
Q7): Two alloys contain 70% copper and 40% copper. How many kg of 70% alloy is needed to make 25 kg of 52% copper alloy?
A) 8 kg
B) 10 kg
C) 12 kg
D) 15 kg
Answer: B) 10 kg
Explanation:
- Step 1: Use alligation on copper %
- Step 2: Lower = 40, Mean = 52, Higher = 70
- Step 3: Ratio (Higher : Lower) = (Mean − Lower) : (Higher − Mean) = (52−40):(70−52)=12:18=2:3
- Step 4: Total parts = 2+3 = 5; total = 25 kg
- Final: 70% alloy = (2/5)×25 = 10 kg, so option B
Q8): A 50 L mixture has 20 L milk and 30 L water. If 10 L mixture is removed and replaced with pure milk, what is the new percentage of milk?
A) 48%
B) 50%
C) 52%
D) 54%
Answer: C) 52%
Explanation:
- Step 1: Milk fraction initially = 20/50 = 0.4
- Step 2: In 10 L removed, milk removed = 0.4×10 = 4 L
- Step 3: Remaining milk = 20 − 4 = 16 L (total now 40 L)
- Step 4: Add 10 L pure milk ⇒ milk = 26 L, total = 50 L
- Final: Milk% = (26/50)×100 = 52%, so option C
Q9): A 12 L solution contains 30% salt. How much pure salt should be added to make the solution 40% salt?
A) 1 kg
B) 1.5 kg
C) 2 kg
D) 2.5 kg
Answer: C) 2 kg
Explanation:
- Step 1: Initial salt = 30% of 12 = 3.6 kg
- Step 2: Add x kg salt ⇒ new salt = 3.6 + x
- Step 3: New total = 12 + x and new % = 40% ⇒ (3.6 + x)/(12 + x) = 0.40
- Step 4: 3.6 + x = 4.8 + 0.4x ⇒ 0.6x = 1.2
- Final: x = 2 kg, so option C
Q10): 36 L of a 32% solution is made by mixing 18% and 42% solutions. How many litres of 42% solution are used?
A) 18 L
B) 20 L
C) 21 L
D) 24 L
Answer: C) 21 L
Explanation:
- Step 1: Use alligation ratio (Lower : Higher) = (Higher − Mean) : (Mean − Lower)
- Step 2: Lower = 18, Mean = 32, Higher = 42
- Step 3: Ratio = (42−32):(32−18) = 10:14 = 5:7
- Step 4: Total parts = 12; higher part = 7
- Final: 42% quantity = (7/12)×36 = 21 L, so option C
Q11): A container has 40 L pure milk. 5 L is removed and replaced with water twice. How much milk remains after two operations?
A) 30.0 L
B) 30.625 L
C) 31.25 L
D) 32.0 L
Answer: B) 30.625 L
Explanation:
- Step 1: After one replacement, milk left factor = (1 − 5/40) = 35/40
- Step 2: After two replacements, milk left = 40×(35/40)²
- Step 3: (35/40)² = (7/8)² = 49/64
- Step 4: Milk left = 40×49/64 = 1960/64 = 30.625
- Final: Milk remaining = 30.625 L, so option B
Q12): A 50 L solution has 20% alcohol. How many litres should be removed and replaced with 60% alcohol solution to make it 30% alcohol?
A) 10 L
B) 12.5 L
C) 15 L
D) 18 L
Answer: B) 12.5 L
Explanation:
- Step 1: Initial alcohol = 20% of 50 = 10 L
- Step 2: Remove x L of 20% ⇒ alcohol removed = 0.2x
- Step 3: Add x L of 60% ⇒ alcohol added = 0.6x
- Step 4: New alcohol = 10 − 0.2x + 0.6x = 10 + 0.4x = 30% of 50 = 15
- Final: 0.4x = 5 ⇒ x = 12.5 L, so option B
Q13): 20 kg of pulses at ₹80/kg is mixed with some pulses at ₹60/kg to get a mixture worth ₹68/kg. Find the quantity of ₹60 pulses.
A) 24 kg
B) 28 kg
C) 30 kg
D) 32 kg
Answer: C) 30 kg
Explanation:
- Step 1: Apply alligation on prices
- Step 2: Cheaper = 60, Mean = 68, Dearer = 80
- Step 3: Ratio (Cheaper : Dearer) = (80−68):(68−60)=12:8=3:2
- Step 4: Dearer quantity = 20 kg corresponds to 2 parts
- Final: 1 part = 10 kg ⇒ cheaper = 3 parts = 30 kg, so option C
Q14): A 24 L acid-water mixture has acid : water = 5 : 3. How much pure acid should be added to make acid 70% of the final mixture?
A) 4 L
B) 5 L
C) 6 L
D) 7 L
Answer: C) 6 L
Explanation:
- Step 1: Total parts = 5+3 = 8 ⇒ acid = (5/8)×24 = 15 L
- Step 2: Add x L pure acid ⇒ acid = 15 + x, total = 24 + x
- Step 3: Required acid% = 70% ⇒ (15 + x)/(24 + x) = 0.70
- Step 4: 15 + x = 16.8 + 0.7x ⇒ 0.3x = 1.8
- Final: x = 6 L, so option C
Q15): A 20 L solution contains 25% salt. How many litres of water must be evaporated to make the solution 40% salt?
A) 6 L
B) 7 L
C) 7.5 L
D) 8 L
Answer: C) 7.5 L
Explanation:
- Step 1: Salt amount stays same in evaporation
- Step 2: Initial salt = 25% of 20 = 5 L (salt equivalent)
- Step 3: For 40% concentration, total must be = 5/0.40 = 12.5 L
- Step 4: Water evaporated = 20 − 12.5 = 7.5 L
- Final: Evaporate 7.5 L water, so option C
Q16): A 12 L solution is diluted by adding 8 L water, and the new concentration becomes 15%. What was the original concentration?
A) 20%
B) 25%
C) 30%
D) 35%
Answer: B) 25%
Explanation:
- Step 1: After adding water, total volume = 12 + 8 = 20 L
- Step 2: New solute amount = 15% of 20 = 3 L
- Step 3: Solute amount was same initially (only water added)
- Step 4: Original % = (3/12)×100 = 25%
- Final: Original concentration = 25%, so option B
Q17): 15 kg tea at ₹240/kg is mixed with some tea at ₹200/kg to get a mixture worth ₹225/kg. How many kg of ₹200 tea is used?
A) 6 kg
B) 8 kg
C) 9 kg
D) 10 kg
Answer: C) 9 kg
Explanation:
- Step 1: Use weighted average: (15×240 + x×200)/(15 + x) = 225
- Step 2: 3600 + 200x = 225(15 + x)
- Step 3: 3600 + 200x = 3375 + 225x
- Step 4: 225 = 25x ⇒ x = 9
- Final: ₹200 tea = 9 kg, so option C
Q18): 40 L of 30% solution is prepared by mixing 20% and 50% solutions. How many litres of 50% solution are required?
A) 10 L
B) 12 L
C) 13.33 L
D) 16 L
Answer: C) 13.33 L
Explanation:
- Step 1: Apply alligation on %
- Step 2: Lower = 20, Mean = 30, Higher = 50
- Step 3: Ratio (Lower : Higher) = (50−30):(30−20) = 20:10 = 2:1
- Step 4: Total parts = 3; higher (50%) part = 1
- Final: 50% quantity = (1/3)×40 = 13.33 L, so option C
Q19): A 30 L mixture contains 80% milk. How many litres should be removed and replaced with water to make milk 60%?
A) 6 L
B) 7 L
C) 7.5 L
D) 8 L
Answer: C) 7.5 L
Explanation:
- Step 1: Initial milk = 80% of 30 = 24 L
- Step 2: Remove x L of mixture ⇒ milk removed = 0.8x
- Step 3: Add x L water ⇒ milk stays 24 − 0.8x
- Step 4: Required milk = 60% of 30 = 18 ⇒ 24 − 0.8x = 18
- Final: 0.8x = 6 ⇒ x = 7.5 L, so option C
Q20): A 25 L mixture has alcohol and water in the ratio 3:2 (alcohol is pure). How many litres of 40% alcohol solution should be added to make the final alcohol concentration 50%?
A) 20 L
B) 22.5 L
C) 25 L
D) 27.5 L
Answer: C) 25 L
Explanation:
- Step 1: Initial pure alcohol = (3/5)×25 = 15 L
- Step 2: Add x L of 40% solution ⇒ alcohol added = 0.4x
- Step 3: New total = 25 + x and new alcohol = 15 + 0.4x
- Step 4: Required concentration 50% ⇒ (15 + 0.4x)/(25 + x) = 0.5
- Final: 15 + 0.4x = 12.5 + 0.5x ⇒ 2.5 = 0.1x ⇒ x = 25 L, so option C