Q1): A 60 L container has pure milk. 10 L is removed and replaced with water. This operation is done 3 times. How much milk remains (approx.)?
A) 33.33 L
B) 34.72 L
C) 35.50 L
D) 37.50 L
Answer: B) 34.72 L
Explanation:
- Step 1: Formula: Milk left = V × (1 − x/V)^n
- Step 2: Here V = 60, x = 10, n = 3
- Step 3: Milk left = 60 × (1 − 10/60)^3 = 60 × (50/60)^3
- Step 4: = 60 × (5/6)^3 = 60 × (125/216) = 34.72 (approx.)
- Final: Milk remaining ≈ 34.72 L, so option B
Q2): A 40 L solution has 25% acid. If 8 L is removed and replaced with 50% acid solution, what is the new acid percentage?
A) 28%
B) 30%
C) 32%
D) 35%
Answer: B) 30%
Explanation:
- Step 1: Initial acid = 25% of 40 = 10 L
- Step 2: Acid removed = 25% of 8 = 2 L
- Step 3: Acid added = 50% of 8 = 4 L
- Step 4: New acid = 10 − 2 + 4 = 12 L in 40 L
- Final: New % = (12/40)×100 = 30%, so option B
Q3): A 60 L mixture of 18% solution is made using 10% solution, 25% solution, and water. If quantity of 25% solution is twice the quantity of 10% solution, how much water is used?
A) 4 L
B) 6 L
C) 8 L
D) 10 L
Answer: B) 6 L
Explanation:
- Step 1: Let 10% = a L, then 25% = 2a L
- Step 2: Water = 60 − (a + 2a) = 60 − 3a
- Step 3: Total solute = 0.10a + 0.25(2a) = 0.10a + 0.50a = 0.60a
- Step 4: Required solute = 18% of 60 = 10.8 ⇒ 0.60a = 10.8 ⇒ a = 18
- Final: Water = 60 − 3×18 = 6 L, so option B
Q4): Pulses costing ₹80/kg and ₹60/kg are mixed and sold at ₹84/kg with 20% profit. Find the ratio of ₹80 pulses to ₹60 pulses.
A) 1 : 2
B) 2 : 1
C) 1 : 1
D) 3 : 2
Answer: C) 1 : 1
Explanation:
- Step 1: Selling price = ₹84 with 20% profit ⇒ Cost price of mixture = 84/1.20 = ₹70
- Step 2: Now mean price = 70 between 60 and 80
- Step 3: Alligation: ₹80 : ₹60 = (70 − 60) : (80 − 70) = 10 : 10
- Step 4: Simplify = 1 : 1
- Final: Ratio is 1:1, so option C
Q5): A 30 L solution has 20% salt. 5 L water evaporates, then x L water is added to make it 15% salt. Find x.
A) 10 L
B) 12 L
C) 15 L
D) 18 L
Answer: C) 15 L
Explanation:
- Step 1: Salt amount stays same in evaporation and water addition
- Step 2: Initial salt = 20% of 30 = 6 L
- Step 3: After evaporation, total = 30 − 5 = 25 L, salt = 6 L
- Step 4: After adding x water: 6/(25 + x) = 0.15 ⇒ 25 + x = 40
- Final: x = 15 L, so option C
Q6): A 50 L solution has 30% sugar. 10 L is removed and replaced with water, then again 10 L is removed and replaced with 60% sugar solution. Find final sugar percentage.
A) 30.8%
B) 31.2%
C) 32.0%
D) 33.6%
Answer: B) 31.2%
Explanation:
- Step 1: Initial sugar = 30% of 50 = 15 L
- Step 2: Remove 10 L ⇒ sugar removed = (15/50)×10 = 3 ⇒ sugar left = 12
- Step 3: Remove 10 L again (now 12/50 sugar fraction) ⇒ removed sugar = (12/50)×10 = 2.4 ⇒ left = 9.6
- Step 4: Add 10 L of 60% solution ⇒ add sugar = 6 ⇒ total sugar = 15.6
- Final: % = (15.6/50)×100 = 31.2%, so option B
Q7): A solution is diluted by adding 20 L water, and concentration becomes 20%. Then 10 L of this diluted solution is removed and replaced with 10 L pure solute (100%). Concentration becomes 25%. Find the original volume of solution.
A) 120 L
B) 130 L
C) 140 L
D) 150 L
Answer: C) 140 L
Explanation:
- Step 1: After adding 20 L water, total volume = V + 20 and solute = 20% of (V + 20)
- Step 2: So solute after dilution = 0.20(V + 20)
- Step 3: Removing 10 L removes solute = 20% of 10 = 2 L; adding 10 L pure adds 10 L solute
- Step 4: Final solute = 0.20(V + 20) − 2 + 10 = 0.20(V + 20) + 8 = 25% of (V + 20)
- Final: 0.20(V + 20) + 8 = 0.25(V + 20) ⇒ 8 = 0.05(V + 20) ⇒ V + 20 = 160 ⇒ V = 140, so option C
Q8): Tea of ₹120/kg, ₹150/kg, and ₹180/kg are mixed in the ratio 1:2:3. At what price per kg should the mixture be sold to earn 20% profit?
A) ₹180
B) ₹192
C) ₹196
D) ₹200
Answer: B) ₹192
Explanation:
- Step 1: Mixture CP = weighted average = (1×120 + 2×150 + 3×180) / (1+2+3)
- Step 2: = (120 + 300 + 540) / 6 = 960/6 = ₹160
- Step 3: Profit 20% ⇒ SP = 160 × 1.20
- Step 4: SP = ₹192
- Final: Sell at ₹192/kg, so option B
Q9): A 30 L solution has 40% acid. x L is removed and replaced with water. Acid becomes 28%. Again x L is removed and replaced with water. Acid becomes 19.6%. Find x.
A) 6 L
B) 8 L
C) 9 L
D) 10 L
Answer: C) 9 L
Explanation:
- Step 1: In replacement with water, concentration multiplies by (1 − x/V) each time
- Step 2: After first operation: 40 → 28 ⇒ factor = 28/40 = 0.7
- Step 3: So 1 − x/30 = 0.7 ⇒ x/30 = 0.3
- Step 4: x = 9 L (and second gives 28×0.7 = 19.6, matches)
- Final: x = 9 L, so option C
Q10): An 80 L mixture of 45% solution is prepared using 10%, 50%, and 70% solutions such that quantities of 10% and 70% are equal. How many litres of 50% solution are used?
A) 30 L
B) 35 L
C) 40 L
D) 45 L
Answer: C) 40 L
Explanation:
- Step 1: Let 10% = a L and 70% = a L, then 50% = 80 − 2a
- Step 2: Total solute = 0.10a + 0.70a + 0.50(80 − 2a)
- Step 3: = 0.80a + 40 − a = 40 − 0.20a
- Step 4: Required solute = 45% of 80 = 36 ⇒ 40 − 0.20a = 36 ⇒ a = 20
- Final: 50% used = 80 − 2×20 = 40 L, so option C
Q11): 60 L of 30% alcohol is mixed with 40 L of 45% alcohol. From this 100 L mixture, 20 L is removed and replaced with water. Find final alcohol percentage.
A) 26.4%
B) 27.6%
C) 28.8%
D) 30.0%
Answer: C) 28.8%
Explanation:
- Step 1: Alcohol after mixing = 60×0.30 + 40×0.45 = 18 + 18 = 36 L
- Step 2: Mixture volume = 100 L ⇒ concentration = 36%
- Step 3: Remove 20 L ⇒ alcohol removed = 36% of 20 = 7.2 L
- Step 4: Alcohol left = 36 − 7.2 = 28.8 L; add 20 L water ⇒ volume = 100 L
- Final: % = (28.8/100)×100 = 28.8%, so option C
Q12): Dry fruits costing ₹500/kg and ₹800/kg are mixed and sold at ₹840/kg with 20% profit. Find the ratio of ₹500 to ₹800 in the mixture.
A) 2 : 1
B) 1 : 2
C) 3 : 1
D) 2 : 3
Answer: B) 1 : 2
Explanation:
- Step 1: SP = 840 with 20% profit ⇒ CP of mixture = 840/1.20 = ₹700
- Step 2: Now mean price = 700 between 500 and 800
- Step 3: Alligation (₹500 : ₹800) = (800 − 700) : (700 − 500) = 100 : 200
- Step 4: Simplify = 1 : 2
- Final: Ratio is 1:2, so option B
Q13): A 40 L milk-water mixture undergoes two operations: each time 10 L is removed and replaced with pure milk. After the second operation, milk becomes 77.5%. What was the initial milk percentage?
A) 55%
B) 60%
C) 62.5%
D) 65%
Answer: B) 60%
Explanation:
- Step 1: Let initial milk fraction be p in 40 L
- Step 2: After one replacement: new fraction p1 = (3p + 1)/4
- Step 3: After second replacement: p2 = (3p1 + 1)/4 and given p2 = 0.775
- Step 4: Work backward: p1 = (4p2 − 1)/3 = (4×0.775 − 1)/3 = 0.7
- Final: p = (4p1 − 1)/3 = (4×0.7 − 1)/3 = 0.6 = 60%, so option B
Q14): An 80 L solution has 25% acid. How many litres should be removed and replaced with 40% acid solution to make it 30% acid?
A) 24 L
B) 25 L
C) 26.67 L
D) 28 L
Answer: C) 26.67 L
Explanation:
- Step 1: Initial acid = 25% of 80 = 20 L
- Step 2: If x L removed, acid removed = 0.25x
- Step 3: Acid added with 40% solution = 0.40x
- Step 4: New acid = 20 − 0.25x + 0.40x = 20 + 0.15x = 30% of 80 = 24
- Final: 0.15x = 4 ⇒ x = 26.67 L, so option C
Q15): A 50 L solution has 10% salt. 5 L water evaporates, then 5 L of 30% salt solution is added. Find the new salt percentage.
A) 12%
B) 12.5%
C) 13%
D) 14%
Answer: C) 13%
Explanation:
- Step 1: Initial salt = 10% of 50 = 5 L
- Step 2: After evaporation, volume = 45 L but salt remains 5 L
- Step 3: Add 5 L of 30% solution ⇒ salt added = 0.30×5 = 1.5 L
- Step 4: New salt = 5 + 1.5 = 6.5 L; new volume = 50 L
- Final: % = (6.5/50)×100 = 13%, so option C
Q16): A 50 kg rice mixture is made using ₹40/kg and ₹70/kg. Quantity of ₹70 rice is 10 kg less than ₹40 rice. Find the cost price per kg of the mixture.
A) ₹50
B) ₹52
C) ₹54
D) ₹56
Answer: B) ₹52
Explanation:
- Step 1: Let ₹40 rice = x kg, then ₹70 rice = x − 10 kg
- Step 2: Total: x + (x − 10) = 50 ⇒ 2x = 60 ⇒ x = 30
- Step 3: So ₹40 rice = 30 kg and ₹70 rice = 20 kg
- Step 4: Mean price = (30×40 + 20×70)/50 = (1200 + 1400)/50
- Final: Mean = 2600/50 = ₹52, so option B
Q17): Solutions of 10%, 25%, and 40% are mixed in the ratio 2:3:5 to make 50 L mixture. How much water should be added to make the final concentration 20%?
A) 20 L
B) 22.5 L
C) 23.75 L
D) 25 L
Answer: C) 23.75 L
Explanation:
- Step 1: Weighted concentration = (2×10 + 3×25 + 5×40) / (2+3+5)
- Step 2: = (20 + 75 + 200)/10 = 295/10 = 29.5%
- Step 3: Solute in 50 L = 29.5% of 50 = 14.75 L
- Step 4: Add x water ⇒ 14.75/(50 + x) = 0.20 ⇒ 50 + x = 73.75
- Final: x = 23.75 L, so option C
Q18): A 60 L solution has 30% acid. First 10 L is removed and replaced with water. Then 15 L is removed and replaced with water. Find final acid percentage.
A) 17.5%
B) 18.0%
C) 18.75%
D) 19.5%
Answer: C) 18.75%
Explanation:
- Step 1: After first replacement: C1 = 30% × (1 − 10/60)
- Step 2: C1 = 30% × (50/60) = 25%
- Step 3: After second replacement: C2 = 25% × (1 − 15/60)
- Step 4: C2 = 25% × (45/60) = 18.75%
- Final: Final concentration = 18.75%, so option C
Q19): A 48% alcohol mixture is formed by mixing 72% alcohol, water (0%), and 36% alcohol. If water and 36% alcohol are in the ratio 1:2, find the ratio of 72% alcohol to water.
A) 2 : 1
B) 3 : 1
C) 4 : 1
D) 5 : 2
Answer: B) 3 : 1
Explanation:
- Step 1: Let water = x, then 36% solution = 2x
- Step 2: Let 72% solution = y
- Step 3: Total alcohol = 0.36(2x) + 0.72y = 0.72x + 0.72y
- Step 4: Total volume = x + 2x + y = 3x + y and % is 48% ⇒ 0.72x + 0.72y = 0.48(3x + y)
- Final: 0.72x + 0.72y = 1.44x + 0.48y ⇒ 0.24y = 0.72x ⇒ y:x = 3:1, so option B
Q20): A 50 L solution has 20% salt. x L of 50% salt solution is added so that concentration becomes 30%. Then 10 L of this new mixture is removed and replaced with water. Final concentration becomes 26%. Find x.
A) 20 L
B) 22.5 L
C) 25 L
D) 27.5 L
Answer: C) 25 L
Explanation:
- Step 1: Initial salt = 20% of 50 = 10 L
- Step 2: After adding x of 50%: salt = 10 + 0.5x, volume = 50 + x, and concentration is 30%
- Step 3: So salt after mixing = 30% of (50 + x) = 0.3(50 + x)
- Step 4: After replacement with water, salt reduces by 30% of 10 = 3 L, so final salt = 0.3(50 + x) − 3
- Final: Final concentration 26% ⇒ 0.3(50 + x) − 3 = 0.26(50 + x) ⇒ 0.04(50 + x) = 3 ⇒ 50 + x = 75 ⇒ x = 25, so option C