Exams › SSC CGL (Prelims) › General › Mixture and Alligation
21 questions with worked solutions.
Answer: 31.25 litres
In each step, the fraction of milk left is \(\frac{54-9}{54}=\frac{5}{6}\). After 3 repetitions, milk left = \(54\left(\frac{5}{6}\right)^3\). This equals 31.25 litres.
Answer: 19:6
Initially, milk = 60 L and water = 24 L. Removing 21 L of the same mixture removes 15 L milk and 6 L water, leaving 45 L milk and 18 L water. Adding 12 L milk makes it 57 L milk and 18 L water, so the ratio is 19:6.
Answer: ₹ 25/kg
Since the mixture is sold at ₹46/kg with 15% profit, its cost price is $46/1.15 = ₹40$ per kg. Using the ratio 1:2:3:4, the weighted average cost is $(30+2\cdot45+3\cdot60+4x)/10 = 40$, which gives $x=25$.
Answer: ₹ 31.6/kg
Selling price is ₹64/kg with 20% profit, so cost price of the mixture is ₹64 ÷ 1.2 = ₹53.33/kg. Using the weighted average, (2×50 + 3×70 + 2×x)/7 = 53.33, which gives x = ₹31.6/kg.
Answer: 30 litres
Initially, water = 16% of 50 = 8 litres. If x litres of oil are added, water remains 8 litres and total mixture becomes \(50+x\). We need \(8/(50+x)=10\%\), so \(8=0.1(50+x)\), giving \(x=30\).
Answer: 52.5%
Each container has the same volume. After filling with milk, the first contains 65% milk and the second contains 40% milk. Since the volumes are equal, the mixture in the larger vessel has milk percentage equal to the average of these two percentages: $(65+40)/2=52.5\%$.
Answer: 49: 15
Each time 15 liters are removed from 120 liters, the fraction of juice remaining becomes 105/120 = 7/8. After two such operations, juice left = \(120\times(7/8)^2 = 73.125\) liters. Water = 120 - 73.125 = 46.875 liters, giving the ratio 49:15.
Answer: ₹ 171
In a no-profit-no-loss mixture, the selling price equals the cost price of the mixture, so the weighted average cost is ₹200 per kg. Using alligation with prices ₹240 and x in the ratio 5:7, we get \((240-200):(200-x)=7:5\). Solving gives \(40:(200-x)=7:5\), so \(200-x=200/7\approx 28.57\), hence \(x\approx 171.43\), nearest integer ₹171.
Answer: 287
Let the initial mixture be \(x\) litres. Then strawberry, blueberry, and raspberry quantities are \(\frac{6x}{20}, \frac{9x}{20}, \frac{5x}{20}\). After removing 40 litres, the remaining amounts are reduced proportionally, and then 15 litres of strawberry and 10 litres of raspberry are added. Using the condition that blueberry exceeds strawberry by 22 litres gives \(\frac{9x}{20}-18 = \left(\frac{6x}{20}-12\right)+15+22\), which yields \(x\approx 286.67\), so the initial quantity is about 287 litres.
Answer: 1:1
Each container has the same volume, so we can directly combine the ratios by parts. The first mixture contributes 4 parts juice and 1 part water, while the second contributes 1 part juice and 4 parts water. Together, juice = 5 parts and water = 5 parts, so the ratio is 1:1.
Answer: 80 L
Initially, juice = 100 - 40 = 60 litres and water = 40 litres. For the ratio juice:water = 1:2, water must be 120 litres when juice is 60 litres. So, water to be added = 120 - 40 = 80 litres.
Answer: Rs. 52
The mixture is in the ratio 3:2, so the average cost per liter is the weighted mean of the two prices. Compute \((3\times 60 + 2\times 40)/5 = (180+80)/5 = 260/5 = 52\).
Answer: 48%
Initially, acid = $\frac{3}{5}\times 200=120$ L. In 40 L of mixture, acid removed = $\frac{3}{5}\times 40=24$ L, so remaining acid = 96 L. Final volume is still 200 L, so percentage of acid = $\frac{96}{200}\times 100=48\%$.
Answer: 15 L
In 60 L, milk = \(60\times \frac{7}{12}=35\) L and water = 25 L. To make the ratio 7:8, if milk is 35 L, water must be \(35\times \frac{8}{7}=40\) L. So, 15 L water must be added.
Answer: 10 L
In 50 litres with ratio 3:2, acid = 30 L and water = 20 L. To make the ratio 1:1, water must also become 30 L, so 10 L of water must be added.
Answer: Add 50 L of water, then 30 L of salt
Initially, salt = 60 L and water = 40 L. To make concentration 40%, the total must be \(60/0.4 = 150\) L, so 50 L water is added. Then to make concentration 50% with 60 L salt, total must be 120 L, so 30 L pure salt must be added.
Answer: 71.2%
Initially, acid = \(100 \times \frac{4}{5} = 80\) litres. After removing 20 litres of the mixture, acid removed = \(20 \times \frac{4}{5} = 16\), so acid left = 64 litres; then 20 litres of water is added, making acid still 64 litres in 100 litres. Next, 20 litres is removed again, so acid removed = \(20 \times \frac{64}{100} = 12.8\), leaving 51.2 litres acid; adding 20 litres pure acid gives 71.2 litres acid, i.e. 71.2%.
Answer: 30 L
Initially, oil = 20 L and kerosene = 10 L. If x litres of kerosene are added, the new ratio becomes 1:2, so 20 : (10 + x) = 1 : 2. Solving gives 10 + x = 40, hence x = 30 L.
Answer: 13:7
Initially, acid = 30 L and water = 10 L. Removing 8 L of a 3:1 mixture removes 6 L acid and 2 L water, leaving 24 L acid and 8 L water. Replacing with 8 L of a 1:3 mixture adds 2 L acid and 6 L water, giving 26 L acid and 14 L water, i.e. 13:7.
Answer: 20 litres
In 50 litres with ratio 3:2, juice = 30 litres and water = 20 litres. If $x$ litres of water are added, then the new ratio becomes $30:(20+x)=3:4$. Solving gives $120=60+3x$, so $x=20$ litres.
Answer: 40 litres
Initially, milk = 70 L and water = 10 L. If \(x\) litres of mixture are removed, water left = \(10 - \frac{x}{8}\) and milk left = \(70 - \frac{7x}{8}\); then \(x\) litres of pure milk are added. The final ratio becomes 15:1, which gives \(x=40\).