Question 1

Diluting a 15-L mixture by adding water: A 15-liter mixture contains 20% alcohol (and 80% water). If 3 liters of water are added, what is the new percentage of alcohol?

Accepted Answer

Introduction / Context: Adding water does not change the absolute alcohol quantity, only the total volume. Therefore, the new percentage is alcohol mass divided by the increased total volume, expressed as a percentage. Given Data / Assumptions: Initial alcohol = 20% of 15 L = 3 L. Added water = 3 L. New total volume = 18 L. Concept / Approach: New alcohol % = (unchanged alcohol) / (new total) * 100%. Substitute the values and reduce the fraction to a mixed number if needed. Step-by-Step Solution: Alcohol remains 3 L.New total = 15 + 3 = 18 L.Alcohol % = 3/18 * 100% = 1/6 * 100% = 16 2/3%. Verification / Alternative check: Decimal form: 0.1666… × 100% = 16.666…%, which matches 16 2/3%. Why Other Options Are Wrong: They assume changed alcohol amount or use incorrect total volume after addition. Common Pitfalls: Reducing alcohol amount by 20% of 3 L incorrectly; the alcohol content is unchanged by adding pure water. Final Answer: 16 2/3%

Question 2

Profit by adulteration (water to milk): A milk seller pays ₹ 500 per kiloliter (i.e., ₹ 0.50 per liter) for pure milk. He adds water and sells the mixture at ₹ 0.56 per liter, making an overall profit of 40%. What is the ratio of water to milk suppl…

Accepted Answer

Introduction / Context: Since water is free, the cost of 1 L of mixture equals the cost of the milk fraction alone. The given selling price and profit percentage let us compute the cost per liter of mixture; from that we deduce the milk fraction and hence the water:milk ratio. Given Data / Assumptions: Milk cost = ₹ 0.50 per L. Selling price = ₹ 0.56 per L. Profit = 40% ⇒ SP = 1.4 * cost_per_liter_of_mixture. Concept / Approach: cost_per_liter_mixture = 0.56 / 1.4 = ₹ 0.40. If milk fraction is m, then cost_per_liter_mixture = 0.50 * m (water is free). Solve for m, then water fraction = 1 − m, and ratio water:milk follows. Step-by-Step Solution: 0.50 * m = 0.40 ⇒ m = 0.80.Water fraction = 1 − 0.80 = 0.20.Water : Milk = 0.20 : 0.80 = 1 : 4. Verification / Alternative check: Check profit: cost 0.40 ⇒ profit 40% ⇒ 0.16 ⇒ SP = 0.56, matching the statement. Why Other Options Are Wrong: Other ratios correspond to milk fractions different from 0.80 and would not satisfy the 40% profit at ₹ 0.56/L. Common Pitfalls: Treating ₹ 500 per kiloliter incorrectly (it is ₹ 0.50 per liter); forgetting water’s zero cost. Final Answer: 1 : 4

Question 3

Mixing bananas of two price types to hit a target price: How many bananas at 5 for ₹ 1.20 (₹ 0.24 each) should be mixed with 300 bananas at 6 for ₹ 2.10 (₹ 0.35 each) so that the mixture is worth ₹ 3.60 per dozen (₹ 0.30 each)?

Accepted Answer

Introduction / Context: This is a weighted-average price problem. The target price per unit dictates the total value for the combined quantity. Let the number of cheaper bananas be x, write total value and total items, and enforce the target average price per banana (₹ 0.30). Given Data / Assumptions: Cheaper: ₹ 0.24 each; quantity = x. Costlier: ₹ 0.35 each; quantity = 300. Target average = ₹ 0.30 each. Concept / Approach: (0.24x + 0.35*300) / (x + 300) = 0.30. Solve for x to determine how many cheaper bananas to add. Step-by-Step Solution: 0.24x + 105 = 0.30x + 90105 − 90 = 0.30x − 0.24x ⇒ 15 = 0.06xx = 250. Verification / Alternative check: Total items = 550; total value = 0.24*250 + 0.35*300 = 60 + 105 = ₹ 165; 165/550 = ₹ 0.30 each ⇒ ₹ 3.60 per dozen. Why Other Options Are Wrong: Any other x changes the average above or below ₹ 0.30; the equation balances only at x = 250. Common Pitfalls: Mixing “per dozen” with “per banana” inconsistently; always convert to a per-unit basis for equations. Final Answer: 250

Question 4

Converting fresh fruit to dry fruit by preserving solids: Fresh fruit contains 72% water (28% solids). Dry fruit contains 20% water (80% solids). From 100 kg of fresh fruit, how many kilograms of dry fruit can be obtained (solids conserved)?

Accepted Answer

Introduction / Context: Drying removes water but keeps the solids mass unchanged. Therefore, equate the solids in the fresh fruit to the solids in the final dry fruit to compute the final dry mass. Given Data / Assumptions: Fresh fruit = 100 kg with 28% solids ⇒ 28 kg solids. Dry fruit has 80% solids (since water = 20%). Concept / Approach: Let W be the dry fruit mass. Then solids conservation gives 0.80 * W = 28. Solve for W directly. Step-by-Step Solution: 0.80W = 28 ⇒ W = 28 / 0.80 = 35 kg. Verification / Alternative check: Water in dry fruit = 20% of 35 = 7 kg; solids 28 kg, equal to fresh solids, consistent. Why Other Options Are Wrong: 30–33 kg would imply solids below 28 kg at 80% solids content; 32 kg corresponds to only 25.6 kg solids. Common Pitfalls: Averaging percentages or changing solids content; the only invariant is solids mass. Final Answer: 35 kg

Question 5

Successive dilution fraction removed after five replacements: From a vessel of alcohol, one-fifth of the contents is removed and replaced with water. This operation is repeated five times. What fraction of the original alcohol has been removed?

Accepted Answer

Introduction / Context: Each cycle retains a fixed fraction of the alcohol. If one-fifth is removed, four-fifths remain. After n cycles, the remaining fraction is (4/5)^n, so the removed fraction is 1 − (4/5)^n. Apply this with n = 5. Given Data / Assumptions: Removal each time = 1/5 of whatever is present at that time. n = 5 successive replacements. Concept / Approach: Remaining fraction = (4/5)^5 = 1024/3125. Therefore removed fraction = 1 − 1024/3125 = 2101/3125. Step-by-Step Solution: (4/5)^5 = 1024/3125.Removed = 1 − 1024/3125 = (3125 − 1024)/3125 = 2101/3125. Verification / Alternative check: Decimal check: 1024/3125 ≈ 0.32768 ⇒ removed ≈ 0.67232; aligns with five successive 20% removals. Why Other Options Are Wrong: 1024/3125 is the remaining fraction, not removed; 1024/2101 is not meaningful in this context; other numerators/denominators do not match (4/5)^5. Common Pitfalls: Subtracting 1/5 each time linearly instead of compounding the retention factor. Final Answer: 2101/3125

Question 6

Mixing sugar syrups to target 10% concentration: Solution A is 15% sugar; solution B is 5% sugar. How many liters of solution B must be added to 20 L of solution A to obtain a 10% sugar mixture?

Accepted Answer

Introduction / Context: Use sugar mass balance: sugar from A plus sugar from B must equal 10% of the total final volume. One linear equation in the unknown liters of B is sufficient to solve the problem cleanly and exactly. Given Data / Assumptions: A: 20 L at 15% ⇒ sugar = 3 L. B: x L at 5% ⇒ sugar = 0.05x L. Final: (20 + x) L at 10% ⇒ sugar = 0.10(20 + x) L. Concept / Approach: Set 3 + 0.05x = 0.10(20 + x) and solve for x. This equates total sugar before and after mixing. Step-by-Step Solution: 3 + 0.05x = 2 + 0.10x1 = 0.05x ⇒ x = 20 L. Verification / Alternative check: Final volume 40 L; sugar = 3 + 1 = 4 L ⇒ 4/40 = 10%, as required. Why Other Options Are Wrong: 10 L or 15 L would yield sugar fractions above 10%; 5 L is too small; only 20 L balances at exactly 10%. Common Pitfalls: Averaging percentages instead of doing mass balance; forgetting to convert percent to fractions in equations. Final Answer: 20 L

Question 7

Combining two wine–water mixtures and adding extra water: Cask 1 has 48 L with wine:water = 13 : 7. Cask 2 has 42 L with wine:water = 18 : 17. After mixing their contents and adding 20 L of water, what is the final wine:water ratio?

Accepted Answer

Introduction / Context: Convert ratios to absolute amounts using the given volumes, then add the wine and water from both casks. Finally add 20 L water and reduce the resulting wine:water pair to the simplest integer ratio. Given Data / Assumptions: Cask 1 (48 L): wine = 13/20*48 = 31.2 L; water = 16.8 L. Cask 2 (42 L): wine = 18/35*42 = 21.6 L; water = 20.4 L. Extra water added = 20 L. Concept / Approach: Sum wine and water separately, then express as a reduced ratio. The added water changes only the denominator (water amount), not wine. Step-by-Step Solution: Total wine = 31.2 + 21.6 = 52.8 L.Total water before addition = 16.8 + 20.4 = 37.2 L.After adding 20 L water ⇒ water = 57.2 L.Wine : Water = 52.8 : 57.2 = 528 : 572 = 12 : 13. Verification / Alternative check: Divide both by 44: 528/44 = 12; 572/44 = 13, so the ratio simplifies cleanly. Why Other Options Are Wrong: 13 : 12 is the inverse; 21 : 31 and 31 : 21 are unrelated to the computed totals. Common Pitfalls: Averaging ratios directly; always compute actual amounts before reducing to a ratio. Final Answer: 12 : 13

Question 8

Adjusting milk:water ratio by adding water: A 729-ml mixture has milk:water = 7 : 2. How much additional water must be added to make the ratio 7 : 3?

Accepted Answer

Introduction / Context: Ratios convert to absolute amounts when total volume is known. Compute current milk and water, then find the target water amount given a fixed milk quantity and new target ratio, and finally the extra water required. Given Data / Assumptions: Total = 729 ml; milk:water = 7 : 2 ⇒ total parts = 9. Milk = 7/9 * 729 = 567 ml. Current water = 729 − 567 = 162 ml. Target ratio milk:water = 7 : 3 with the same milk amount (no milk added). Concept / Approach: For ratio 7 : 3 and milk = 567 ml, water must be (3/7)*567 = 243 ml. The additional water is target minus current. Step-by-Step Solution: Target water = (3/7)*567 = 243 ml.Extra water needed = 243 − 162 = 81 ml. Verification / Alternative check: New total = 567 + 243 = 810 ml; ratio 567:243 reduces to 7:3, correct. Why Other Options Are Wrong: Offered numerical choices (600, 710, 520) are far from the computed 81 ml; hence “None of these” is correct for the provided list. Common Pitfalls: Trying to maintain total volume 729 ml; note that adding water increases total volume but does not change milk amount. Final Answer: None of these

Question 9

Two-price currants blended to a target price: Currants at ₹ 50/kg are mixed with currants at ₹ 90/kg to make 17 kg of mixture worth ₹ 70/kg. How many kilograms of each type are used?

Accepted Answer

Introduction / Context: This is a direct weighted average. Let x be the mass at ₹ 50/kg, then (17 − x) is at ₹ 90/kg. Enforce the total value equal to 17 * ₹ 70, and solve x. The solution reveals whether the mix is equal or skewed to either side. Given Data / Assumptions: Total mass = 17 kg; target price = ₹ 70/kg ⇒ total value = ₹ 1190. Costs: ₹ 50/kg and ₹ 90/kg. Unknowns: x kg at ₹ 50/kg; 17 − x kg at ₹ 90/kg. Concept / Approach: Equation: 50x + 90(17 − x) = 1190. Solve for x. The complementary mass is 17 − x. Step-by-Step Solution: 50x + 1530 − 90x = 1190−40x = −340 ⇒ x = 8.5 kg.Hence both are 8.5 kg each. Verification / Alternative check: Value = 8.5*50 + 8.5*90 = 425 + 765 = 1190; average = 1190/17 = ₹ 70/kg, correct. Why Other Options Are Wrong: Other splits do not yield ₹ 1190 total value at 17 kg; the average would shift away from ₹ 70/kg. Common Pitfalls: Using ratio 50:90 directly on masses without enforcing the 17-kg total and the ₹ 70/kg target average. Final Answer: 8 1/2 kg of each

Question 10

The ratio of milk to water is 5:4. If two litres of water is added, the ratio becomes 10:9, then find the new amount of water in the mixture.

Accepted Answer

Let initial quantity of milk & water in the mixture are 5x & 4x 2 litres water is added. Hence, New quantity of milk & water will be 5x & (4x+2) ATQ, 5x/(4x+2) = 10/9 x/(4x+2) = 2/9 9x = 8x + 4 x=4 Hence, New amount of water in the mixture = 4x+2 = (4*4) + 2 = 18 litres.

Question 11

A vessel, full of water, weighs 24 kg. When the vessel is 1/4 full, it weighs 9 kg. Find the weight of empty vessel.

Accepted Answer

Let the weight of vessel be x Weight of water = 24 – x ⇒ (24 – x)/4 + x = 9 ⇒ 24 – x + 4x = 36 ⇒ 3x = 12 ⇒ x = 4 kg

Question 12

Three bottles of equal capacity containing mixture of milk and water in ratio 2 : 5, 3 : 4 and 4 : 5 respectively. These three bottles are emptied into a large bottle. What will be the ratio of milk and water respectively in the large bottle?

Accepted Answer

73 : 116

Question 13

In a mixture of milk and water, there is 19% milk. If the quantity of milk in the mixture is 437 mililitres, then what is the total quantity (in millilitres) of mixture?

Accepted Answer

2300

Question 14

In a mixture of 25 litres, the ratio of milk to water is 4:1. Another 3 litres of water is added to the mixture. The ratio of milk to water in the new mixture is

Accepted Answer

5:2

Question 15

Two bottles contain acid and water in the ratio 2 : 3 and 1 : 2 respectively. These are mixed in the ratio 1 : 3. What is the ratio of acid and water in the new mixture ?

Accepted Answer

7:13

Question 16

In two types of brass, the ratios of Copper to Zinc are 8:3 and 15:7 respectively. If the two types of brass be melted and mixed in the ratio 5:2 a new type of brass is obtained. The ratio of Copper to Zinc in this new type of brass is

Accepted Answer

5:2

Question 17

A jar contains a blend of a fruit juice and water in the ratio 5 : x. When 1 litre of water is added to 4 litres of the blend the ratio of fruit juice to water becomes 1 : 1. What is the value of x?

Accepted Answer

3

Question 18

If a dairy mixes cow's milk which contains 10% fat with buffalo's milk which contains 20% fat, then the resulting mixture has fat (120/7)% of fat. What ratio was the cow's milk mixed with buffalo's milk?

Accepted Answer

2 : 5

Question 19

In what ratio should tea costing Rs 300/kg be mixed with tea costing Rs 200/kg so that the cost of the mixture is Rs 225/kg?

Accepted Answer

1 : 3

Question 20

In a container, there is 960 ltr of pure milk from which 48 ltr of milk is replaced with 48 ltr of water, again 48 ltr milk is replaced by same amount of water, as this process is done once more. Now, what is the amount of pure milk?

Accepted Answer

Amount of pure milk= a(1 –b/a)n(n = 3, a = pure milk and b = amount replaced)= 960 (1 –48/960)^3= 960(1-1/20)^3= 960 * 19/20 * 19/20*19/20= 823.08 lt.

Alligation or Mixture Questions