Question 1

Three mixtures to yield 1:1 syrup–water: From boxes with ratios 2:3 and 3:4, 10 L and 21 L are taken respectively. How much from a 7:5 box must be added to make final ratio 1:1?

Accepted Answer

Introduction / Context: Combine mixtures by tracking absolute amounts of syrup and water. Goal ratio is 1:1 (equal amounts). Given Data / Assumptions: Box1 2:3 (syrup:water), 10 L taken. Box2 3:4, 21 L taken. Box3 7:5, take x L. Concept / Approach: Compute syrup and water from each, sum them, and set syrup = water to solve x. Step-by-Step Solution: Box1 syrup = 2/5 * 10 = 4; water = 6.Box2 syrup = 3/7 * 21 = 9; water = 12.Box3 syrup = 7/12 * x; water = 5/12 * x.Totals: syrup = 4 + 9 + 7x/12 = 13 + 7x/12; water = 6 + 12 + 5x/12 = 18 + 5x/12.Set 13 + 7x/12 = 18 + 5x/12 ⇒ 2x/12 = 5 ⇒ x = 30.But this gives syrup = water + 5, re-check arithmetic: We must equalize totals; actually 13 + 7x/12 = 18 + 5x/12 ⇒ (7x − 5x)/12 = 5 ⇒ 2x/12 = 5 ⇒ x = 30. Now verify totals with x=30:Syrup = 13 + (7/12)*30 = 13 + 17.5 = 30.5; Water = 18 + (5/12)*30 = 18 + 12.5 = 30.5 (balanced). So x = 30 L. Correction and Options: The computed x is 30 L, which is not the given key; therefore we select the closest correct among options: 30 Litres is option D, which matches the derivation. Why Other Options Are Wrong: 20 L, 25 L, 35 L lead to mismatched syrup and water totals. Common Pitfalls: Equating ratios directly instead of equating absolute amounts. Final Answer: 30 Litres

Question 2

Alloy copper:tin = 3:2. After adding 250 g copper, copper becomes twice tin: Find tin (g) in the original alloy.

Accepted Answer

Introduction / Context: This is a ratio adjustment problem: adding a metal shifts the copper:tin relation to a specified post-addition ratio. Given Data / Assumptions: Initial copper:tin = 3:2 ⇒ let copper = 3k, tin = 2k (g). Add 250 g copper; then copper becomes twice tin. Concept / Approach: Form equation: (3k + 250) = 2*(2k) = 4k to solve for k, then compute tin = 2k. Step-by-Step Solution: 3k + 250 = 4k ⇒ k = 250.Tin = 2k = 500 g. Verification / Alternative check: Original copper = 750 g, tin = 500 g; after +250 g copper → copper = 1000 g ⇒ 1000 = 2*500 (true). Why Other Options Are Wrong: 250, 750, 1000 correspond to k or copper amounts; correct tin is 500 g. Common Pitfalls: Setting final ratio as 2:1 of total weight rather than copper to tin only. Final Answer: 500

Question 3

Successive replacement in an 80 L ethanol tank: Remove 20 L and replace with water, repeat once more. How much water is now in the tank?

Accepted Answer

Introduction / Context: We start with pure ethanol and repeatedly remove a fixed volume and replace with water. After two operations, compute ethanol remaining then infer water. Given Data / Assumptions: Total V = 80 L; draw d = 20 L; n = 2 operations. Initially, 100% ethanol. Concept / Approach: Remaining ethanol = V*(1 − d/V)^n = 80*(1 − 20/80)^2 = 80*(3/4)^2. Step-by-Step Solution: Remaining ethanol = 80*(9/16) = 45 L.Therefore, water = 80 − 45 = 35 L. Verification / Alternative check: Sequential: after first step ethanol = 60 L; after second, remove 1/4 of the 60 (15), leaving 45 → water 35. Why Other Options Are Wrong: 40 and 44 misapply single-step removal; 45 is ethanol, not water. Common Pitfalls: Reporting remaining ethanol instead of the asked water volume. Final Answer: 35

Question 4

Alloy cost from metals A and B: A costs Rs 2000/kg, B costs Rs 400/kg, mixed in 3:1 (A:B). Find the cost of 8 kg of alloy.

Accepted Answer

Introduction / Context: Weighted average cost per kg = (3 parts of A at 2000 + 1 part of B at 400)/4 parts. Given Data / Assumptions: Mix ratio A:B = 3:1. Concept / Approach: Cost per kg = (3*2000 + 1*400)/4 = (6000 + 400)/4 = 1600 Rs/kg. Step-by-Step Solution: Cost per kg = Rs 1600.Total for 8 kg = 8 * 1600 = Rs 12800. Note: The correct total is Rs 12800. Options show 6400 also; ensure 8 kg is recognized. Since the computed total is Rs 12800, we select 12800. Why Other Options Are Wrong: 6400 equals 4 kg at Rs 1600/kg; 9800 not aligned; 1600 is per-kg, not total. Common Pitfalls: Stopping at per-kg cost or multiplying wrongly by 4 instead of 8. Final Answer: 12800

Question 5

Acid–water ratio becomes 2:5: A beaker has acid:water = 1:x. When 300 ml of this mixture is mixed with 50 ml water (only), the new ratio is 2:5. Find x.

Accepted Answer

Introduction / Context: Start from a mixture with acid:water = 1:x. Taking 300 ml preserves the ratio inside that sample. Then 50 ml pure water is added; set up the new ratio. Given Data / Assumptions: Initial acid:water = 1:x. From that, 300 ml is taken: acid = 300*(1/(1+x)); water = 300*(x/(1+x)). Add 50 ml water. Concept / Approach: New mixture components: acid stays the same; water increases by 50 ml. New ratio = 2:5. Step-by-Step Solution: Acid A = 300/(1+x).Water W = 300x/(1+x) + 50.Given A:W = 2:5 ⇒ 5A = 2W.5*(300/(1+x)) = 2*(300x/(1+x) + 50)1500/(1+x) = (600x/(1+x)) + 100Multiply both sides by (1+x): 1500 = 600x + 100 + 100x1500 = 700x + 100 ⇒ 700x = 1400 ⇒ x = 2. Recheck: Wait: plugging x=2 gives A = 100; W = 200 + 50 = 250 → 100:250 = 2:5 (works). Option A shows 2, but the original set lists 4 as a candidate too. Correct is x = 2. Why Other Options Are Wrong: 1, 3, 4 do not satisfy the derived proportion with the 50 ml extra water. Common Pitfalls: Forgetting the added water is pure water, not mixture. Final Answer: 2

Question 6

Milk–water mixture parts 11:2: After adding 35 L water, the mixture contains twice as much milk as water. Find original pure milk (L).

Accepted Answer

Introduction / Context: Given a parts ratio and an addition of water, enforce the final condition milk = 2*water to find original quantities. Given Data / Assumptions: Original milk:water = 11:2 ⇒ milk = 11k, water = 2k. Add 35 L water ⇒ final water = 2k + 35. Final condition: milk = 2*(final water). Concept / Approach: 11k = 2*(2k + 35) ⇒ solve k, then compute milk = 11k. Step-by-Step Solution: 11k = 4k + 70 ⇒ 7k = 70 ⇒ k = 10.Original milk = 11k = 110 L. Verification / Alternative check: Original water 20 L; after addition 55 L; milk 110 L is indeed twice water 55 L. Why Other Options Are Wrong: 55 is water after addition; 70, 220 mismatch the given condition. Common Pitfalls: Doubling the original water rather than the final water (which includes the added 35 L). Final Answer: 110

Question 7

In a coloured picture, blue and yellow colours are used in the overall ratio of 4 : 3 (blue : yellow). In the upper half of the picture, the ratio of blue to yellow is 2 : 3. Assuming that the upper and lower halves are of equal area, what is the ra…

Accepted Answer

Introduction: This question tests your understanding of ratio, proportion, and mixing of two components in different parts of a whole. The overall ratio of blue and yellow colours is known, one half has a particular ratio, and you must deduce the ratio for the other half so that the combined picture maintains the overall ratio. Given Data / Assumptions: Overall ratio of blue : yellow in the entire picture = 4 : 3. In the upper half, blue : yellow = 2 : 3. Upper half and lower half have equal area and therefore equal total amount of colour units. Let the ratio in the lower half be x : y (blue : yellow), which we need to find. Concept / Approach: We treat the amount of colour as proportional to area. If each half has the same area, then the total colour units in the upper and lower halves are equal. Let us use a common scaling factor to represent quantities in each half and then combine them to match the overall ratio of 4 : 3. This is a standard mixture and proportion problem. Step-by-Step Solution: Step 1: Let the total number of colour units in the upper half be 5k (since 2 + 3 = 5 parts). Step 2: Then in the upper half, blue = 2k and yellow = 3k. Step 3: Because the lower half has equal area, let its total colour units also be 5k for convenience. Let its blue and yellow be B and Y respectively such that B + Y = 5k. Step 4: Overall blue = 2k + B and overall yellow = 3k + Y, and their ratio must be 4 : 3: (2k + B) : (3k + Y) = 4 : 3. Step 5: Translate this into an equation: (2k + B) / (3k + Y) = 4/3. Step 6: Cross multiply: 3(2k + B) = 4(3k + Y). Step 7: Expand: 6k + 3B = 12k + 4Y. Step 8: Rearrange to group B and Y: 3B − 4Y = 6k. Step 9: Use the relation B + Y = 5k. Solve the two equations: From B + Y = 5k, we get B = 5k − Y. Substitute into 3B − 4Y = 6k: 3(5k − Y) − 4Y = 6k ⇒ 15k − 3Y − 4Y = 6k ⇒ 15k − 7Y = 6k. Step 10: Simplify: 15k − 6k = 7Y ⇒ 9k = 7Y ⇒ Y = 9k/7 and B = 5k − 9k/7 = (35k − 9k)/7 = 26k/7. Step 11: Therefore, the ratio in the lower half is B : Y = 26k/7 : 9k/7 = 26 : 9. Verification / Alternative check: Using upper half (2k, 3k) and lower half (26k/7, 9k/7), total blue = 2k + 26k/7 = (14k + 26k)/7 = 40k/7 and total yellow = 3k + 9k/7 = (21k + 9k)/7 = 30k/7. The overall ratio is 40 : 30 = 4 : 3 as required, so the result is correct. Why Other Options Are Wrong: 1 : 1 or 2 : 1 would give a much higher proportion of blue and would not maintain the overall 4 : 3 ratio. 9 : 26 would produce more yellow than blue in the lower half and disturb the overall blue dominance. 3 : 2 is also inconsistent with the required combined ratio. Common Pitfalls: A common error is to average the ratios directly, which is incorrect since ratios must be combined using weighted sums, not simple averaging. Another mistake is ignoring that the areas of the upper and lower halves are equal, which is crucial for setting up the equations correctly. Final Answer: The ratio of blue to yellow in the lower half is 26 : 9.

Question 8

Tea worth of Rs. 135/kg & Rs. 126/kg are mixed with a third variety in the ratio 1: 1 : 2. If the mixture is worth Rs. 153 per kg, the price of the third variety per kg will be____?

Accepted Answer

Rs. 175.50

Question 9

8 litres are drawn from a cask full of wine and is then filled with water. This operation is performed three more times. The ratio of the quantity of wine now left in cask to that of the water is 16 : 65. How much wine the cask hold originally?

Accepted Answer

Let the quantity of the wine in the cask originally be x litres Then, quantity of wine left in cask after 4 operations = x 1 - 8 x 4 litres ∴ x 1 - 8 x 4 x = 16 81 ⇒ 1 - 8 x 4 = 2 3 4 ⇒ x = 24

Question 10

A can contains a mixture of two liquids A and B in the ratio 7 : 5. When 9 litres of mixture are drawn off and the can is filled with B, the ratio of A and B becomes 7 : 9. How many litres of liquid A was contained by the can initially?

Accepted Answer

Suppose the can initially contains 7x and 5x litres of mixtures A and B respectively Quantity of A in mixture left = 7 x - 7 12 × 9 = 7 x - 21 4 litres. Quantity of B in mixture left = 5 x - 5 12 × 9 = 5 x - 15 4 litres. 7 x - 21 4 5 x - 15 4 + 9 = 7 9 ⇒ 28 x - 21 20 x + 21 = 7 9 ⇒ x = 3 So, the can contained 21 litres of A.

Question 11

A merchant has 1000 kg of sugar part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. The Quantity sold at 18% profit is

Accepted Answer

600 kg

Question 12

A vessel is filled with liquid, 3 parts of which are water and 5 parts of syrup. How much of the mixture must be drawn off and replaced with water so that the mixture may be half water and half syrup?

Accepted Answer

Suppose the vessel initially contains 8 litres of liquid. Let x litres of this liquid be replaced with water. Quantity of water in new mixture = 3 - 3 x 8 + x litres. Quantity of syrup in new mixture = 5 - 5 x 8 litres. 3 - 3 x 8 + x = 5 - 5 x 8 => 5x + 24 = 40 - 5x => 10x = 16 => x = 8/5 So, part of the mixture replaced = 8 5 × 1 8 = 1/5.

Question 13

The ratio of expenditure and savings is 3 : 2 . If the income increases by 15% and the savings increases by 6% , then by how much percent should his expenditure increases?

Accepted Answer

21

Question 14

A mixture of 150 liters of wine and water contains 20% water. How much more water should be added so that water becomes 25% of the new mixture?

Accepted Answer

Number of liters of water in 125 liters of the mixture = 20% of 150 = 1/5 of 150 = 30 liters Let us Assume that another 'P' liters of water are added to the mixture to make water 25% of the new mixture. So, the total amount of water becomes (30 + P) and the total volume of the mixture becomes (150 + P) Thus, (30 + P) = 25% of (150 + P) Solving, we get P = 10 liters

Question 15

How many kilograms of sugar costing Rs. 9 per kg must be mixed with 27 kg of sugar costing Rs. 7 per Kg so that there may be a gain of 10 % by selling the mixture at Rs. 9.24 per Kg ?

Accepted Answer

63 Kg

Question 16

The cost of Type 1 rice is Rs. 15 per kg and Type 2 rice is Rs.20 per kg. If both Type 1 and Type 2 are mixed in the ratio of 2 : 3, then the price per kg of the mixed variety of rice is

Accepted Answer

Rs. 18

Question 17

The diluted wine contains only 8 liters of wine and the rest is water. A new mixture whose concentration is 30%, is to be formed by replacing wine. How many liters of mixture shall be replaced with pure wine if there was initially 32 liters of water…

Accepted Answer

Wine Water 8L 32L 1 : 4 20 % 80% (original ratio) 30 % 70% (required ratio) In ths case, the percentage of water being reduced when the mixture is being replaced with wine. so the ratio of left quantity to the initial quantity is 7:8 Therefore , 7 8 = 1 - K 40 => K = 5 Lit

Question 18

The ratio of petrol and kerosene in the container is 3:2 when 10 liters of the mixture is taken out and is replaced by the kerosene, the ratio become 2:3. Then total quantity of the mixture in the container is:

Accepted Answer

pool : kerosene 3 : 2(initially) 2 : 3(after replacement) R e m a i n i n g Q u a n t i t y I n i t i a l Q u a n t i t y = 1 - R e p l a c e d Q u a n t i t y T o t a l Q u a n t i t y (for petrol) 2 3 = 1 - 10 k => K = 30 Therefore the total quantity of the mixture in the container is 30 liters.

Question 19

A container contains 40 litres of milk.From this container 4 litres of milk was taken out and replaced by water. This process was repeated further two times. How much milk is now contained by the container.

Accepted Answer

Amount of milk left after 3 operations 40 1 - 4 40 3 = 40 * 9 10 * 9 10 * 9 10 = 29 . 16 litres

Question 20

A milk man sells the milk at the cost price but he mixes the water in it and thus he gains 9.09%. The quantity of water in the mixture of 1 liter is :

Accepted Answer

Profit (%) = 9.09 % = 1/11 Since the ratio of water and milk is 1 : 11, Therefore the ratio of water is to mixture = 1:12 Thus the quantity of water in mixture of 1 liter = 1000 x (1/12) = 83.33 ml

Alligation or Mixture Questions