📘 Quiz

Test your skills & challenge yourself 🚀

Question 1 / 20
1:00
1
Mixture 1 has wine and water in a 3:2 ratio, and Mixture 2 has them in a 4:5 ratio. If 3 litres of Mixture 1 are used, how many litres of Mixture 2 must be added so that the final combined mixture has equal quantities of wine and water?
0:00
Solution: Step 1: Calculate the quantities of wine and water in 3 litres of Mixture 1 (ratio 3:2): * Wine in Mixture 1 = (3 / (3+2)) * 3 = (3/5) * 3 = 9/5 litres. * Water in Mixture 1 = (2 / (3+2)) * 3 = (2/5) * 3 = 6/5 litres. Step 2: Let 'x' be the quantity of Mixture 2 (ratio 4:5) to be added. * Wine in Mixture 2 = (4 / (4+5)) * x = 4x/9 litres. * Water in Mixture 2 = (5 / (4+5)) * x = 5x/9 litres. Step 3: The resulting mixture should have equal quantities of wine and water. Set up the equation for total wine and total water: * Total Wine = (9/5) + (4x/9) * Total Water = (6/5) + (5x/9) * (9/5) + (4x/9) = (6/5) + (5x/9) Step 4: Solve for x: * 9/5 - 6/5 = 5x/9 - 4x/9 * 3/5 = x/9 * x = (3/5) * 9 = 27/5 litres. Step 5: Convert the answer to a mixed fraction: 27/5 litres = 5 and 2/5 litres.
2
In what proportion should a grocer mix two varieties of tea, priced at Rs. 60 per kg and Rs. 65 per kg, so that he earns a 10% profit by selling the mixture at Rs. 68.20 per kg?
0:00
Solution: Step 1: Calculate the Cost Price (C.P.) of the mixture. Selling Price (S.P.) of 1 kg of mixture = Rs. 68.20. Profit = 10%. C.P. = (100 / (100 + Profit%)) * S.P. C.P._mixture = (100 / (100 + 10)) * 68.20 = (100 / 110) * 68.20 = (10 / 11) * 68.20. C.P._mixture = 682 / 11 = Rs. 62 per kg. Step 2: Apply the Rule of Alligation. Cost of 1st kind of tea = Rs. 60 per kg. Cost of 2nd kind of tea = Rs. 65 per kg. Mean C.P. of the mixture = Rs. 62 per kg. Cost 1 (Rs. 60) Cost 2 (Rs. 65) \ / Mean (Rs. 62) / \ Difference 2 Difference 1 Difference 1 = |Mean C.P. - Cost 2| = |62 - 65| = |-3| = 3. Difference 2 = |Mean C.P. - Cost 1| = |62 - 60| = 2. Step 3: Determine the ratio of quantities of the 1st kind to the 2nd kind. The required ratio of quantities of the 1st kind to the 2nd kind is (Difference 1) : (Difference 2). Ratio = 3 : 2. So, the teas must be mixed in the ratio 3 : 2.
3
A 32-litre solution contains acid and water in a 5:3 ratio. If 12 litres of this solution are drawn out and 7.5 litres of water are then added, what will be the acid-to-water ratio in the final solution?
0:00
Solution: Step 1: Calculate initial quantities of acid and water. Total volume = 32 litres Ratio (Acid:Water) = 5:3 (Total 8 parts) Initial Acid = (5/8) * 32 = 20 litres Initial Water = (3/8) * 32 = 12 litres Step 2: Calculate quantities of acid and water removed when 12 litres of solution are taken out. Acid removed = (5/8) * 12 = 7.5 litres Water removed = (3/8) * 12 = 4.5 litres Step 3: Calculate the remaining quantities of acid and water in the vessel. Remaining Acid = 20 - 7.5 = 12.5 litres Remaining Water = 12 - 4.5 = 7.5 litres Step 4: Add 7.5 litres of water to the remaining solution. New Acid quantity = 12.5 litres (acid quantity remains unchanged by adding water) New Water quantity = 7.5 + 7.5 = 15 litres Step 5: Determine the ratio of acid to water in the resulting solution. Required Ratio = New Acid : New Water = 12.5 : 15 Step 6: Simplify the ratio. Multiply by 2 to remove decimals: 25 : 30 Divide by 5: 5 : 6 The ratio of acid and water in the resulting solution is 5 : 6.
4
Vessel A contains a mixture of spirit and water in a 5 : 2 ratio, while Vessel B contains them in a 7 : 6 ratio. In what proportion must the mixtures from Vessels A and B be combined to produce a new mixture in Vessel C that has spirit and water in an 8 : 5 ratio?
0:00
Solution: Step 1: Determine the fraction of spirit in each vessel's mixture and the target mixture. Vessel A (C1): Spirit fraction = 5 / (5 + 2) = 5/7. Vessel B (C2): Spirit fraction = 7 / (7 + 6) = 7/13. Target Mixture C (Cm): Spirit fraction = 8 / (8 + 5) = 8/13. Step 2: Apply the Rule of Alligation using these spirit fractions (or water fractions; the ratio will be the same). C1 (5/7) C2 (7/13) \ / Mean (8/13) / \ (8/13 - 7/13) = 1/13 (5/7 - 8/13) = (65-56)/91 = 9/91 Step 3: The ratio of quantities taken from Vessel A (Q1) to Vessel B (Q2) is the cross-difference ratio: Q1 : Q2 = (Cm - C2) : (C1 - Cm) Q1 : Q2 = (1/13) : (9/91) Step 4: To simplify the ratio, find a common denominator. The least common multiple of 13 and 91 is 91. Q1 : Q2 = (7/91) : (9/91) Step 5: The required mixing ratio of mixtures from Vessel A to Vessel B is 7 : 9.
5
A trader blends two types of lentils priced at Rs. 170 per kg and Rs. 180 per kg with a third type in a 1:2:3 ratio. If the final mixture costs Rs. 173 per kg, what is the price per kg of the third type of lentils?
0:00
Solution: Step 1: Let the price of the third variety of lentils = x per kg Step 2: The first variety of Lentils = Rs.170/kg, second = Rs.180/kg Step 3: All three varieties are mixed in the ratio 1 : 2 : 3 Step 4: The mixture cost is Rs.173/kg. Step 5: Assume each part in the ratio is 1 kg. Step 6: Set up equation: [170(1)+180(2)+x(3)] / 6 = 173 Step 7: Simplify: 170 + 360 + 3x = 173 * 6 Step 8: Further simplify: 530 + 3x = 1038 Step 9: Solve for x: 3x = 508, x = 169.33
6
A container was 2/3 full with a substance. Person A removed 18 units of the substance, and Person B added 12 units. After these changes, the container was 3/5 full. What is the total capacity of the container in units?
0:00
Solution: Step 1: Let the total capacity of the container = x units Step 2: Initial amount of substance = (2/3)x Step 3: After removal by Person A: (2/3)x - 18 Step 4: After addition by Person B: (2/3)x - 18 + 12 = (2/3)x - 6 Step 5: Final amount = (3/5)x Step 6: Set up equation: (2/3)x - 6 = (3/5)x Step 7: Multiply through by 15 to clear fractions: 10x - 90 = 9x Step 8: Simplify: 10x - 9x = 90 Step 9: x = 90 units
7
Determine the mixing ratio for a 30% alcohol solution and a 50% alcohol solution to yield a final mixture with 45% alcohol strength.
0:00
Solution: Step 1: Identify the concentrations of the two components: First component (C1) = 30% alcohol. Second component (C2) = 50% alcohol. Step 2: Identify the desired mean concentration (Cm) = 45% alcohol. Step 3: Apply the Rule of Alligation. Step 4: Difference 1 = `Cm - C1 = 45 - 30 = 15`. Step 5: Difference 2 = `C2 - Cm = 50 - 45 = 5`. Step 6: The ratio of quantities (Q1 : Q2) = `Difference 2 : Difference 1 = 5 : 15`. Step 7: Simplify the ratio: `5 : 15 = 1 : 3`.
8
A 729-litre mixture of milk and water has a milk-to-water ratio of 7:2. To achieve a new mixture with a 7:3 milk-to-water ratio, what quantity of water must be introduced?
0:00
Solution: Step 1: Calculate initial quantities of milk and water. Step 2: Total mixture = 729 litres. Milk:Water ratio = 7:2. Step 3: Sum of ratio parts = `7 + 2 = 9`. Step 4: Quantity of milk = `(7/9) * 729 = 7 * 81 = 567` litres. Step 5: Quantity of water = `(2/9) * 729 = 2 * 81 = 162` litres. Step 6: Let `x` litres of water be added. Step 7: The new milk quantity remains 567 litres. The new water quantity becomes `162 + x` litres. Step 8: The new ratio of milk to water is 7:3. Step 9: Set up the proportion: `567 / (162 + x) = 7 / 3`. Step 10: Cross-multiply: `567 * 3 = 7 * (162 + x)`. Step 11: `1701 = 1134 + 7x`. Step 12: `7x = 1701 - 1134 = 567`. Step 13: `x = 567 / 7 = 81`. Step 14: Therefore, 81 litres of water must be added.
9
Stainless steel of two varieties has chromium-to-steel ratios of 2:11 and 5:21. In what ratio should these two varieties be combined to produce a new mixed type where the chromium-to-steel ratio is 7:32?
0:00
Solution: Step 1: Convert the ratios of chromium to steel into fractions of chromium in the mixture. Type 1 chromium fraction = 2 / (2 + 11) = 2/13 Type 2 chromium fraction = 5 / (5 + 21) = 5/26 Desired mixture chromium fraction = 7 / (7 + 32) = 7/39 Step 2: Apply the rule of alligation. Let the proportion of Type 1 to Type 2 be P1 : P2. P1 is proportional to |Desired - Type 2| = |7/39 - 5/26| P2 is proportional to |Type 1 - Desired| = |2/13 - 7/39| Step 3: Perform fraction subtraction. |7/39 - 5/26| = |(14 - 15) / 78| = 1/78 |2/13 - 7/39| = |(6 - 7) / 39| = 1/39 Step 4: Determine the ratio P1 : P2. P1 : P2 = (1/78) : (1/39) Multiply by 78 to simplify: P1 : P2 = 1 : 2 Step 5: Conclude. The two types should be mixed in the proportion 1 : 2.
10
A 60-litre mixture of milk and water contains 10% water. How much additional water should be introduced into this mixture so that the water content rises to 25%?
0:00
Solution: Method 1: Using quantities directly. Step 1: Calculate the initial quantities of water and milk. Total mixture = 60 litres Initial water = 10% of 60 = 6 litres Initial milk = 60 - 6 = 54 litres Step 2: Let 'W' be the quantity of water added. New total water = 6 + W litres New total mixture = 60 + W litres Milk quantity remains constant = 54 litres Step 3: Formulate an equation based on the new water percentage (25%). The new percentage of water is 25% of the new total mixture, meaning milk is 75%. So, 54 / (60 + W) = 75 / 100 = 3 / 4 Step 4: Cross-multiply and solve for W. 4 * 54 = 3 * (60 + W) 216 = 180 + 3W 3W = 216 - 180 3W = 36 W = 12 litres Step 5: Conclude. 12 litres of water should be added.
11
A container holds a liquid mixture with a water to solvent ratio of 5:7. After spilling 9 liters of the mixture, the container is refilled with water equal to the spilled amount. The new water to solvent ratio becomes 9:7. What was the initial quantity of solvent in the container?
0:00
Solution: Step 1: Let initial mixture be 5x (water) + 7x (solvent) = 12x liters Step 2: After spilling 9 liters, remaining mixture = 12x - 9 Step 3: Water added = 9 liters, new total = 12x - 9 + 9 = 12x Step 4: New water = 5x - (5/12)*9 + 9 = (5x - 4.5 + 9) = 5x + 4.5 Step 5: New ratio 9:7 => (5x + 4.5) / 7x = 9/7 Step 6: Cross-multiply: 7(5x + 4.5) = 9(7x) Step 7: Expand: 35x + 31.5 = 63x Step 8: Rearrange: 31.5 = 28x Step 9: Solve: x = 31.5 / 28 = 1.125 Step 10: Initial solvent = 7x = 7 * 1.125 = 7.875 * 3 = 21 liters
12
A dishonest milkman created a 36-liter mixture by adding 1 liter of water for every 3 liters of milk. If he then adds an additional 15 liters of milk to this mixture, what will be the new ratio of milk to water?
0:00
Solution: Step 1: Determine the initial ratio of water to milk. '1 liter of water for every 3 liters of milk' means Water : Milk = 1 : 3. Step 2: Calculate the initial quantities of water and milk in the 36-liter mixture. Total parts in ratio = 1 (water) + 3 (milk) = 4 parts. Value of one part = 36 liters / 4 parts = 9 liters/part. Initial water = 1 part * 9 liters/part = 9 liters. Initial milk = 3 parts * 9 liters/part = 27 liters. Step 3: Account for the added milk. 15 liters of milk are added, so new milk quantity = 27 + 15 = 42 liters. (The quantity of water remains 9 liters). Step 4: Form the new ratio of milk to water (as requested in the question): New ratio = 42 (milk) : 9 (water). Step 5: Simplify the ratio by dividing both parts by their greatest common divisor (3). 42 / 3 = 14 9 / 3 = 3 Step 6: The new ratio of milk to water is 14:3.
13
A 30-litre mixture contains milk and water in a ratio of 7 : 3. How many litres of water must be added to this mixture so that the resulting solution contains 40% water?
0:00
Solution: Step 1: Calculate the initial quantities of milk and water in the 30-litre mixture. Total ratio parts = 7 + 3 = 10. Initial quantity of milk = (7/10) * 30 litres = 21 litres. Initial quantity of water = (3/10) * 30 litres = 9 litres. Step 2: Let 'a' be the amount of water (in litres) to be added to the mixture. Step 3: After adding 'a' litres of water: Quantity of milk remains 21 litres. Quantity of water becomes (9 + a) litres. Total volume of the new mixture = (30 + a) litres. Step 4: The resultant mixture has 40% water. This means (Quantity of water) / (Total mixture volume) = 40/100 = 2/5. Step 5: Set up the equation: (9 + a) / (30 + a) = 2/5 Step 6: Cross-multiply: 5 * (9 + a) = 2 * (30 + a) Step 7: Expand both sides: 45 + 5a = 60 + 2a Step 8: Gather 'a' terms on one side and constants on the other: 5a - 2a = 60 - 45 3a = 15 Step 9: Solve for 'a': a = 15 / 3 = 5 litres. Step 10: Therefore, 5 litres of water should be added to the mixture.
14
An 80-litre mixture contains milk and water in a 7:3 ratio. How many litres of water must be added to this mixture to change the milk-to-water ratio to 2:1?
0:00
Solution: Step 1: Calculate initial quantity of milk: (7 / (7+3)) * 80 litres = (7/10) * 80 = 56 litres. Step 2: Calculate initial quantity of water: (3 / (7+3)) * 80 litres = (3/10) * 80 = 24 litres. Step 3: Let 'x' litres of water be added. The quantity of milk remains 56 litres. The new quantity of water becomes (24 + x) litres. Step 4: The new ratio of milk to water is 56 / (24 + x) = 2 / 1. Step 5: Cross-multiply the equation: 56 * 1 = 2 * (24 + x). Step 6: Simplify and solve for x: 56 = 48 + 2x => 2x = 56 - 48 => 2x = 8 => x = 4. Step 7: Therefore, 4 litres of water must be added.
15
A vendor mixes a certain quantity of liquid A with 1.5 liters of liquid B. The mixture is sold at Rs.28 per liter, while pure liquid A costs Rs.36 per liter. Determine the original quantity of liquid A.
0:00
Solution: Step 1: Let the quantity of liquid A = x liters Step 2: Set up equation: 36x = 28(x + 1.5) Step 3: Expand: 36x = 28x + 42 Step 4: Simplify: 8x = 42 Step 5: Solve for x: x = 42 / 8 = 5.25 liters
16
A milk vendor has two cans of milk. The first can contains 25% water (and the rest milk), while the second can contains 50% water. How many litres should he mix from each can to obtain 12 litres of a mixture where the ratio of water to milk is 3 : 5?
0:00
Solution: Step 1: Determine the proportion of milk in each can and the desired final mixture. Can 1: 25% water means 75% milk = 3/4 milk. Can 2: 50% water means 50% milk = 1/2 milk. Desired final mixture: Water : Milk = 3 : 5. Total parts = 8. Desired fraction of milk = 5/8. Step 2: Apply the Rule of Alligation using the fraction of milk. Fraction of milk in Can 1: 3/4 Fraction of milk in Can 2: 1/2 Mean fraction of milk (desired): 5/8 Differences for alligation: Difference 1 (Can 2 to Mean) = |5/8 - 1/2| = |5/8 - 4/8| = 1/8. Difference 2 (Can 1 to Mean) = |3/4 - 5/8| = |6/8 - 5/8| = 1/8. Step 3: Determine the ratio of quantities from Can 1 to Can 2. Ratio (Quantity from Can 1) : (Quantity from Can 2) = Difference 1 : Difference 2 = 1/8 : 1/8 = 1 : 1. Step 4: Calculate the quantity from each can for a total of 12 litres. Since the ratio is 1:1, the 12 litres will be equally divided. Quantity from Can 1 = (1 / (1 + 1)) * 12 = 1/2 * 12 = 6 litres. Quantity from Can 2 = (1 / (1 + 1)) * 12 = 1/2 * 12 = 6 litres.
17
A merchant possesses 1000 kg of sugar. He sells a portion of it at an 8% profit and the remaining at an 18% profit. If he realizes a 14% profit on the entire stock, what quantity of sugar was sold at an 18% profit?
0:00
Solution: Step 1: Identify the profit percentages for the two parts and the overall mean profit. Profit on 1st part = 8%. Profit on 2nd part = 18%. Mean profit on the whole = 14%. Step 2: Apply the Rule of Alligation. Profit 1 (8%) Profit 2 (18%) \ / Mean (14%) / \ Difference 2 Difference 1 Difference 1 = |Mean Profit - Profit 2| = |14 - 18| = |-4| = 4. Difference 2 = |Mean Profit - Profit 1| = |14 - 8| = 6. Step 3: Determine the ratio of the quantities of the 1st part to the 2nd part. Ratio (Quantity at 8% profit) : (Quantity at 18% profit) = Difference 1 : Difference 2 = 4 : 6. Step 4: Simplify the ratio. Divide both sides by 2: 2 : 3. Step 5: Calculate the quantity sold at 18% profit. Total quantity of sugar = 1000 kg. The ratio of quantities is 2:3, so there are 2 + 3 = 5 total parts. Quantity sold at 18% profit = (3 / (2 + 3)) * 1000 kg. Quantity sold at 18% profit = (3 / 5) * 1000 kg. Quantity sold at 18% profit = 3 * 200 kg = 600 kg.
18
A dishonest milkman claims to sell milk at its cost price, yet he achieves a 25% profit by adulterating it with water. What is the percentage of water present in this mixture?
0:00
Solution: Step 1: Assume the Cost Price (CP) of 1 litre of pure milk is Rs. 1. Step 2: The milkman sells the mixture at the cost price of milk, so the Selling Price (SP) of 1 litre of the mixture is Rs. 1. Step 3: He earns a profit of 25%. We need to find the Cost Price (CP) of 1 litre of this mixture. CP of mixture = SP * (100 / (100 + Profit%)) CP of mixture = 1 * (100 / (100 + 25)) = 100 / 125 = 4/5 Rs. Step 4: Now, consider the two components of the mixture: pure milk and water. Cost of pure water (C1) = Rs. 0 per litre (as it's free). Cost of pure milk (C2) = Rs. 1 per litre. Mean cost of the mixture (Cm) = Rs. 4/5 per litre. Step 5: Apply the Rule of Alligation to find the ratio of quantities of water to milk: C1 (0) C2 (1) \ / Mean (4/5) / \ (1 - 4/5) = 1/5 (4/5 - 0) = 4/5 Step 6: The ratio of Water to Milk (Q1 : Q2) is (C2 - Cm) : (Cm - C1). Water : Milk = 1/5 : 4/5 Step 7: Simplify the ratio: Water : Milk = 1 : 4. Step 8: The total parts in the mixture are 1 (water) + 4 (milk) = 5 parts. Step 9: Percentage of water in the mixture = (Water parts / Total parts) * 100. Percentage of water = (1/5) * 100 = 20%.
19
A container initially holds 40 litres of milk. 4 litres of milk are removed and replaced with water. This process is repeated two more times. What is the remaining quantity of milk in the container?
0:00
Solution: Step 1: Identify the initial quantity, amount removed, and number of repetitions. Initial quantity of milk (V) = 40 litres. Amount removed and replaced (x) = 4 litres. Total number of times the process is repeated (n) = 1 (initial process) + 2 (further two times) = 3 times. Step 2: Use the formula for remaining quantity after repeated replacement. The quantity of pure liquid left after 'n' operations is given by: Q_final = Q_initial * (1 - x/V)^n. Where: Q_initial = 40 litres, x = 4 litres, V = 40 litres, n = 3. Step 3: Substitute the values into the formula. Q_final = 40 * (1 - 4/40)^3. Q_final = 40 * (1 - 1/10)^3. Q_final = 40 * (9/10)^3. Step 4: Calculate the final quantity. Q_final = 40 * (9 * 9 * 9) / (10 * 10 * 10). Q_final = 40 * (729 / 1000). Q_final = (4 * 729) / 100. Q_final = 2916 / 100. Q_final = 29.16 litres.
20
Vessels A and B contain acid and water in ratios of 4:3 and 5:3, respectively. What ratio of these mixtures should be combined to create a new mixture in Vessel C with an acid-to-water ratio of 3:2?
0:00
Solution: Step 1: Express the acid concentration as a fraction for each vessel and the desired mixture. Vessel A (Acid fraction) = 4 / (4 + 3) = 4/7 Vessel B (Acid fraction) = 5 / (5 + 3) = 5/8 Desired Mixture C (Acid fraction) = 3 / (3 + 2) = 3/5 Step 2: Apply the rule of alligation. Let the proportion of liquid from A to B be P_A : P_B. P_A is proportional to |Desired Acid - Vessel B Acid| = |3/5 - 5/8| P_B is proportional to |Vessel A Acid - Desired Acid| = |4/7 - 3/5| Step 3: Perform fraction subtraction. |3/5 - 5/8| = |(24 - 25) / 40| = 1/40 |4/7 - 3/5| = |(20 - 21) / 35| = 1/35 Step 4: Determine the ratio P_A : P_B. P_A : P_B = (1/40) : (1/35) Multiply by the LCM of 40 and 35, which is 280. P_A : P_B = (280/40) : (280/35) = 7 : 8 Step 5: Conclude. The mixtures should be combined in the ratio 7 : 8.
📊 Questions Status
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20