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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
A mixture is said to be symmetrical if it contains liquids P and Q in ratio 1 ∶ 2. 40 liters of a mixture of P and Q is taken in which P and Q are in the ratio 2 ∶ 3. What should be the ratio of P and Q in 20 liters of mixture that when mixed with first mixture results in a symmetrical mixture?1). 1 ∶ 22). 1 ∶ 33). 1 ∶ 44). 1 ∶ 5 |
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Answer» 40 LITERS of a MIXTURE of P and Q is TAKEN in which P and Q are in the ratio 2 ? 3. Amount of P in this mixture = (2/(2+3)) × 40 = 16 liters Amount of Q in this mixture = (3/(2+3)) × 40 = 24 liters Let amounts of P and Q in second mixture be T liters and (20 – T) liters, respectively. To form a symmetrical mixture, ratio of P and Q in resultant mixture should be 1 ? 2. ⇒ (16 + T)/(24 + 20 – T) = ½ ⇒ T = 12/3 = 4 ∴ ratio of P and Q = T/(20 – T) = 4/16 = 1 ? 4 |
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| 2. |
Quantity B: In a 100 L mixture of petrol and spirit, the spirit is only 2% then find amount of spirit to be added to make spirit concentration 30%.1). Quantity A > Quantity B2). Quantity A < Quantity B3). Quantity A ≥ Quantity B4). Quantity A ≤ Quantity B |
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Answer» QUANTITY A: Let AMOUNT of water mixed be ‘x’ L. ⇒ So amount of water before mixing = 10 × 40/100 = 4 L ⇒ Amount of water after mixing = 4 + x ⇒ Amount % = 50 = (4 + x) × 100/ (40 + x) ⇒ x = 32 L Quantity B: Let amount of spirit mixed is ‘x’ L. ⇒ So amount of spirit before mixing = 2 × 100/100 = 2 L ⇒ Amount of spirit after mixing = 2 + x ⇒ Amount % = 30 = (2 + x) × 100/ (100 + x) ⇒ x = 40 L ∴ Quantity A < Quantity B |
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| 3. |
In a vessel, there is a mixture of apple, orange and mango juices in the ratio of 3 : 5 : 4 respectively. A quantity of 12 litres from the mixture is replaced by 8 litres of apple juice. Thereafter the quantities of apple and orange juices in the resultant mixtures become same. Find out the initial quantity of mixture in the vessel.1). 90 litres2). 70 litres3). 60 litres4). 80 litres |
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Answer» Ratio of apple, orange and mango juices = 3 : 5 : 4 Let 12x is the initial quantity, so, ⇒ 3X – 3 + 8 = 5x – 5 ⇒ 2X = 10 ⇒ X = 5 ∴ Initial quantity = 12 × 5 = 60 LITRES |
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| 4. |
A milkman bought milk at Rs. 2.50 per litre. He mixed 4 litres of water with the milk and sold it at Rs. 2.50 per litre, earning a profit of 5%. What was the quantity of milk that he had?1). 65 litres2). 80 litres3). 45 litres4). 56 litres |
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Answer» LET the quantity of milk BOUGHT be x litres Cost of milk = RS. 2.50 per litre ∴ Total cost of milk = 2.50 × x = Rs. 2.5x Water added = 4 litres ∴ Total quantity of the mixture = (4 + x) litres ∴ Total selling PRICE of the mixture = 2.50 × (4 + x) = Rs. (10 + 2.5x) Now, gain = 5% ∴ Gain = (Gain % × Cost price)/100 ⇒ Selling price – Cost price = (5 × 2.5x)/100 ⇒ (10 + 2.5x) – 2.5x = 2.5x/20 ⇒ 10 = 2.5x/20 ⇒ 200 = 2.5x ⇒ x = 80 litres |
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| 5. |
A piece of alloy weighs 8 kg. It contains tin and copper and their weights are in the ratio of 3 ∶ 1. If the amount of tin is reduced by half, what is the new weight of the piece?1). 7 kg2). 6 kg3). 5 kg4). 4 kg |
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Answer» Let the weight of TIN and copper be 3X kg and 1x kg respectively. 3x + 1x = 8 ⇒ x = 2 ∴ Weight of tin = 6 kg and weight of copper = 2 kg New quantity of tin = 3 kg ∴ New weight of the PIECE = 3 kg + 2 kg = 5 kg |
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| 6. |
In what ratio must a mixture of 15% spirit strength be mixed with that of 25% spirit strength to get a mixture of 21% spirit strength? 1). 1 : 32). 2 : 3 3). 3 : 4 4). 4 : 5 |
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Answer» Solution: Let the RATIO of the MIXING be x:y 15% of x and 25% of y MUST be mixed to give 21% of (x+y). (15x/100) + (25y/100) = (21/100)(x + y) (15x + 25y)/100 = 21(x +y)/100 15x + 25y = 21x + 21y 21x - 15x = 25y - 21y 6x = 4y x/y = 4/6 x : y = 2 : 3 So, the correct OPTION is 2). 2:3 |
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| 7. |
There are two mixtures of honey and water in which the ratio of honey and water is 1 : 3 and 3 : 1 respectively. Two litres are drawn from the first mixture and 3 litres from the second mixture, and it is mixed to form another mixture. What is the ratio of honey and water in it?1). 106 : 692). 103 : 723). 89 : 864). 11 : 9 |
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Answer» The percentage of honey in the FIRST mixture = 1/4 × 100 = 25% The percentage of honey in the second mixture = 3/4 × 100 = 75% Given that, 2 litres are drawn from the first mixture and 3 litres from the second mixture, then Percentage of honey in the NEW mixture $(= \frac{{2\; \times \;25\; + \;3\; \times \;75}}{5}\; = \frac{{275}}{5} = 55\% )$ Then, the ratio of honey and water in the new mixture would be 55 : (100 – 55) = 55 : 45 = 11 : 9 |
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| 8. |
1). 12 litres2). 14 litres3). 10 litres4). 13 litres |
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Answer» Given, TWO vessels CONTAIN mixtures of milk and water in the ratio of 8 : 1 and 1 : 5 respectively. Amount of water in 1st vessel = 1/9 Amount of water in 2nd vessel = 5/6 Final mixture has half milk and half water Amount of water in 3rd vessel = ½ Let the volume taken from 1st vessel be x, then the volume of mixture is taken from 2nd vessel = 26 – x $(\begin{array}{l} \therefore \;\frac{1}{9}\; \times \;x\; + \;\frac{5}{6}\left( {26 - x} \right)\; = \;\frac{1}{2}\; \times \;26\\ \Rightarrow \;\frac{{5\; \times \;26}}{6}\; + \;\frac{x}{9} - \frac{{5x}}{6}\; = \;\frac{{26}}{2}\\ \Rightarrow \;\frac{{13x}}{{18}}\; = \;\frac{{26}}{3} \end{array})$ ⇒ x = 12 LITRES Volume drawn from 2nd vessel = 26 – 12 = 14 litres |
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| 9. |
A medicine is made of chemical and water. Quantity of the chemical in the medicine is \(11\frac{1}{9}{\rm{\% \;}}\) is more than that of the water. The chemical is made from two different liquids A and B. Quantity of liquid A in the chemical is \(11\frac{1}{9}{\rm{\% }}\) less than that of liquid B. Find the ratio of liquid A, liquid B and water in the medicine.1). 35 ∶ 45 ∶ 1112). 87 ∶ 78 ∶ 1503). 80 ∶ 90 ∶ 1534). 70 ∶ 80 ∶ 153 |
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Answer» Let the QUANTITY of the water in the medicine be 9x So, quantity of the chemical in the medicine = 9x × (1 + 1/9) = 10x Ratio of chemical to water in the medicine = 10x? 9x = 10 ? 9 Let the quantity of the liquid B in the chemical be 9Y So, quantity of the liquid A in the chemical = 9y × 8/9 = 8Y Ratio of liquid A to liquid B in the chemical = 8y? 9y = 8 ? 9 Ratio of chemical to water = 170 ? 153 Ratio of liquid A to liquid B = 80 ? 90 ∴ ratio of liquid A, liquid B and water in the medicine = 80 ? 90 ? 153 |
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| 10. |
1200 ml of mixture contains milk and water in the ratio of 5 : 3. What amount of water should be added to make the ratio 5 : 4?1). 150 ml2). 185 ml3). 135 ml4). 250 ml |
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Answer» 1200 ml mixture contains milk and water in ratio of 5:3 ∴ Amount of milk in mixture $(= \frac{5}{{5 + 3}} \times 1200 = \frac{5}{8} \times 1200 = 750\;ml)$ ∴ Amount of water in mixture = 1200 – 750 = 450 ml Let’s assume that ‘x’ ml of water is ADDED to 1200 ml of mixture so that the ratio of milk to water becomes 5:4. The TOTAL amount of milk does not change. Total amount of water = 450 + x ? the required ratio of milk to water is 5:4, $(\Rightarrow \frac{{750}}{{450\; + \;x\;}} = \frac{5}{4})$ ⇒ 4 × (750) = 5 × (450 + x) ⇒ 3000 = 2250 + 5x ⇒ 750 = 5x ⇒ x = 750/5 = 150 ml. ∴ 150 ml of water has to be added. |
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| 11. |
The ratio of petrol and kerosene in 3 vessels of equal capacity is 4 ∶ 3, 2 ∶ 3 and 2 ∶ 5 respectively. Find the ratio of petrol and kerosene in the final fixture, if we mix the solutions of 3 vessels in the ratio of 1 ∶ 2 ∶ 1.1). 23 ∶ 472). 29 ∶ 413). 13 ∶ 174). 27 ∶ 43 |
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Answer» Suppose the capacity of each vessel is 70 litres; Since the solutions of 3 vessels are mixed in the ratio of 1 ? 2 ? 1; ∴ QUANTITIES taken from 3 vessels to prepare final mixture must be 35L, 70L and 35L respectively. Since the ratio of PETROL and kerosene in 3 vessels of equal capacity is 4 ? 3, 2 ? 3 and 2 ? 5 respectively; ∴ QUANTITY of petrol in the final vessel = (35 × 4/7) + (70 × 2/5) + (35 × 2/7) = 20 + 28 + 10 = 58 L Quantity of kerosene in the final vessel = (35 × 3/7) + (70 × 3/5) + (35 × 5/7) = 15 + 42 + 25 = 82 L ∴ Ratio of petrol and kerosene in the final fixture = 58 ? 82 = 29 ? 41 |
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| 12. |
1). \(7\frac{4}{5}\)%2). \(5\frac{4}{5}\)%3). \(3\frac{5}{7}\)%4). \(7\frac{9}{5}\)% |
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Answer» Given, two litres of a mixture of spirit and water contains 12% water. Volume of water = 12% of 2 = 0.24 litres Now, this is added to 3 litres of another mixture of spirit and water CONTAINING 5% of water. Total volume of water = 0.24 + 5% of 3 = 0.39 litres % of water in the mixture $(= \;\frac{{0.39}}{5} \times 100\% \; = \;\frac{{39}}{5}\% \; = \;7\frac{4}{5}\% )$ |
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| 13. |
20 kg of a mixture of wheat and husk contains 5% husk. How many kg more of husk must be added to make the husk content 20% in the new mixture?1). 2.752). 3.753). 4.754). 5.75 |
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Answer» GIVEN, Husk in 20 KG of mixture = 5% of 20 = 1 kg Let x kg husk is been ADDED According to the question, ⇒ 1 + x = 20% of (20 + x) ⇒ 1 + x = (1/5) × (20 + x) ⇒ 5 + 5x = 20 + x ⇒ 4x = 15 ⇒ x = 15/4 = 3.75 ∴ 3.75 kg of husk has been added |
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| 14. |
Manoj is making lemonade at the side of the street. If he wants to earn 15% profit on selling the juice at the cost price of the lemon, what should be the ratio of lemon juice and water?1). 3 ∶ 202). 10 : 113). 20 ∶ 234). 23 ∶ 20 |
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Answer» Let the cost price of pure LEMON JUICE be Rs. 1 per liter. Cost price of water is Rs. 0 per liter. Selling price of the mixture is also Rs. 1 per liter. Gain = 15% Let X be the cost price of the mixture. (1 – x)/x × 100 = 15 ⇒ x = 1/1.15 = 100/115 = 20/23 ⇒ 3/23 ? 20/23 = 3 ? 20 |
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| 15. |
1). Rs. 482). Rs. 503). Rs. 494). Rs. 47 |
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Answer» ⇒ Total cost of TEA = (45 × 10) + (50 × 8) ⇒ Total cost of tea = Rs. 850 ⇒ Total WEIGHT of tea = 18 kg ⇒ Profit = Rs. 32 ⇒ Selling Price of 18 kg tea = 850 + 32 ⇒ Selling Price of 18kg tea = Rs. 882 ⇒ Selling Price of 1kg tea = 882/18 ∴ Selling Price of 1kg tea = Rs. 49 |
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| 16. |
1). 8 litres2). 10 litres3). 14 litres4). 22 litres |
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Answer» RATIO of new mixture = 2 ? 1 Quantity of FIRST mixture in 30 LITRES of new mixture = 2/3 × 30 = 20 litres Ratio of first mixture = 4 ? 1 Quantity of LIQUID A in 20 litres of first mixture = 4/5 × 20 = 16 litres Quantity of second mixture in 30 litres of new mixture = 1/3 × 30 = 10 litres Ratio of second mixture = 3 ? 2 Quantity of liquid A in 10 litres of second mixture = 3/5 × 10 = 6 litres ∴ Quantity of liquid A in 30 litres of new mixture = 16 + 6 = 22 litres |
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| 17. |
20 L of a mixture contains alcohol and water in the ratio 2 : 3. If 4 L of water is mixed in it, the percentage of alcohol in the new mixture will be1). 502). 253). 204). 33.33% |
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Answer» Amount of ALCOHOL in the mixture = 2/5 × 20 = 8 L And amount of water in the mixture = 20 – 8 = 12 L Given, 4 L of water is added. HENCE, Amount of water in the new mixture = 16 L Hence, the percentage of alcohol in the new mixture = $(\frac{8}{{24}} \TIMES 100 = 33\frac{1}{3}\%)$ |
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| 18. |
A shopkeeper makes varying profits on selling certain amount of sugar weighing 84 kgs. On certain part of the total quantity, he makes a profit of 9% profit and rest at 16% profit. If the overall gain for the shopkeeper is 12% profit on the whole quantity, then find how much is sold at 16% profit?1). 36 kgs2). 48 kgs3). 66 kgs4). None of these |
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Answer» USING the rule of allegation, if two quantities are mixed, then ⇒ Quantity of FIRST part/Quantity of SECOND part = (Profit percentage of second part – Mean Profit)/(Mean profit percentage – Profit percentage of first part) ⇒ Substituting the values, Quantity of first part/Quantity of second part = (16 – 12)/(12 – 9) ⇒ Quantity of first part/Quantity of second part = 4/3 ⇒ Ratio is 4 : 3 ⇒ Quantity sold at 16% profit = [3/(4 + 3)] × 84 kg ⇒ (3/7) × 84 kg ⇒ 36 kg ∴ Quantity sold at 16% profit = 36 kgs |
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| 19. |
Sagar sold total 240 groceries. He sold bottles at 13% loss and lunchbox at profit of 35% thus he gains 15% on the whole. How much bottles did he has sold?1). 802). 1203). 1004). 140 |
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| 20. |
A vessel of 120 litres is filled with milk and water.80% of milk and 40% of water is taken out of the vessel. It is found that the vessel is vacated by 65%.What was the ratio of milk to water in the original mixture? 1). 5 ∶ 32). 6 ∶ 53). 3 ∶ 54). 4 ∶ 3 |
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Answer» Let the volume of milk be = x Then the volume of WATER = 120 - x Percent of vessel empty after drawing = 65% Thus according to the question, we can write the equation ⇒ 0.8x + 0.4(120 - x) = 0.65 × 120 ⇒ 0.8x + 48 - 0.4x = 78 ⇒ 0.4x = 78 - 48 ⇒ 0.4x = 30 ⇒ x = 300/4 = 75 liters ∴ Volume of water = 120 – 75 = 45 liters ∴ RATIO of milk to water =$(\FRAC{{75}}{{45}} = \frac{5}{3})$ Hence, the ratio of milk to water is 5 : 3. |
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| 21. |
1). Rs. 30 per kg2). Rs. 42 per kg3). Rs. 36 per kg4). Rs. 28 per kg |
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Answer» Total PRICE of type 1 wheat = 312 × 72 = RS. 22,464 Total price of type 2 wheat = 138 X x = Rs. 138x Total kg = 312 + 138 = 450 kg Average price = Rs. 59.12 Total price of the MIXTURE = 59.12 × 450 = Rs. 26,604 ∴ Price of type 2 wheat per kg = x = $(\frac{{26604\; - \;22464}}{{138}})$ = Rs. 30 per kg |
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| 22. |
In 1 kg mixture of sand and iron, 23% is iron. How much sand should be added so that the proportion of iron becomes 13%1). 769.2 gms2). 800 gms3). 760.92 gms4). Data insufficient |
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Answer» Thus initially amount of iron in the MIXTURE = 23% if 1000 ∴ initially amount of iron = 230 gm ∴ initially amount of SAND = 770 gm After adding ‘X’ gms of sand the new ratio of iron : sand will be 13:87 $(\therefore \;\frac{{230}}{{770 + x}} = \;\frac{{13}}{{87}})$ ⇒ 10010 + 13x = 20010 ⇒ x = 769.2 gms |
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| 23. |
In an alloy, lead and tin are in the ratio of 2 : 3. In the second alloy, the ratio of same elements is 3 : 4. If equal quantities of these two alloy are mixed to form a new alloy, then what will be the ratio of these two elements in the new alloy?1). 1 : 32). 29 : 413). 25 : 374). 31 : 43 |
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Answer» LET ‘x’ QUANTITY of each of two alloy be mixed Quantity of lead in the new alloy = (2/5)x + (3/7)x = 29x/35 Quantity of TIN in the new alloy = (3/5)x + (4/7)x = 41x/35 ∴ REQUIRED ratio = (29x/35) ? (41x/35) = 29 ? 41 |
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| 24. |
In what ratio wheat costing Rs. 17/ kg should be mixed with another wheat costing Rs. 12/kg to get a mixture costing Rs. 13.5/ kg?1). 1 : 22). 2 : 53). 3 : 74). 4 : 9 |
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Answer» Applying Alligation formula, (Quantity of higher PRICE/ Quantity of Lesser Price) = (Average Price – Lesser Price)/(Higher Price – Average Price) ∴ Required Ratio = 3 : 7 |
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| 25. |
1). 94%2). 93%3). 95%4). 92% |
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Answer» Let the weight of both ALLOYS be 100 kg In first alloy, AMOUNT of iron = 95 kg Amount of carbon = 3 kg Amount of chromium = 2 kg In second alloy, Let the amount of iron = x kg ⇒ Amount of carbon = (100 - x) kg Total weight of MIXTURE = 100 + 100 = 200 kg ∴ Amount of chromium = 1% of 200 kg = 2 kg ⇒ Amount of iron = 94% of 200 = 188 kg Total iron present = iron from first alloy + iron from second alloy ∴ 188 = 95 + x ⇒ x = 93 ∴ Percentage of iron in second alloy = 93% |
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| 26. |
The ratio of the volumes in water an glycerin in 320 cc of a mixture is 1 : 3. The quantity of water (in cc) that should be added to the mixture so that the new ratio of the volumes of water and glycerin becomes 2 : 3 is1). 552). 803). 62.54). 64 |
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Answer» ⇒ Amount of water in previous MIXTURE = 320 × (1/4) = 80 ⇒ After ADDING x cc amount of water will be (80 + x) ⇒ Amount of glycerin = 320 × (3/4) = 240 ⇒ New ratio = (80 + x)/240 ⇒ (80 + x)/240 = 2/3 ⇒ (80 + x) = 160 ∴ x = 80 |
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| 27. |
6 kg sugar costing Rs. 10/kg is mixed with 4 kg sugar costing Rs. 15/kg. What is the average cost of the mixture per kilogram?1). Rs. 122). Rs. 153). Rs. 104). Rs. 13 |
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Answer» Price of sugar PER kg = Rs. 10 ∴ TOTAL cost = 10 × 6 = Rs. 60 Another quantity of sugar = 4 kg Price of sugar per kg = Rs. 15 ∴ Total cost = 15 × 4 = Rs. 60 ⇒ Total cost of 10 kg of sugar = 60 + 60 = Rs. 120 ⇒ Cost per kilogram = 120/10 = Rs. 12 per kilogram |
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| 28. |
A and B are two alloys of tin and lead prepared by mixing metals in the ratios 3 : 5 and 6 : 11 respectively. Equal quantities of these alloys are melted to form a third alloy C. The ratio of tin and lead in the alloy C is1). 95 : 1742). 42 : 913). 99 : 1734). 24 : 20 |
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Answer» From the given DATA, Tin present in 1 GM of A = 3/8 gm Tin present in 1 gm of B = 6/17 gm Tin present in 2 gm of C = 3/8 + 6/17 = 99/136 gm Lead present in 1 gm of A = 5/8 gm Lead present in 1 gm of B = 11/17 gm Lead present in 2 gm of C = 5/8 + 11/17 = 173/136 gm ⇒ Ratio of tin and Lead in C = (99/136) : (173/136) = 99 : 173 |
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| 29. |
1). 5 : 72). 7 : 53). 7 : 64). 6 : 7 |
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Answer» ⇒ Let the quantity of milk and water in 1st and 2nd VESSEL be 2X, x and x, x respectively. ⇒ Milk in the first mixture = 2x/(2x + x) = 2/3 ⇒ Water in the first mixture = x/(2x + x) = 1/3 ⇒ Milk in the Second mixture = x/(x + x) = 1/2 ⇒ Water in the second mixture = x/(x + x) = 1/2 ⇒ RATIO of Milk and water in new mixture = (2/3 + 1/2)/(1/3 + 1/2) ⇒ Ratio of Milk and water in new mixture = {(2 × 2 + 1 × 3)/6}/{(2 × 1 + 3 × 1)/6} ⇒ Ratio of Milk and water in new mixture = (7)/(5) ∴ Ratio of Milk and water in new mixture = 7 : 5 |
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| 30. |
1). 7 : 52). 6 : 53). 7 : 44). 1 : 1 |
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Answer» Let the amounts of FIRST and second mixture taken be M and N, respectively. ⇒ Amount of milk in first mixture = 6M/11 ⇒ Amount of water in first mixture = 5M/11 ⇒ Amount of milk in second mixture = 7N/11 ⇒ Amount of water in second mixture = 4N/11 Amount of final mixture = (M + N) ⇒ Amount of milk in final mixture = 7(M + N)/12 ⇒ 6M/11 + 7N/11 = 7(M + N)/12 ⇒ 72M + 84N = 77M + 77N ⇒ 5M = 7N ⇒ M/N = 7/5 ∴ Ratio should be 7 : 5 |
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| 31. |
1). Rs. 26 per kg2). Rs. 29 per kg3). Rs. 34 per kg4). Rs. 36 per kg |
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Answer» TOTAL price of type 1 wheat = 78 × 24 = Rs. 1,872 Total price of type 2 wheat = 56 × 32 = Rs. 1,792 Total price of type 3 wheat = 26 × x = Rs. 26x Total kg = 78 + 56 + 26 = 160 kg Average price = Rs. 28.75 Total price of the MIXTURE = 28.75 × 160 = Rs. 4,600 Price of type 3 wheat per kg = x = $(\frac{{4600\; - \;\left( {1872\; + \;1792} \right)}}{{26}}\;)$= Rs. 36 per kg |
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| 32. |
1). 19 litres2). 23 litres3). 22 litres4). 26 litres |
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Answer» Let orange juice and water be 9x and 14X Now, by adding three LITRES the ratio becomes 6 ? 7 (9x + 3) ? 14x = 6 ? 7 ⇒ (9x + 3)7 = (14x)(6) ⇒ 63X + 21 = 84x ⇒ x = 1 ∴ Total amount = 9 + 14 = 23 litres |
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| 33. |
In one vessel there is the mixture of milk and water in the ratio 5 : 2 and in another the ratio of milk and water is 4 : 3. The quantity of mixture in the other vessel is double the quantity of the mixture in the first vessel. If both the mixtures are poured in a single vessel, what will be the ratio of milk and water in the new mixture?1). 2 : 12). 11 : 53). 3 : 24). 13 : 8 |
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Answer» ⇒ Ratio in first VESSEL = 5 : 2 (sum of quantity = 5 + 2 = 7) ⇒ Ratio in second vessel = 4 : 3 (sum of quantity = 4 + 3 = 7) But the quantity of the second vessel is double THEREFORE we will multiply it by 2. ⇒ New ratio in second vessel = 8 : 6 Now, if these are mixed the ratio will be = (5 + 8) : (2 + 6) ∴ 13 : 8 |
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| 35. |
1). 27 ∶ 172). 19 ∶ 53). 37 ∶ 214). 13 ∶ 2 |
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| 36. |
In a particular type of alloy the ratio of iron to carbon is 2 : 3. The amount of iron that should be added to 15 kg of this material to make the ratio of the contents1 : 1 is 1). 2 kg2). 3 kg3). 6 kg4). 9 kg |
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Answer» In 15 kg of this ALLOY, we have AMOUNT of iron $(= \frac{2}{5} \times 15 = 6\;kg)$ amount of carbon $(= \frac{3}{5} \times 15 = 9\;kg)$ Hence, the amount of iron to be added to this mixture to make the ratio of the contents 1:1 is 3 kg of iron. |
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| 37. |
In a mixture of 45 litres, the ratio of liquid A and liquid B is 7 ∶ 2. If 11 litres of liquid B is added to the mixture, then what will be the ratio of liquid A and liquid B in the new mixture?1). 7 ∶ 42). 8 ∶ 53). 6 ∶ 54). 5 ∶ 3 |
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Answer» As, LIQUID A ? Liquid B = 7 ? 2 Quantity of Liquid A in 45 LITERS of mixture = {7/(7 + 2)} × 45 = (7/9) × 45 = 35 liters Quantity of Liquid B in 45 liters of mixture = 45 – 35 = 10 liters New Quantity of Liquid B = 10 + 11 = 21 liters ∴ New Ratio = 35 ? 21 = 5 ? 3 |
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| 38. |
Nitesh wants to make a wine solution of 50% concentration by mixing 5L of wine solution with 2L of another wine solution of 70% concentration but by mistake he used 2L of water in place of 2L wine solution of 70% concentration. Now how much wine solution of 60% concentration be mixed with the solution formed by mistake to get wine solution of 50% concentration.1). 12L2). 14L3). 16L4). 18L |
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Answer» Let concentration 5L of wine solution be x% ⇒ 5 × (x/100) + 2 × (70/100) = 7 × (50/100) ⇒ x = 42 ∴ Concentration of 5L of wine solution is 42% Let concentration of solution formed by mixing 5L of wine solution of 42% with WATER be y% ⇒ 5 × (42/100) + 2 × (0/100) = 7 × (y/100) ⇒ y = 30 Let z L of wine solution of 60% concentration is MIXED with wine solution of 30% concentration formed by mixing water and 5l of wine solution ⇒ 7 × (30/100) + z × (60/100) = (7 + z) × (50/100) ⇒ 210 + 60z = 350 + 50z ⇒ z = 14 ∴ 14L of wine solution of 60% concentration is mixed to FORM wine solution of 50% concentration. |
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| 39. |
Rahul adds 2 litres of alcohol in 6 litres of water and Dinesh adds 1 litre of alcohol in 9 litres of water. What is the ratio of the percentage of alcohol in the two mixtures?1). 7 ∶ 42). 2 ∶ 13). 5 ∶ 24). 8 ∶ 3 |
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Answer» PERCENTAGE of alcohol in Rahul’s mixture = [2/(2 + 6)] × 100 = 25% Percentage of alcohol in DINESH’s mixture = [1/(1 + 9)] × 100 = 10% ∴ Required RATIO = 25 ? 10 = 5 ? 2 |
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| 40. |
1). 10 litres2). 20 litres3). 30 litres4). 40 litres |
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Answer» Total amount = 85 litres Milk ? Water = 11 ? 6 Amount of milk =litres Amount of water = 85 - 55 = 30 litres Let X amount of water be added,such that 55 ? (30 + x) = 11 ? 8 ⇒ 55 × 8 = (30 + x) × 11 ∴ x = 10 litres Thus,10 litres of water should be added to make the RATIO 11 ? 8. |
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| 41. |
Three cans A, B and C are having mixtures of syrup and water in the ratio 1 ∶ 5 , 3 ∶ 5 and 5 ∶ 7 respectively. If the capacities of the cans are in the ratio 5 ∶ 4 ∶ 5, then find the ratio of syrup to water, if the mixtures of all the three cans are mixed together.1). 44 ∶ 1192). 24 ∶ 1113). 46 ∶ 1434). 53 ∶ 115 |
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Answer» Ratio of milk and water = [(1/6) × 5 + (3/8) × 4 + (5/12) × 5] ? [(5/6) × 5 + (5/8) × 4 + (7/12) × 5] = 53 ? 115 |
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| 42. |
In what ratio the milk costing Rs. 24 per liter must be mixed with water to make mixture cost Rs. 18 per liter taking cost of pure water to be 0 ?1). 3 : 72). 2 : 33). 3 : 14). 1 : 3 |
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Answer» Let QUANTITY of milk = x liters Price of pure milk = RS. 24/liter ∴ Cost of milk in mixture = 24x Let quantity of water = y liters Price of water = 0 Total mixture = x + y Total cost of mixture = 24x + 0 = 24x As per the question, Total cost of mixture: 18 × (x + y) ∴ 24x = 18 × (x + y) ⇒ 24x = 18x + 18Y ⇒ 24x – 18x = 18y ⇒ 6x = 18y $(\Rightarrow \;\FRAC{x}{y} = \;\frac{{18}}{6}\; \Rightarrow \;\frac{x}{y} = \;\frac{3}{1}\;)$ ∴ Required RATIO = 3 ? 1 |
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| 43. |
1). 16 : 492). 9 : 83). 10 : 34). 49 : 16 |
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| 44. |
What quantity of water should be added to 5 L of 20% solution of salt, so that it becomes a 10% salt solution?1). 3.5 L2). 5.7 L3). 5 L4). Cannot be determined |
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| 45. |
There are two vessels A & B of capacities 12 litres and 18 litres. They have different concentrations of alcohol. Equal quantity of mixture is taken from both the vessels at the same time and poured into one another, then the concentration of alcohol become equal in both the vessels. Find the quantity of mixture taken from each vessel.1). 4.8 litres2). 6 litres3). 7.2 litres4). 8.4 litres |
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| 46. |
An alloy is made by mixing metal A costing Rs. 2000/kg and metal B costing Rs. 400/kg in the ratio A ∶ B = 3 ∶ 1. What is the cost (in Rs.) of 8 kilograms of this alloy?1). 16002). 98003). 64004). 12800 |
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Answer» Since RATIO in which A and B are MIXED = 3 ? 1 ⇒ Weight of METAL A in 8 kg of alloy = 3/4 × 8 = 6 kg ⇒ Weight of metal B in 8 kg of alloy = 1/4 × 8 = 2 kg ∴ Cost of 8 kg of the alloy = 2000 × 6 + 400 × 2 = Rs. 12800 |
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| 47. |
A big container contains 300 liters of Milk. The owner of the container replaces 30 liters of milk with water. After, sometime out of greed he again replaces 60 liters of milk with water. Meanwhile a thief came and stole 90 liters of milk and replaces it with water. Find what is the quantity of water in the final mixture.1). 148.8 liters2). 128.8 liters3). 130.8 liters4). 120.8 liters |
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Answer» Some PERCENTAGE of milk is replaced with water THREE times ⇒ Remaining milk after replacement of 30 liters of milk with water = 300 × (1 – 30/300 × 100) ⇒ Remaining milk = 300 × (1 – 10%) ⇒ Remaining milk = 270 liters ⇒ Remaining milk after replacement of 60 liter of milk with water = 270 × (1 – 20%) ⇒ Remaining milk = 270 × 80% = 216 liters ⇒ Remaining milk after replacement of 90 liter of milk with water ⇒ Remaining milk = 216 × (1 – 30%) = 216 × 70% = 151.2 liters ∴ Quantity of water in the final mixture = 300 – 151.2 = 148.8 liters |
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| 48. |
1). 5/42). 5/23). 3/44). 4/3 |
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Answer» ⇒ LET x LITERS of pure ALCOHOL added to mixture ⇒ Quantity of alcohol in 10 liters of mixture = 15% of 10 = 1.5 litre ⇒ (10 + x) x 25% = (1.5 + x) ⇒ 2.5 + 0.25x = 1.5 + x ⇒ 1 = 0.75x ∴ x = 4/3 |
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| 49. |
1). 825 kg2). 990 kg3). 780 kg4). 690 kg |
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| 50. |
1). 51 : 1152). 52 : 1153). 53 : 1154). 54 : 115 |
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Answer» ⇒ Let the QUANTITY be 5X, 4x and 5x litres RESPECTIVELY in the 3 containers. ⇒ Quantity of MILK in 1ST container = 5x x (x/6x) = 0.833x ⇒ Quantity of milk in 2nd container = 4x x (3x/8x) = 1.5x ⇒ Quantity of milk in 3rd container = 5x x (5x/6x) = 2.0833x ⇒ Total quantity of milk is = 0.833x + 1.5x + 2.0833x = 4.4166x litres ⇒ The amount of water in the mixture = 14x - 4.4166x = 9.5833x ⇒ Hence, the ratio of milk to water = 4.4166x/9.5833 ∴ Ratio of milk to water = 53 : 115 |
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