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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. |
1). 5 litres2). 2 litres3). 3 litres4). 4 litres |
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Answer» First of all, we use GIVEN quantities and place them in the formula, to know the required quantities. Cost of $(2\frac{1}{3}{\rm{\;or}}\frac{7}{3})$ LITRES MILK = Rs. 42 ⇒ Cost of 1 litre milk = Rs. $(\frac{{42 \times 3}}{7})$ = Rs. 18 Cost of dearer = d = Rs. 18 Cost of 1 litre water = Cost price of cheaper = c = Rs. 0 Desired cost of 1 litre of mixture = Mean price = m = $({\rm{Rs}}.{\rm{\;}}17\frac{1}{3} = Rs.\frac{{52}}{3})$ Quantity of cheaper = d – m $(= 18 - \frac{{52}}{3} = \frac{{54 - 52}}{3} = \frac{2}{3})$ Quantity of dearer = m – c $(= \frac{{52}}{3} - 0 = \frac{{52}}{3})$ Required rate $(= \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{d - m}}{{m - c}} = \frac{{2 \times 3}}{{3 \times 52}} = \frac{1}{{26}})$ ∴ Quantity of water to be added $(= \frac{1}{{26}} \times 52\;litres = 2\;litres)$ |
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| 2. |
1). 3 : 52). 1 : 33). 2 : 34). 4 : 3 |
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Answer» FIRST of all, we use given quantities and place them in the formula, to know the required quantities. Cost of 1 kg fruit of 1st quality = Cost price of dearer = d = Rs. 90 Cost of 1 kg fruit of 2nd quality = Cost price of cheaper = C = Rs. 65 Desired cost of 1 kg of the mixture = Mean price = m = Rs. 75 $(\begin{ARRAY}{l} {\rm{Required\;rate}} = \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{{\rm{d}} - {\rm{m}}}}{{{\rm{m}} - {\rm{c}}}}\\ \Rightarrow \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{90 - 75}}{{75 - 65}} = \frac{{15}}{{10}} = \frac{3}{2} \end{array})$ ∴ The fruit seller should mix the mixture in the ratio 2 : 3 |
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| 3. |
Soham mixes 100 kg of sugar at Rs. 20 per kg with 40 kg of tea at cost price of Rs. 50 per kg. In order to earn a profit of 30% what should be the sale price of the mixed tea?1). Rs. 23.752). Rs. 223). Rs. 37.054). Rs. 19.20 |
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Answer» Given that, Soham mixes 100 kg of sugar at Rs. 20 PER kg with 40 kg of tea at cost PRICE of Rs. 50 per kg. According to the question, Cost price of the mixture $(= \frac{{\left( {100 \TIMES 20} \right) + \left( {40 \times 50} \right)}}{{100 + 40}} = \frac{{2000 + 2000}}{{140}} = \frac{{4000}}{{140}})$ = Rs. 28.5 So, SELLING price will be $(\frac{{100 + 30}}{{100}} \times 28.5)$ = Rs. 37.05 ∴ The sale price of the mixed tea should be Rs. 37.05. |
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| 4. |
1). 44 kg2). 48 kg3). 52 kg4). 56 kg |
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| 5. |
80% milk solution is mixed with 30% milk solution to produce 45 liter of 50% milk solution. If the price of 45 liter of 50% milk solution is Rs.1,080 and the price of 30% milk solution is Rs. 20 per liter, then what is the price of 80% milk solution in the final mixture?1). Rs.6402). Rs.6503). Rs.4004). Rs.250 |
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Answer» Say there is X liters of 30% milk solution and REMAINING (45 – X) liters of 80% milk. AMOUNT of milk in the solution = 0.30X + 0.80(45 – X) Amount of milk in 45 liters of 50% solution = 0.50 × 45 = 22.5 ∴ 0.30X + 36 – 0.8X = 22.5 ⇒ 0.5X = 13.5 ⇒ X = 27 liters Volume of 80% milk in the mixture = 45 – X = 18 liters. Price of 30% milk at Rs.20 PER liter = 20 × 27 = Rs.540 Price of 80% milk = 1080 – 540 = Rs.540 |
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| 6. |
A mixture of oils is prepared by mixing two mixtures A and B in the ratio 2 ∶ 1. Mixture A is made by mixing olive oil and jasmine oil in ratio 3 ∶ 1, while mixture B is made by mixing coconut oil and almond oil in ratio 2 ∶ 3. How much jasmine oil and coconut oil is required to prepare 750 ml of the final mixture of oils respectively?1). 100 ml, 100 ml2). 100 ml, 125 ml3). 125 ml, 100 ml4). 125 ml, 125 ml |
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Answer» RATIO of final mixture = 2 ? 1 Quantity of mixture A in 750 ML final mixture = 2/3 × 750 = 500 ml Ratio of mixture A = 3 ? 1 ⇒ Quantity of jasmine oil in 500 ml mixture A = 1/4 × 500 = 125 ml Similarly, Quantity of mixture B in 750 ml final mixture = 1/3 × 750 = 250 ml Ratio of mixture B = 2 ? 3 ⇒ Quantity of coconut oil in 250 ml mixture B = 2/5 × 250 = 100 ml ∴ To prepare 750 ml of final mixture of OILS, 125 ml jasmine oil and 100 ml coconut oil is required |
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| 7. |
1). 182). 203). 244). 30 |
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Answer» TOTAL price of MILK in mixture = 30 × Rs. 40 = Rs. 1200 Total price of water in mixture = N × Rs. 1 = Rs. N ⇒ Average price of mixture = (1200 + N)/(30 + N) = 24.4 ⇒ 1200 + N = 732 + 24.4N ⇒ N = 468/23.4 = 20 |
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| 8. |
4 litres of milk is removed from a 40 litre container of milk and it is then replaced with 4 litres of water. This procedure is repeated two more times. How much milk is now present in the container?1). 29.16 litres2). 28 litres3). 27.36 litres4). 26.34 litres |
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Answer» Original quantity of milk in the container = 40 litres At first, the quantity of liquid removed from the container = 4 litres Quantity of milk left in the container after first procedure = 40 – 4 = 36 litres Let us try to write this quantity in terms of 40 litres to make it easier to do bigger calculations. 36 litres = $(40\;\left[ {1 - \FRAC{4}{{40}}} \right])$ Quantity of milk left in the container after SECOND procedure $(\begin{array}{l} = \left\{ {40\;\left[ {1 - \frac{4}{{40}}} \right]} \right\} - \left\{ {\frac{4}{{40}}of\;40\;\left[ {1 - \frac{4}{{40}}} \right]} \right\}\\ = \left\{ {40\;\left[ {1 - \frac{4}{{40}}} \right]} \right\} \times \left[ {1 - \frac{4}{{40}}} \right]\\ = \left[ {40{{\left( {1 - \frac{4}{{40}}} \right)}^2}} \right]\END{array})$ Similarly, quantity of milk left in the container after third procedure $(\begin{array}{l} = \left[ {40{{\left( {1 - \frac{4}{{40}}} \right)}^3}} \right]\\ = \left[ {40{{\left( {\frac{{40 - 4}}{{40}}} \right)}^3}} \right]\\ = \left[ {40{{\left( {\frac{{36}}{{40}}} \right)}^3}} \right]\\ = \left[ {40{{\left( {\frac{9}{{10}}} \right)}^3}} \right]\\ = \left[ {40 \times \frac{9}{{10}} \times \frac{9}{{10}} \times \frac{9}{{10}}} \right]\end{array})$ = 29.16 litres ∴ Milk now present in the container is 29.16 litres |
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| 9. |
A container contains some amount of milk. A milkman adds 200 ml of water for each one litre of milk in the container. 12 litres of the mixture is sold from the container and 20 litres of milk is added to the remaining mixture. If now the ratio of milk to water in container is 25 : 3, find the initial quantity of milk in the container.1). 35 liters2). 24 liters3). 40 liters4). 30 liters |
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Answer» LET initial QUANTITY of milk = 10x liters, For each 1 liter, 200 ml of water is added, So for 10x liters 2x liters of water is added RATIO of milk and water = 10x ? 2x = 5 ? 1 So after adding water, quantity of mixture become = 12x liters Now 12 liters of mixture is sold, and 20 liters of milk is added So remaining quantity is (12x – 12 + 20) = (12x + 8) Quantity of milk sold = $(12{\rm{\;}} \times {\rm{\;}}\frac{5}{6} = 10)$ In this final quantity, milk = 10x – 10 + 20 = (10x + 10) The ratio of milk to water in container is 25 : 3 $(\frac{{10{\rm{x}} + 10}}{{12{\rm{x}} + 8}} = {\rm{\;}}\frac{{25}}{{28}})$ ⇒ 28 × (10x + 10) = 25 × (12x + 8) ⇒ 280x + 280 = 300x + 200 ⇒ 20X = 80 ⇒ x = 4So initial quantity of milk = 10x = 40 liters |
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| 10. |
Zinc and copper are in the ratio of 5 : 3 in 400 gram of an alloy. How much gram of copper should be added to make the ratio of zinc to copper as 3 : 5?1). \(133\frac{1}{3}\)2). 2/33). \(266\frac{2}{3}\;\)4). 260 |
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Answer» Zinc and copper are in the ratio 5 : 3 Zinc in 400 gm of ALLOY = 400 × 5/8 = 250 gm Copper in 400 gm of alloy = 400 – 250 = 150 gm x gram of copper is ADDED to make the required ratio, $(\Rightarrow \frac{{250}}{{150\; + x}} = \frac{3}{5})$ ⇒ 1250 = 450 + 3x ⇒ 3x = 800 $(\Rightarrow x = \frac{{800}}{3} = 266\frac{2}{3})$ |
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| 11. |
1). 2 litres2). 3 litres3). 4 litres4). 6 litres |
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Answer» Quantity of WATER in first mixture = (2/5) × 10 = 4 litres Quantity of water in SECOND mixture = (1/5) × 10 = 2 litres ∴ Quantity of water in new mixture = 4 + 2 = 6 litres |
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| 12. |
In what ratio should a seller mix oil of Rs. 95 per litre with oil at Rs. 89 per litre, so that the mixture would worth Rs. 92.50 per litre?1). 3 : 22). 7 : 53). 4 : 74). 3 : 1 |
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Answer» First of all, we use GIVEN quantities and PLACE them in the formula, to know the required quantities. COST of 1 litre oil of first TYPE = Cost price of dearer = d = Rs. 95 Cost of 1 litre oil of second type = Cost price of cheaper = c = Rs. 89 Desired cost of 1 litre of the mixture = Mean price = m = Rs. 92.50 $(Required\;rate = \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{d - m}}{{m - c}})$ $(\Rightarrow \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{95 - 92.50}}{{92.50 - 89}} = \frac{{2.50}}{{3.50}} = \frac{5}{7})$ ∴ The ratio of the mixture should be 7 : 5 |
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| 13. |
The ratio of males and females in a village is 8 : 5 respectively and the percentage of children among males and females is 30% and 25% respectively. If the number of adult males in the village is 274400, what is the number of female children in the village?1). 675002). 612503). 694004). 76500 |
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Answer» Let’s assume that the total population of the village is X ? RATIO of male & female in village is = 8 : 5 ∴ Number of males in village $(= \frac{8}{{8 + 5}}\; \times \;X = \frac{{8X}}{{13}})$ & Number of FEMALES in village $(= \frac{5}{{5 + 8}}\; \times \;X\; = \frac{{5X}}{{13}})$ Number of male children is 30% among all males ∴ number of male children = (8/13)X × 0.30 = (2.4/13)X ∴ number of adult males is = (8/13)X - (2.4/13)X = (5.6/13)X Given that number of adult males = 274400 ∴ (5.6/13)X = 274400 ⇒ X = 637000 Number of female children is 25% among all females ∴ Number of female children = (5/13)X × 0.25 = (1.25/13)X ⇒ Number of female children = (1.25/13)× 637000 ⇒ Number of female children = 61250 |
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| 14. |
1). 6 kg2). 4 gram3). 500 gram4). 3 kg |
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Answer» In 1 kg mixture of iron FILINGS and sand, 25% is sand ⇒ Quantity of sand in the mixture = 250 grams Let’s assume x grams iron filings is added, then mixture BECOMES (1000+ x) grams ∴ 5% of (1000 + x) = 250 ⇒ 1000 + x = 5000 ⇒ x = 4000 GRAM = 4 kg Thus, 4 kg iron filings should be added, So that the proportion of sand becomes |
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| 15. |
A shopkeeper has 1000 kg of salt part of which he sells at 8% profit and the rest at 18% profit. He gains 14% on the whole. What is the total Selling Price of the salt on which he gained 18% if the Selling Price of each kg is Rs. 2?1). 12002). 4003). 6004). 500 |
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Answer» We know that, $(\frac{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;cheaper\;substance}}\left( {\rm{c}} \right)}}{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;dearer\;substance}}\left( {\rm{d}} \right)}} = \frac{{{\rm{Mean\;price}} - {\rm{c\;}}}}{{{\rm{d}} - {\rm{Mean\;price}}}})$ For 8% profit salt, 18 - 14 = 4 For 18% profit salt, 14 - 8 = 6 Quantity of 8% profit salt ? Quantity of 18% profit salt = 4 ? 6 = 2 ? 3 Given that the TOTAL quantity is 1000 kg. So the quantity sold at 18% profit = $(1000 \times \frac{3}{5} = 600\;{\rm{kg}})$ The total selling price of this part of salt = 2 × 600 = Rs. 1200 ∴ The total selling price of the 18% profit salt is Rs. 1200 |
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| 16. |
Jar-A contained a mixture of milk and water in the respective ratio of 5 ∶ 1.The difference between quantities of milk and water in Jar-A was 32 litre. If 12 litres of mixture was taken out and 1.5 litres of water was added to Jar-A. What will be the new respective ratio between quantities of milk and water in Jar-A?1). None2). 4 ∶ 13). 9 ∶ 24). 3 ∶ 1 |
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Answer» The ration of MILK and water in Jar-A = 5 ? 1 Let the QUANTITY of milk and water in Jar-A be 5x and 1x The difference between the quantity of milk and water in Jar-A = 32 ⇒ 5x - x = 32 ⇒ 4x = 32 ⇒ x = 8 The quantity of milk = 5x = 5 × 8 = 40 The quantity of water = 1x = 8 Now, If 12 LITERS of the mixture was taken out and 1.5 liters of water was ADDED to Jar-A, we get $4 : 1$ |
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| 17. |
1). 6 L2). 6.5 L3). 5.5 L4). 5 L |
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Answer» Amount of water INITIALLY = (10/100) × 40 = 4 L Let ‘x’ be the amount of water to be added. ⇒ [(4 + x)/( 40 + x)] × 100 = 20 ⇒ 40 + 10X = 80 + 2x ⇒ 8X = 40 ⇒ x = 40/8 = 5 ∴ 5 L of water is to be added to the mixture. |
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| 18. |
In what proportion must water be added to hydrochloric acid to gain 25% by selling it at the cost price?1). 2 : 32). 1 : 43). 3 : 44). 4 : 7 |
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| 19. |
Two bottles A and B contain mixture of milk and water in the ratio of 5 ∶ 3 and 1 ∶ 5 respectively. Find the ratio in which these two mixtures be mixed to obtain a new mixture containing milk and water in the ratio 3 ∶ 4.1). 5 ∶ 62). 3 ∶ 83). 4 ∶ 34). 2 ∶ 3 |
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Answer» Let Amount of mixture CONTAINED by both bottles A and B be 1 litre Bottle A contains milk and water mixture in the ratio 5 ? 3 ⇒ Amount of milk in the bottle A = 5/8 litre ⇒ Amount of water in the bottle A = 3/8 litre Bottle B contains milk and water mixture in the ratio 1 ? 5 ⇒ Amount of milk in the bottle B = 1/6 litre ⇒ Amount of water in the bottle B = 5/6 litre Let ‘x’ and ‘y’ LITRES from bottle A and B respectively are mixed ⇒ Amount of milk in the NEW mixture in litre = 5/8x + 1/6y ⇒ Amount of milk in the new mixture in litre = 3/8x + 5/6y ? New mixture contains milk and water in the ratio 3 ? 4 ⇒ (5/8x + 1/6y) / (3/8x + 5/6y) = 3/4 ⇒ (15x + 4y) / (9x + 20y) = 3/4 ⇒ 60x + 16y = 27x + 60y ⇒ 33x = 44y ⇒ x/y = 4/3 |
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| 20. |
A beaker contains 30 litres of juice. From this beaker 3 litres of juice are taken out and replaced by water. This process was repeated further 2 times. How much juice is now contained by the container?1). 21 litres2). 24.3 litres3). 21.87 litres4). 24 litres |
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Answer» Amount of JUICE LEFT after 3 removals = 30(1 – 3/30)3 = 30 × (9/10)3 = 30 × 9/10 × 9/10 × 9/10 LITRES = 21.87 litres Detailed Solution: Quantity of juice remained after FIRST removal = (30 – 3) litres = 27 litres Quantity of juice remained after second removal = (27 – 27/30 × 3) litre = (27 – 2.7) litres = 24.3 litres Quantity of juice remained after third removal = (24.3 – 24.3/30 × 3) litre = (24.3 – 2.43) litres = 21.87 litres |
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| 21. |
1). 4 ∶ 32). 5 ∶ 23). 2 ∶ 54). 3 ∶ 4 |
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| 22. |
It is given that 1st type of sugar costing Rs. 6.60 per kg was mixed with 56 kg of 2nd type of sugar costing Rs. 9.60 per kg, to get a mixture worth Rs. 8.20 per kg. What is the difference in quantities of the 1st type of sugar and 2nd type of sugar present in the mixture?1). 52). 93). 34). 7 |
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Answer» We know that, $(\FRAC{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;cheaper\;SUBSTANCE}}\left( {\rm{c}} \right)}}{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;dearer\;substance}}\left( {\rm{d}} \right)}} = \frac{{{\rm{MEAN\;price}} - {\rm{c}}}}{{{\rm{d}} - {\rm{Mean\;price}}}})$ For CP of 1st variety sugar, 9.6 - 8.2 = 1.4 For CP of 2nd variety sugar, 8.2 - 6.6 = 1.6 Quantity of 1st type of sugar ? Quantity of 2nd type of sugar = 1.4 ? 1.6 = 7 ? 8 Quantity of 1st type of sugar ? 56 = 7 ? 8 Quantity of 1st type of sugar = 49 The difference between the both quantities = 56 - 49 = 7 ∴ the required difference between 1st type of sugar and 2nd type of sugar is 7 |
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| 23. |
Six litres of a 35% solution of alcohol in water are mixed with 4 litres of a 45% solution of alcohol in water. The % alcoholic strength of the mixture is1). 492). 393). 364). 48 |
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Answer» Percentage of alcohol in 6 litre solution = 35 Percentage of alcohol in 4 litre solution = 45 Let x% of alcoholic MIXTURE is present Also, ACCORDING to the question (35% of 6) + (45% of 4) = x% of 10 ⇒ 2.1 + 1.8 = x/10 ⇒ x = 39% ∴ The % alcoholic STRENGTH of the mixture is 39. |
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| 24. |
A and B are two alloys of copper and tin prepared by mixing metals in the ratio of 4 : 3 and 5 : 7 respectively. These alloys are melted in the ratio 1 : 2 to form a third alloy C. The ratio of tin and copper in the alloy C is1). 65 : 532). 67 : 593). 65 : 674). 68 : 55 |
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Answer» Let 1 kg of the alloy A and 2 kg of alloy B are MIXED together In 1 kg of alloy A, Quantity of COPPER = $(\frac{4}{7})$ kg Quantity of tin = $(\frac{3}{7})$ kg In 2 kg alloy B, Quantity of copper = $(2\; \times \;\frac{5}{{12}} = \;\frac{5}{6})$ kg Quantity of tin = $(2\; \times \;\frac{7}{{12}} = \;\frac{7}{6})$ kg ∴ Required ratio of = $(\left( {\frac{3}{7} + \;\frac{7}{6}} \right) \sim \left( {\frac{4}{7} + \;\frac{5}{6}} \right) = \;\frac{{67}}{{42}} \sim \frac{{59}}{{42}} = \frac{{67}}{{59}})$ ∴ Required ratio = 67 ? 59 |
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| 25. |
A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaced is:1). \(\frac{1}{3}\)2). \(\frac{2}{3}\)3). \(\frac{2}{5}\)4). \(\frac{3}{5}\) |
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| 26. |
1). 1 ∶ 602). 1 ∶ 503). 1 ∶ 554). 1 ∶ 45 |
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Answer» ? Cost of 150 ml tasty solution = Rs. 27 ∴ Cost of 1 ml tasty solution = Rs. $(\frac{{27}}{{150}})$ ∴ Cost of 1 litre of tasty solution = 1000 × (9/50) = Rs. 180 Cost of 1 kg of salt = Cost price of cheaper = c = Rs. 30 Cost of 1 litre of soup = Cost price of dearer = d = Rs. 183 Cost of 1 litre of tasty solution = MEAN price = m = Rs. 180 By formula, $(\BEGIN{array}{l} \frac{{{\rm{Quantity\;of\;salt}}}}{{{\rm{Quantity\;of\;soup}}}} = \frac{{{\rm{Quantity\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{{\rm{d}} - {\rm{m}}}}{{{\rm{m}} - {\rm{c}}}}\\ \Rightarrow \frac{{{\rm{Quantity\;of\;Salt}}}}{{{\rm{Quantity\;of\;Soup}}}} = \frac{{183 - 180}}{{180 - 30}} = \frac{3}{{150}} \end{array})$ ∴ The required ratio = Salt ? Soup = 1 ? 50. |
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| 27. |
1). 3 : 52). 1 : 33). 5 : 34). 4 : 3 |
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Answer» First of all, we use given QUANTITIES and place them in the FORMULA, to KNOW the required quantities. Cost of 1 kg fruit of 1ST quality = Cost price of dearer = d = Rs. 90 Cost of 1 kg fruit of 2nd quality = Cost price of cheaper = c = Rs. 50 Desired cost of 1 kg of the mixture = Mean price = m = Rs. 65 $(\begin{array}{l} Required\;rate\; = \;\frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}}\; = \;\frac{{d - m}}{{m - c}}\\ \Rightarrow \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}}\; = \;\frac{{65 - 50}}{{90 - 65}}\; = \;\frac{{15}}{{25}}\; = \;\frac{3}{5} \end{array})$ ∴ The fruit seller should mix the mixture in the ratio 3 : 5 |
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| 28. |
1). 3 litres2). 4 litres3). 8 litres4). 12 litres |
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| 29. |
A bottle full of Brandy contains 40% alcohol. A part of this Brandy was replaced by another one having 19% alcohol and the percentage now became 26%. What was the quantity of Brandy replaced?1). 2/32). 1/33). 1/24). 5/6 |
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Answer» Concentration of Alcohol in the FIRST BOTTLE = 40% Concentration of Alcohol in the SECOND bottle = 19% We know that, $(\frac{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;cheaper\;substance}}\left( {\rm{c}} \right)}}{{{\rm{Cost\;price\;of\;unit\;quantity\;of\;dearer\;substance}}\left( {\rm{d}} \right)}} = \frac{{{\rm{Mean\;price\;}} - {\rm{\;c\;}}}}{{{\rm{d\;}} - {\rm{Mean\;price}}}})$ For first bottle, 26 - 19 = 7 For second bottle, 40 - 26 = 14 Hence the RATIO is 1 ? 2. The part of BRANDY replaced = $(\frac{2}{3})$ ∴ The part of Brandy replaced is 2/3 |
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| 30. |
The presence of milk in four different containers is 1/3, 2/5, 2/3 an 3/5. A shopkeeper withdraws 2 liter, 3 liter, 1 liter and 4 liter from each container and mixes them in new container. Find the ratio of milk and water in the new container?1). 34 : 312). 37 : 323). 32 : 354). 37 : 41 |
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Answer» ⇒ Presence of milk = 1/3 : 2/5 : 2/3 : 3/5 ⇒ Quantity of milk from container 1 = 1/3 × 2 = 2/3 liter ⇒ Quantity of milk from container 2 = 2/5 × 3 = 6/5 liter ⇒ Quantity of milk from container 3 = 2/3 × 1 = 2/3 liter ⇒ Quantity of milk from container 4 = 3/5 × 4 = 12/5 liter ⇒ So, total milk in new container = 2/3 + 6/5 + 2/3 + 12/5 = 74/15 liter ⇒ Total water in new container = 10 – 74/15 = 76/15 liter ∴ REQUIRED RATIO = 74/15 : 76/15 or 37 : 38 |
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| 31. |
A dishonest milkman fills up his bucket, which is only 4/7th full of milk, with water. He again removes 3 litres of this mixture from the bucket and adds an equal quantity of water. If milk is now 40% of the mixture, then what is the capacity of the bucket?1). 25 litres2). 40 litres3). 15 litres4). 10 litres |
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Answer» Let the capacity of bucket be X litres ⇒ Quantity of MILK = (4/7)X ⇒ Quantity of water = 1 – (4/7)X = (3/7)X ⇒ If 3 litres mixture is REMOVED, REMAINING mixture = X – 3 ⇒ Now quantity of milk = (4/7) × (X – 3) ⇒ If 3 litres water is added, quantity of water = (3/7) × (X – 3) + 3 ⇒ (4/7) × (X – 3) = 40% of X ⇒ (4/7)X – 12/7 = (40/100)X ⇒ {(4/7) – (40/100)}X = 12/7 ∴ X = 10 litres |
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| 32. |
Hanna bought 20 kg of wheat at the rate of Rs. 8.50 per kg and 35 kg at the rate of Rs. 8.75 per kg. He mixed the two. At what price per kg should he sell the mixture to make 40% profit at the cost price?1). Rs. 122). Rs. 153). Rs. 164). Rs. 20 |
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Answer» CP = 20 × 8.5 + 35 × 8.75 = 476.25 Profit = 40% SELLING Price of mixture per kg = $(\frac{{\left( {100{\RM{\;}} + {\rm{\;Profit\% }}} \right)}}{{100}} \times {\rm{CP}})$ $(= \frac{{100{\rm{\;}} + {\rm{\;}}40}}{{100}} \times 476.25 = 666.75)$ TOTAL quantity = 20 + 35 = 55 kg SP per kg = $(\frac{{666.75}}{{55}} \approx 12)$ ∴ The Selling Price of the mixture is RS. 12 per kg |
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| 33. |
You are given a mixture of 18 litres, where the ratio of milk and water is 3 ∶ 2. If 3 litres of this mixture is replaced by 3 litres of milk, then what will be the ratio of milk and water in the newly formed mixture?1). 12 ∶ 32). 8 ∶ 53). 2 ∶ 14). 1 ∶ 1 |
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Answer» Total quantity of MIXTURE = 18 liters and 3 LITRES of mixture is removed from the container Now, you are left with only 15 litres of mixture in 3 ? 2 ratio. ⇒ Milk in 15 litres mix = 15 × 3/(2 + 3) = 9 litres ⇒ Water in 15 litres mix = 15 × 2/(2 + 3) = 6 litres We add 3 litres milk to this. ⇒ Milk in new mix is = 9 litres + 3 litres = 12 litres ⇒ Water = 6 litres ∴ Ratio of milk ? water = 12 ? 6 = 2 ? 1 |
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| 34. |
Some amount of sugar costing Rs. 9 per kg is mixed with 27 kg of sugar costing Rs. 7 per kg. There is a gain of 10% by selling the mixture at Rs. 9.24 per kg. In what ratio both the qualities of sugar have been mixed?1). 3 ∶ 42). 3 ∶ 53). 7 ∶ 34). 2 ∶ 5 |
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Answer» Selling PRICE of 1 KG of the mixture = Rs. 9.24 Gain % = 10% Let the cost price of 1 kg of the mixture = Rs. x By the formula of Cost price, CP + 10% of CP = SP $(\begin{array}{l} \Rightarrow x + \frac{{10}}{{100}} \times x = 9.24\\ \Rightarrow \frac{{110}}{{100}} \times x = 9.24 \end{array})$ ∴ x = Rs. 8.40 Cost of 1 kg sugar of 1st QUALITY = Cost price of dearer = d = Rs. 9 Cost of 1 kg sugar of 2nd quality = Cost price of cheaper = c = Rs. 7 Cost of 1 kg of the mixture = Mean price = m = Rs.8.40 By formula, $(\begin{array}{l} \frac{{{\rm{QUANTITY\;of\;Cheaper}}}}{{{\rm{Quantity\;of\;Dearer}}}} = \frac{{{\rm{d}} - {\rm{m}}}}{{{\rm{m}} - {\rm{c}}}}\\ \Rightarrow \frac{{9 - 8.40}}{{8.40 - 7}} = \frac{3}{7} \end{array})$ ∴ The ratio of dearer to cheaper or 1st quality : 2nd quality is 7 : 3 |
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| 35. |
One quality of wheat at Rs. 9.30 per kg is mixed with a higher quality wheat in the ratio 8 ∶ 7. If the mixture so formed costs Rs. 10 per kg, what is the rate per kg of the second quality of wheat?1). Rs. 10.302). Rs. 10.603). Rs. 10.804). Rs. 11 |
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Answer» COST of 1 KG wheat of 1st quality = Cost price of cheaper = c = Rs. 9.30 Cost of 1 kg wheat of 2nd quality = Cost price of dearer = d = Rs. x (suppose) Cost of 1 kg of the mixture = Mean price = m = Rs.10 By formula, $(\begin{array}{L} \FRAC{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{d - m}}{{m - c}}\\ \RIGHTARROW \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{x - 10}}{{10 - 9.30}} = \\ \frac{8}{7} \Rightarrow \frac{{x - 10}}{{0.70}} = \frac{8}{7} \end{array})$ ⇒ 7x – 70 = 8 × 0.70 ⇒ 7x = 5.6 + 70 ⇒ 7x = 75.6 ⇒ x = Rs. 10.8 ∴ The rate of the second quality of wheat is Rs. 10.80 per kg. |
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| 36. |
How many litres of water should be added to a 30 litre mixture of milk and water containing milk and water in the ratio of 7 : 3 such that the resultant mixture has 40% water in it?1). 2 litres2). 5 litres3). 7 litres4). 9 litres |
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Answer» 30 litres of the mixture has MILK and WATER in the ratio 7 : 3 ⇒ Solution has 21 litres of milk and 9 litres of water When you add more water, the amount of milk in the mixture remains constant at 21 litres In the first CASE, before addition of further water, 21 litres of milk accounts for 70% by volume After water is added, the new mixture contains 60% milk and 40% water ⇒ The 21 litres of milk accounts for 60% by volume ⇒ 100% volume = 21/0.6 = 35 litres We started with 30 litres and ended up with 35 litres ∴ 5 litres of water was added |
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| 37. |
In a mixture of 40 litres, the ratio of milk and water is 3 : 2. How many litres of milk must be added to make the ratio 4 : 1?1). 452). 403). 354). 30 |
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Answer» Let ‘a’ be the number of litres of milk to be added. Given the ratio of milk and WATER before and after MIXING with milk as 3 : 2 and 4 : 1 respectively. Equating the quantity of milk present in the INGREDIENTS to the FINAL mixture we have, 0.6 × 40 + a = 0.8 × (40 + a) ⇒ 0.2 × a = 0.2 × 40 ⇒ a = 40 ∴ Quantity of milk to be added is 40 litres. |
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| 38. |
1). 18.18%2). 22.22%3). 28.57%4). 33.33% |
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Answer» Quantity of WATER in 16 litres of alcohol-water mixture = 1/4 × 16 = 4 litres Quantity of water in 20 litres of milk-water mixture = 1/5 × 20 = 4 litres Quantity of water in (16 + 20 = ) 36 litres of NEW mixture = 4 + 4 = 8 litres ∴ Required percentage = 8/36 × 100 = 22.22% |
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| 39. |
A grocer purchased 2 kg. of rice at the rate of Rs. 15 per kg. and 3 kg. of rice at the rate of Rs. 13 per kg. At what price per kg should he sell the mixture to earn 33.33% profit on the cost price?1). Rs. 28.002). Rs. 20.003). Rs. 18.404). Rs. 17.40 |
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Answer» Price of 2 kg rice I = Rs. 30 Price of 3 kg rice II = Rs. 39 ⇒ Price of 5 kg mixture = Rs. 30 + Rs. 39 = Rs. 69 Now, 33.33% profit is required ⇒ 100 + 100/3 of CP ⇒ 400/3% of CP = [400/(3 × 100)] × 69 = 4 × 23 = Rs. 92 ⇒ SELLING price of 5 kg mixture of rice = Rs. 92 ∴ Selling price of 1 kg mixture = 92/5 = Rs. 18.40 PER kg |
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| 40. |
In a barrel of 56 litres, the volume of milk is 62.5%. If the mixture requires to make 85.71% of milk, how much more amount of pure milk is needed to be added?1). 91 litres2). 88 litres3). 90 litres4). 85 litres |
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Answer» Total volume of mixture = 56 LITRES Amount of milk in mixture = 62.5% of total volume = (5/8) × 56 = 35 litres Let x litres of pure milk is added in the mixture. To CHANGE the proportion of mixture, Required amount of milk = 85.71% of total volume = (6/7) × (56 + x) ⇒ 35 + x = (6/7) × (56 + x) ⇒ 35 + x = 48 + 6x/7 ⇒ x/7 = 13 ⇒ x = 91 litres ∴ Amount of milk required = 91 litres |
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| 41. |
1). 150 ml2). 225 ml3). 240 ml4). 330 ml |
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Answer» Quantity of liquid B = 70% of 450 = 315 ML Let x ml of liquid B be added. 315 + x = 80% of (450 + x) ⇒ 315 + x = 0.8 × (450 + x) ⇒ 315 + x = 360 + 0.8x ⇒ 0.2x = 45 ⇒ x = 225 |
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| 42. |
A bottle full of juice containing 40% carbohydrate. A part of this juice is replaced by another juice containing 19% carbohydrate and percentage of carbohydrate in the new mixture becomes 26%. The part of the juice replaced is?1). 1/3rd2). 2/3rd3). 2/5th4). 3/5th |
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Answer» Let amount of juice in the BOTTLE is 1 litre ? Juice CONTAINS 40% carbohydrate ⇒ Amount of carbohydrate in the bottle = 0.40 litre Let amount of juice replaced from the bottle is ‘x’ litre After REPLACING x litre with another juice containing 19% carbohydrate, ⇒ Amount of carbohydrate in the NEW MIXTURE = (0.40 – 0.40x + 0.19x) litre ? Percentage of carbohydrate in the new mixture = 26% ⇒ (0.40 – 0.40x + 0.19x) /1 × 100% = 26% ⇒ 0.40 – 0.21x = 0.26 ⇒ x = 14/21 ⇒ x = 2/3 ∴ 2/3rd part of the juice is replaced |
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| 43. |
The milk and water in a mixture are in the ratio 7 ∶ 5. When 15 liters of water is added to it, the ratio of milk and water in the new mixture becomes 7 ∶ 8. The total quantity of water in the new mixture is1). 35 litres2). 40 litres 3). 60 litres4). 96 litres |
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Answer» Suppose initially mixture contains 7M and 5m milk and WATER respectively. QUANTITY of water left in mixture = 5m + 15 $(\Rightarrow \frac{{7m}}{{5m + 15}} = \frac{7}{8})$ ⇒ 56m = 35m + 105 ⇒ 21m = 105 ⇒ m = 5 ∴ Quantity of water in mixture = 5m + 15 = 5 × 5 + 15 = 40 lit. |
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| 44. |
In what ratio must a grocer mix two varieties of pulses costing Rs. 15 and Rs. 20 per kg respectively so as to get a mixture worth Rs. 16.50 per kg?1). 3 ∶ 72). 5 ∶ 73). 7 ∶ 34). 7 ∶ 5 |
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Answer» Cost of 1 kg PULSES of 1st quality = Cost price of cheaper = c = Rs. 15 Cost of 1 kg pulses of 2ND quality = Cost price of DEARER = d = Rs. 20 Cost of 1 kg of the mixture = Mean price = m = Rs. 16.50 By formula, $(\begin{array}{l} \frac{{Quantity\;of\;Cheaper}}{{Quantity\;of\;Dearer}} = \frac{{d - m}}{{m - c}}\\ \RIGHTARROW \frac{{20 - 16.50}}{{16.50 - 15}} = \frac{{3.50}}{{1.50}} = \frac{7}{3} \end{array})$ ∴ The ratio of cheaper to dearer or 1st quality ? 2nd quality is 7 ? 3 |
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| 45. |
A certain amount of mixture contains milk and water in the ratio of 4 : 1.If 10 liters of this mixture is replaced by 10 liters of water, then the resultant ratio becomes 2 : 3.How much pure milk was originally there in the mixture at first.1). 8 liters2). 10 liters3). 12 liters 4). 16 liters |
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Answer» LET the volume of water be = x Then, volume of milk = 4x Volume of MIXTURE replaced = 10 liters Volume of water added = 10 liters Volume of milk replaced = (4/5) × 10 = 8 liters Volume of water replaced = 10 – 8 = 2 liters Ratio of milk to water when mixture is replaced by 10 liters of water = 2 : 3 Initial, volume of milk = $(\frac{{4{\rm{x}} - 8}}{{{\rm{x}} - 2 + 10}} = \frac{2}{3})$ ⇒ 12x – 24 = 2x + 16 ⇒ 10x = 24 + 16 ⇒ x = 40/10 = 4 ∴ Volume of milk initially = 4 × 4 = 16 liters ∴ The initial volume of milk is 16 liters |
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| 46. |
In a mixture of 225 liters, the ratio of acid to water is 4 ∶ 1. Another 35 liters of water is added to the mixture. The ratio of acid to water in the new mixture is:1). 5 : 92). 9 : 53). 3 : 94). 4 : 9 |
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Answer» TOTAL mixture = 225 liters Acid in mixture = $(225\times \frac{4}{5} = 180\;liters)$ Water in mixture = $(225\times \frac{1}{5} = 45\;liters)$ Water added = 35 liters NEW ratio = $(\frac{{180}}{{45 + 35\;}} = \;\frac{{180}}{{80}} = \;\frac{9}{4})$ ∴ The ratio of acid to water in the new mixture is 9 ? 4 |
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| 47. |
The cost of Basmati rice is Rs. 30 per kg and Ponni rice is Rs. 20 per kg. Both quantities are mixed in the ratio 2 ∶ 3. Then the price of 2 kg of the mixed variety of rice is?1). Rs. 242). Rs. 253). Rs. 484). Rs. 20 |
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Answer» Basmati and Ponni RICE are mixed in the ratio of 2 ? 3 ⇒ Amount of Basmati in 1 KG of the mixture = 2/5 kg ⇒ Amount of Ponni rice in 1 kg of the mixture = 3/5 kg ⇒ Price of 2/5 kg of Basmati rice in RS. = 2/5 × 30 = Rs. 12 ⇒ Price of 3/5 kg of Ponni rice in Rs. = 3/5 × 20 = Rs. 12 ⇒ Price of 1 kg of the mixed VARIETY = Rs. (12 + 12) = Rs. 24 ∴ Price of 2 kg of the mixed variety = Rs. (2 × 24) = Rs. 48 |
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| 48. |
1). 36 litres2). 45 litres3). 40 litres4). 25 litres |
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Answer» RATIO of milk and water = 64 ? 17 ∴ Ratio of remaining milk and initial milk = 64 ? (64 + 17) = 64 ? 81 Let, total AMOUNT of container = K liter $(\therefore \frac{{remaining\;milk}}{{initial\;milk}} = {(1 - \frac{{quantity\;taken\;out}}{{total\;amount}})^n})$ [where n = number of attempts] $(\Rightarrow \frac{{64}}{{81}} = {(1 - \frac{4}{k})^2})$ ⇒ 8/9 = 1 - 4/k ⇒ 4/k = 1 - 8/9 ⇒ 4/k = 1/9 ⇒ k = 36 ∴ total amount of the container = 36 LITRES |
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| 49. |
Two mixtures M1 and M2 contain mixture of two types of pulses P1 and P2. M1 and M2 contain P1 and P2 in the ratio of 4 : 5 and 8 : 3 respectively. M1 and M2 are mixed to form a third mixture. Now the ratio of P1 : P2 becomes 8 : 5 in the resultant mixture. Find the amount of pulses P2 in the M2, if the resultant quantity is 364 kg of pulses.1). 62 kg2). 64 kg3). 60 kg4). 68 kg |
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Answer» P2 in M1 = 5/9, in M2 = 3/11 and in resultant mixture = 5/13 By ALLIGATION rule : - 5/9 3/11 5/13 (5/13 – 3/11) = 16/143 (5/9 – 3/11) = 20/117 ⇒ 16/143 : 20/117 = 36 : 55 ∴ Amount of M2 in 364 kg = 55/91 × 364 = 220 kg Now amount of pulses P2 in M2 = (3/11) × 220 = 60 kg |
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| 50. |
1). 18 litres2). 38 litres3). 56 litres4). 84 litres |
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Answer» Milk ? Water = 15 ? 7 AMOUNT = 66 litres Amount of milk = 45 litres Amount of water = 66 - 45 = 21 litres Let x amount of milk be added, such that ratio becomes 3 ? 1 (45 + x) ? 21 = 3 ? 1 ⇒ (45 + x) = 21(3) ⇒ x = 18 litres ∴ New amount = 66 + 18 = 84 litres |
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