Two varieties of sugar, same capacities of milk mixture and other mixture problems with solutions
Two varieties of sugar are mixed in a certain ratio and other selected milk or wheat mixture problems to be solved. Learn to solve from easy solutions.
Contents are,
- 10 sugar, milk or wheat mixture problems to solve in 15 minutes.
- Answers to the problems.
- Quick and easy solution to the 10 problems.
Take the test first and then learn to solve quickly from easy solutions.
Sugar, milk or wheat mixture problems SSC CGL Set 85 - time to solve 15 mins
Problem 1.
A merchant buys 25 litres of milk daily at the rate of Rs.12 per litre. He mixes 5 litres of water in it and sells at the rate of Rs.10.40 per litre. His profit percentage is,
- 8%
- 6%
- 4%
- 2%
Problem 2.
80 litres mixture of milk and water contains 10% milk. How much milk (in litres) must be added to make the water percentage in the mixture as 80%?
- 12
- 10
- 9
- 8
Problem 3.
In what ratio tea at Rs.240 per kg should be mixed with tea at Rs.280 per kg so that on selling the mixture at Rs.324 per kg there is a profit of 20%?
- 1 : 3
- 1 : 1
- 1 : 2
- 1 : 4
Problem 4.
Three boxes of capacities 24 kg, 36 kg and 84 kg are completely filled with three varieties of wheat A, B and C respectively. All the three boxes were emptied and the three types of wheat were thoroughly mixed and the mixture was put back in the three boxes. How many kgs of type A wheat would be there in the third box?
- 12
- 16
- 10
- 14
Problem 5.
A milkman mixes water with milk and sells the mixture at the cost price of the milk. The volume of water in litres to be mixed with each litre of pure milk to get a 25% profit is,
- $1\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{5}$
- Cannot be determined with given information
Problem 6.
Three equal capacity bottles contain mixtures of milk and water in ratios 2 : 5, 3 : 4 and 4 : 5 respectively. The three bottles are emptied into a large vessel. What will be the ratio of milk and water respectively in the large vessel?
- 73 : 113
- 73 : 189
- 73 : 106
- 73 : 116
Problem 7.
In a mixture of 25 litres, the ratio of milk to water is 4 : 1. Another 3 litres of water is added to the mixture. The ratio of milk to water in the new mixture is,
- 5 : 1
- 5 : 2
- 5 : 4
- 5 : 3
Problem 8.
Two alloys contain tin and iron in the ratio 1 : 2 and 2 : 3. If the two alloys are mixed in the proportion 3 : 4 respectively by weight, the ratio of tin and iron in the newly formed alloy is,
- 13 : 22
- 12 : 23
- 10 : 21
- 14 : 25
Problem 9.
A litre pf water weighs a kilogram while a litre of a second liquid weighs 1.340 kilogram. A mixture of the two weighs 1.270 kilogram per litre. The ratio of their volume in a litre of mixture is,
- 7 : 27
- 7 : 17
- 27 : 34
- 17 : 24
Problem 10.
Two varieties of sugar are mixed in a certain ratio. The cost of the mixture per kg is Rs.0.50 less than that of the superior and Rs.0.75 more than the inferior variety. The ratio in which the inferior and the superior varieties of sugar have been mixed is,
- 5 : 2
- 5 : 1
- 3 : 2
- 2 : 3
Answers to the sugar, milk or wheat mixture problems SSC CGL Set 85
Problem 1. Answer: Option c: 4%
Problem 2. Answer: Option b: 10.
Problem 3. Answer: Option a: 1 : 3.
Problem 4. Answer: Option d: 14.
Problem 5. Answer: Option b: $\displaystyle\frac{1}{4}$.
Problem 6. Answer: Option d: 73 : 116
Problem 7. Answer: Option b: 5 : 2.
Problem 8. Answer: Option a: 13 : 22.
Problem 9. Answer: Option a: 7 : 27.
Problem 10. Answer: Option d: 2 : 3.
Solution to the Sugar, milk or wheat mixture problems SSC CGL Set 85 - time to solve was 15 mins
Problem 1.
A merchant buys 25 litres of milk daily at the rate of Rs.12 per litre. He mixes 5 litres of water in it and sells at the rate of Rs.10.40 per litre. His profit percentage is,
- 8%
- 6%
- 4%
- 2%
Solution 1: Problem analysis and solving in mind by basic mix and profit and loss concepts
Total cost for 25 litres bought is Rs. 300.
He sells $25+5=30$ litres costing Rs.300 at $\text{Rs.}10.40\times{30}=\text{Rs.}312$.
Profit is Rs.12 for cost price of Rs.300.
Profit percentage is,
$\displaystyle\frac{12}{300}\times{100}=4$%
Answer: Option c: 4%.
Key concepts used: Profit concept-- Profit percentage concept -- Concept of water is free -- Solving in mind.
Problem 2.
80 litres mixture of milk and water contains 10% milk. How much milk (in litres) must be added to make the water percentage in the mixture as 80%?
- 12
- 10
- 9
- 8
Solution 2: Problem analysis and conceptual solving in mind with awareness that water amount remains unchanged
10% of 80 litres is milk. So, 90%, that is, 72 litres is water.
Milk is added, not water. So we are interested in this amount of water that will remain fixed. After adding of milk this amount of 72 litres will form 80% of the mixture.
So by unitary method, 100% of the new mixture will be,
$\displaystyle\frac{72}{80}\times{100}=90$ litres.
Thus, 10 litres of milk is to be added.
Answer: Option b: 10.
Key concepts used: Percentage concept -- Mathematical reasoning -- Awareness of which component remains fixed and forms the new percentage -- Unitary method -- Solving in mind.
Problem 3.
In what ratio tea at Rs.240 per kg should be mixed with tea at Rs.280 per kg so that on selling the mixture at Rs.324 per kg there is a profit of 20%?
- 1 : 3
- 1 : 1
- 1 : 2
- 1 : 4
Solution 3: Problem analysis and solution in mind by Basic profit and loss concept and Mix ratio evaluation technique
The problem would be solved in two stages. At the first stage, the cost price per kg of the mixture would be evaluated from profit statement, and in the second stage the ratio of the mix to achieve this mix cost price.
For 20% profit, sale price would be 120% of cost price, that is, $\displaystyle\frac{6}{5}$ times of cost price. As sale price is Rs.324 per kg, cost price would be,
$\displaystyle\frac{5}{6}\times{324}=5\times{54}=\text{Rs.}270$ per kg.
Here we will introduce a simple technique of mix ratio evaluation to find the desired ratio of mix.
Mix ratio evaluation technique
Assume total mixed tea $x$ kg and higher priced tea, that is, tea of Rs.280 per kg as $y$ kg. So lower priced tea at Rs.240 per kg would be $(x-y)$ kg.
Equating prices,
$240(x-y)+280y=270x$,
Or, $(270-240)x=(280-240)y$.
Or, $x:y=4:3$.
Total is 4 portions out of which 3 portions is higher priced tea and 1 portion is lower priced tea.
The ratio will be, 1 : 3.
While solving in mind, we won't go through these deductions. We would assume $x$ and $y$ as stated and form the differences as shown to get the ratio of total portions to portions of higher priced tea. That is easy to do.
Answer: Option a: 1 : 3.
Key concepts used: Rich concept of profit, 20% profit means sale price is 1.2 times cost price -- Mix ratio evaluation technique -- Basic ratio concepts -- Basic mixing concepts -- Portions concepts.
Problem 4.
Three boxes of capacities 24 kg, 36 kg and 84 kg are completely filled with three varieties of wheat A, B and C respectively. All the three boxes were emptied and the three types of wheat were thoroughly mixed and the mixture was put back in the three boxes. How many kgs of type A wheat would be there in the third box?
- 12
- 16
- 10
- 14
Solution 4: Problem analysis and solving in mind by basic mixture concept
In the homogeneous mixture of three types of wheat, total weight of mixture is,
$24+36+84=144$ kg and the ratio of three types, $24:36:84$, totalling 144 portions with each portion of value 1 kg.
So when the 84 kg box is filled with the homogeneous mixture, its type A wheat content will be by unitary method,
$\displaystyle\frac{24}{144}\times{84}=14$ kg.
Answer: Option d: 14.
Key concepts used: Homogeneity in a mixture -- Ratio concept -- Portions concept -- Unitary mehod -- Mixing concept -- Solving in mind.
Problem 5.
A milkman mixes water with milk and sells the mixture at the cost price of the milk. The volume of water in litres to be mixed with each litre of pure milk to get a 25% profit is,
- $1\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{4}$
- $\displaystyle\frac{1}{5}$
- Cannot be determined with given information
Solution 5: Problem analysis and solving in mind by rich profit and loss concept and basic mixture concept
Sale price of 1 litre mixture is equal to the cost price of 1 litre of pure milk. This sale price to earn a profit of 25%, it would be $\displaystyle\frac{5}{4}$th of the cost price and so, the amount of pure milk in 1 litre of mixture should be, $\displaystyle\frac{4}{5}$th of 1 litre, that is, 0.8 litres. This would be the actual cost of the diluted milk which he sells at the price of 1 litre of pure milk and earns a profit of 25%.
By unitary method then, with every 0.8 litres of pure milk he mixes 0.2 litres of water, so with each litre of pure milk he would have to mix $\displaystyle\frac{1}{4}$ litre of water.
Answer: Option b: $\displaystyle\frac{1}{4}$.
Key concepts used: Basic mixing concept -- Basic profit and loss concept -- Event mapping -- Context awareness -- Solving in mind.
Problem 6.
Three equal capacity bottles contain mixtures of milk and water in ratios 2 : 5, 3 : 4 and 4 : 5 respectively. The three bottles are emptied into a large vessel. What will be the ratio of milk and water respectively in the large vessel?
- 73 : 113
- 73 : 189
- 73 : 106
- 73 : 116
Solution 6: Problem analysis and solving in mind by equal mix volume technique
By the equal mix volume technique, we will assume that we would mix $2+5=7$ litres from first bottle, $3+4=7$ litres from second bottle and 7 litres also from the third bottle. We have chosen to do this because the bottles are of equal capacity. If we take equal volumes of mixed milk from each bottle then, the ratio of milk to water in this mix will be the same as when the full bottles are mixed (because of proportionality and homogeneity of mixed liquid).
For the first two bottles, the amount of milk would be 2 litres and 3 litres, that is 5 litres in total. This is because the total volume of 7 litres ensured that portion value of each ratio is 1 litre.
For the third bottle though the portion value is different and we have 4 litres of milk for evey 9 litres of mixture. By unitary method then, in 7 litres of mixed milk from the third bottle the milk amount will be,
$7\times{\displaystyle\frac{4}{9}}=\displaystyle\frac{28}{9}$ litres.
In 21 litres of mixture that we made, the milk amount would then be,
$5+\displaystyle\frac{28}{9}=\displaystyle\frac{73}{9}$ litres.
In every litre of mixture then the milk will be,
$\displaystyle\frac{73}{9\times{21}}=\frac{73}{189}$ litres, and the amount of water will be the rest,
$1-\displaystyle\frac{73}{189}=\frac{116}{189}$ litres.
Milk to water ratio will be,
$73 : 116$.
Answer: Option d: 73 : 116.
Key concepts used: Mixing liquid concept -- Homogeneity of mixed liquid -- Equal mix volume technique -- Ratio concept -- Portion concept -- Unitary method -- Solving in mind.
The equal mix volume technique reduces number of fraction additions and so saves time. We have taken the opportunity of total number of portions as 7 for both the first two ratios and fixed the equal volume as 7 litres to be mixed from each bottle. We had to do fraction calculation only for the third bottle.
With clarity in concepts and the quick solving method, the problem could be solved easily in mind as the calculations are straightforward.
Problem 7.
In a mixture of 25 litres, the ratio of milk to water is 4 : 1. Another 3 litres of water is added to the mixture. The ratio of milk to water in the new mixture is,
- 5 : 1
- 5 : 2
- 5 : 4
- 5 : 3
Solution 7: Problem analysis and solving in mind by portion value concept and mixing concept
By the given ratio, in 25 litres of liquid, milk is 4 portions that is 20 litres and rest 5 litres is water. Adding 3 litres of water makes the total volume as 28 litres and water volume 8 litres. Milk volume remains same as 20 litres.
So in the new mix, the milk to water ratio would be $20 : 8 = 5 :2$.
Answer: Option b: 5 : 2.
Key concepts used: Homogeneity in mixture -- liquid addition -- Basic mixing concept, one liquid volume remains fixed -- Event mapping -- Solving in mind.
Problem 8.
Two alloys contain tin and iron in the ratio 1 : 2 and 2 : 3. If the two alloys are mixed in the proportion 3 : 4 respectively by weight, the ratio of tin and iron in the newly formed alloy is,
- 13 : 22
- 12 : 23
- 10 : 21
- 14 : 25
Solution 8: Problem analysis and solving in mind by Multiple of LCM of total portion volume technique
To avoid time taking fraction calculation as much as possible we will assume the volumes of two alloys to be mixed as the multiples of the LCM of total number of portions of the two ratios, that is 3 times LCM 15 of 3 and 5, that is 45 kgs for first alloy and 4 times 15, that is 60 kgs for the second alloy. This will ensure in 45 kgs of first alloy tin and iron ratio to remain unchanged as 1 : 2 and same for the second alloy with its ratio of tin and iron to remain as 2 : 3 in 60 kgs. In addition it will ensure the ratio of 3 : 4 in the mix of two alloys.
The advantage with this selection of volumes to be mixed is—the amount of tin and iron in both the alloys being mixed will be pure integers free from fractions. Fraction calculation takes time. We want to speed up the solution process.
Total volume will be 105 kgs in which tin will be $15+24=39$ kgs and iron $30+36=66$ kgs.
Ratio of tin and iron in the new alloy will then be,
$39 : 66=13:22$.
Answer: Option a: 13 : 22.
Key concepts used: Basic mixing concept -- Homogeneity of a mix concept -- Multiple of LCM of total portion volume technique -- Solving in mind.
The technique is based on advanced concepts on ratio, portions, mixing and homogeneity of a mix. But once you are clear about how the technique works, applying it takes very little time.
In the conventional solution you calculate per kg tin in first alloy to be, $\displaystyle\frac{1}{3}$ kg and in the second alloy to be, $\displaystyle\frac{2}{5}$ kg. Take 3 kgs of first alloy with tin as 1 kg and 4 kgs of second alloy with tin as $\displaystyle\frac{8}{5}$ kg.
So in total of 7 kgs of new alloy you get tin as,
$1+\displaystyle\frac{8}{5}=\displaystyle\frac{13}{5}$kgs,
And the rest iron as,
$7-\displaystyle\frac{13}{5}=\displaystyle\frac{22}{5}$ kgs.
The ratio of tin and iron in the new alloy,
$13 : 22$.
With the advanced technique you add integers and in the conventional solution carefully manipulate fractions that is generally more time consuming.
Problem 9.
A litre of water weighs a kilogram while a litre of a second liquid weighs 1.340 kilogram. A mixture of the two weighs 1.270 kilogram per litre. The ratio of their volume in a litre of mixture is,
- 7 : 27
- 7 : 17
- 27 : 34
- 17 : 24
Solution 9: Problem analysis and solving in mind by Mix ratio evaluation technique
As a higher density liquid is mixed with the lower density water, the density of the mix, that is, weight per litre will be higher than water but lower then the higher density liquid.
The technique we use in this situation is what we call Mix ratio evaluation technique.
Mix ratio evaluation technique
We assume the volume $y$ litres as the volume of higher density liquid that is mixed. If total volume is $x$ in the mix, lighter liquid water volume will be $(x-y)$ litres.
The weight of $x$ litres of the mix will be,
$(x-y)\times{1}+y\times{1.340}=x\times{1.270}$
Or, $y\times{(1.340-1)}=x\times{(1.270-1)}$.
We do not deduce these steps, but directly use this relation to get finally,
$0.340y=0.270x$,
Or, $y=\displaystyle\frac{0.270}{0.340}x$.
This is,
$\text{Volume of higher density liquid}$
$=\displaystyle\frac{\text{mix density}- \text{lighter liquid density}}{\text{heavier liquid density}-\text{lighter liquid density}}\times{\text{Total mix volume}}$
This results in ratio of heavier liquid volume to total mix volume as,
$\displaystyle\frac{y}{x}=\frac{0.270}{0.340}$,
And the ratio of volumes of lighter liquid water to total mix volume as,
$\displaystyle\frac{x-y}{x}=1-\displaystyle\frac{y}{x}=\displaystyle\frac{0.340-0.270}{0.340}=\frac{0.70}{0.340}$.
The required ratio of water and heavier liquid will then be,
$\displaystyle\frac{x-y}{y}=\frac{7}{27}$
This technique is a general one and can be applied for mixing of any two matter with different costs, or weights by volume.
Once you understand the technique, you can solve such problems mentally and very quickly.
We call it as Mix ratio evaluation technique that can be applied for mixing solids into alloys, mixing two types of tea into blended tea or mixing liquids of different densities.
Answer: Option a: 7 : 27.
Key concepts used: Basic mixing concept -- Mixing of two liquids of different densities in a specific ratio to get a given density -- Mix ratio evaluation technique -- Solving in mind.
Problem 10.
Two varieties of sugar are mixed in a certain ratio. The cost of the mixture per kg is Rs.0.50 less than that of the superior and Rs.0.75 more than the inferior variety. The ratio in which the inferior and the superior varieties of sugar have been mixed is,
- 5 : 2
- 5 : 1
- 3 : 2
- 2 : 3
Solution 10: Problem analysis and solving in mind by Mix ratio evaluation technique
This problem can be solved very quickly as the problem description gives the difference between the cost of mix and cost of inferior quality sugar as 0.75 which is the numerator in our relation,
$y=\displaystyle\frac{0.75}{\text{higher quality sugar cost}-\text{inferior quality sugar cost}}\times{x}$,
where $x$ is the total weight of mixture and $y$ the weight of higher quality sugar in the mixture.
The denominator is just the sum of the two given differences, that is, 1.25.
So,
$y=\displaystyle\frac{0.75}{1.25}x$,
Or, $(x-y)=\displaystyle\frac{0.50}{1.25}x$
And the ratio of inferior quality and superior quality sugar as,
$(x-y):y=0.50 : 0.75=2:3$.
This makes sense as the mix cost is nearer to the superior quality sugar cost compared to the inferior quality sugar cost implying that more of superior quality sugar is present in the mixture.
Answer: Option d: 2 : 3.
Note: Unless you are conceptually clear about the technique as well as having good practice in using the technique, it is safer to use the more time consuming conventional method.
In the conventional method, instead of using the final result of the technique you deduce the relation.
Key concepts used: Basic mixing concepts -- Homogeneity of a mix concept -- Mix ratio evaluation technique -- Solving in mind.
Overall concepts and techniques used: Homogeneity in a mixture, Basic percentage concepts, Solving in mind, Mathematical reasoning, Basic ratio concepts, Event mapping, Portion use technique, Unitary method, Basic mixing concepts, Cost of a mixture, Profit and loss concept, Profit percentage concept, Concept of water is free, Mix ratio evaluation technique, Rich concept of profit and loss.
Note: Though this set of 10 mixture problems is not easy, all the problems can be quickly solved in mind by applying concept based techniques.
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