Arithmetic Chapter 6 Mixture
If two or more articles of different values and in different quantities or volume are mixed to produce an article of a different value and quantity. This is known as mixture.
Arithmetic Chapter 6 Mixture Examples
Example 1. If the ratio of measurements of urea and potash in 25 kg of mixed fertilizer is 3:2, then find the amount of potash.
Solution: The ratio of measurements of urea and potash = 3:2
The proportional part of the quantity of potash = \(\frac{2}{3+2}=\frac{2}{5}\)
The amount of potash is 25 kg of mixed fertilizer is \(\left(25 \times \frac{2}{5}\right)\) kg = 10 kg
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Example 2. If 2 kg of Darjeeling tea costing ₹300 per kg mixed with 3 kg of Assam tea kg of mixture tea. costing ₹200 per kg, then find the price of 1 kg of mixture tea.
Solution: The price of 2 kg of Darjeeling tea at the rate of ₹300 per kg is ₹(2 x 300) or ₹600.
The price of 3 kg Assam tea is ₹(3 x 200) or ₹600.
The price of (2+3) kg or 5 kg of mixtured tea is ₹(600+ 600) or ₹1200.
∴ The price of 1 kg of mixtured tea is ₹\(\frac{1200}{5}\) or ₹240.
Example 3. In 48 litres of mixture the ratio of measurements of water and glycerine is 2 1. What value of glycerine should be added to mixture so that the ratio of measurements of water and glycerine becomes 3: 2?
Solution: The quantity of water is \(\left(48 \times \frac{2}{2+1}\right)\) litres = \(\left(48 \times \frac{2}{3}\right)\) litres = 32 litres.
The quantity of glycerine is (48 – 32) litres = 16 litres.
Let x litres of glycerine be added to the mixture.
According to questions,
⇒ \(\frac{32}{16+x}=\frac{3}{2}\)
⇒ 48 + 3x = 64
⇒ 3x = 64 – 48 = 16
⇒ x = \(\frac{16}{3}\) = 5\(\frac{1}{3}\)
∴ 5\(\frac{1}{3}\) litre of glycerine should be added.
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Example 4. The ratio of measurements of two types of tea in a mixture is 2: 3. If 2 kg of first type of tea is mixed with 55 kg of blended tea then, find the ratio of measurements of two types of tea.
Solution: The quantity of 1st type of tea = \(\left(55 \times \frac{2}{2+3}\right) \mathrm{kg}=\left(55 \times \frac{2}{5}\right) \mathrm{kg}=22 \mathrm{~kg}\)
The quantity of 2nd type of tea = (55-22) kg = 33 kg.
If 2 kg of 1st types of tea is mixed, then the ratio of the measurement of two types of tea is (22 + 2): 33 = 24: 33 = 8: 11
8: 11 ratio of measurements of two types of tea.
Example 5. The ratio of measurements of copper, zinc and nickel in German silver is 4: 3: 2 respectively. Find out what weight of zinc should be added to 54 kg of German silver, so that the ratio of measurements become 6:5: 3.
Solution: In 54 kg of German silver the quantity of zinc = (54 x \(\frac{3}{4+3+2}\))kg
= \(\left(54 \times \frac{3}{9}\right)\) kg = 18 kg.
Let x kg of zinc be added.
∴ The quantity of zinc in (54 + x) kg of German silver is (54 + x) x \(\frac{5}{6+5+3}\) kg = \(\frac{5}{14}\)(54 + x) kg.
According to question, x + 18 = \(\frac{5}{14}\)(54 + x)
⇒ 14 (x + 18) = 5(54 + x)
⇒ 14x + 252 = 270+ 5x
⇒ 14x – 5x = 270 – 252
⇒ 9x = 18
⇒ x =\(\frac{18}{9}\) = 2
∴ 2 kg of zinc be added.
Example 6. In a vessel of beverage the ratio of measurement of syrup and water is 4:1. Find out what part of the drink should removed and replaced by water so that the volume of syrup and water becomes equal.
Solution: Let x unit of beverage is there in a vessel. y unit of it is removed and replaced by the same volume of water.
The quantity of syrup in x units of beverage is \(\left(\frac{4}{4+1} \times x\right)\) units = \(\frac{4x}{5}\) units and the quantity of water = \(\left(\frac{1}{4+1} \times x\right)\) units = \(\frac{x}{5}\) units
Again the quantity of syrup in y units = \(\frac{4y}{5}\) units and water = \(\frac{y}{5}\) units
In new beverage the quantity of syrup is \(\left(\frac{4 x}{5}-\frac{4 y}{5}\right)\) units and water is \(\left(\frac{x}{5}-\frac{y}{5}+y\right)\) units.
According to question,
\(\frac{4 x}{5}-\frac{4 y}{5}=\frac{x}{5}-\frac{y}{5}+y \Rightarrow \frac{4 x}{5}-\frac{x}{5}=\frac{4 y}{5}-\frac{y}{5}+y\)
⇒ \(\frac{4 x-x}{5}=\frac{4 y-y+5 y}{5} \Rightarrow \frac{3 x}{5}=\frac{8 y}{5} \Rightarrow 3 x=8 y \Rightarrow y=\frac{3 x}{8}\)
∴ \(\frac{3}{8}\) part of the beverage will be removed.
Example 7. In what ratio tea of price ₹35 per kg and ₹28 per kg be mixed, so that there will be neither loss or gain on selling mixed tea at the rate of 30 per kg?
Solution: Let x kg tea of price ₹35 per kg and y kg tea price ₹28 per kg be mixed.
Total price of mixture tea is ₹(35x + 28y).
According to the question:
35x + 28y = 30 (x + y)
⇒ 35x – 30x = 30y – 28y ⇒ 5x = 2y
⇒ \(\frac{x}{y}\) = \(\frac{2}{5}\)
⇒ x: y = 2:5
∴ The required ratio is 2: 5.
Example 8. \(\frac{1}{2}\) and \(\frac{1}{5}\) parts two similar vessels contain fruit juice. The remaining empty parts of these vessels are filled with water and poured the juice mixed with water of the two vessels into another big vessels. Find out the ratio of measurements of fruit and water in the new vessel.
Solution: In the 1st vessel contains \(\frac{1}{2}\) parts fruit juice and (1-\(\frac{1}{2}\)) parts or \(\frac{1}{2}\) parts water.
In the second vessel contains \(\frac{1}{5}\) parts fruit juice and (1-\(\frac{1}{5}\)) or \(\frac{4}{5}\) parts water.
The ratio of fruir juice and water in new vessel is (\(\frac{1}{2}\)+\(\frac{1}{5}\)): (\(\frac{1}{2}\)+\(\frac{4}{5})\) = \(\frac{7}{10}\): \(\frac{13}{10}\) = 7: 13
Example 9. Two different types of stainless steel contain chromium and steel in the ratio of measurement of 3 : 8 and 5: 17 respectively. Find out in what proportion these two types of steel should be mixed, so that the ratio of measurement of chromium and steel becomes 8: 25.
Solution: Let x unit of 1st type of stainless steel is mixed with y unit of 2nd type of stainless steel.
The quantity of chromium in x unit of 1st type stainless steel is (x+\(\frac{3}{3+8}\)) unit or \(\frac{3x}{11}\) unit and quantity of steel is (x x \(\frac{8}{3+8}\)) unit or \(\frac{8x}{11}\) unit.
The quantity of chromium in y unit of 2nd type stainless steel is (y x \(\frac{5}{5+17}\)) unit or \(\frac{5y}{22}\)
In the new mixture the quantity of chromium is (\(\frac{3x}{11}\)+\(\frac{5y}{22}\)) unit and steel is (\(\frac{8x}{11}\)+\(\frac{17y}{22}\)) unit.
According to the questions,
\(\frac{\frac{3 x}{11}+\frac{5 y}{22}}{\frac{8 x}{11}+\frac{17 y}{22}}=\frac{8}{25} \Rightarrow \frac{\frac{6 x+5 y}{22}}{\frac{16 x+17 y}{22}}=\frac{8}{25}\)
⇒ \(\frac{6 x+5 y}{16 x+17 y}=\frac{8}{25}\)
⇒ 150x + 125y = 128x + 136y
⇒ 150x – 128x = 136y – 125y
⇒ 22x = 11y ⇒ \(\frac{x}{y}\)= \(\frac{11}{2}\)
⇒ x: y = 1: 2
∴ The two types of stainless steel are mixed up in the ratio of measurements of 1: 2.
Example 10. In what ratio is water to be mixed with milk at ₹30 per litre to make the price of the mixture ₹25 per litre?
Solution: Let x litre of water is mixed with y litre of milk.
The price of y litre of milk is ₹30y.
The price of (x + y) litre of mixture is ₹25 (x + y)
According to question,
25(x+y)= 30 y
⇒ 25x + 25y = 30y
⇒ 25x = 30y – 25y
⇒ 25x = 5y
⇒ \(\frac{x}{y}\)= \(\frac{5}{25}\)
⇒ x: y = 1:5
∴ The ratio of measurement of water and milk is 1: 5.
Example 11. Choose the Correct Answer:
1. If the ratio of measurement of milk and water in 30 litres is 4: 1, the measurement of milk is
- 5 litres
- 6 litres
- 24 litres
- 25 litres
Solution: The proportional part of the quantity of milk = \(\frac{4}{4+1}\) = \(\frac{4}{5}\)
The quantity of milk in 30 litres of mixture = (\(\frac{4}{5}\) x 30) litres = 24 litres
∴ So the correct answer is 3. 24 litres
The measurement of milk is 24 litres
2. 2/5 parts of a vessel contain syrup. If the vessel filled with water, then the ratio of syrup and water becomes
- 2: 5
- 3: 5
- 2: 3
- 4: 5
Solution: 2/5 parts of a vessel contain syrup. If the vessel filled with water, then the amount of water is (1- \(\frac{2}{5}\)) part or 3/5 part of that vessel.
The ratio of syrup and water = \(\frac{2}{5}\): \(\frac{3}{5}\) = 2:3
∴ The correct answer is 3. 2: 3
The ratio of syrup and water becomes 2: 3
3. In a alloy iron contains 20% copper contains 30% and the rest is nickel. The ratio of measurements of iron, copper and nickel is
- 5: 2: 3
- 2: 3: 5
- 3: 2: 5
- 5: 3: 2
Solution: The measurements of nickel is {100 (20+30)} % = 50%
∴ The ratio of measurements of iron, copper and nickel is 20%: 30%: 50% = \(\frac{20}{100}\): \(\frac{30}{100}\): \(\frac{50}{100}\) = 2: 3: 5
∴ The correct answer is 2. 2: 3: 5
The ratio of measurements of iron, copper and nickel is 2: 3: 5
Example 12. Write ‘True’ or ‘False’:
1. In 60 litres of a mixture of syrup and water contains 30% water. If 30 litres of syrup be added to the mixture, then the new mixture contains 20% water.
Solution: The quantity of water in 60 litres of a mixture is (60x\(\frac{30}{100}\)) litres or 18 litres.
So the quantity of syrup (60 – 18) litres = 42 litres.
If 30 litres of syrup be added to the mixture the total volume is (60+30) litres or 90 litres.
∴ The quantity of water = (\(\frac{18}{90}\)x100)% = 20%
∴ So the statement is true.
2. In 50 c.c. of concentrated sulphuric acid contains 2% water. The ratio of measurements of acid and water is 25: 1.
Solution: The quantity of water in 50 c.c. of concentrated sulphuric acid is
(50x\(\frac{2}{100}\)) = 1 c.c.
The quantity of acid is (50 – 1) c.c. = 49 c.c.
The ratio of acid and water is 49: 1.
∴ So the statement is false.
Example 13. Fill in the blanks:
1. If the ratio of measurements of sulphur, charcoal and potassium nitrate in 240 gm of gun powder is 1 : 3: 4, then the quantity of sulphur is ______ gm.
Solution: The quantity of sulphur = (240 x \(\frac{1}{1+3+4}\)) gm = (240 x \(\frac{1}{8}\)) gm = 30 gm
2. In air nitrogen contains 77.16%, oxygen contains 20.60%, carbon dioxide contains 0.04%, water vapour contains 1.40% and the rest is inert gas. The ratio of measurements of carbon dioxide and inert gas is ________
Solution: The quantity of inert gas is
{100 (77.16 + 20.60 + 0.04 + 1·40)}% = (100 – 99.20)% = 0.80%
The ratio of measurements of carbon dioxide and inert gas is
0.04%: 0.80% = 4: 80 = 1: 20.
The ratio of measurements of carbon dioxide and inert gas is 1: 20.