In the previous section we saw how to calculate the given percentage of a quantity. We also saw some solved examples. In this section, we will see more solved examples.
Solved example 8.15
A basket contains 240 apples. 10% of this is Golden apples 15% is Green apples. The rest is ordinary apples. What is the number of ordinary apples?
Solution:
• Total number of apples = 240
• Number of Golden apples = 10% of 240 = 240 × 0.1 = 24
• Number of Green apples = 15% of 240 = 240 × 0.15 = 36
• Let the number of ordinary apples (the unknown which we are required to calculate) be x
• Then we can write: 24 +36 + x =240 ⇒ 60 + x =240
∴ x = 240 -60 =180
Solved example 8.16
An iron bar is 150 cm in length. 32% of this length is to be welded to a steel post. Can you draw the finished drawing after the weld is made?
Solution:
• Length of iron bar = 150 cm
• Portion of this 150 cm, which is to be welded = 32% of 150 = 150 × 0.32 = 48 cm
• So the final drawing will be as shown in the fig.8.2(c) below:
Thus we have seen the method to calculate any percentage of any quantity. Some times We will have to do the process in a 'reverse order'. Let us see some examples:
Solved example 8.17
A person saves ₹ 3600 from his salary. If this is 12% of his salary, what is his salary?
Solution:
• Let the salary (the unknown which we are required to calculate) be equal to x
• Then x × 0.12 = 3600 ⇒ x × 12⁄100 = 3600⁄1 ⇒ 12x⁄100 = 3600⁄1
• Taking cross products we get: 12x × 1 = 3600 × 100 ⇒12x = 360000
∴ x = 360000⁄12 = 30000
• So the salary = 30000
Solved example 8.18
A student scored 108 marks in an exam. If this is 72% of the total marks, then, what is the total marks of the exam?
Solution:
• Let the total marks (the unknown which we are required to calculate) be x
• Then x × 0.72 =108 ⇒ x × 72⁄100 = 108⁄1 ⇒ 72x⁄100 = 108⁄1
• Taking cross products we get: 72x × 1 = 108 × 100 ⇒same as 72x = 10800
∴ x = 10800⁄72 = 150
• So the total marks = 150.
Solved example 8.19
A person spends 25% of his income on food, and 7% as house rent. If he spends ₹750 on these two items, what is his income?
Solution:
• Let the income be equal to x
• Amount spent on food = 25% of x = x × 0.25 = 0.25x
• Amount spent on rent = 7% of x = x × 0.07 = 0.07x
• Total amount spent on these two items = 0.25x + 0.07 x = 0.32x
• But this total amount is given as ₹4800
• So we can write 0.32x = 4800 ⇒ 32x⁄100 = 4800⁄1
• Taking cross products we get 32x = 4800 × 100 ⇒32x = 480000
∴ x = 480000⁄32 = 15000
• So the income = ₹15000
Solved example 8.20
Mrs A spends 15% of her salary on house rent. Out of the remaining amount, 45% is spent on food, and 35% on other expenditure. If she manages to save ₹3400, what is her salary?
Solution:
• Let the salary be x.When she receives her salary, the first thing that she does, is to keep aside 15%. This is for house rent. So we can write:
• House rent = 0.15x. And the remaining = (100 - 15)% of x ⇒ 85% of x = 0.85x
• 45% of this remaining is spent for food. So we can write:
Food expenditure = 45% of 0.85x = 0.45 × 0.85x = 0.3825x
•35% of this remaining is spent for other expenditure. So we can write:
Other expenditure = 35% of 0.85x = 0.35 × 0.85x = 0.2975x
• So total expenditure (out of the remaining 0.85x) = 0.3825x + 0.2975x = 0.68x
∴ the remaining = 0.85x - 0.68x = 0.17x
• This is the amount which is saved, and it is given as ₹3400/-. So we can write:
0.17x = 3400 ⇒ 17x⁄100 = 3400 ⇒17x = 340000
∴ x = 340000⁄17 = 20000
• So the salary is equal to ₹20000/-
Solved example 8.21
A man lost 15% of his money. He spent 30% of what he had left. After this spending, the money that he had was ₹2800. How much money did he originally have?
Solution:
• Let the money that was originally had be x. 20% of this was lost. So the remaining is 80%. So we can write:
• Money in hand after loosing 20% = 80% of x = 0.80x
• He spent 30% of this remaining. So 70% of the remaining will be in hand after spending. So we can write:
• Money in hand after spending = 70% of 0.80x = 0.7 × 0.80 x = 0.56x
• This amount is given as ₹2800. So we can write: 0.56x = 2800
∴ x = 2800⁄0.56 = 5000
• So the amount of money that he originally had was ₹5000/-
In the next section we will see the relation between percentage and ratios.
Solved example 8.15
A basket contains 240 apples. 10% of this is Golden apples 15% is Green apples. The rest is ordinary apples. What is the number of ordinary apples?
Solution:
• Total number of apples = 240
• Number of Golden apples = 10% of 240 = 240 × 0.1 = 24
• Number of Green apples = 15% of 240 = 240 × 0.15 = 36
• Let the number of ordinary apples (the unknown which we are required to calculate) be x
• Then we can write: 24 +36 + x =240 ⇒ 60 + x =240
∴ x = 240 -60 =180
Solved example 8.16
An iron bar is 150 cm in length. 32% of this length is to be welded to a steel post. Can you draw the finished drawing after the weld is made?
Solution:
• Length of iron bar = 150 cm
• Portion of this 150 cm, which is to be welded = 32% of 150 = 150 × 0.32 = 48 cm
• So the final drawing will be as shown in the fig.8.2(c) below:
 |
| Fig.8.2 |
Thus we have seen the method to calculate any percentage of any quantity. Some times We will have to do the process in a 'reverse order'. Let us see some examples:
Solved example 8.17
A person saves ₹ 3600 from his salary. If this is 12% of his salary, what is his salary?
Solution:
• Let the salary (the unknown which we are required to calculate) be equal to x
• Then x × 0.12 = 3600 ⇒ x × 12⁄100 = 3600⁄1 ⇒ 12x⁄100 = 3600⁄1
• Taking cross products we get: 12x × 1 = 3600 × 100 ⇒12x = 360000
∴ x = 360000⁄12 = 30000
• So the salary = 30000
Solved example 8.18
A student scored 108 marks in an exam. If this is 72% of the total marks, then, what is the total marks of the exam?
Solution:
• Let the total marks (the unknown which we are required to calculate) be x
• Then x × 0.72 =108 ⇒ x × 72⁄100 = 108⁄1 ⇒ 72x⁄100 = 108⁄1
• Taking cross products we get: 72x × 1 = 108 × 100 ⇒same as 72x = 10800
∴ x = 10800⁄72 = 150
• So the total marks = 150.
Solved example 8.19
A person spends 25% of his income on food, and 7% as house rent. If he spends ₹750 on these two items, what is his income?
Solution:
• Let the income be equal to x
• Amount spent on food = 25% of x = x × 0.25 = 0.25x
• Amount spent on rent = 7% of x = x × 0.07 = 0.07x
• Total amount spent on these two items = 0.25x + 0.07 x = 0.32x
• But this total amount is given as ₹4800
• So we can write 0.32x = 4800 ⇒ 32x⁄100 = 4800⁄1
• Taking cross products we get 32x = 4800 × 100 ⇒32x = 480000
∴ x = 480000⁄32 = 15000
• So the income = ₹15000
Solved example 8.20
Mrs A spends 15% of her salary on house rent. Out of the remaining amount, 45% is spent on food, and 35% on other expenditure. If she manages to save ₹3400, what is her salary?
Solution:
• Let the salary be x.When she receives her salary, the first thing that she does, is to keep aside 15%. This is for house rent. So we can write:
• House rent = 0.15x. And the remaining = (100 - 15)% of x ⇒ 85% of x = 0.85x
• 45% of this remaining is spent for food. So we can write:
Food expenditure = 45% of 0.85x = 0.45 × 0.85x = 0.3825x
•35% of this remaining is spent for other expenditure. So we can write:
Other expenditure = 35% of 0.85x = 0.35 × 0.85x = 0.2975x
• So total expenditure (out of the remaining 0.85x) = 0.3825x + 0.2975x = 0.68x
∴ the remaining = 0.85x - 0.68x = 0.17x
• This is the amount which is saved, and it is given as ₹3400/-. So we can write:
0.17x = 3400 ⇒ 17x⁄100 = 3400 ⇒17x = 340000
∴ x = 340000⁄17 = 20000
• So the salary is equal to ₹20000/-
Solved example 8.21
A man lost 15% of his money. He spent 30% of what he had left. After this spending, the money that he had was ₹2800. How much money did he originally have?
Solution:
• Let the money that was originally had be x. 20% of this was lost. So the remaining is 80%. So we can write:
• Money in hand after loosing 20% = 80% of x = 0.80x
• He spent 30% of this remaining. So 70% of the remaining will be in hand after spending. So we can write:
• Money in hand after spending = 70% of 0.80x = 0.7 × 0.80 x = 0.56x
• This amount is given as ₹2800. So we can write: 0.56x = 2800
∴ x = 2800⁄0.56 = 5000
• So the amount of money that he originally had was ₹5000/-
In the next section we will see the relation between percentage and ratios.
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