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Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1 - #NCSOLVE 📚

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Students often refer to Class 8 Maths Notes and Part 2 Chapter 1 Fractions in Disguise Class 8 Notes during last-minute revisions.

Class 8 Maths Chapter 1 Fractions in Disguise Notes

Class 8 Fractions in Disguise Notes

→ Percentages are widely used in our daily lives.

→ Percentages are fractions with a denominator of 100.

→ Percentages are denoted using the symbol ʻ%ʼ, pronounced ʻper centʼ.
x% = \(\frac {x}{100}\)

→ Fractions can be converted to percentages and vice versa. Decimals, too, can be converted to percentages and vice versa.
For example, \(\frac {4}{10}\) = 0.4 = 40%.
Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1 1

→ We have learnt to find the exact number when a certain percentage of the total quantity is given.

→ When parts of a quantity are given to us as ratios, we have seen how to convert them to percentages.

Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1

→ The increase or decrease in a certain quantity can also be expressed as a percentage.

→ The profits or losses incurred in transactions, and tax rates, can be expressed in terms of percentages.

→ We have seen how a quantity or a number grows when compounded. Interest rates are a common example of compounding.
If p is the principal, r is the rate of interest, and t is the number of terms, then the total amount after the maturity period is
Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1 2

→ A situation or a problem can often be solved by describing it using a rough diagram. We have learnt to estimate and do mental computations to solve problems related to percentages.

Fractions As Percentages
A way of comparing quantities is a percentage. Percentages are numerators of fractions with a denominator of 100. Per cent means per hundred.
62 out of 100 is written as 62%.

To write a fraction as a percent, we multiply the fraction by 100, simplify, and attach the symbol of %.
\(\frac {2}{5}\) can be written in percent as \(\frac {2}{5}\) × 100 = 40%.

To write a percent as a fraction, we drop the % symbol and multiply the number by \(\frac {1}{100}\).
25% can be written as a fraction as 25 × \(\frac {1}{100}\) = \(\frac {1}{4}\)

Percentage of Some Quantity
To find a% of a number, we multiply the number by \(\frac {a}{100}\). [ a% of n = \(\frac {a}{100}\) × n]
We can find the exact number when a certain percentage of the total quantity is given.
Suppose a% of x is b, then x = b × \(\frac {100}{a}\)
Freehand computation:

  • To find 50%, divide by 2
  • To find 25%, divide by 4
  • To find 20%, divide by 5
  • To find 12\(\frac {1}{2}\)%, divide by 8
  • To find 10%, divide by 10
  • To find 5%, divide by 20
  • To find 2%, divide by 50
  • To find 1 %, divide by 100

To find n% of a number, first find 1% and then multiply by ‘n’.

Using Percentages
Percentage increase = \(\frac{\text { increase in quantity }}{\text { original quantity }} \times 100 \%\)

Percentage decrease = \(\frac{\text { decrease in quantity }}{\text { original quantity }} \times 100 \%\)

To increase a number by a% multiply it by \(\frac{100+a}{100}\)

Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1

To decrease a number by a% multiply it by \(\frac{100-a}{100}\)

Cost Price (CP): The price at which an article is bought.

Selling Price (SP): The price at which an article is sold.

Profit or Gain: If SP > CP, then the amount by which SP exceeds CP.

Loss: If CP > SP, then the amount by which CP exceeds SP.

SP > CP (profit or gain)

Profit = SP – CP

Profit % = \(\frac{\mathrm{SP}-\mathrm{CP}}{100} \times 100 \%\) or \(\frac{\text { Profit }}{\text { CP }} \times 100 \%\)

If profit is P %, then,
SP = \(\frac{(100+\mathrm{P})}{100} \times \mathrm{CP}\)
CP = \(\frac{100}{(100+\mathrm{P})} \times \mathrm{SP}\)
Profit = \(\frac{\mathrm{P}}{100} \times \mathrm{CP}\)

SP < CP (Loss)

Loss = CP – SP

Loss % = \(\frac{\mathrm{CP}-\mathrm{SP}}{\mathrm{CP}} \times 100 \%\) or \(\frac{\text { Loss }}{\mathrm{CP}} \times 100 \%\)

If loss is L %, then,
SP = \(\frac{(100-L)}{100} \times C P\)
CP = \(\frac{100}{(100-L)} \times S P\)
Loss = \(\frac{\mathrm{L}}{100} \times \mathrm{CP}\)

MP (marked price): the price written on the label of the object ( MP is also called list price)

SP (selling price): the price at which an item is sold. It is obtained after deducting the discount from the marked price.

Discount: The reduction made on the marked price of the item. If no discount is given, then the selling price is the same as the marked price.

Discount = MP – SP

SP = MP – discount

MP = SP + discount

If the discount given is D%, then
SP = \(\frac{100-\mathrm{D}}{100} \times \mathrm{MP}\)
MP = \(\frac{100}{100-D} \times S P\)
D % = \(\frac{\mathrm{MP}-\mathrm{SP}}{\mathrm{MP}} \times 100 \%\)

Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1

Goods and Services Tax (GST): Tax to be paid to the government on the sale of a particular item. It is calculated as a given percentage of the list price.
Rates of GST in India are 5%, 12%, 18% and 28%.
However, GST on gold is 3%.
If the rate of sales tax is r%, then GST = \(\frac {r}{100}\) × SP.

Growth and Compounding
Simple interest (no compounding):
I = \(\frac {PRT}{100}\)
A = P + SI
Note: T is always taken in years.
If time is given in months, then to convert it into year we divide it by 12.
If time is given in days, then to convert it into years, we divide it by 365.
Also: P = \(\frac{100 \times \mathrm{I}}{\mathrm{P} \times \mathrm{T}}\), T = \(\frac{100 \times I}{P \times R}\), and R = \(\frac{100 \times I}{P \times T}\)

Compound Interest: Money is said to be at compound interest if, at the end of the year or a fixed period, the interest that becomes due is not paid but added to the principal, and the amount obtained becomes the new principal. The process is repeated till the amount for the last period is determined. Now the difference between the final amount and the principal is called compound interest. Compound interest is always more than the simple interest, provided the principal, rate, and time are the same.
A = \(\mathrm{P}\left(1+\frac{r}{100}\right)^n\)
CI = A – P
Where, P: principal, r: effective rate, n: number of conversion periods, A: amount

If same principal P is lent at compound interest compounded annually as well as simple interest at the same rate of r% per annum then for the first year Cl and SI are same, but for two year CI exceeds SI by \(\mathrm{P}\left(\frac{r}{100}\right)^2\)

Decline
An increase in price or quantity with the passage of time is called appreciation.
Appreciated value is always more than the original value.
A decrease in price or quantity with the passage of time is called depreciation.
Depreciated value is always less than the original value.

Original Value (P): Value or quantity at the beginning of the period.
Rate of Appreciation and Depreciation (r): The rate at which the value or quantity increases or decreases. Unless otherwise specified, it is generally taken as per annum.
n: number of conversion periods
Final Value (A): Value or quantity at the end of the period.
For appreciation A = \(\mathrm{P}\left(1+\frac{r}{100}\right)^n\)
Net increase = A – P

Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1

For depreciation A = \(\mathrm{P}\left(1-\frac{r}{100}\right)^n\)
Net decrease = P – A

Tricky Percentages
If successive discounts of D1 % and D2 % are offered then SP = \(\frac{100-D_1}{100} \times \frac{100-D_2}{100} \times M P\)
Successive discounts are also called a discount series.
For “buy x get y free” the percentage discount = \(\frac{y}{x+y} \times 100 \%\)

The post Fractions in Disguise Class 8 Notes Maths Part 2 Chapter 1 appeared first on Learn CBSE.



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