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Title: FORMULAS
Description: CONTAIN IMPORTANT FORMULAS

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Must Know Concepts
Percentages:
1

1

1
...
Percentage change =

𝐹𝑖𝑛𝑎𝑙 𝑉𝑎𝑙𝑢𝑒−𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒
𝐼𝑛𝑖𝑡𝑖𝑎𝑙 𝑉𝑎𝑙𝑢𝑒

x 100 (If positive then it’s an increase and if negative

then it’s a decrease
...
If final value becomes ‘k’ times the initial then there’s an increase of [(k-1) x 100]%
1

b
...
To increase a value ‘X’ by r% = X x (
Short-cut: X x

𝑏+𝑎
𝑏

100+𝑟
)
100

𝑎

where, r% = 𝑏 (in fraction)
100−𝑟
)
100

4
...
Reverse percentage change:
Short-cut:
𝑎
𝑏

a
...


𝑘−1
x
𝑘

100)%

(The same can be used if ‘A’ is increased by r% to find by how much % it should be decreased to get
the original value
...
67%
...
By what percentage should they
reduce the price now?
Ans: 14
...
If ‘A’ is 𝑏 th less than ‘B’, then ‘B’ is 𝑏−𝑎th more than ‘A’
...
)
6
...
C) = a + b + 100
(Here, keep a note to use negative sign for decrease
...

Example: Area of a rectangle = Length x Breadth; Expenditure = Price x Consumption;
Distance = Speed x Time
Application of successive percentage change:
a
...
C & to find b% (or a%) (If R
...
)
Quick Check-2: If the price of the petrol is decreased by 11
...
By how much should you increase
your consumption to keep the expenses constant?
Ans: 12
...
By how much should you reduce your
consumption so that the expenses increases by 10%?
Ans: -12%
b
...
C=0 & to
find the absolute value for which absolute change is given
...

Ans: 28

Profit & Loss:
Cost Price (C
...
Add all expenses while purchasing to get the
final C
...


Selling Price (S
...
Any commission paid to agent for selling an article
will be deducted from the S
...

Marked Price/List Price (M
...

Case-I: C
...
P then Profit
1
...
P – C
...
Profit% =

𝑃𝑟𝑜𝑓𝑖𝑡
𝐶
...
S
...
P x

(100 + 𝑃𝑟𝑜𝑓𝑖𝑡%)
100

(Use fractions conversion of percentages as and when possible)

4
...
P = S
...
Margin =

𝑃𝑟𝑜𝑓𝑖𝑡
𝑆
...
P > S
...
Loss = C
...
P
2
...
𝑃

x 100

3
...
P = C
...
C
...
P x

100
(100 − 𝐿𝑜𝑠𝑠%)

(Use reverse percentage change as and when possible)

Case-III: Discount
1
...
P – S
...
Discount% =

𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡
𝑀
...
Mark Up Price = M
...
P
4
...
𝑃

x 100

5
...
P = M
...
M
...
P x

100
(Use reverse percentage change
(100 − 𝐷𝑖𝑠𝑐𝑜𝑢𝑛𝑡%)

fractions conversion of percentages as and when possible)
as and when possible)

Short-Cut:
1
...
P of two articles is same and on one there’s a profit of ‘k%’ and on the other there’s a loss
(𝒌)𝟐

of ‘k%’, then there’s always a loss of 𝟏𝟎𝟎 %
2
...
If an article is Marked Up by ‘M%’, available at a Discount of ‘D%’ and yields a Profit of ‘P%’ then
𝑴𝑫

P = (M + D + 𝟏𝟎𝟎) (Use negative sign for D as it is a decrease)

𝒎

4
...
If 𝐶
...
If a shopkeeper uses a false weight of ‘x kg’ instead of ‘y kg’ (where x < y) and professes to sell the
𝒚−𝒙
article at the C
...
If a shopkeeper uses a false weight of ‘x kg’ instead of ‘y kg’ (where x is ‘k%’ less than y) and
(𝒑 + 𝒌)

professes to sell the article at ‘p%’ profit, then his Actual Profit% = (𝟏𝟎𝟎 − 𝒌) x 100

Simple Interest & Compound Interest:
Principal (P): Money lent or borrowed
Rate of Interest (R): Percentage of Principal paid as interest
...
e
...

Amount (A): Total money paid to clear the principal as well as interest
...
I): When the interest is calculated only over the principal
...
S
...
A = P + S
...
If an amount becomes ‘k’ times of itself when given at ‘R%’ rate of interest, then
R = (k-1) x

𝟏𝟎𝟎
𝒌

Compound Interest (C
...

1
...
C
...
If the amount for the nth year and the (n+k)th year be ‘x’ and ‘y’ respectively, then
𝒚 𝟏

R = [(𝒙)𝒌 - 1] x 100
Hence, if the amount for two consecutive years be ‘p’ and ‘q’ then R =
Difference between C
...
I:
1
...
𝐼)2 𝑦𝑟 = 𝑃(100)2
𝑅

𝑅

3
Title: FORMULAS
Description: CONTAIN IMPORTANT FORMULAS