## Introduction

This system is based upon the place value and face value of a digit in a number. We have learnt that a natural number can be written as the sum of the place values of all digits of the numbers. For example

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
Such a form of a natural number is known as its expanded form.

The expanded form of a number can also be expressed in terms of powers of 10 by using

$${10}^{0}$$ = 1, $${10}^{1}$$ = 10, $${10}^{2}$$ = 100, $${10}^{3}$$ = 1000 etc.
For example,

3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
=>    3256 = 3 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 5 x $${10}^{1}$$ + 6 x $${10}^{0}$$

Clearly, each digit of the natural number is multiplied by $${10}^{n}$$, where n is the number of digits to its right and then they are added.

## Illustrative Examples

Example 1:    Write the following numbers in the expanded exponential forms:

(i) 32005    (ii) 56719    (iii) 8605192    (iv) 2500132
Solution.

(i) 32005 = 3 x $${10}^{4}$$ + 2 x $${10}^{3}$$ + 0 x $${10}^{2}$$ + 0 x $${10}^{1}$$ + 5 x $${10}^{0}$$

(ii) 560719 = 5 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 7 x $${10}^{2}$$ + 1 x $${10}^{1}$$ + 9 x $${10}^{0}$$

(iii) 8605192 = 8 x $${10}^{6}$$ + 6 x $${10}^{5}$$ + 0 x $${10}^{4}$$ + 5 x $${10}^{3}$$ + 1 x $${10}^{2}$$ + 9 x $${10}^{1}$$ + 2 x $${10}^{0}$$

(iv) 2500132 = 2 x $${10}^{6}$$ + 5 x $${10}^{5}$$ + 0 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 1 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 2 x $${10}^{0}$$

Example 2:    Find the number from each of the following expanded forms:

(i) 5 x $${10}^{4}$$ + 4 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 5 x $${10}^{0}$$

(ii) 7 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 9 x $${10}^{0}$$

(iii) 9 x $${10}^{5}$$ + 4 x $${10}^{2}$$ + 1 x $${10}^{1}$$

Solution.

(i) 5 x $${10}^{4}$$ + 4 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 5 x $${10}^{0}$$

= 5 x 10000 + 4 x 1000 + 2 x 100 + 3 x 10 + 5 x 1

= 50000 + 4000 + 200 + 30 + 5

= 54235

(ii) 7 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 9 x $${10}^{0}$$

=  7 x 100000 + 6 x 10000 + 0 + 9 x 1

= 700000 + 60000 + 9

= 760009

(iii) 9 x $${10}^{5}$$ + 4 x $${10}^{2}$$ + 1 x $${10}^{1}$$

= 9 x 100000 + 4 x 100 + 1 x 10

= 900000 + 400 + 10

= 900410

## Formulas

Profit % = ($$\frac{ Profit}{ C.P.}$$ ) x 100

Loss % = ($$\frac{ Loss}{ C.P.}$$) x 100

C.P. = $$\frac{ 100 X S.P.}{ (100 – Loss \%)}$$

C.P. = $$\frac{ 100 X S.P.}{ (100 + Profit \%)}$$

## Illustrative Examples

Example 1: By selling a fan for Rs 649, Anil earns a profit of 18%. Find its cost price.

Solution. S.P. of the fan = Rs 649, profit = 18%

Therefore, Rs 649 = ( 1 + $$\frac{18}{100}$$) of C.P.

=> Rs 649 = $$\frac{118}{100}$$ of C.P.

=>  C.P. = Rs (649 x $$\frac{100}{118}$$) = Rs 550

Hence the cost price of the fan = Rs 550.

Example 2: By selling a chair for Rs 391, Ali suffers a loss of 15%. Find its cost price.

Solution. S.P. of the chair = Rs 391, Loss = 15%

Therefore,    Rs 391 = ( 1- $$\frac{15}{100}$$) of C.P.

Rs 391 = $$\frac{85}{100}$$ of C.P.

=>  C.P. = (Rs 391 x $$\frac{100}{85}$$ )= Rs (23 x 20) = Rs 460

Hence the cost price of the chair = Rs 460.

Example 3: A man sells his scooter for Rs 18000 making a profit of 20%. How much did the scooter cost him?

Solution. Let the cost price of the scooter be Rs 100. Then, Profit = Rs 20

S.P. = C.P. + Profit = Rs 100 + Rs 20= Rs 120
Thus, if the S.P. is Rs 120, then C.P. = Rs 120 – 20 = Rs 100

If the S.P. is Rs 18000, then C.P. = Rs ($$\frac{100}{120}$$ x 18000) = Rs 15000

Hence, the cost of the scooter = Rs 15000

## When the Interest is Compounded Annually

Formula

Let principal = P, rate = R% per annum and time = n years.
Then, the amount A is given by the formula
A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$ .

Illustrative Examples

Example 1: Find the amount of Rs 8000 for 3 years, compounded annually at 10% per annum. Also,find the compound interest.

Solution.    Here, P = Rs 8000, R =10% per annum and n =3 years.

Using the formula A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$ , we get

Amount after 3Years = {$${ 8000 \times (1+\frac { 10 }{ 100 } ) }^{ 3 }$$ }

= Rs (8000 x  $$\frac{11}{10}$$ x $$\frac{11}{10}$$ x $$\frac{11}{10}$$)

= Rs 10648.

Thus, Amount after 3 years = Rs 10648.

And, compound interest = Rs (10648 – 8000) = Rs 2648

Example 2: Rakesh lent Rs 8000 to his friend for 3 years at the rate of 5% per annum compound interest. What amount does Rakesh get after 3 years?

Solution.    Here, P = Rs 8000, R = 5% per annum and n =3.

Amount after 3 year = P$${ (1+\frac { R }{ 100 } ) }^{ n }$$

= Rs 8000 x $${ (1+\frac { 5 }{ 100 } ) }^{ 3 }$$

= Rs 8000 x $${ (1+\frac { 1 }{ 20 } ) }^{ 3 }$$

= Rs 8000 x $${ (\frac { 21 }{ 20 } ) }^{ 3 }$$

= Rs 8000 x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$

= Rs 9261

Example 3: Find the amount and compound interest on Rs 5000 for 2 years at 10%, interest being pay yearly.

Solution.    Here, P= Rs 5000, R = 10%, n = 2 years

Using the formula, A (Amount) = P  $${ (1+\frac { R }{ 100 } ) }^{ n }$$, we have

Therefore,    A = Rs 5000  $${ (1+\frac { 10 }{ 100 } ) }^{ 2 }$$

= Rs 5000 x $$\frac{110}{100}$$ x $$\frac{110}{100}$$
= Rs 6050
Therefore, Compound Interest = A – P = Rs 6050 – Rs 5000 = Rs 1050.

## When the Interest is Compounded Half-Yearly

Formula

If the interest is paid half-yearly, then in the formula A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$, for R we take $$\frac{R}{2}$$ , because R% p.a. means $$\frac{R}{2}$$ % half-yearly and for n we take 2n, because n years is equal to 2n half-years.

Therefore,                 A = P $${ (1+\frac { R }{ 200 } ) }^{ 2n }$$

Illustrative Examples

Example 1: compute the compound interst on Rs 10000 for 2 years at 10% per annum when compounded half-yearly.

Solution.

Here, Principal P = Rs 10000, R = 10% per annum, and n = 2 years

Amount after 2 years

= P $${ (1+\frac { R }{ 200 } ) }^{ 2n }$$

= Rs 10000 x P $${ (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 }$$

= Rs 10000 x  P $${ (1+\frac { 1 }{ 20 } ) }^{ 4 }$$
= Rs 10000 x P $${ (\frac { 21 }{ 20 } ) }^{ 4 }$$
= Rs 10000 x $$\frac{21}{20 }$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$
= Rs 10000 x $$\frac{194481}{160000}$$
= Rs 12155.06
Therefore, Compound Interest = Ra(12155.06 – 10000) = Rs 2155.065

Example 2: How much will Rs 256 amount to in one year at $$12\frac{1}{2}$$% per annum, when the interest is compounded half-yearly.

Solution.   P= Rs 256, 1 year= 2 half years, n = 2, Annual rate = 12 %

= $$12\frac{1}{2}$$%

Therefore, Half-yearly rate = $$\frac{1}{2}$$($$\frac{25}{2}$$ %) = $$\frac{25}{4}$$%

Thus Amount(A) =  $${ (1+\frac { R }{ 100 } ) }^{ n }$$

= Rs 256 $$(1+{ \frac { \frac { 25 }{ 4 } }{ 100 } ) }^{ 2 }$$

= Rs 256 $${ (1+\frac { 1 }{ 16 } ) }^{ 2 }$$
= Rs 256 x $$\frac{17}{16}$$ x $$\frac{17}{16}$$
= Rs 289

Example 3:  How much would a sum of Rs 16000 amount to in 2 years time at 10% per annum compounded interest, interest being payable half-yearly?

Solution.    Here, P = Rs 16000, R = 10% per annum and n = 2 years.

Amount after 2 years

= P$${ (1+\frac { R }{ 200 } ) }^{ 2n }$$

= Rs 16000 x $${ (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 }$$

= Rs 16000 x $${ (1+\frac { 1 }{ 20 } ) }^{ 4 }$$

= Rs 16000 x $${ (\frac { 21 }{ 20 } ) }^{ 4 }$$

= Rs 16000 x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$

= Rs 19448.10

Hence, a sum of Rs 16000 amounts to Rs 19448. 10 in 2 years.

## When the Interest is Compounded Quarterly

Formula

If P = Principal, R = Interest rate percent per annum and  n = number of years, then

A = $${ (1+\frac { R }{ 400 } ) }^{ 4n }$$

C.I. = A – P

Illustrative Examples

Example 1: Find the compound interest on Rs 360000 for one year at the rate of 10% per annum, if the interest is compounded quarterly.

Solution.    Here, P = Rs 360000, R = 10%  per annum and n= 1 year

Amount after 1 year

= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$

= Rs 360000 x $${ (1+\frac { 10 }{ 400 } ) }^{ 4 \times 1 }$$

= Rs 360000 x $${ (1+\frac { 1 }{ 40 } ) }^{ 4 }$$

= Rs 360000 x $${ (\frac { 41 }{ 40 } ) }^{ 4 }$$

= Rs 360000 x $$\frac{41}{40}$$ x $$\frac{41}{40}$$ x $$\frac{41}{40}$$ x $$\frac{41}{40}$$

= Rs 397372.64

Therefore, Compound Interest = Rs 397372.64 – Rs 360000

= Rs 37372.64

Example 2: Sharukh deposited in a bank Rs 8000 for 6 months at the rate of 10% interest compounded quarterly. Find the amount he received after 6 months.

Solution.    Here, P = Rs 8000, R = 10%  per annum and n= 6 months

= $$\frac{6}{12}$$

= $$\frac{1}{2}$$ year

Amount after 6 months

= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$

= Rs 8000 x $$(1+{ \frac { 10 }{ 400 } ) }^{ 4\times \frac { 1 }{ 2 } }$$

= Rs 8000 x $${ (1+\frac { 1 }{ 40 } ) }^{ 2 }$$

= Rs 8000 x $${ (\frac { 41 }{ 40 } ) }^{ 2 }$$

Example 3: Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months?

Solution.    Here, P = Rs 7500, R = 10%  per annum and n = 9 months

= $$\frac{9}{12}$$

= $$\frac{3}{4}$$ year

Amount after 9 months

= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$

= Rs 7500 x $$(1+{\frac { 12 }{ 400 } ) }^{ 4 \times \frac{3}{4} }$$

= Rs 8000 x $${ (1+\frac { 3 }{ 100 } ) }^{ 3 }$$

= Rs 8000 x $${ (\frac { 103 }{ 100 } ) }^{ 2 }$$

= Rs 8000 x $$\frac{103}{100}$$ x $$\frac{103}{100}$$ x $$\frac{103}{100}$$

= Rs 8195.45

= Rs 8000 x $$\frac{41}{40}$$ x $$\frac{41}{40}$$

= Rs 8405

## When the Rate of Interest for Successive years are Different

Formula

If the rate of interest is different for every year say, $${ R }_{ 1 }$$, $${ R }_{ 2 }$$, $${ R }_{ 3 }$$…$${ R }_{ n }$$ for the first, second, third year… nth year then the amount is given by

A = P $$(1+\frac { { R }_{ 1 } }{ 100 } )$$$$(1+\frac { { R }_{ 2 } }{ 100 } )$$$$(1+\frac { { R }_{ 3 } }{ 100 } )$$….$$(1+\frac { { R }_{ n } }{ 100 } )$$

Illustrative Examples

Example 1: Find the amount of Rs 50000 after 2 years, compounded annually; the rate interest being 8% p.a. during the first year and 9% p.a. during the second ear. Also,find the compound Interest.

Solution.    Here, P = Rs 50000,  $${ R }_{ 1 }$$= 8% p.a. and $${ R }_{ 2 }$$ = 9% p.a.

Using the formula A = P $$(1+\frac { { R }_{ 1 } }{ 100 } )$$$$(1+\frac { { R }_{ 2 } }{ 100 } )$$ we have:

Amount after 2 years

= Rs 50000 $$(1+\frac { 8 }{ 100 } )$$ $$(1+\frac { 9 }{ 100 } )$$

= Rs 50000 $$\frac{27}{25}$$ x $$\frac{109}{100}$$

= Rs 58860

Thus, amount after 2 years = Rs 58860.

And, compound interest = Rs (58860 – 50000) = Rs 8860.

Example 2: Find the compound interest on Rs 80,000 for 3 years if the rates 4%, 5% and 10% respectively.

Solution.    Here, P = Rs 80000, $${ R }_{ 1 }$$ = 4%, $${ R }_{ 2 }$$ = 5% and $${ R }_{ 3 }$$ = 10%

amount after 3 years

= Rs 80000 $$(1+\frac { 4 }{ 100 } )$$ $$(1+\frac { 5 }{ 100 } )$$ $$(1+\frac { 10 }{ 100 } )$$

= Rs 80000 x $$\frac{104}{100}$$ x $$\frac{105}{100}$$ x $$\frac{110}{100}$$

= Rs 96096
Therefore,    Compound interest = Rs (96096 — 80000) = Rs 16096.

## When Interest is Compounded Annually but Time is a Fraction

Formula

If P = Principal, R = Rate % per annum and Time = $$3\frac{3}{4}$$ years (say), then

A = P$$(1+{ \frac { R }{ 100 } ) }^{ 3}$$ x $$(1+\frac { \frac { 3 }{ 4 } \times R }{ 100 } )$$

Illustrative Examples

Example 1: Find the compound interest on Rs 31250 at 8% per annum for $$2\frac{3}{4}$$ years.

Solution.    Amount after $$2\frac{3}{4}$$ years

= Rs 31250 x $$(1+{ \frac { 8 }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 3 }{ 4 } \times 8 }{ 100 } )$$

= Rs 31250 x $$(1+{ \frac { 27 }{ 25 } ) }^{ 2}$$ x $$(\frac { 53 }{ 50 } )$$

= 31250 x $$(1+\frac { 27 }{ 25 } )$$ x  $$(1+\frac { 27 }{ 25 } )$$ x  $$(1+\frac { 53 }{ 50 } )$$

= Rs 38637

Therefore,    Amount = Rs 38637.

Hence,    compound interest = Rs (38637 — 31250) = Rs 7387.

Example 2: Find the compound interest on Ra 24000 at 15% per annum for $$2\frac{1}{3}$$  years.

Solution.

Here, P = Rs 24000, R =15% per annum and Time = $$2\frac{1}{3}$$ years.

Amount after $$2\frac{1}{3}$$years

= P$$(1+{ \frac { R }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 1 }{ 3 } \times R }{ 100 } )$$

= Rs 24000 x $$(1+{ \frac { 15 }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 1 }{ 3 } \times 15 }{ 100 } )$$

= Rs 24000 x $$(1+{ \frac { 3 }{ 20 } ) }^{ 2}$$ x $$(1+\frac { 1 }{ 20 } )$$

= Rs 24000 x $$({ \frac { 23 }{ 20 } ) }^{ 2}$$ x $$(\frac { 21 }{ 20 } )$$

= Rs 33327

Therefore,    Compound interest = Rs (33327 — 24000) = Rs 9327

## Standard Form

A number written as ( m x $$10^n$$ ) is said to be in standard form if m is a decimal number such that 1 $$\le$$ m $$<$$10 and n is either a positive or a negative integer.

The standard form of a number is also known as Scientific notation.

## Expressing Very Large Numbers in Standard Form

In order to write large numbers in the standard form,following steps must be followed:

STEP I– Obtain the number and move the decimal point to the left till you get just one digit to the left of the decimal point.

STEP II– Write the given number as the product of the number so obtained and $$10^n$$ , where n is the number of places the decimal point has been moved to the left. If the given number is between 1 and 10, then write it as the product of the number itself and $$10^0$$ .

Illustrative Examples

Example 1: Express the following numbers in the standard form:

(i) 3,90,878    (ii) 3,186,500,000    (iii) 65,950,000

Solution.

(i) We have,

3,90,878 = 390878.00
Clearly, the decimal point is moved through five places to obtain a number in which there is just one digit to the left of the decimal point.

Therefore,    390878.00 = 3.90878 x $$10^5$$

(ii) We have,

3,186,500,000 = 3.186500000 x $$10^9$$
= 3.1865 x $$10^9$$
(iii) We have,

65,950,000 = 65,950,000.00

= 6.5950000 x $$10^7$$
= 6.595 x $$10^7$$

Example 2: The distance between sun and earth is (1.496 x $${10}^{11}$$) m and the distance between earth and moon is (3.84 x $$10^8$$) m. During solar eclipse moon comes in between earth and sun. At that time what Is the distance between moon and sun?

Solution.    Required distance

= {(1.496 x $${10}^{11}$$) – (3.84 x $$10^8$$) } m

= {$$\frac { 1496\times { 10 }^{ 11 } }{ { 10 }^{ 3 } }$$ – (3.84 x $$10^8$$)} m

= {1496 x $$10^8$$) – (3.84 x $$10^8$$)} m

= {(1496 – 3.84) x $$10^8$$)} m

= (1492.16 x $$10^8$$) m

Hence, the distance between moon and sun is (1492.16 x $$10^8$$ ) m.

Example 3: Write the following numbers in the usual form:

(i) 7.54 x $$10^6$$    (ii)2.514 x $$10^7$$

Solution.    We have

(i) 7.54 x $$10^6$$

= $$\frac{754}{100}$$ x $$10^6$$

= $$\frac{754 \times {10}^{6} }{{10}^{2}}$$

= 754 x $${10}^{(6-2)}$$

= (754 x $${10}^{4}$$ )

= (754 x 10000) = 7540000

(ii) 2.514 x $$10^7$$

= $$\frac{2514}{1000}$$ x $$10^7$$

= $$\frac{2514 \times {10}^{7} }{{10}^{3}}$$

= 2514 x $${10}^{(7-3)}$$

= (2514 x $${10}^{4}$$ )

= (2514 x 10000) = 25140000

## Expressing Very Small Numbers in Standard Form

In order to write very small numbers in the standard form,following steps must be followed:

STEP I- Obtain the number and count the number of decimal values after the decimal point. Consider it as n.

STEP II- Divide the number by $${10}^{n}$$). If the number is between 1 and 10, then write it as the product of the number itself and $${10}^{-n}$$

Example 1: Write the following numbers in the standard form:

(i) 0.000000059    (ii) 0.00000000526
Solution. We may write:

(i)    0.000000059

= $$\frac{59}{{10}^{9}}$$

=  $$\frac{5.9 \times 10}{{10}^{9}}$$

= $$\frac{5.9}{{10}^{8}}$$ = (5.9 x $${10}^{-8}$$)

(ii) 0.00000000526

= $$\frac{526}{{10}^{11}}$$

= $$\frac{5.26 \times 100}{{10}^{11}}$$

= $$\frac{5.26 \times {10}^{2}}{{10}^{11}}$$

= $$\frac{5.26}{{10}^{(11 – 2)}}$$

= $$\frac{5.26}{{10}^{9}}$$ = (5.26 x $${10}^{-9}$$)

Example 2: The size of a red blood cell is 0.000007 m and that of a plant cell Is 0.00001275 m. Show that a red blood cell is half of plant cell in size.

Solution.    We have,

Size of a red blood cell = 0.000007 m = $$\frac{7}{{10}^{6}}$$ m = (7 x $${10}^{-6}$$)

Size of a plant cell

= 0.00001275 m

= $$\frac{1275}{{10}^{8}}$$ m

= $$\frac{1.275 \times {10}^{3}}{{10}^{8}}$$ m

= $$\frac{1.275}{{10}^{(8-3)}}$$ m

= $$\frac{1.275}{{10}^{5}}$$ m = (1.275 x $${10}^{-5}$$) m

$$\frac{Size of a red blood cell}{Size of a plant cell}$$

= $$\frac{7 \times {10}^{-6}}{1.275 \times {10}^{-5}}$$

= $$\frac{7 \times {10}^{-6 + 5}}{1.275}$$

= $$\frac{7 \times {10}^{-1}}{1.275}$$

= $$\frac{7}{1.275 \times 10}$$

= $$\frac{7}{12.75}$$

= $$\frac{7}{13}$$ (nearly)

= $$\frac{1}{2}$$  (approximately)

Therefore,    size of a red blood cell = $$\frac{1}{2}$$ x (size of a plant cell)

Example 3: Express the following numbers in usual form:

(i) 3 x $${10}^{-3}$$    (ii) 2.34 x $${10}^{-4}$$
Solution.    We have,

(i) 3 x $${10}^{-3}$$

=  $$\frac{3}{{10}^{3}}$$

= $$\frac{3}{1000}$$ = 0.003

(ii) 2.34 x $${10}^{-4}$$

= $$\frac{234}{100}$$ x $$\frac{1}{{10}^{4}}$$

= $$\frac{234}{{10}^{2} \times {10}^{4}}$$

= $$\frac{234}{{10}^{6}}$$

= $$\frac{234}{1000000}$$ = 0.000234

## Formulas

Profit % = $$\frac{Profit}{ C.P.}$$ x 100

Loss % = $$\frac{ Loss}{C.P.}$$ x 100

S.P. = ($$\frac{100 + Profit \%}{ 100}$$) x C.P.

S.P. = ($$\frac{100 – Loss \%}{ 100}$$) x C.P.

## Illustrative Examples

Example 1: Kirpal bought a certain number of apples at Rs 75 per score and sold them at a profit of 40%. Find the selling price per apple.

Solution. C.P. of one score i.e.,  20 apples = Rs 75, profit = 40%
S.P. of 20 apples= ( 1 + $$\frac{40}{100}$$ )of Rs 75

= Rs ( $$\frac{140}{100}$$ x 75 ) = Rs 105

S.P. of one apple = Rs $$\frac{105}{20}$$

= Rs $$\frac{21}{4}$$ = Rs 5.25

Example 2: Bashir bought an article for Rs 1215 and spent Rs 35 on its transportation. At what price should he sell the article to have a gain of 16%?

Solution. The effective cost price of the article is equal to the price at which it was bought plus the transportation charge.

C.P. of the given article = Rs (1215 + 35) = Rs 1250
Gain percent = 16%

Gain = 16% of cost price = Rs ($$\frac{16}{100}$$ x 1250) = Rs 200

S.P. = C.P. + Gain = Rs 1250 + Rs 200 = Rs 1450

Example 3: Krishnamurti bought oranges at Rs 5 a dozen. He had to sell them at a loss of 4%. Find the selling price of one orange.

Solution. We have, C.P. of one dozen oranges = Rs 5.

Loss percent = 4%

Loss = 4 % of Rs 5 = Re($$\frac{4}{100}$$ x 5) = Re ($$\frac{1}{5}$$)

S.P. = C.P. — Loss = Rs (5 – $$\frac{1}{5}$$) = Rs $$\frac{24}{5}$$

Thus, S.P. of one dozen oranges = Rs $$\frac{24}{5}$$

Therefore, S.P.of an orange = Re ($$\frac{24}{5}$$ x $$\frac{1}{12}$$)
= Re $$\frac{2}{5}$$
= $$\frac{2}{5}$$ x 100 paise
= 40 paise

## Definitions

Compound Interest: If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half year or a quarter of a year etc) so that the amount ( = Principal + Interest) at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals, calculated in this way is called the compound interest and is abbreviated as C.I.

Clearly, compound interest at the end of certain specified period is equal to the difference between the amount at the end of the period and the original principal i.e.
C.I. = Amount — Principal

Conversion Period: The fixed interval of time at the end of which the interest is calculated and added to the principal at the beginning of the interval is called the conversion period.

In other words, the period at the end of which the interest is compounded is called the conversion period.

When the interest is calculated and added to the principal every six months, the conversion period is six months. Similarly, the conversion period is 3 months when the interest is calculated and added quarterly.

## Finding CI When Interest Is Compounded Annually

when Interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.

Example 1: Maria invests Rs 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate:
(i) The amount standing to her credit at the end of second year.
(ii) The interest for the third year.
Solution.    (i) We have,
Principal for the first year Rs 93750

Rate of interest = 9.6% per annum.

Therefore,          Interest for the first year = Rs ($$\frac{93750 \times 9.6 \times 1}{100}$$) = Rs 9000

Amount at the end of the first year = Rs 93750 +Rs 9000

= Rs 102750

Principal for the second year = Rs 102750

Interest for the second year = Rs ($$\frac{102750 \times 9.6 \times 1}{100}$$) = Rs 9864

Amount at the end of second year = Rs 102750 + Rs 9864

= Rs 112614

(ii) Principal for the third year = Rs 112614

Interest for the third year =Rs ($$\frac{112614 \times 9.6 \times 1}{100}$$ ) = Rs 10810.94

Example 2: Find the compound interest on Rs 25000 for 3 years at 10% per annum, compounded annually.

Solution.    Principal for the first year = Rs 25000

Interest for the first year  = ($$\frac{25000 \times 10 \times 1}{100}$$) = Rs 2500

Amount at the end of the first year = (25000 + 2500) = Rs 27500

Principal for the second year = Rs 27500

Interest for the second year = ($$\frac{27500 \times 10 \times 1}{100}$$) = Rs 2750

Amount at the end of the second year = (27500 + 2750) = Rs 30250.

Principal for the third year = Rs 30250.

Interest for the third year = ($$\frac{30250 \times 10 \times 1}{100}$$) = Rs 3025

Amount at the end of the third year = (30250 + 3025) = Rs 33275.

Therefore, compound interest = (33275 — 25000) = Rs 8275

## Finding CI When Interest Is Compounded Half-Yearly

If the rate of Interest is R% per annum then it is clearly ($$\frac{R}{2}$$)% per half-year.
The amount after the first half-year becomes the principal for the next half-year, and so on.

Example 3: Find the compound interest on Rs 5000 for 1 year at 8% per annum, compound half-yearly.

Solution.    Rate of interest             = 8% per annum = 4% per half-year.

Time                                          = 1 year = 2 half-years.

Original principal                    = Rs 5000.

Interest for the first half-year = ($$\frac{5000 \times 4 \times 1}{100}$$) = Rs 200.

Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = Rs 5200.

Interest for the second half-year = (($$\frac{5200 \times 4 \times 1}{100}$$)) = Rs 208.

Amount at the end of the second half-year = Rs (5200 + 208) = Rs 5408.
Therefore,        compound interest = Rs (5408 — 5000) = Rs 408.

Example 4: Find the compound interest on Rs 8000 for $$1\frac{1}{2}$$ years at 10% per annum, interest being payable half-yearly.

Solution.    We have,

Rate of interest                = 10% per annum = 5% per half-year.

Time                                  = $$1\frac{1}{2}$$ years = $$\frac{3}{2}$$ x 2 = 3 half- years

Original principal             = Rs 8000

Interest for the first half-year = Rs ( $$\frac{8000 \times 5 \times 1}{100}$$) = Rs 400

Amount at the end of the first half-year = Rs 8000 + R 400 = Rs 8400
Principal for the second half-year = Rs 8400

Interest for the second half-year = Rs ($$\frac{8400 \times 5 \times 1}{100}$$) = Rs 420

Amount at the end of the second half-year = Rs 8400 + Rs 420 = Rs 8820
Principal for the third half-year = Rs 8820

Interest for the third half-year = Rs ($$\frac{8820 \times 5 \times 1}{100}$$) = Rs 441

Amount at the end of third half-year = Rs 8820 + Rs 441 = Rs 9261
Therefore,        Compound interest = Rs 9261 — Rs 8000 = Rs 1261

## Finding CI When Interest Is Compounded Quarterly

If the rate of interest is R % per annum and the interest is compounded quarterly, then it is $$\frac{R}{4}$$ %  per quarter.

Example 5: Find the compound interest on Rs 10000 for 1 year at 20% per annum compounded quarterly.

Solution.    We have,

Rate of interest             = 20% per annum = $$\frac{1}{5}$$% = 5% per quarter

Time                         = 1 year = 4 quarters.

Principal for the first quarter = Rs 10000

Interest for the first quarter = Rs ($$\frac{10000 \times 5 \times 1}{100}$$) = Rs 500

Amount at the end of first quarter = Rs 10000 + Rs 500 = Rs 10500

Principal for the second quarter = Rs 10500

Interest for the second quarter = Rs ($$\frac{10500 \times 5 \times 1}{100}$$) = Rs 525

Amount at the end of second quarter = Rs 10500 + Rs 525 = Rs 11025
Principal for the third quarter = Rs 11025

Interest for the third quarter = Rs ($$\frac{11025 \times 5 \times 1}{100}$$) = Rs 551.25

Amount at the end of the third quarter = Rs 11025 + Rs 551.25

= Rs 11576.25

Principal for the fourth quarter = Rs 11576.25

Interest for the fourth quarter = Rs ($$\frac{11576.25 \times 5 \times 1}{100}$$) = Rs 578.8125

Amount at the end of fourth quarter = Rs 11576.25 + Rs 578.8125

= Rs 12155.0625

Therefore,        Compound interest = Rs 12155.0625 — Rs 10000

= Rs 2155.0625

= Rs 2155.06

## Formulas

Profit = S.P. –C.P.

Loss = C.P. – S.P.

Profit % = ( $$\frac{Profit}{C.P.}$$ x 100) %

Loss % = ( $$\frac{Loss}{C.P.}$$ x 100) %

## Illustrative Examples

Example 1: John bought a watch for Rs 540 and sold it for Rs 585. Find his profit and profit percentage.

Solution. C.P. of the watch = Rs 540, S.P. of the watch = Rs 585

Profit = S.P. — C.P.
= Rs 585 — Rs 540 = Rs 45

Profit percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %

= ($$\frac{45}{540}$$ X 100 ) %

=  $$\frac{100}{12}$$ %

= $$\frac{25}{3}$$ %
= $$8\frac{1}{3}$$ %

Example 2: By selling a bike for Rs 22464, Ansari incurs a loss of Rs 1536. Find his loss percenta

Solution. S.P. of the bike = Rs 22464, loss = Rs 1536
C.P. of the bike = S.P. + loss = Rs 22464 + Rs 1536 = Rs 24000

Loss percentage = ( $$\frac{Loss}{C.P.}$$ x 100) %

= ($$\frac{1536}{24000}$$ X 100 ) %

=  $$\frac{1536}{240}$$ %
= $$\frac{64}{10}$$ %
= 6.4 %

Example 3: B ijoy bought bananas at the rate of 5 for Rs 4 and sold them at the rate of 4 for Rs 5. Calculate his gain percentage.

Solution.

C.P. of 5 bananas = Rs 4

C.P. of 1 banana = Rs $$\frac{4}{5}$$ = Rs 0.80

S.P. of 4 bananas = Rs 5

S.P. of 1 banana = Rs $$\frac{5}{4}$$ = Rs 1.25

Therefore,        Gain on the sale of one banana = S.P. — C.P. = Rs 1.25 — Rs 0.80 = Rs 0.45

Gain percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %

= ($$\frac{0.45}{0.80}$$ X 100 ) %

=  $$\frac{145}{80}$$ %
= $$\frac{225}{4}$$ %
= 56.25 %

## Cubes

The cube of a number is the number raised to the power 3. Thus,
cube of 2 = $${ 2 }^{ 3 }$$ = 2 x 2 x 2 = 8,
cube of 5 = $${ 2 }^{ 3 }$$ = 5 x 5 x 5 = 125.

## Perfect Cube

We know that $${ 2 }^{ 3 }$$ = 8, $${ 3 }^{ 3 }$$ = 27, $${ 26}^{ 3 }$$ = 216, $${ 7 }^{ 3 }$$ = 343, $${ 10 }^{ 3 }$$= 1000.
The numbers 8, 27, 216, 343, 1000, … are called perfect cubes. A natural number is said to be a perfect cube, if it is the cube of some natural number, i.e.,
A natural number n is a perfect cube if there exists a natural number m such that  m X m X m = n.

Procedure :

Step I- Obtain the natural number.

Step II- Express the given natural number as a product of prime factors.

Step III- Group the factors in triples in such a way that all the three factors in each triple are equal.

Step IV- If no factor is left over in grouping in step III, then the number is a perfect cube, otherwise not.

Illustrative Examples :

Example 1: Show that 729 is a perfect cube.
Solution. Resolving 729 into prime factors, we have
729 =3 x 3 x 3 x 3 x 3 x 3
Here, we find that the prime factor 3 of the given number can be grouped into triplets and no factor is left out. Hence, 729 is a perfect cube.
Also, 729 is the cube of 3 x 3, i.e., 729 = $${ (9) }^{ 3 }$$•

Example 2: What is the smallest number by which 1323 may be multiplied so that the product is a perfect cube?
Solution. Resolving 1323 into prime factors, we have
1323 = 3 x 3 x 3 x 7 x 7
Since one more 7 is required to make a triplet of 7, the smallest number by which 1323 should be multiplied to make it a perfect cube is 7.

Example 3: What is the smallest number by which 1375 should be divided so that the quotient may be a perfect cube?
Solution. Resolving 1375 into prime factors, we have
1375 = 5 x 5 x 5 x 11
The factor 5 makes a triplet, and 11 is left out. So, clearly 1375 should be divided by 11 to make it a perfect cube.

## Properties of Cubes of Numbers

1. Cubes of all odd natural numbers are odd. Thus  $${ 3 }^{ 3 }$$ = 27, $${ 5 }^{ 3 }$$ = 125, $${ 7 }^{ 3 }$$ = 343, $${ 9 }^{ 3 }$$ = 729, etc.

2. Cubes of all even natural numbers are even. Thus $${ 2 }^{ 3 }$$ = 8, $${ 4 }^{ 3 }$$ = 64, $${ 6 }^{ 3 }$$ = 216, $${ 8 }^{ 3 }$$ = 512, etc.

3. The cube of a negative integer is always negative
e.g.,  $${ (-1) }^{ 3 }$$ = (—1)x (—1) x (—1) = (1) x (—1) = —1.
$${ (-2) }^{ 3 }$$ = -2 x -2 x -2 =(-2 X -2) X -2 = 4 X —2 = -8.

4. For any rational number $$\frac { a } { b }$$, we have $${ (\frac { a } { b } ) }^{ 3 }=(\frac { { a }^{ 3 } } { { b }^{ 3 } } )$$.
Thus, $${ (\frac { 2 }{ 3 } ) }^{ 3 }=(\frac { { 2 }^{ 3 } }{ { 3 }^{ 3 } } )$$ = $$\frac { 8 }{ 27 }$$ .

$${ (\frac { -4 }{ 5 } ) }^{ 3 }=(\frac { { (-4) }^{ 3 } }{ { 5 }^{ 3 } } )$$ = $$\frac { -64 }{ 125 }$$ .

5. The sum of the cubes of first n natural numbers is equal to the square of their sum. That is,

$${ 1 }^{ 3 }$$ + $${ 2 }^{ 3 }$$ + $${ 3 }^{ 3 }$$ … + $${ n }^{ 3 }$$ = $${ (1 + 2 + 3+ … + n) }^{ 2 }$$

6. Cubes of the numbers ending in digits 1, 4, 5, 6 and 9 are the numbers ending in the same digit. Cubes of numbers ending in digit 2 ends in digit 8 and the cube of numbers ending in digit 8 ends in digit 2. The cubes of the numbers ending in digits 3 and 7 ends in digit 7 and 3 respectively.

## Steps involved in finding a percent of a given number

Step I– Obtain the number, say x.

Step II– Obtain the required percent, say P %.

Step III– Multiply x by P and divide by 100 to obtain the required P % of x
i.e.,                        P% of x = $$\frac{ P}{ 100}$$ * x

## Illustrative Examples

Example 1: Find
(i)  30% of Rs 180            (ii) 13% of Rs 6500            (iii) 16% of 25 litres
Solution.   We know that P% of x is equal to $$\frac{ P }{ 100}$$ * x. So, We have

(i) 30% of Rs 180 = Rs ( $$\frac{ 30 }{ 100}$$ x 180) = Rs 54

(ii) 13% of Rs 6500 = Rs ( $$\frac{ 13 }{ 100}$$ x 6500 ) = Rs 845

(iii) 16% of 25 litres = Rs ( $$\frac{ 16 }{ 100}$$ x 25) = 4 litres

Example 2: If 23% of a is 46, then find a.

Solution. We have,

23% of a = $$\frac{ 23 }{ 100}$$ x a.

But, 23% of a is given as 46.

Therefore,   $$\frac{ 23 }{ 100}$$ x a => a = 46 x $$\frac{ 100 }{ 23 }$$

=> a = 200

Example 3: 72% of 25 students are good at Mathematics. How many are not good at it?

Solution. We have,

Number of students who are good at Mathematics

= 72% of 25

= $$\frac{ 72 }{ 100}$$ x 25 = 18

## Steps involved in conversion of a given per cent into decimal form

STEP I– Obtain the per cent which is to be converted into decimal.

STEP II– Express the given per cent as a fraction with denominator as 100.

STEP III– Write the fraction obtained in step II in decimal form.

Illustration 1: Express each of the following as a decimal:

(i) 25%         (ii) 12%         (iii) 5.6%         (iv) 0.5%
Solution. We have,

(i) 25% = $$\frac{ 25}{ 100}$$ = 0.25

(ii) 12% = $$\frac{ 12 }{ 100}$$ = 0.12

(iii) 5.6% = $$\frac{ 5.6}{ 100}$$ = 0.056

(iv) 0.5% = $$\frac{ 0.5 }{100}$$ =  $$\frac{ 5 }{ 1000}$$  = 0.005

## Steps involved in conversion of decimal into a per cent

STEP I– Obtain the number in decimal form.

STEP II– Convert it into a fraction by removing the decimal point. In order to remove decimal, divide by 10 or 100 or 1000 according to the number of digits on the right side of the decimal point 1 or 2 or 3 respectively.

STEP III– Multiply by 100 and put% sign.

Illustration 2: Express each of the following as percent:

(i) 0.073                       (ii) 0.001                     (iii) 2.4
Solution. We have,

(i) 0.073 = $$\frac{73}{ 1000}$$ = ($$\frac{ 73}{ 1000}$$ x 100)% = 7.3%

(ii) 0.001 = $$\frac{ 1}{ 1000}$$ = ($$\frac{ 1 }{1000}$$ x 100) = 0.1%

(iii) 2.4 = $$\frac{ 24}{ 10 }$$ = ($$\frac{ 24}{10}$$ x 100) = 240%

## Percentage Increase

To increase a quantity by a percentage find the percentage of the quantity and add it to the original quantity.

Example 1: Increase 320 by 20%.

Solution. 20% of 320 = $$\frac{ 20}{ 100}$$ x 320 = 64

Therefore,            Increased amount = Rs 320 + Rs 64 = Rs 384.

## Percentage Decrease

To decrease a quantity by a percentage, find the percentage of the quantity and subtract it from the original quantity.

Example 2: Decrease 120 by $$12\frac{1}{2}$$ %.

Solution. $$12\frac{1}{2}$$% of 120

= Rs $$\frac{ 12.5}{ 100}$$ of 120

= Rs $$\frac{ 125 }{ 10 * 100}$$ x 120

= Rs 15

Therefore,       Decreased amount = Rs 120 — Rs15 = Rs 105.

## Percentage Change

 Percentage change = ( $$\frac{ Actual change (Increase or Decrease)}{ Original quantity}$$ x 100) % Percentage error = ($$\frac{ Error}{Actual Value}$$ x 100) %

Example 3: Sabita’s weight decreased from 80 kg to 40 kg. Find the percentage decrease.

Solution. Decrease in weight = (80 — 40) kg = 40 kg.

Therefore,        % decrease = $$\frac{ 40}{ 80}$$ x100% = 50%.

Example 4: The distance between two places was 200 km. It was measured as 300 km. Find the percentage error.

Solution. Error = 300 km — 200 km = 100 km.

Therefore,     %error = $$\frac{ error}{ actual value}$$ x 100%

= $$\frac{ 100}{ 200}$$ x 100 = 50%.