Finding The S.P. When C.P. and Profit Or Loss Percent Are Given

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

Percentage Increase and Decrease And Percentage Change

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%.

Conversion Of A Percent Into A Fraction And Vice Versa

Steps involved in conversion of a per cent into a fraction

STEP I– Obtain the given per cent. Let it be x%.

STEP II– Drop the per cent sign (i.e %) and divide the number by 100. Thus, x% =  \(\frac{ x }{ 100 }\)

Illustration 1: Express the following per cents as fractions in the simplest forms:
(i) 57%            (ii) 36%             (iii) 115%
Solution. We have,

(i) 57% =  \(\frac{ 57 }{ 100 }\)
(ii) 36%=  \(\frac{ 36 }{ 100 }\) =  \(\frac{ 9 }{ 25 }\)
(iii) 115% =  \(\frac{ 115 }{ 100 }\) =  \(\frac{ 23 }{ 20 }\)

Illustration 2: Express each of the following per cents as fractions in the simplest
(i) 0.375%            (ii) 0.4%            (iii) 16%
Solution. We have,

(i) 0.375% = \(\frac{ 0.375 }{ 100 }\) =  \(\frac{ 375 }{ 100000 }\) =  \(\frac{ 3 }{ 800 }\)

(ii) 0.4% =  \(\frac{ 0.4 }{ 100 }\) =  \(\frac{ 4 }{ 1000 }\) =  \(\frac{ 1 }{ 250 }\)

(iii) \(16\frac{2}{3}\) = \(\frac{ 50 }{ 3 }\)% = \(\frac { \frac { 50 }{ 3 }  }{ 100 } \) = \(\frac{ 50}{ 3 }\) x \(\frac{ 1 }{ 100 }\) = \(\frac{ 1 }{ 6 }\)

Steps involved in conversion of a fraction into a percent

STEP I– Obtain the fraction. Let it be \(\frac{ a }{ b }\)

STEP II- Multiply the fraction by 100 and put the per cent sign% to obtain the required  percent. Thus, \(\frac{ 4 }{ 5 }\) = ( \(\frac{ 4 }{ 5 }\) x 100)%

Illustration 1: Express each of the following fractions as per cents:

(i) \(\frac{ 4 }{ 5 }\)    (ii) \(\frac{ 9 }{ 20 }\)    (iii) \(5\frac{1}{4}\)
Solution. We have,
(i) \(\frac{ 4 }{ 5 }\) = (\(\frac{ 4 }{ 5 }\)  x 100)% = 80%
(ii) \(\frac{ 9 }{ 20 }\) = (\(\frac{ 9 }{ 20 }\)  x 100)% = 45%
(iii) \(5\frac{1}{4}\) = \(\frac{ 21 }{ 4 }\) = (\(\frac{ 21 }{ 4 }\) x 100)% = 525%

Illustration 2: Express each of the following into per cents:

(i) 0.375         (ii) 0.005         (iii) 2.45
Solution. We have,

(i)  0.375 =  \(\frac{ 375 }{ 1000 }\)% = (\(\frac{ 375 }{ 1000 }\) x 100) = 37.5%

(ii)  0.005 = \(\frac{ 5 }{ 1000 }\) = (\(\frac{ 5 }{ 1000 }\) x 100)% = 0.5%

(iii)  2.45 = \(\frac{ 245 }{ 100 }\) = (\(\frac{ 245 }{ 100 }\) x 100)% = 245%

Laws of Exponents

First Law

If a is any non-zero rational number and m, n are natural numbers, then

\( { a }^{ m } \) x \( { a }^{ n } \) = \( { a }^{ m + n } \)
Generalised form of above law:

If a is a non-zero rational number and m, n, p are natural numbers, then,

\( { a }^{ m } \) x \( { a }^{ n } \) x \( { a }^{ p } \) = \( { a }^{ m + n + p } \)

Illustration : Simplify and write the answer of each of the following in exponential form:

(i) \( { 4 }^{ 2 } \) x \( { 4 }^{ 3 } \)        (ii) \( { 2 }^{ 2 } \) x \( { 2 }^{ 3 } \) x \( { 2 }^{ 4 } \)
(iii) \( { 6 }^{ x } \) x \( { 6 }^{ 3 } \)     (iv) \( { (\frac { 3 }{ 2 } ) }^{ 3 } \) x \( { (\frac { 3 }{ 2 } ) }^{ 6 } \)

Solution. Using first law of exponents, We have

(i) \( { 4 }^{ 2 } \) x \( { 4 }^{ 3 } \) = \( { 4 }^{ 2 + 3 } \) = \( { 4 }^{ 5 } \)

(ii) \( { 2 }^{ 2 } \) x \( { 2 }^{ 3 } \) x \( { 2 }^{ 4 } \) = \( { 2 }^{ 2 + 3 + 4 } \) = \( { 2 }^{ 9 } \)

(iii) \( { 6 }^{ x } \) x \( { 6 }^{ 3 } \) = \( { 6 }^{ x + 3 } \)

(iv) \( { (\frac { 3 }{ 2 } ) }^{ 3 } \) x \( { (\frac { 3 }{ 2 } ) }^{ 6 } \)= \( { (\frac { 3 }{ 2 } ) }^{ 6 + 3 } \) = \( { (\frac { 3 }{ 2 } ) }^{ 9 } \)

Second Law

If a is any non-zero rational number and m and n are natural numbers such that m > n, then

\( { a }^{ m } \) \( \div \) \( { a }^{ n } \) = \( { a }^{ m – n } \) or \( \frac { { a }^{ m } }{ { a }^{ n } } \) = \( { a }^{ m – n } \)

Illustration : Simplify and write each of the following in exponential form:

(i) \( { 8 }^{ 6 } \) \( \div \) \( { 8 }^{ 3 } \)        (ii) \( { (-5) }^{ 10 } \) ÷ \( { (-5) }^{ 4 } \)
(iii) \( { (\frac { -3 }{ 5 } ) }^{ 6 } \) ÷ \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)
Solution. Using second law of exponents, We have

(i) \( { 8 }^{ 6 } \) \( \div \) \( { 8 }^{ 3 } \)

= \( \frac { { 8 }^{ 6 } }{ { 8 }^{ 3 } } \)
=  \( { 8 }^{ 6 – 3 } \)
=  \( { 8 }^{ 3 } \)
(ii) \( { (-5) }^{ 10 } \) ÷ \( { (-5 )}^{ 4 } \)

= \( \frac { { (-5) }^{ 10 } }{ {(-5) }^{ 4 } } \)

= \( { (-5) }^{ 10 – 4 } \)
= \( { (-5) }^{ 6 } \)
(iii) \( { (\frac { -3 }{ 5 } ) }^{ 6 } \) ÷ \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)

= \( { \frac { { (\frac { -3 }{ 5 }  })^{ 6 } }{ ({ \frac { -3 }{ 5 } ) }^{ 3 } }  } \)
= \( { (\frac { -3 }{ 5 } ) }^{ 6 – 3 } \)
= \( { (\frac { -3 }{ 5 } ) }^{ 3 } \)

Third Law

If a is any rational number different from zero and m, n are natural numbers, then

\( { { (a }^{ m }) }^{ n }\quad =\quad { a }^{ m\quad \times \quad n }\quad =\quad { { (a }^{ n }) }^{ m } \)

Illustration: Simplify and write each of the following in exponential form:

(i) \( { { (3 }^{ 2 }) }^{ 4 }\quad \)    (ii) \( { { ((-2) }^{ 4 }) }^{ 2 }\quad \)
(iii) \( { { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\}  }^{ 4 } \)    (iv) \( { ({ 4 }^{ 2 }) }^{ 3 }\quad \) x \( ({ { 4}^{ 4 }) }^{ 2 }\quad \)
Solution. Using third law of exponents, We have

(i) \( { { (3 }^{ 2 }) }^{ 4 }\quad \)

= \( { 3 }^{ 2 \times 4 } \)
= \( { 3 }^{ 8 } \)
(ii) \( { { ((-2) }^{ 4 }) }^{ 2 }\quad \)

= \( { (-2) }^{ 4 \times 2 } \)

= \( { (-2) }^{ 8 } \)
(iii) \( { { \{ (\frac { 3 }{ 5 } ) }^{ 3 }\}  }^{ 4 } \)

= \( { (\frac { 3 }{ 5 } ) }^{ 3 \times 4 } \)
= \( { (\frac { 3 }{ 5 } ) }^{ 12 } \)

(iv) \( { ({ 4 }^{ 2 }) }^{ 3 }\quad \)x \( { ({ 4}^{ 4 }) }^{ 2 }\quad \)

= \( ({ 4 }^{ 2 \times 3}) \) x \( ({ 4 }^{ 4 \times 2 }) \)
= \( { 4 }^{ 6 } \) x \( { 4 }^{ 8 } \)
= \( { 4 }^{ 6 + 8 } \)
= \( { 4 }^{ 14 } \)

Fourth Law

If a, b are non-zero rational numbers and n is a natural number, then

\( { a }^{ n } \) x \( { b }^{ n } \) = \( { (ab) }^{ n } \)
Generalised form of above law:

If a, b, c are non-zero rational numbers and n is a natural number, then

\( { a }^{ n } \) x \( { b }^{ n } \) x \( { c }^{ n } \) = \( {(abc)}^{ n } \)

Illustration: Express each of the following products of powers as the exponent of a rational number:

(i) \( { 2 }^{ 4 } \) x \( { 5 }^{ 4 } \)    (ii) \( {(-3) }^{ 3 } \) x \( { (-2) }^{ 3 } \)
(iii) \( { 3 }^{ 2 } \) x \( { x }^{ 2 } \) x \( { y }^{ 2 } \)    (iv) \( { (\frac { 3 }{ 2 } ) }^{ 2 } \) x \( { (\frac { 2 }{ 5 } ) }^{ 2 } \)
Solution. We have,

(i) \( { 2 }^{ 4 } \) x \( { 5 }^{ 4 } \)

= \( { (2 \times 5 )}^{ 4 } \)
= \( { 10 }^{ 4 } \)
(ii) \( {(-3) }^{ 3 } \) x \( { (-2) }^{ 3 } \)

= \( { ((-3) \times (-2) )}^{ 3 } \)

= \( { 6 }^{ 3 } \)

(iii) \( { 3 }^{ 2 } \) x \( { x }^{ 2 } \) x \( { y }^{ 2 } \)

= \( {( 3 \times x \times y  )}^{ 2 } \)
= \( { (3xy) }^{ 2 } \)

(iv) \( { (\frac { 3 }{ 2 } ) }^{ 2 } \) x \( { (\frac { 2 }{ 5 } ) }^{ 2 } \)

= \( { (\frac { 3 }{ 2 }  }\times \frac { 2 }{ 5 } )^{ 2 } \)
= \( { (\frac { 3 }{ 5 } ) }^{ 2 } \)

Fifth Law

If a and b are non-zero rational numbers and n is a natural number, then

\( \frac { { a }^{ n } }{ { b }^{ n } } \) = \( { (\frac { a }{ b } ) }^{ n } \)

Illustration: Write each of the following in the form \(\frac{p}{q}\)

(i) \( { (\frac { 2 }{ 5 } ) }^{ 3 } \)    (ii) \( { (\frac { -3 }{ 2 } ) }^{ 4 } \)
Solution. We have,

(i) \( { (\frac { 2 }{ 5 } ) }^{ 3 } \)

= \( \frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } } \)
= \(\frac{2 \times 2 \times 2}{5 \times 5 \times 5 }\)
= \(\frac{8}{125}\)
(ii) \( { (\frac { -3 }{ 2 } ) }^{ 4 } \)

= \( \frac { { (-3) }^{ 4 } }{ { 2 }^{ 4 } } \)

= \(\frac{(-3) \times (-3) \times (-3) \times (-3)}{2 \times 2 \times 2 \times 2}\)
= \(\frac{81}{16}\)

Illustrative Examples

Example 1: Use the laws of exponents to simplify the following :

(i) \( { [{ (2^{ 3 }) }^{ 4 }] }^{ 5 } \)    (ii) \( [{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 } \)
(iii) \( { 81 }^{ -1 } \) x \( { 3 }^{ 5 } \)    (iv) \( { (\frac { 2 }{ 3 } ) }^{ 0 } \) + \( { (\frac { 2 }{ 3 } ) }^{ -2 } \)
Solution. We have,

(i) \( { [{ (2^{ 3 }) }^{ 4 }] }^{ 5 } \)

= \( { [{ 2^{ 3 \times 4 }}] }^{ 5 } \)
= \( { [{ 2^{ 12 }}] }^{ 5 } \)
= \( { [{ 2^{ 12 \times 5}}] } \)
= \( { 2 }^{ 60 } \)
(ii) \( [{ { 3 }^{ 6 }\quad \div \quad { 3 }^{ 4 }] }^{ 3 } \)

= \( { [ 3^{ 6 – 4 }]^3} \)

= \( { [{ 3^{ 2 }}] }^{ 3 } \)
= \( { 3 }^{ 2 \times 3 } \)
= \( { 3 }^{ 6 } \)

(iii) \( { 81 }^{ -1 } \) x \( { 3 }^{ 5 } \)

= \( { { (3 }^{ 4 }) }^{ -1 }\quad \) x \( { 3 }^{ 5 } \)
= \( { 3 }^{ -4 + 5 } \)
= \( { 3 }^{ 1 } \)
= 3

(iv) \( { (\frac { 2 }{ 3 } ) }^{ 0 } \) + \( { (\frac { 2 }{ 3 } ) }^{ -2 } \)

= 1 + \( { \frac { 1 }{ ({ \frac { 2 }{ 3 } ) }^{ 2 } }  } \)
= \( { \frac { 1 }{ \frac { { 2 }^{ 2 } }{ { 3 }^{ 2 } }  }  } \)
= \( { \frac { 1 }{ \frac { 4 }{ 9 }  }  } \)
= 1 + \(\frac{9}{4}\)
= \(\frac{13}{4}\)

Example 2: Simplify and write the answer in the exponential form:

(i) \( { ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad  } \)    (ii) \( { (-4) }^{ 3 } \) x \( { (5) }^{ -3 } \) x \( { (-5) }^{ -3 } \)

Solution.

(i) We have,

\( { ({ { 2 }^{ 5 }\quad \div \quad { 2 }^{ 8 }) }^{ 5 }\quad \times \quad { 2 }^{ -5 }\quad  } \)
= \( { (\frac { { 2 }^{ 5 } }{ { 2 }^{ 8 } } ) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { { (2 }^{ 5 – 8 }) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { { (2 }^{ -3 }) }^{ 5 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -3 \times 5 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -15 } \) x \( { (2) }^{ -5 } \)
= \( { (2) }^{ -15-5 } \) = \( { (2) }^{ -20 } \)

(ii) We have,

\( { (-4) }^{ 3 } \) x \( { (5) }^{ -3 } \) x \( { (-5) }^{ -3 } \)
= \( { [-4 \times 5 \times (-5)] }^{ -3 } \)
= \( { (100) }^{ -3 } \)
= \( { { (10 }^{ 2 }) }^{ -3 }\quad \)
= \( { (10) }^{ 2 \times -3 } \) = \( { (10) }^{ -6 } \)

Example 3: Simplify

(i) \( { ({ { 4 }^{ -1 } \div  { 3 }^{ -1 }) }^{ -2 }} \)    (ii) \( { ({ { 5 }^{ 3 } \times  { 3 }^{ -1 }) }^{ -1 }} \) ÷ \( { 6 }^{ -1 } \)

Solution.

(i) \( { ({ { 4 }^{ -1 } \div  { 3 }^{ -1 }) }^{ -2 }} \)

=  \( { (\frac { 1 }{ 4 } \times \frac { 3 }{ 1 } ) }^{ -2 } \)

=  \( { (\frac { 3 }{ 4 })^{-2}} \)

= \( { (\frac { 4 }{ 3 })^{2}} \)

= \(\frac{16}{9}\)

(ii) \( { ({ { 5 }^{ -1 } \times  { 3 }^{ -1 }) }^{ -1 }} \) ÷ \( { 6 }^{ -1 } \)

= \( { (\frac { 1 }{ 5 } \times \frac { 1 }{ 3 } ) }^{ -1 } \) ÷ \(\frac{1}{6}\)

= \( { (\frac { 1 }{ 15 })^ {-1} } \) ÷ \(\frac{1}{6}\)

= \(\frac{15}{1}\) ÷ \(\frac{1}{6}\)

= \(\frac{15}{1} \times \frac{6}{1}\)

= 90

Example 4: Find the value of m if \( { (\frac { 2 }{ 9 })^ {3} } \) x \( { (\frac { 2 }{ 9 })^ {-6} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)

Solution.

\( { (\frac { 2 }{ 9 })^ {3} } \) x \( { (\frac { 2 }{ 9 })^ {-6} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)

=>    \( { (\frac { 2 }{ 9 })^ {3+(-6)} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)

=>    \( { (\frac { 2 }{ 9 })^ {-3} } \) = \( { (\frac { 2 }{ 9 })^ {2m-1} } \)

In an equation, when bases on both sidess are equal, their powers must also be equal.

Therefore, 2m – 1 = -3 or 2m = -3 + 1

=>    2m = -2

=>    m = \(\frac{-2}{2}\) = –1

Example 5: What number should \( { (\frac { 2 }{ 3 })^ {-2} } \) be multiplied so that the product is \( { (\frac { 4 }{ 27 })^ {-1} } \) ?

Solution.

\( x \) x \( { (\frac { 2 }{ 3 })^ {-2} } \)  = \( { (\frac { 4 }{ 27 })^ {-1} } \)

=>    \( x \) x  \( { (\frac { 3 }{ 2 })^ {2} } \) = \(\frac{27}{4}\)

=>    \( x \) x \(\frac{9}{4}\) = \(\frac{27}{4}\)

=>    \( x \) = \(\frac{27}{4}\) x \(\frac{4}{9}\)

=>    \( x \) = 3.

Hence, the required number is 3.

Example 6: Simplify

\( \frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 } \)
Solution.    We have,

\( \frac { { 12 }^{ 4 }\times { 9 }^{ 3 }\times 4 }{ { 6 }^{ 3 }\times { 8 }^{ 2 }\times 27 } \)
= \( \frac { ({ 2 }^{ 2 }\times 3)^{4} \times ({ 3 }^{ 2 }) ^ {3} \times {2}^{2} }{ { {2}^ {3}\times 3 }^{ 3 }\times ({ 2 }^{ 3 })^{2}\times {3}^{3}) } \)
= \( \frac { { ({ 2 }^{ 2 }) }^{ 4 }\times { 3 }^{ 4 }\times { ({ 3 }^{ 2 }) }^{ 3 }\times { 2 }^{ 2 } }{ ({ 2 }^{ 3 }\times { 3 }^{ 3 })\times { { (2 }^{ 3 }) }^{ 2 }\times { 3 }^{ 3 } } \)
= \( \frac { { 2 }^{ 8 }\times { 3 }^{ 4 }\times {3}^{6} \times {2}^{2} }{ { 2 }^{ 3 }\times { 3 }^{ 3 }\times {2}^{6} \times {3}^{3} } \)
= \( \frac { ({ 2 }^{ 8 }\times { 2 }^{ 2 } )\times ({3}^{6} \times {3}^{4}) }{ ({ 2 }^{ 3 }\times { 2 }^{ 6 }\times ({3}^{3} \times {3}^{3}) } \)
= \( \frac { { 2 }^{ 8+2 }\times { 3 }^{ 4+6 } }{ { 2 }^{ 3+6 }\times { 3 }^{ 3+3 } } \)
= \( \frac { { 2 }^{ 10 }\times { 3 }^{ 10 } }{ { 2 }^{ 9 }\times { 3 }^{ 6 } } \)
= \( \frac { { 2 }^{ 10 } }{ { 2 }^{ 9 } } \) x \( \frac { { 3 }^{ 10 } }{ { 3 }^{ 6 } } \)
= \( { 2 }^{ 10-9 } \) x \( { 3 }^{ 10-6 } \)
= \( { 2 }^{ 1 } \) x \( { 3 }^{ 4 } \)
= 2 x 81 = 162

Example 7: Find the values of n in each of the following:

(i) \( { ({ 2 }^{ 2 }) }^{ n } \) = \( { ({ 2 }^{ 3 }) }^{ 4 } \)    (ii) \( { 2 }^{ 5n } \) ÷ \( { 2 }^{ n } \) = \( { 2 }^{ 4 } \)

Solution.

(i) We have,

\( { ({ 2 }^{ 2 }) }^{ n } \) = \( { ({ 2 }^{ 3 }) }^{ 4 } \)
=>    \( { 2 }^{ 2n } \) = \( { 2 }^{ 3 \times 4 } \)

=>    \( { 2 }^{ 2n } \) = \( { 2 }^{ 12 } \)

=>    2n = 12        [On equating the exponents]

=>    n = \(\frac{12}{2}\) = 6

(ii) We have,

\( { 2 }^{ 5n } \) ÷ \( { 2 }^{ n } \) = \( { 2 }^{ 4 } \)
=>    \( \frac { { 2 }^{ 5n } }{ { 2 }^{ n } } \) = \( { 2 }^{ 4 } \)

=>    \( { 2 }^{ 5n-n } \) = \( { 2 }^{ 4 } \)    [since, \( \frac { { a }^{ m } }{ { a }^{ n } } \) = \( { a }^{ m-n } \) ]

=>    \( { 2 }^{ 4n } \) = \( { 2 }^{ 4 } \)

=>     4n = 4         [On equating the exponents]

=>    n = \(\frac{4}{4}\) = 1

Example 8: If \({25}^{n-1}\) + 100 = \({5}^{2n-1}\), find the value of n.

Solution.    We have,

\({25}^{n-1}\) + 100 = \({5}^{2n-1}\)

=>    \({5}^{2n-1}\) – \({5}^{2n-1}\) = 100

=>    \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { { 25 }^{ n } }{ { 25} } \) = \({10}^{2}\)

=>    \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { ({ 5 } ^{2})^{ n } }{ { 25} } \) = \( ({2 \times 5}) ^{2}\)

=>    \( \frac { { 5 }^{ 2n } }{ { 5} } \) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \(\frac{1}{5}\) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \((\frac{1}{5})\) – \( \frac { { 5 }^{ 2n } }{ { 25} } \) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \((\frac{5 – 1}{25})\) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \((\frac{4}{25})\) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \( \frac { { 2 }^{ 2 } }{ { 5 }^{ 2 } } \) = \({2}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) x \({2}^{2}\) = \({2}^{2} \times {5}^{2} \times {5}^{2}\)

=>    \({5}^{2n}\) = \( \frac { { 2 }^{ 2 } \times {5}^{2} \times {5}^{2} }{ { 2 }^{ 2 } } \)

=>    \({5}^{2n}\) = \({2}^{2 – 2}\) x \({5}^{2 + 2}\)

=>    \({5}^{2n}\) = \({2}^{0}\) x \({5}^{4}\)

=>    \({5}^{2n}\) = \({5}^{4}\)

=>    2n = 4     [On equating the exponents]

=>    n = \(\frac{4}{2}\) = 2

Simple Interest

Definitions

Principal: The money borrowed (lent or invested) is called principal.

Interest: The additional money paid by the borrower to the moneylender in lieu of the money used by him is called interest.

Amount: The total money paid by the borrower to the moneylender is called amount.

Thus, amount = principal + interest

Rate: It is the interest paid on Rs 100 for specified period.

For example:
(i) Rate of 6% per annum means that the interest paid on Rs 100 for one year is Rs 6

(ii) Rate of 1.25% per month means that the interest paid on Rs 100 for one month is Rs 1.25

(iii) Rate of 2.5% per quarterly means that the interest paid on Rs 100 for 3 months is Rs 2.5

However, if the time period for the interest rate is not given, then we shall take the time period as one year.

Time: It is the time for which the money is borrowed (or invested).

Simple interest: It is the interest calculated on the original money (principal) at given rate of interest for any given time.

Following example explains the above terms :

Vijay borrowed Rs 10,000 from a credit society for 1 year. At the end of 1 year, he was required to return Rs 11,200 instead of Rs 10,000 which he borrowed.

Why did Vijay not return the same amount of money i.e., Rs 10,000 which he borrowed?

Why was he required to pay Rs 1200 extra to the credit society?

Vijay was required to pay Rs 1200 extra to the credit society because he used the credit society’s money (Rs 10,000) for 1 year. The extra money is the charge for using the credit society’s money which Vijay borrowed. It is called the interest charged by the credit society.

• The money that Vijay borrowed (Rs 10,000) is called the Principal.

• The duration (1 year) for which he borrowed the money is called the Period.

• The charges (Rs 1200) for using the money is called the Interest.

• The total money which Vijay returned at the end of 1 year

(Rs 10,000 + Rsl200 = Rs 11,200) is called the Amount.

Formulas

If P denotes the principal, R the rate of interest, T the time for which the money is borrowed (or invested), I (or S.I.) the simple interest and A the amount, then

I = \(\frac{(P * R * T)}{100}\)
P = \(\frac{I * 100}{R * T}\), R = \(\frac{I * 100}{P * T}\), T = \(\frac{I * 100}{P * R}\)
A = P + I = P + \(\frac{P * R * T}{100}\) = (1 + \(\frac{R * T}{100}\)) P

Points To Remember:

-> For counting the time between two given dates, only one of the two dates counted (either first or last). Usually, we exclude the date of start and include the date of return.

-> For converting the time in days into years, always divide by 365, whether it is leap year or not.

-> The time must be taken in accordance with the interest rate percent. Thus, if interest rate is per month then time must be taken in months.

Illustrative Examples

Example1: Find the simple Interest on Rs 7200 at 5% per annum for 8 months. Also, find the amount.

Solution. Principal = Rs 7200, rate = 5% p.a. and time = \(\frac{8}{12}\) year = \(\frac{2}{3}\) year.

Therefore,        P = Rs 7200, R = 5% p.a. and T = \(\frac{2}{3}\) year.

=>    SI = \(\frac{P * T * R}{100}\)

= Rs (7200 x 5 x \(\frac{2}{3}\) x \(\frac{1}{100}\)) = Rs 240.

=>  amount = (principal + SI)

= Rs (7200 + 240) = Rs 7440.

Therefore,              SI = Rs 240 and amount = Rs 7440.

Example 2: Find the simple interest on Rs 4500 at 8% per annum for 73 days. Also, find the amount.

Solution.    Principal = Rs 4500, rate = 8% p.a. and time = \(\frac{73}{365}\) year = \(\frac{1}{5}\) year.

Therefore,        P = Rs 4500, R = 8% p.a. and T = \(\frac{1}{5}\) year.

=> SI = \(\frac{P * T * R}{100}\)

= Rs (4500 x 8 x \(\frac{1}{5}\) x \(\frac{1}{100}\)) = Rs 72.

=> amount = (principal + SI)

= Rs (4500 + 72) = Rs 4572.

Therefore,            SI = Rs 72 and amount = Rs 4572.

Example 3: How long will it take for Rs 5660 invested at 10% per annum simple interest to amount to Rs 7641?

Solution.    Here, P = Rs 5660, A = Rs 7641, R = 10% p.a.

I = A — P = Rs 7641 — Rs 5660 = Rs 1981

Let T years be the required time.

Using    T = \(\frac{I * 100}{P * R}\),

we get T = \(\frac{1981 * 100}{5660 * 10}\)
= \(\frac{1981}{566}\)
= \(\frac{7}{2}\) = \(3\frac{1}{2}\)

Hence the required time = \(3\frac{1}{2}\) years = 3 years 6 months.

Example 4: In what time will the simple interest on a certain sum of money at 6% per annum be of itself?

Solution.    Let the sum of money (principal) be Rs P, then

Interest = \(\frac{3}{8}\) of Rs P = Rs \(\frac{3}{8}\)P

Rate of simple interest = \(6\frac{1}{4}\) p.a. = \(\frac{25}{4}\) % p.a.

Let T years be the required time

Using, T = \(\frac{I * 100}{P * R}\)

we get T = \( \frac { \frac { 3 }{ 8 } \times P\times 100 }{ P\times \frac { 25 }{ 4 }  } \)

= \(\frac{3}{8}\) x 100 x \(\frac{4}{25}\)
= 6
Hence the required time = 6 years.

Example 5: If the interest charged for 9 months be 0.09 times the money borrowed, find the rate of simple interest per annum.

Solution.    Let the money borrowed (principal) be Rs P, then

I (simple interest) = 0.09 of Rs P = Rs \(\frac{9}{100}\)  P

Time = 9 months = \(\frac{9}{12}\) years = \(\frac{3}{4}\) years

Let R be the rate percent of simple interest per annum.

Using R = \(\frac{I \times 100}{P \times T}\),
we get R = \( \frac { \frac { 9 }{ 100 } \times P\times 100 }{ P\times \frac { 3 }{ 4 }  } \)

= 9 x  \(\frac{4}{3}\)
= 12

Hence the rate of simple interest per annum 12%.

Example 6: At what rate percent simple interest will a sum of money will amount to \(\frac{5}{3}\) of itself in 6 years 8 months?

Solution.    Let the money borrowed (principal) be Rs.P, then

amount = \(\frac{5}{3}\) P = Rs  \(\frac{5}{3}\) P

I (simple interest) = Amount – Principal = Rs  \(\frac{5}{3}\) P – Rs P
= ( \(\frac{5}{3}\) P – P)
= Rs ( \(\frac{5}{3}\) – 1) P
= Rs  \(\frac{2}{3}\) P

Time = 6 years 9 months = \( 6\frac{8}{12}\) years = \(6\frac{2}{3}\) years =  \(\frac{20}{3}\) years

Let R be the rate percent of simple interest per annum.

Using R = \(\frac{I \times 100}{P \times T}\),
we get R = \( \frac { \frac { 2 }{ 3 } \times P\times 100 }{ P\times \frac { 20 }{ 3 }  } \)

= \(\frac{200}{3}\) x  \(\frac{3}{20}\)
= 10

Hence the rate of simple interest per annum 10%.

Example 7: Sudhir borrowed Rs 3,00,000 at 12% per annum from a money-lender. At the end of 3 years, he cleared the account by paying Rs 2,60,000 and a gold necklace. Find the cost of the necklace.

Solution.            SI = \(\frac{P * T * R}{100}\)

Here        P = Rs 3,00,000, R = 12%, T= 3 years
Therefore,        S.I. = Rs \(\frac{300000 \times 12 \times 3}{100}\) = Rs 108000

So, amount to be returned by Sudhir

= Principal + Interest = Rs300000 + Rs 108000 = Rs 408000
Amount returned by Sudhir = Rs 2,60,000
So, cost of the necklace = Rs 4,08,000— Rs2,60,000 = Rs 1,48,000.

Finding A Percentage Of A Number

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

Conversion Of Per cent Into Decimal And Vice Versa

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%

Finding The C.P. When S.P. and Profit Or Loss Percent Are Given

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

Decimal Number System

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

Compound Interest

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

Exponents

Definition

If a is any real number and n is a natural number, then \( { a }^{ n } \) = a x a x a…. n times

where a is called the base, n is called the exponent or index and  \( { a }^{ n } \) is the exponential expression. \( { a }^{ n } \) is read as ‘a raised to the power n’ or ‘a to the power n’ or simply ‘a power n’.

For zero power, we have :

\( { a }^{ 0 } \) = 1 (where a \( \neq \) 0)

For example :

(i) \( { 7 }^{ 0 } \) =  1    (ii) \( { (-\frac { 2 }{ 3 } ) }^{ 0 } \) = 1    (iii) \(( { \sqrt { 7 }  })^{ 0 } \) = 1

For negative powers, we have :

\( \sqrt [ n ]{ { a }} \) = \( \frac { 1 }{ { a }^{ n } } \) and \( \frac { 1 }{ { a }^{ -n } } \) = \( { a }^{ n } \)

For example:
(i) \( { 5 }^{ -2 } \) = \( \frac { 1 }{ { 5 }^{ 2 } } \)
(ii) \( { -2 }^{ -3 } \) = \( \frac { 1 }{ { -2 }^{ 3 } } \)
(iii) \( \frac { 1 }{ { 2 }^{ -5 } } \) = \( { 2 }^{ 5 } \)

For fractional indices,  we have :

\( { \sqrt { a}  }^{ n } \) = \( { a }^{ \frac { 1 }{ n }  } \) and \( \sqrt [ n ]{ { a }^{ m } } \) = \( { a }^{ \frac { m }{ n }  } \)

For example:

(i) \( { \sqrt { 3 }  }\) = \( { 3 }^{ \frac { 1 }{ 2 }  } \)
(ii) \( { \sqrt { 8 }  }^{ 3 } \) = \( { 8 }^{ \frac { 1 }{ 3}  } \)
(iii) \( \sqrt [ 4 ]{ { 5 }^{ 3 } } \) = \( { 5 }^{ \frac { 3 }{ 4 }  } \)

Finding the value of the Number given in the Exponential Form

Example 1: Find the value of each of the following:

(i) \( { 12 }^{ 2 } \)    (ii) \( { 8 }^{ 3 } \)    (iii) \( { 4 }^{ 4 } \)

Solution.

(i) We have,

\( { 12 }^{ 2 } \) = 12 x 12 = 144
(ii) We have,

\( { 8 }^{ 3 } \) = 8 x 8 x 8

= (8 x 8) x 8
= 64 x 8
= 512
(iii) We Have,

\( { 4 }^{ 4 } \)= 4 x 4 x 4 x 4

= (4 x 4 ) x 4 x 4
= (16 x 4) x 4
= 64 x 4
= 256

Example 2: Simplify:

(i) 2 x \( { 10 }^{ 3 } \)    (ii) \( { 5 }^{ 2 } \) x \( { 4 }^{ 2 } \)    (iii) \( { 3 }^{ 3 } \) x 4

Solution.

(i) We have,

2 x \( { 10 }^{ 3 } \) = 2 x 1000 = 2000     [since \( { 10 }^{ 3 } \)=10 x10 x 10 = 1000]

(ii) We have,

\( { 5 }^{ 2 } \) x \( { 4 }^{ 2 } \)

= 25 x 16 = 400

(iii) We have,

\( { 3 }^{ 3 } \) x 4  = 27 x 4 = 108

Expressing Numbers in Exponential Form

Example 1: Express each of the following in exponential form:

(i) (-4) x (-4) x (-4) x (-4) x (-4)    (ii) \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\)
Solution. We have,

(i)  (-4) x (-4) x (-4) x (-4) x (-4) = \( { -4}^{ 5 } \)

(ii) \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) x \(\frac{2}{5}\) = \( ({ \frac { 2 }{ 5 } ) }^{ 4 } \)

Example 2: Express each of the following in exponential form:

(i) 3 x 3 x 3 x a x a    (ii) a x a x a x a x a x a x b x b x b c x c x c x c
(iii) b x b x b x \(\frac{2}{5}\) x \(\frac{2}{5}\)
Solution. We have,

(i) 3 x 3 x 3 x a x a = \( { 3 }^{ 3 } \) x \( { a }^{ 2 } \)

(ii) a x a x a x a x a x a x b x b x b x c x c x c x c = \( { a }^{ 6 } \) x \( { b }^{ 3 } \) x \( { c }^{ 4 } \)

(iii) b x b x b x \(\frac{2}{5}\) x \(\frac{2}{5}\) = \( { a }^{ 3 } \) x \( ({ \frac { 2 }{ 5 } ) }^{ 2 } \)

Example 3: Express each of the following numbers in exponential form:

(i) 128    (ii) 243    (iii) 3125
Solution.

(i) We have,

128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
128 = \( { 2 }^{ 7 } \)

Express-In-Exponential-Form-example3(i)

(ii) We have,

243 = 3 x 3 x 3 x 3 x 3
243 = \( { 3 }^{ 5 } \)

Express-In-Exponential-Form-example3(ii)
(iii) We have,

625 = 5 x 5 x 5 x 5

625 = \( { 5 }^{ 4 } \)

Express-In-Exponential-Form-example3(iii)

Positive Integral Exponent of a Rational Number

Let \(\frac{a}{b}\) be any rational number and n be a positive integer. Then,

\( {(\frac { a } { b })^{ n } } \) = \(\frac{a}{b}\) x \(\frac{a}{b}\) x \(\frac{a}{b}\)…n times
= \( \frac { a\quad \times \quad a\quad \times \quad a….n\quad times }{ b\quad \times \quad b\quad \times \quad b….n\quad times } \)
= \( \frac { { a }^{ n } }{ { b }^{ n } } \)
Thus, \( {(\frac { a }{ b })^{ n } } \) = \( \frac { { a }^{ n } }{ { b }^{ n } } \) for every positive integer n.

Example : Evaluate:

(i) \( {(\frac { 3 } { 7 })^{ 3 } } \)    (ii) \( {(\frac { -2 }{ 5 })^{ 3 } } \)

Solution.

(i) \( {(\frac { 3 } { 7 })^{ 3 } } \) = \( \frac { { 3 }^{ 3 } }{ { 7 }^{ 3 } } \) = \(\frac{127}{343}\)
(ii) \( {(\frac { -2 } { 5 }) ^{ 3 } } \) = \( \frac { { (-2) }^{ 3 } }{ { 5 }^{ 3 } } \)
= \( -\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } } \)
= \(-\frac{8}{125}\)

Negative Integral Exponent of a Rational Number

Let \(\frac{a}{b}\) be any rational number and n be a positive integer.

Then, we define,  \( {(\frac { a }{ b })^{ -n } } \) = \( {(\frac { b }{ a })^{ n } } \)

Example : Evaluate:

(i) \( {(\frac { 1 } { 2 })^{ -3 } } \)    (ii) \( {(\frac { 2 } { 7 })^{ -2 } } \)

Solution.

(i) \( {(\frac { 1 }{ 2 })^{ -3 } } \)
= \( {(\frac { 2 } { 1 })^{ 3 } } \) = \( \frac { { 2}^{ 3 } }{ { 1 }^{ 3 } } \)
= 8
(ii) \( {(\frac { 2 } { 7 })^{ -2 } } \)
= \( {(\frac { 7 } { 2 })^{ 2 } } \)

= \( \frac { { 7}^{ 2 } }{ { 2 }^{ 2 } } \)
= \(\frac{49}{4}\)