Category: Maths

Growth and Decay

The formula A = P\((1+{ \frac { R }{ 100 } ) }^{ n} \) is called the compound interest law, and applies to any quantity which increases or decreases so that the amount at the end of each period of constant length bears a constant ratio to amount at the beginning of that period. This ratio is called the growth factor, if it is greater than 1, and the decay factor, if less than 1.

For example, if the population of a town increases steadily by 2% p.a. of the population at the beginning each year, the yearly growth factor is \( (1+\frac { 2 }{ 100 } ) \) i.e., 1.02, and the population after n years is \( { (1.02) }^{ n } \) times population at the beginning of that period. If the population decreases by 2%, then the yearly decay factor is \( (1-\frac { 2 }{ 100 } ) \) , i.e., 0.98.

Illustrative Examples

Example 1: If the population of a town decreases 2.5% annually and the present population is 3,26,40,000, find its population after 3 years.

Solution. Required population = P\((1+{ \frac { R }{ 100 } ) }^{ n} \), Here d = 2.5

Therefore,    Population after 3 years

= 32640000 \((1+{ \frac { R }{ 100 } ) }^{ n} \)

= 32640000 \((1+{ \frac { 2.5 }{ 100 } ) }^{ 2} \)

= 32640000 \((1+{ \frac { 25 }{ 1000 } ) }^{ 2} \)

= 32640000 x \( (\frac { 15 }{ 16 } ) \) x \( (\frac { 15 }{ 16 } ) \) x \( (\frac { 15 }{ 16 } ) \)

= 31028400

Example 2: The population of a certain city is 1,25,000. If the annual birth rate is 3.3% and the annual death rate is 1.3%, calculate the population after 3 years.

Solution.    Present population of the city (P) = 125000

Time (n) = 3 years, Rate of birth \( ({ R }_{ 1 } ) \) = 3.3%,

Rate of death \( ({ R }_{ 2 } ) \) = 1.3%.

So the net rate of increase (\( { R }_{ 1 } \) — \( { R }_{ 2 } \))

= 3.3 — 1.3 = 2%

Therefore,    Population after 3 years

= 1250001\((1+{ \frac { 2}{ 100 } ) }^{ 3} \)

= 125000 x \( (\frac { 51 }{ 50 } ) \) x \( (\frac { 51 }{ 50 } ) \) x \( (\frac { 51 }{ 50 } ) \)

= 51 x 51 x 51

= 132651.

Appreciation and Depreciation

When the value of an article increases with the passage of time, the article is said to appreciate. When the value of an article decreases with the passage of time, the article is said to depreciate.

For example, if a man buys a car and uses it for two years, it is obvious that the car will not be worth the e as a new one. The car will thus have depreciated in value.

On the other hand, if a man buys a piece of land, he will probably find that in a few years he will be able to get a better price for it than the price he paid for it. The value of the land will thus have appreciated. When things are difficult to obtain, they have a rarity value and appreciate.

Illustrative Examples

Example 1: In a certain experiment the count of bacteria was increasing at the rate of 2.5% per hour. Initially, the count was 512000. Find the bacteria at the end of 2 hours.

Solution.    Using the formula P\((1+{ \frac { R }{ 100 } ) }^{ n} \)

Bacteria at the end of 2 hours

= { 512000 x \((1+{ \frac { 5 }{ 2 \times 100 } ) }^{ 2} \) }

= (512000 x \( (\frac { 41 }{ 40 } ) \) x \( (\frac { 41 }{ 40 } ) \))

= 537920

Hence, the bacteria at the end of 2 hours = 537920.

Example 2: The value of a residential flat constructed at a cost of Rs 1,00,000 is appreciating at the rate 10% per year annum. What will be its value 3 years after construction?

Solution.

Present value of the flat (P) = Rs 100000, rate of appreciation = 10%, Time (n) = 3 years.

Therefore, Value of the flat after 3 years

= Rs 100000 \((1+{ \frac { 10 }{ 100 } ) }^{ 3} \)

= Rs 100000 \((1+{ \frac { 1 }{ 10 } ) }^{ 3} \)

= l00000 x \( (\frac { 11}{ 10 } ) \) x \( (\frac { 11}{ 10 } ) \) x \( (\frac { 11}{ 10 } ) \)

= Rs 1,33,100.

Example 3: A motorcycle Is bought at Rs 160000. Its value depreciates at the rate of 10% per annum. Find its value after 2 years.

Solution.

Value of the motorcycle after 2 year

= Rs { 160000 x \((1-{ \frac { 10 }{ 100 } ) }^{ 2} \) }

= Rs { 160000 x \((1-{ \frac { 1 }{ 10 } ) }^{ 2} \) }

= Rs (160000 x \( (\frac { 9 }{ 10 } ) \) x \( (\frac { 9 }{ 10 } ) \) )

= Rs 129600

Therefore, value after 1 year = Rs 129600

Example 4: Raghu purchased a boat for Rs 16,000. If the cost of the boat is depreciating at the rate of 5% annum, calculate its value after 2 years.

Solution.

Present value of the boat (P) = Rs 16,000, Rate (r) of depreciation = — 5%, Time (n) = 2 years.

Therefore,    The value of the boat after 2 years

= 16000 x \( (1-{ \frac { 5 }{ 100 } ) }^{ 2} \)

= Rs 16000 x \( (1-{ \frac { 1 }{ 20 } ) }^{ 2} \)

= Rs 16000 x \( (\frac { 19 }{ 20 } ) \) x \( (\frac { 19 }{ 20 } ) \)

= Rs 14,440.

Read More:

RELIANCE Pivot Point Calculator

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.

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

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%.

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%

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

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.

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%