Category: Maths

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

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

(ii) We have,

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

(iii) We have,

625 = 5 x 5 x 5 x 5

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

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) \( {(\frac { 2 } { 7 })^{ -2 } } \)
= \( {(\frac { 7 } { 2 })^{ 2 } } \)

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

We have :  \( { (a+b) }^{ 3 } \) = \( { a }^{ 3 } \) + 3 \( { a }^{ 2 } \) b+ 3 a \( { b }^{ 2 } \) + \( { b }^{ 3 } \).
Method : For finding the cube of a two-digit number with the tens digit = a and the units digit b, we make four columns, headed by
\( { a }^{ 3 } \), 3 \( { a }^{ 2 } \) b, 3 a \( { b }^{ 2 } \) and \( { b }^{ 3 } \).
The rest of the procedure is the same as followed in squaring a number by the column method.

We simplify the working as :

Example 1: Find the value of \( { (29) }^{ 3 } \) by the short-cut method.
Solution. Here,            a=2 and b=9

Therefore,        \( { (29) }^{ 3 } \) = 24389.

Example 2: Find the value of \( { (71) }^{ 3 } \) by the short-cut method.
Solution. Here, a = 7 and b = 1.

Therefore,     \( { (71) }^{ 3 } \) = 357911.

Definitions

Sometimes to increase the sale or to dispose off the old stock, a dealer offers his goods at reduced prices. The reduction in price offered by the dealer is called discount.

Marked Price: The printed price or the tagged price of an article is called the marked price (M.P.). It is also called the list price.

Discount: The deduction allowed on the marked price is called discount. Discount is generally given as per cent of the marked price.

Net Price: The selling price at which the article is sold to the customer after deducting the discount from the marked price is called the net price.

Formulas

(i) S.P. = M.P. – Discount

(ii) Rate of discount = Discount % =  \(\frac{Discount}{M.P.}\) x 100

(iii) S.P. = M.P. x (\(\frac{100 – Discount \%}{100}\))

(iv) M.P. = \(\frac{100 * S.P.}{100 – Discount \%}\)

Illustrative Examples

Example 1: Find S.P. if M.P. = Rs 650 and Discount = 10%

Solution. (i) We have,

M.P. = Rs 650, Discount = 10%

Discount = 10% of Rs 650 = Rs(\(\frac{10}{100}\) x 650) = Rs 65

Hence, S.P. = M.P. — Discount = Rs 650 — Rs 65 = Rs 585

Alternative Solution–    We have,

M . P. = Rs 650, Discount % =10

S.P. = M.P. x \(\frac{(100 – Discount \%)}{100}\)

=>    S.P. = Rs {650 x \(\frac{(100 -10)}{100}\)} = Rs(65 x 9) = Rs 585

Example 2: Find the rate of discount when M.P. = Rs 600 and S.P. = Rs 510.

Solution.    M.P. = Rs 600, S.P. = Rs 510

Therefore,    Discount = M.P. — S.P. = Rs 600 — Rs 5l0 = Rs 90

Therefore,    Rate of discount, i.e., discount% = \(\frac{Discount}{M.P.}\) x 100 = \(\frac{90}{600}\) x 100% = 15%.

Example 3: Find the M.P. When S.P. = Rs 9,000 and discount = 10%.

Solution.    S.P. = 9000, discount = 10%
Let the M.P. be Rs 100. Since discount = 10%, So S.P. = Rs 90.

When S.P. is 90, M.P. is 100.

When S.P. is Rs 1, M.P. is Rs \(\frac{100}{90}\)
When S.P. is Rs 9000, M.P. is Rs \(\frac{100}{90}\) x 9000 = Rs 10,000

Example 4: A garment dealer allows his customers 10% discount on a marked price of the goods and still g a profit of 25%. What is the cost price if the marked price of a shirt is Rs 1250?

Solution.    M.P. = 1250, Discount = 10%

When M.P. is 100, S.P. is 90

When M.P. is 1250, S.P. is Rs \(\frac{900}{100}\) = Rs 1125

Profit = 25%, So C.P. = \(\frac{100}{(100 + Profit \%)}\) x S.P.
= Rs \(\frac{100}{(100 + 25)}\) x 1125

= Rs \(\frac{100}{125}\) x 1125
= Rs(100 x 9) = Rs 900

Successive Discounts

Example 5: A car is marked at 4,00,000. The dealer allows successive discounts of 5%, 3% and \(2\frac{1}{2}\)% on it. What is the net selling price ?

Solution.    Marked price of the car = Rs 4,00,000

First discount = 5% of 4,00,000 = (\(\frac{5}{100}\) x 400000) = Rs 20,000

Net price after first discount = (4,00,000 — 20,000) = 3,80,000

Second discount = 3% of Rs 3,80,000 = (\(\frac{3}{100}\) x 380000) = Rs 11,400

Net price after second discount = (3,80,000 — 11,400) = Rs 3,68,600

Third discount = Rs ( \( \frac { 2\frac { 1 }{ 2 }  }{ 100 } \) x 3, 68, 600)
= (\(\frac{5}{200}\) x 368600) = Rs 9215

Net selling price = Rs (3,68,600 — 9215) = Rs 3,59,385.

Example 6: Find a single discount equivalent to two successive discounts of 20% and 5%.

Solution.    Let the marked price be Rs 100.

First discount = Rs 20

Net price after first discount = Rs (100 — 20) = Rs 80

Second discount 5% of Rs 80 = Rs (\(\frac{5}{100}\) x 8O) = Rs 4

Net price after second discount = Rs (80 — 4) = Rs 76

Total discount allowed = Rs (100 — 76) = Rs 24

Hence, the required single discount = 24%.

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 %

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

Steps involved in conversion of a ratio into per cent

STEP I– Obtain the ratio, say, a : b.

STEP II– Convert the given ratio into the fraction \(\frac{ a }{ b }\)

STEP III– Multiply the fraction obtained in step II by 100 and put per cent sign %.

Illustration 1: Express the following as per cents:
(i)  14 : 25                (ii) 5 : 6                (iii) 111 : 125

Solution. We have :

(i) 14 : 25 =\(\frac{ 14 }{ 25 }\) = (\(\frac{ 14 }{ 25 }\) x 100)% = 56%.
(ii) 5 : 6 = \(\frac{ 5 }{ 6 }\) = (\(\frac{ 5 }{ 6 }\) x 100)% = \(\frac{ 250 }{ 3 }\)% = \(83\frac{1}{3}\)%
(iii) 111 : 125 = \(\frac{ 111 }{ 125 }\) = (\(\frac{ 111 }{ 125 }\) x 100)% = 88.88%

Steps involved in conversion of a per cent into ratio

STEP I– Obtain the per cent.

STEP II– Convert the given per cent into a fraction by dividing it by 100 and removing per cent sign %.

STEP III– Express the fraction obtained in step II in the simplest form.

STEP IV– Express the fraction obtained in step III as a ratio.

Illustration 2 : Express each of the following per cents as a ratio in the simplest form:

Solution.  We have:

(i) 36% = \(\frac{ 36 }{ 100 }\) = 0.36
(ii) 5.4% = \(\frac{ 5.4 }{ 100 }\) = \(\frac{ 54 }{ 1000 }\) = 0.054
(iii) 0.25% = \(\frac{ 0.25 }{ 100 }\) = \(\frac{ 25 }{ 10000 }\) = 0.0025
(iv) 135% = \(\frac{ 135 }{ 100 }\) = 1.35

Procedure

Sometimes we are given two quantities and we want to find what per cent of one quantity is of the other quantity. In other words, we want to find how many hundredths of one quantity should be taken so that it is equal to the second quantity. In such type of problems, we proceed as discussed below:

Let a and b be two numbers and we want to know: what per cent of a is b ?

Let x% of a be equal to b. Then,

\(\frac{ x}{ 100}\)    x a = b

=> x = b x \(\frac{ 100}{ a }\)
=> x = \(\frac{ b}{ a}\) x lOO

Thus, b is (\(\frac{ b}{ a }\) x 100)% of a.

Illustrative Examples:

Example 1: What per cent of 25 kg is 3.5 kg?

Solution. We have,

Required per cent = ( \(\frac{ 3.5 kg}{ 25 kg}\) x 100) = \(\frac{ 3.5 * 100}{ 25}\)

= \(\frac{ 35 * 100}{ 250}\)

=\(\frac{ 35 * 2}{ 5}\)

= 7x 2

= 14
Hence, 3.5 kg is 14% of 25 kg.

Alternative Solution-

Let x% of 25 kg be 3.5 kg. Then,
x% of 25kg = 3.5kg
=> \(\frac{ x}{ 100}\) x 25 = 3.5
=> x = \(\frac{ 3.5 * 100}{ 25}\)     [Multiplying both sides by \(\frac{ 100}{ 25}\) ]
=> x = \(\frac{ 35 * 100}{ 250}\) = \(\frac{ 35 * 2}{ 5}\) = 7 x 2 = 14.

Example 2: Express 75 paise as a per cent of Rs 5.

Solution. We have, Rs 5 = 500 paise.

Let x% of Rs 5 be 75 paise. Then,

x% of Rs 5 = 75 paise

=> x% of 500 paise = 75 paise

=> \(\frac{ x}{ 100}\) x 500 = 75

=> x = \(\frac{ 75 * 100}{ 500}\)

=> x = 15.

Hence, 15% of Rs 5 is 75 paise.

Alternative Solution-  The required per cent = ( \(\frac{ 75}{ 500}\) x 100) % = 15%

Example 3 : Find 10% more than Rs 90.
Solution. We have,

10% of Rs 90 = Rs ( \(\frac{ 10 }{ 100}\) x 90 ) = Rs 9
Therefore, 10% more than Rs 90 = Rs 90 + Rs 9 = Rs 99

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.

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