Category: Maths

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

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

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

When the Interest is Compounded Annually

Formula

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

Illustrative Examples

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

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

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

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

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

= Rs 10648.

Thus, Amount after 3 years = Rs 10648.

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

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

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

Amount after 3 year = P\( { (1+\frac { R }{ 100 } ) }^{ n } \)
= Rs 8000 x \( { (1+\frac { 5 }{ 100 } ) }^{ 3 } \)

= Rs 8000 x \( { (1+\frac { 1 }{ 20 } ) }^{ 3 } \)
= Rs 8000 x \( { (\frac { 21 }{ 20 } ) }^{ 3 } \)
= Rs 8000 x \(\frac{21}{20}\) x \(\frac{21}{20}\) x \(\frac{21}{20}\)

= Rs 9261

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

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

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

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

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

When the Interest is Compounded Half-Yearly

Formula

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

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

Illustrative Examples

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

Solution.

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

Amount after 2 years

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

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

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

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

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

= \(12\frac{1}{2}\)%

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

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

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

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

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

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

Amount after 2 years

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

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

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

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

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

= Rs 19448.10

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

When the Interest is Compounded Quarterly

Formula

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

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

C.I. = A – P

Illustrative Examples

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

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

Amount after 1 year

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

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

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

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

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

= Rs 397372.64

Therefore, Compound Interest = Rs 397372.64 – Rs 360000

= Rs 37372.64

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

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

= \(\frac{6}{12}\)

= \(\frac{1}{2}\) year

Amount after 6 months

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

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

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

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

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

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

= \(\frac{9}{12}\)

= \(\frac{3}{4}\) year

Amount after 9 months

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

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

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

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

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

= Rs 8195.45

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

= Rs 8405

When the Rate of Interest for Successive years are Different

Formula

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

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

Illustrative Examples

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

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

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

Amount after 2 years

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

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

= Rs 58860

Thus, amount after 2 years = Rs 58860.

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

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

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

amount after 3 years

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

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

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

When Interest is Compounded Annually but Time is a Fraction

Formula

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

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

Illustrative Examples

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

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

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

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

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

= Rs 38637

Therefore,    Amount = Rs 38637.

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

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

Solution.

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

Amount after \(2\frac{1}{3}\)years

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

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

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

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

= Rs 33327

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

Try More:

Nr7 Candle

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