phani cbse

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

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.

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