Question

Find the compound interest on 4500 at 10% per annum for 5 years, compounded half-yearly,
Find the compound interest on 2000 at the rate of 8% per annum for 18 months, when the interest is
calculated half-yearly.​

Answer 1

S O L U T I O N :

$\underline{\bf{Given\::}}$

• Principal, (P) = Rs.4500
• Rate, (R) = 10% p.a
• Time, (n) = 5 years .

$\underline{\bf{Explanation\::}}$

As we know that formula of the compounded half - yearly;

$\boxed{\bf{Amount = Principal\bigg(1+\frac{R/2}{100} \bigg)^{2n}}}$

A/q

$\mapsto\tt{A =P\bigg(1+ \dfrac{R/2}{100} \bigg)^{2n}}$

$\mapsto\tt{A =4500\bigg(1+ \dfrac{10/2}{100} \bigg)^{2\times 5}}$

$\mapsto\tt{A =4500\bigg(1+ \dfrac{\cancel{10/2}}{100} \bigg)^{10}}$

$\mapsto\tt{A =4500\bigg(1+ \dfrac{5}{100} \bigg)^{10}}$

$\mapsto\tt{A =4500\bigg(1+ \cancel{\dfrac{5}{100}} \bigg)^{10}}$

$\mapsto\tt{A =4500\bigg(1+ \dfrac{1}{20} \bigg)^{10}}$

$\mapsto\tt{A =4500\bigg( \dfrac{20+1}{20} \bigg)^{10}}$

$\mapsto\tt{A =4500\bigg( \dfrac{21}{20} \bigg)^{10}}$

$\mapsto\tt{A =\cancel{4500} \times \dfrac{21}{\cancel{20}} \times \dfrac{21}{\cancel{20}} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times\dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} }$

$\mapsto\tt{A =11.25 \times \dfrac{21}{1} \times \dfrac{21}{1} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times\dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} \times \dfrac{21}{20} }$

$\mapsto\tt{A =\dfrac{187648661004761.3}{25600000000} }$

$\mapsto\tt{A =\cancel{\dfrac{187648661004761.3}{25600000000} }}$

$\mapsto\bf{A = Rs.7330.02}$

Now, as we know that compound Interest;

→ C.I. = Amount - Principal

→ C.I. = Rs.7330.02 - Rs.4500

→ C.I. = Rs.2830.02

Thus,

The compound Interest will be Rs.2830.02 .

Again,

$\underline{\bf{Given\::}}$

• Principal, (P) = Rs.2000
• Rate, (R) = 8% p.a
• Time, (n) = 18 months  [18/12 = 3/2 years ]

$\underline{\bf{Explanation\::}}$

A/q

$\mapsto\tt{A =P\bigg(1+ \dfrac{R/2}{100} \bigg)^{2n}}$

$\mapsto\tt{A =2000\bigg(1+ \dfrac{8/2}{100} \bigg)^{2\times 3/2}}$

$\mapsto\tt{A =2000\bigg(1+ \dfrac{\cancel{8/2}}{100} \bigg)^{\cancel{2} \times 3/\cancel{2}}}$

$\mapsto\tt{A =2000\bigg(1+ \dfrac{4}{100} \bigg)^{3}}$

$\mapsto\tt{A =2000\bigg(1+ \cancel{\dfrac{4}{100}} \bigg)^{3}}$

$\mapsto\tt{A =2000\bigg(1+ \dfrac{1}{25} \bigg)^{3}}$

$\mapsto\tt{A =2000\bigg(\dfrac{25+1}{25} \bigg)^{3}}$

$\mapsto\tt{A =2000\bigg(\dfrac{26}{25} \bigg)^{3}}$

$\mapsto\tt{A =\cancel{2000} \times \dfrac{26}{\cancel{25}} \times \dfrac{26}{\cancel{25}} \times \dfrac{26}{25} }$

$\mapsto\tt{A =3.2 \times \dfrac{26}{1} \times \dfrac{26}{1} \times \dfrac{26}{25} }$

$\mapsto\tt{A =\cancel{\dfrac{56243.2}{25} }}$

$\mapsto\bf{A =Rs.2249.72}$

Now, as we know that compound Interest;

→ C.I. = Amount - Principal

→ C.I. = Rs.2249.72 - Rs.2000

→ C.I. = Rs.249.72

Thus,

The compound Interest will be Rs.249.72 .

Answer 2

Step−by−stepexplanation:

Let 'a' be any positive integer and b = 4.

Using Euclid Division Lemma,

a = bq + r [ 0 ≤ r < b ]

⇒ a = 3q + r [ 0 ≤ r < 4 ]

Now, possible value of r :

r = 0, r = 1, r = 2, r = 3

CASEI:

If we take, r = 0

⇒ a = 4q + 0

⇒ a = 4q

On cubing both sides,

⇒ a³ = (4q)³

⇒ a³ = 4 (16q³)

⇒ a³ = 9m [16q³ = m as integer]

CASEII:

If we take, r = 1

⇒ a = 4q + 1

On cubing both sides ;

⇒ a³ = (4q + 1)³

⇒ a³ = 64q³ + 1³ + 3 * 4q * 1 ( 4q + 1 )

⇒ a³ = 64q³ + 1 + 48q² + 12q

⇒ a³ = 4 ( 16q³ + 12q² + 3q ) + 1

⇒ a³ = 4m + 1 [ Take m as some integer ]

CASEIII:

If we take r = 2,

⇒ a = 4q + 2

On cubing both sides ;

⇒ a³ = (4q + 2)³

⇒ a³ = 64q³ + 2³ + 3 * 4q * 2 ( 4q + 2 )

⇒ a³ = 64q³ + 8 + 96q² + 48q

⇒ a³ = 4 ( 16q³ + 2 + 24q² + 12q )

⇒ a³ = 4m [Take m as some integer]

CASEIV:

If we take, r = 3

⇒ a = 4q + 3

On cubing both the sides;

⇒ a³ = (4q + 3)³

⇒ a³ = 64q³ + 27 + 3 * 4q * 3 (4q + 3)

⇒ a³ = 64q³ + 24 + 3 + 144q² + 108q

⇒ a³ = 4 (16q³ + 36q² + 27q + 6) + 3

⇒ a³ = 4m + 3 [Take m as some integer]

Hence, the cube of any positive integer is in the form of 4m, 4m+1 or 4m+3.

__________________

Identity used ;

( a + b )³ = a³ + b³ + 3ab ( a + b )