Question

$\rm\huge\blue{\underline{ Question :-}}$

Q. Find the amount of ₹50,000 after 2 years, compounded annually; the rate of interest being 8% p.a. during first year and 9% p.a. during the second year. Also, find the compound interest.

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Answer 1

$\sf\red{\underline{ <u>Question</u> :- }}$

• Q. Find the amount of ₹50,000 after 2 years, compounded annually; the rate of interest being 8% p.a. during first year and 9% p.a. during the second year. Also, find the compound interest.

$\sf \pink{\underline{ <u>Given</u> :- }}$

• (P) Principal = ₹50,000.
• (r) rate = 8% p.a. (first year) and, 9% p.a. (second year).
• (n) Time = 2 years.

$\sf \orange{\underline{ <u>To</u> <u>\</u><u>:</u><u> </u><u>find</u> :- }}$

• The amount after being compounded first year as S. I. and second year at C. I.

$\sf \green{\underline{ <u>Solu</u><u>tion</u> :- }}$

• ✧ Let's take out the amount of first year :-

Where,

• P (principal)= ₹50,000,
• r (rate)= 8%. and,
• n (time) = 1 year (as it is first year).

By using the formula of S. I. :-

$\bf S. I.\: = \frac{P \times r \times t}{100} \\ \bf \: = \frac{500 \cancel{00} \times 8 \times 1}{1 \cancel{00}} \\ \bf \: = { \fcolorbox{pink}{orange}{₹4000}}$

$\bf \therefore \: amount \: for \: the \: first \: year \: = (50000 + 4000 ) \\ \bf \: = { \fcolorbox{orange}{pink} { ₹54000}}$

✯Now,

• ✧ For the second year :-

Where,

• P (principal) = ₹54,000.
• n (time) = 1 year (as it is second year).
• r (rate) = 9% (given).

Using the formula of C. I. :-

$\bf \therefore \: A = p(1 + \frac{r}{100} {)}^{n} \\ \bf \implies \: A = \: 54000 {(1 + \frac{9}{100}) }^{1} \\ \bf \implies \: A = 540{ \cancel{00}} \times \frac{109} {\cancel{100}} \\ \bf \implies \: A(amount) =( 540 \times 109) \\ \bf \implies{ \fcolorbox{pink}{orange}{ \: A =₹58860}}$

$\bf \therefore \: C. I. = ₹( 58886 - 50000) ={ \fcolorbox{orange}{pink}{ ₹8860 \:\: ➜\:\: ans.}}$

$\rm\bigstar \purple{\underline{\boxed{\rm <u>Alternative</u> <u>\</u><u>:</u><u>method</u> }}}{\bigstar}$

✧ By using the formula ✧

$\bf{ \bigstar} { \underline { \boxed{\sf \bold a = \: p(principal) \times {(1 + \frac{p}{100} )} \times (1 + \frac{q}{100} )}}}{ \bigstar}$

• Here, p (rate of first year), and q (rate of second year).

$\sf a = 50000 \times (1 + \frac{9}{100} ) \times (1 + \frac{8}{100} ) \\ \bf </p><p></p><p>\implies \: a \: = 5{ \cancel{0000}} \times ( \frac{109}{ \cancel{100}} \times \frac{108}{ \cancel{100} )} \\ \sf </p><p>\implies \: a = 5 \times 108 \times 109 \\ \sf </p><p>\implies \: { \fcolorbox{orange}{pink}{a = ₹58860}} \\ \bf </p><p>\therefore \: C. I. = \: ₹(58860 - 50000) = { \fcolorbox{orange}{pink}{₹8860 \:\: ➜\:\: ans.}}$

Answer 2

Answer:

Answer is 8860

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