Question

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Find the compound interest at the rate of 5% per annum for 3 years on that sum which in 3 years at the rate of 5% per annum gives 1200 as simple interest.​

Answer 1

Answer:

➡️⏩➡️⏩➡️⏩➡️⏩➡️⏩➡️⏩➡️

given that

S.I = 1️⃣2️⃣0️⃣0️⃣

T = 3️⃣ years

so,

by using the formula

S.I = PRT/1️⃣0️⃣0️⃣

P = S.I × 1️⃣0️⃣0️⃣/RT

P = 1️⃣2️⃣0️⃣0️⃣×1️⃣0️⃣0️⃣

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5️⃣× 3️⃣

P = 8️⃣0️⃣0️⃣0️⃣0️⃣

now,

C.I= P{(1️⃣➕R/1️⃣0️⃣0️⃣)^n ➖1️⃣}

C.I ={2️⃣1️⃣}^3 ➖8️⃣0️⃣0️⃣0️⃣

C.I = 9️⃣2️⃣6️⃣1️⃣➖8️⃣0️⃣0️⃣0️⃣

C.I= 1️⃣2️⃣6️⃣1️⃣. RS

hope it helps u

Answer 2

$\huge{\bigstar{\underline{\red{\mathfrak{Question}}}}}$

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ғɪɴᴅ ᴛʜᴇ ᴄᴏᴍᴘᴏᴜɴᴅ ɪɴᴛᴇʀᴇsᴛ ᴀᴛ ᴛʜᴇ ʀᴀᴛᴇ ᴏғ 5% ᴘᴇʀ ᴀɴɴᴜᴍ ғᴏʀ 3 ʏᴇᴀʀs ᴏɴ ᴛʜᴀᴛ sᴜᴍ ᴡʜɪᴄʜ ɪɴ 3 ʏᴇᴀʀs ᴀᴛ ᴛʜᴇ ʀᴀᴛᴇ ᴏғ 5% ᴘᴇʀ ᴀɴɴᴜᴍ ɢɪᴠᴇs 1200 ᴀs sɪᴍᴘʟᴇ ɪɴᴛᴇʀᴇsᴛ.

$\huge{\bigstar{\underline{\red{\mathfrak{Answer}}}}}$

$\sf{S.I. = 1200}$

Rate of Interest = 5% p.a.

$\sf{Time = 3 years}$

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$\sf{P = \frac{S.I. \times 100}{R \times T}}$

$\sf{ = \frac{1200 \times 100}{5 \times 3} }$

$\sf{ = 400 \times 20}$

$\sf{ = 8000}$

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$\sf{=> C.I. = P [(1 + \frac{R}{100} {)}^{n} - 1]}$

$\sf{= 8000[(1+ { \frac{5}{100} )}^{3} - 1]}$

$\sf{= 8000[(1+ { \frac{1}{20} })^{3} - 1]}$

$\sf{= 8000[ \frac{{21}^{3} - {20}^{3} }{8000} ]}$

$\sf{= 8000[ \frac{9261 - 8000}{8000} ]}$

$\sf{= 8000[\frac{1216}{8000}]}$

$\sf{= 8000 \times \frac{1216}{8000}}$

$\sf{= 1261}$

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