Question

What sum of money will get an interest of RS 1200 at 6% per annum for 2 years ?

Answer 1

Answer:

SI = rs. 1200

Time =2years

Rate p.a =6%

S.I. = P×R×T/100

1200 = P × 6 ×2 /100

120000= P×12

P= 120000/12

P= rs. 10000

so, sum of money= rs. 10000

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Answer 2

Information provided with us:

➪ Simple Interest ( S.I )

$\implies \sf \:1200\: Rs$

➪Time ( T )

$\implies \sf \: 2 \: years$

➪ Rate of Interest ( R )

$\sf\implies \: 6 \: \% \: per \: annum$

What we have to calculate༄

➪The required Principal or sum of money

✰Assumption :

➪ Consider that Principle be P

✰Formula used ༄

$\clubsuit \: \rm \:Simple \: Interest :$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\orange{ \bigg(\dfrac{P \times R \times T}{100}\bigg)}}}\: \: \: \bigstar$

✰Where ༄

➡ P = Principle

➡ R = Rate of Interest

➡ T = Time

✰Now ༄

➡ Substitute the given values in above formula and solve

$\longrightarrow\bigstar \: \: \sf \boxed{\bold{\green{1200 = \bigg( \dfrac{P\times 6 \times 2}{100}\bigg)}}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\red{P = \dfrac{1200 \times 100}{6 \times 2}\bigg)}}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\blue{P = \dfrac{120000}{12} }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\pink{P = 10,000 \:Rs }}}\: \: \: \bigstar$

✰Therefore༄

$\rm \therefore \: Thus \: required \: sum \: of \:money \: is$

$\implies\bigstar \: \: \sf\boxed{\bold{\gray{10,000 \:Rs }}}\: \: \: \bigstar$

✰Verification ༄

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\orange{\:1200 = \dfrac{10000 \times \: 6 \times 2 }{100} }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\red{\:1200 = \dfrac{120000 }{100} }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\purple{\:1200 = 1200 }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\blue{\:L.H.S = R.H.S }}}\: \: \: \bigstar$

$\longrightarrow\bigstar \: \: \sf\boxed{\bold{\purple{Hence \: verified }}}\: \: \: \bigstar$