Question

a man lent a part of ₹30000 at 12% and the remaining at 10% simple intrest. the total intrest he received after 2 years is ₹36480.calculate the amount he lent at 12%. Plz help me with this question​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Amount\:lent\:at\:12\%=12000\:rupees}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Amount \: lent = 30000 \: rupees \\ \\ \tt: \implies Rate(r_{1}) = 12\% \\ \\ \tt: \implies Rate(r_{2}) = 10\% \\ \\ \tt: \implies Time(t) = 2 \: years \\ \\ \tt: \implies Amount \: after \: two \: years = 36480 \: rupees \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Amount \: lent \: at \: 12\% = ?$

• According to given question :

$\tt \circ \: Let \: amount \: lent \: at \: 12\% = x \\ \\ \tt \circ \: Amount \: lent \: at \: 10\% = 30000 - x \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies S.I = \frac{p \times r_{1}\times t}{100} \\ \\ \tt: \implies S.I= \frac{x \times 12 \times 2}{100} \\ \\ \green{\tt: \implies S.I = 0.24x} \\ \\ \bold{Similarly \: for \: r_{2} } \\ \tt: \implies S.I_{1} = \frac{(30000 - x) \times 10 \times 2}{100} \\ \\ \tt: \implies S.I_{1} = \frac{600000 - 20x}{100} \\ \\ \green{\tt: \implies S.I_{1} =6000 - 0.2x} \\ \\ \bold{For \: total \: Simple \: Interest} \\ \tt: \implies S.I_{t}=36480 - 30000 \\ \\ \tt: \implies S.I + S.I_{1} = 6480 \\ \\ \tt: \implies 0.24x + 6000 - 0.2x = 6480 \\ \\ \tt: \implies 0.04x = 6480 - 6000 \\ \\ \tt: \implies 0.04x = 480 \\ \\ \tt: \implies x = \frac{480}{0.04} \\ \\ \green{\tt: \implies x = 12000 \: rupees} \\ \\ \green{\tt \therefore Amount \: lent \: at \: 12\% \: is \: 12000 \: rupees}$

Answer 2

Given:-

• Rate, r = 12 %

• S.I = 10 %

• Time, n = 2 years

• Total interest received = ₹ 36480

To find:-

The amount he lent at 12%.

Solution:-

Let the sum lent at the rate of 12 % per annum be ₹ x.

So, the lent at 10 % p.a. is ₹ (3000-x).

S.I at the end of 2 years =  ₹ (36480 - 30000)

= ₹ 6480

So,

[(x*12*2)/100] + {[(3000-x)*10*2]/100} = 6480

24x + 600000 - 20x = 648000

⇒4x = 48000

$\boxed{x=Rs.\ 12000}$