Question

A sum of Rs.1536 put at compound interest amount s to Rs.1632 in one year .How much would it amount to in the second year

Answer 1

A = P[1+(R/100)]^T

$1632 = 1536 {(1 + \frac{r}{100} )}^{1} \\ \frac{1632}{1536} = \frac{100 + r}{100} \\ 1.0625 = \frac{100 + r}{100} \\ 106.25 = 100 + r \\ r = 6.25$

NOW FINDING AMOUNT FOR 2ND YEAR.

$a = p {(1 + \frac{r}{100} )}^{2} \\ a = 1536 {( \frac{100 + 6.25}{100}) }^{2} \\ a = 1536 \times \frac{10625 \times 10625}{10000 \times 10000} \\ a = 1536 \times 1.12890625 \\ a = 1734$

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WE CAN ALSO FIND THIS BY ANOTHER METHOD OF [{SIMPLE INTEREST}]

WE KNOW THAT COMPOUND INTEREST FOR 1 YEAR ON THE SAME RATE EQUALS TO THE SIMPLE INTEREST FOR 1 YEAR AT THE SAME RATE.

SO,

TAKING,

PRINCIPAL = 1632

RATE = 6.25

TIME = 1 YEAR

SI = PRT/100

SI = 1632×6.25×1/100

SI = 1632×2.5×2.5/100

SI = 1632×(1/16)

SI = 102

AMOUNT = PRINCIPAL + SI

AMOUNT = 1632 + 102

AMOUNT = 1734

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Answer 2

Step-by-step explanation:

Given,

P= ₹1536, A=₹1632, t=1 yr.

therefore SI= A-P= ₹(1632-1536)=₹96

So, SI=₹96, P=₹1536, t=1 yr, r=?

r= SI×100/P×t= 96×100/1536×1

=6•25%

therefore r=6•25%

For 2nd year

P=₹1632

t=1 year

r=6•25%

SI= Prt/100

SI= 1632×6•25×1/100

SI= ₹102

Amount for 2nd year =₹(1632+102)=₹1734 Ans.