Question

ashish deposits certain sum of money every month in a recurring account for a period of 12 months if the bank repays at the rate of 11℅ per annum and ashish gets 12,175 at the maturity value of this account did what sum did he pay every month?​

Answer 1

Answer:⭐Rs.1000⭐

S.I = PRT/100

S.I = P× n(n+1)/2×12×R/100

S.I= P× 12(12+1)/2×12×11/100

S.I = 0.715 P

Maturity value: (P×12)+0.715P

12,715 = 12.715P

P = 12715/ 12.715

P = Rs.1000

Answer 2

$\underline{ \underline{{ \huge{ \mathtt{Q}}} \mathtt{UESTION -}}}$

Ashish deposits certain sum of money every month in a recurring account for a period of 12 month if the bank repays at the rate of 11% p.a. and Ashish get ₹ 12,715 at the maturity value of this account. What sum did he pay every month?

$\underline{ \underline{{ \huge{ \mathtt{G}}} \mathtt{IVEN -}}}$

• number of months (n) = 12 months
• Rate of interest (r) = 11%
• Maturity value (M.V) = ₹ 12,715

$\underline{ \underline{{ \huge{ \mathtt{T}}} \mathtt{O} \: \: \: {{ \huge{ \mathtt{F}}} \mathtt{ IND -}}}}$

• Sum that he pay every month (principal amount).

$\underline{ \underline{{ \huge{ \mathtt{F}}} \mathtt{ORMULA } \: \: \: {{ \huge{ \mathtt{U}}} \mathtt{SED -}}}}$

$\boxed{ \underline{ \mathtt{ \gray{M.V = P \times n + \frac{P \times n(n + 1)}{2 \times 12} \times \frac{r}{100}} }} }$

where,

P is the principal amount.

n is the number of months.

r is the rate of interest.

M.V is the maturity value.

$\underline{ \underline{{ \huge{ \mathtt{S}}} \mathtt{OLUTION- }}}$

$\mathtt{M.V = P \times n + \frac{P \times n(n + 1)}{2 \times 12 } \times \frac{r}{100} } \\ \\ \mathtt{\implies{₹ \: 12,715 =P \times 12 + \frac{P \times 12(12 + 1)}{2 \times 12} } \times \frac{r}{100} } \\ \\ \mathtt{ \implies{12715 =P \times 12 + \frac{P \times 12 \times 13}{2 \times 12} \times \frac{11}{100} }} \\ \\ \mathtt{ \implies{12715 =P \times 12 + \frac{P \times \cancel{12 }\times 13}{2 \times \cancel{ 12}} \times \frac{11}{100} }} \\ \\ \mathtt{ \implies{12715 = 12 \: P + \frac{143 \: P}{200} }} \\ \\ \mathtt{ \implies{12715 = \frac{2400 \: P + 143 \: P}{100} }} \\ \\ \mathtt{ \implies{12715 = \frac{2543 \: P}{200} }} \\ \\ \mathtt{ \implies { \frac{ \cancel{12715} ^{5} \times 200}{ \cancel{2543}} = P }} \\ \\ \boxed{ \purple{ \mathtt{ \therefore \: {P =₹ \: 1,000 }}}}$

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@MissSolitary✌️

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