Question

Mrs.goswami deposits 1000 every month in a recurring deposit account for 3 years at 8% interest per annum .Find the matured value.​

Answer 1

AnswEr :

Rs.40440

$\bf{\red{\underline{\underline{\bf{Given\::}}}}}$

Mrs. goswami deposits 1000 every month in a regarding deposits account for 3 years at 8% Interest per annum.

$\bf{\red{\underline{\underline{\bf{To\:find\::}}}}}$

The matured value.

$\bf{\red{\underline{\underline{\bf{Explanation\::}}}}}$

Formula use :

S.I. = P × n(n+1)/2 × 12 × r/100

• Mrs. Goswami deposit (P) = Rs.1000
• Time (n) = 3 years = (12 × 3 ) = 36 months
• Rate (r) = 8%

A/q

S.I. = 1000 × 36(36+1)/2 × 12 × 8/100

S.I. = 1000 × 36(37)/24 × 8/100

S.I. = 1000 × 1332/3 × 100

S.I. = Rs.(10 × 444)

S.I. = Rs.4440.

So,

Matured Value = Principal × Time + Interest

Matured Value = 1000 × 36 + 4440

Matured Value = 36000 + 4440

Matured Value = Rs.40440.

Answer 2

$\huge \mathfrak \red{answer}$

➹40440

✥Question:✥

Mrs.goswami deposits 1000 every month in a recurring deposit account for 3 years at 8% interest per annum .Find the matured value.

★step to step explanation★

$\sf{given \: from \: the \: question}$

➹$\rm{p = 1000}$

➹$\rm{t = 3yrs}$

➹$\rm{n = 12 \times t}$

➹$\sf{ = 12 \times 3 = 36}$

➹$\rm{r = 8\%}$

❁________________________________❁

problem solving:✙

$\tt{p = \frac{36(36 + 1)}{2} \times 1000}$

$\tt{ = \frac{36 \times 37 \times 1000 \times 8}{2 \times 12 \times 100}}$

$\tt{12 \times 37 \times 10 = 4440}$

✤______________________________✤

$\tt{matured \: value = 36000 + 4440}$

$\bf{ \huge{ \boxed{ \red{ \tt{40440 \: }}}}}$