(a) ₹ $7,500$ at $12%$ p.a. in $3$ years.
Solution:-
Principal = P
Rate = R
Time = T
Given :
Principal, P = ₹ $7,500$, Rate, r = $12% p.a.,$ Time, t = 3 years
For the first year, t = $1$ year
S.I. = (P × r × t)/$100$
= $(7,500\times 12\times 1)/100$.
= ₹ $900$
A = P + S.I.
= $7,500 + 900$
= ₹ $8,400$
New principal is ₹ 8,400.
For the second year, t = $1$ year, p = ₹ $8,400$
S.I. = (P × r × t)/$100$
= $(8,400\times 12\times 1)/100$
= ₹ $1,008$
A = P + S.I.
= $8,400 + 1,008$
= ₹ $9,408$
New principal is ₹ $9,408$.
For the Third year, t = $1$ year, p = ₹ $9,408$
S.I. = (P × r × t)/$100$
= $(9,408\times 12\times 1)/100$
= ₹ $1,128.96$
A = P + S.I.
= $9,408 + 1,128.96$
= ₹ $10,536.96$
C.I. = Interest in first year + interest in second year + interest in third year
= ₹ $(900 + 1,008 + 1,128.96)$
= ₹ $3,036.96$
Solution:-
P = ₹ $13,500$, r = $10% p.a.$, t = $2$ years
For the first year, t = $1$ year
S.I. = (P × r × t)/$100$
= $(13,500\times 10\times 1)/100$
= ₹ $1,350$
A = P + S.I.
= $13,500 + 1,350$
= ₹ $14,850$
New principal is ₹ $14,850$.
For the second year, t = $1$ year, p = ₹ $14,850$
S.I. = (P × r × t)/$100$
= = $(14,850\times 10\times 1)/100$
= ₹ $1,485$
A = P + S.I.
= $14,850 + 1,485$
= ₹ $16,335$
C.I. = Interest in first year + interest in second year
= ₹ $(1,350 + 1,485)$
= ₹ $2,835$