Solution:- Given Prashant borrowed ₹ $35,000$ at $12$% p.a. p = ₹ $35,000$, t = years = $1½$ = $1.5$ years, r = $12$% p.a. Amount = P${{(1+r/100)}^{2t}}$ Amount = $35,000\text{ }{{\left( 1\text{...
Solution:- The amount received by him had he chosen the duration of the deposit to be $2$ years, P2 = $34,496$
Mr. Mohan invested ₹ $12,500$ at $16$% p.a. Given p = ₹ $12,500$, t = $1.5$ years, r = $16$% p.a. Calculation Amount $=P{{(1+r/100)}^{t}}$ Amount =$12,500{{(1+(16/100))}^{1.5}}$ =...
Given, Pradeep gave $₹ 16,000$ to a friend for 1.5 years at $1% p.a$. p $= ₹ 16,000$, t $= 1.5$ years, r $= 15% p.a$. Calculation, Amount = $P{{(1+r/100)}^{2t}}$ Amount =...
Solution:- Given, Amita needed = $₹40,000$ Where, p =$₹40,000$, t$=1$ $1/2$years $=1.5$years, r $=12%$ Amount = $p{{(1+r/100)}^{2t}}$ Amount = $35,000{{(1+(12/200))}^{3}}$ Amount = $₹48,620.25$...
Solution : Given Neha loaned ₹ $27,500$ to a friend for $1\frac{3}{4}$ years at $8$% p.a. Where, P = ₹ $27,500$, t = $1\frac{3}{4}$ years = $1.75$ years, r = $8$% p.a. Amount = P${{(1+r/100)}^{2t}}$...
Solution:- Given Shekhar had a fixed deposit of ₹ $24,000$ for $3$ years. P = ₹ $24,000$, t = $3$ years, r = $10$% p.a. Amount = $p{{(1+r/100)}^{t}}$ Amount = $p{{(1+(10/100))}^{3}}$ = ₹...
Solution:- Given Prerna borrowed ₹ $16,000$ from a friend at $15$% p.a. C1 = (P × r × t)/$100$ $=(16,000\times 15\times 1)/100$ = ₹ $2,400$ P1 = $16,000+2,400$ = ₹ $18,400$ C2 = (P ×...
Solution:- Given Gayatri invested ₹ $25,00$ for $3$ years and $6$ months in a bank which paid $10$% p.a. Calculation C1 = (P × r × t)$/100$ $=(25,000\times 10\times 1)/100$ = ₹ $2,500$...
Solution:- C1 = (P × r × t)/$100$ $=(60,000\times 15\times 1)/100$ = ₹ $9,000$ P1 = $60,000+9,000$ = ₹ $69,000$ C2 = (P × r × t)/$100$ $=(69,000\times 15\times 1)/100$ = ₹ $10,350$...
Solution:- The amount received by him had he chosen the duration of the deposit to be $2$ years, P2 = $34,496$
Given Harjyot deposited ₹ $27,500$ in a deposit scheme paying $12$%p.a. Time, t = $3$years C1 = (P × r × t)/$100$ =$(27,500\times 12\times 1)/100$ = ₹ $3,300$ P1 = $27,500+3,300$ =...
Given P = Principal R = Rate T = Time P = ₹ $24,000$, r = $10$%p.a., t = $2.5$ years Solution:- C1 = (P × r × t)/$100$ = $\left( 24,000\times 10\times 1 \right)/100$ = ₹ $2,400$ P1 =...
Given P = Principal R = Rate T = Time P = ₹ $36,000$, r = $10$%p.a., t = $1.5$year Solution:- C1 = (P × r × t)/$100$ = $(36,000\times 10\times 1)/100$ = ₹ $3,600$ P1 =...
Solution:- C3 = (P × r × t)/$100$ = $(87,480\times 8\times 1)/100$ = ₹ $6,998.4$ Solution:- C4 = (P × r × t)/$100$ = $(94,478.4\times 8\times 1)/100$ = ₹...
Given P = Principal R = Rate T = Time Alisha invested ₹ $75,000$ for $4$ years at $8$%p.a. P = ₹ $75,000$, r = $8$%p.a., t = $4$ years C1 = (P × r × t)/$100$ = $(75,000\times 8\times 1)/100$ =...
Solution:- C1 + C2 = $5,200+5,616$ = ₹ $10,816$ Solution:- C2 = (P*R*T)$/100$ = $(75,816*8*1)/100$ = ₹ $6,065.28$
Solution:- C1 + C2 = $5,200+5,616$ = ₹ $10,816$ C2 = (P × r × t)/$100$ =$(75,816\times 8\times 1)/100$ = $6,065.28$
P = Principal R = Rate T = Time P = ₹ \[65,000\], r = \[8%\]p.a., t = \[3\] years Solution:- C1 = (P × r × t)\[/100\] = \[\left( 65,000\times 8\times 1 \right)/100\] = ₹ \[5,200\] Then,...
Solution:- Given : P = ₹ $22,500$, r = $12%$ p.a., t =\[\mathbf{1}{\scriptscriptstyle 3\!/\!{ }_4}\]years For the first year, t = $1$ year We know that, S.I. = (P × r × t)/$100$ = $(22,500 ×...
Solution:- Given : P = ₹ $25,000$, r = $5\frac{1}{4}$% p.a. = $21/4$%, Time, t = $1\frac{1}{3}$ years For the first year, t = $1$ year S.I. = (P × r × t)/$100$ = $(40,000 × 21 × 1)/(100 × 4)$ = ₹...
Solution:- Given : P = ₹ $10,000$, r = $8$% p.a., t = $2\frac{1}{3}$ years For the first year, t = 1 year S.I. = (P × r × t)/$100$ = $(20,000 × 9 × 1)/100$ = ₹ $1,800$ A = P + S.I. = $20,000 +...
Solution:- Given : P = ₹ $30,000$, r = $8%$ p.a., t = $2½$ years For the first year, t = $1$ year S.I. = (P × r × t)/$100$ = ₹ $2,400$ A = P + S.I. = $30,000 + 2,400$ = ₹ $32,400$ New principal is ₹...
Solution:- Given : P = ₹ $17,500$, r = $12% p.a.$, t = $3$ years For the first year, t = $1$ year S.I. = (P × r × t)/$100$ = $(17,500\times 12\times 1)/100$. = ₹ $2,100$ A = P + S.I. = $17,500 +...
(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...