Question 13 Marks
Find the simple interest and the amount when:
Principal $= Rs. 2650,$ rate $= 8\%$ p.a. and time $=2\frac12\text{ years.}$
AnswerPrincipal $(P) = Rs. 2650$
Rate $(r) = 8\%$ p.a.
$\text{Time(t)}=2\frac12=\frac52\text{ years}$
$\therefore\text{S.I}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{2650\times8\times5}{100\times2}$
$=\text{Rs. }530$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }2650+530$
$=\text{Rs. }3180$
View full question & answer→Question 23 Marks
Find the time when: Principal $= Rs. 8250, S.I. = Rs. 1100$ and time $= 2$ years.
AnswerPrincipal $(P) = Rs. 8250$
$S.I. = Rs. 1100$
Time $(t) = 2$ years $\text{Rate}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1100\times100}{8250\times2}$
$=\frac{20}{3}\%=6\frac23\%\text{ p.a.}$
View full question & answer→Question 33 Marks
Find the time when:
In what time will $Rs. 5600$ amount to $Rs. 6720$ at $8\%$ per annum$?$
AnswerPrincipal $(P) = Rs. 5600$
Amount $(A) = Rs. 6720$
$S.I. = A - P$
$= Rs. 6720 - 5600$
$= Rs. 1120$
Rate $(r) = 8\%$ p.a.
$\therefore\text{Time}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1120\times100}{5600\times8}$
$=\frac52$
$=2\frac12\text{ years}$
View full question & answer→Question 43 Marks
Find the time when:
The simple interest on a certain sum for $3$ years at $10\%$ per annum is $Rs. 829.50.$ Find the sum.
Answer$S.I. = Rs. 829.50$
Rate $(r) = 10\%\ p.a.$
Time $(t) = 3$ years
$\therefore\text{Sum(P)}=\frac{\text{S.I.}\times100}{\text{r}\times\text{t}}$
$=\frac{82950\times100}{100\times10\times3}$
$=\text{Rs. }2765$
View full question & answer→Question 53 Marks
At what rate percent per annum simple interest will $Rs. 3625$ amount to $Rs. 4495$ in $2$ years$?$
Answer$P = Rs. 3625$
$A = Rs. 4495$
$T = 2$ years
$S.I. = A - P$
$= Rs. 4495 - Rs. 3625 = Rs. 870$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow870=\frac{3625\times\text{R}\times2}{100}$
$\Rightarrow\text{R}=\frac{870\times100}{3625\times2}$
$\Rightarrow\text{R}=\frac{87000}{7250}=12\%$
View full question & answer→Question 63 Marks
Find the time when:
Principal $= Rs. 5200, S.I. = Rs. 975$ and time $=2\frac12\text{ years}.$
AnswerPrincipal $(P) = Rs. 5200$
$S.I. = Rs. 975$
$\text{Time(t)}=2\frac12=\frac52\text{ years}$
$\therefore\text{Rate}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{975\times100\times2}{5200\times5}$
$=\frac{15}{2}\%$
$=7\frac12\%\text{ p.a.}$
View full question & answer→Question 73 Marks
Find the time when:
Principal $= Rs. 3560,$ Amount $= Rs. 4521.20$ and time $= 3$ years.
AnswerPrincipal $(P) = Rs. 3560$
Amount $(A) = Rs. 4521.20$
$S.I. = A - P$
$= Rs. 4521.20 - 3560$
$= Rs. 961.20$
Time $(t) = 3$ years.
$\therefore\text{Rate}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{961.20\times100}{3560\times3}$
$=9\%\text{ p.a.}$
View full question & answer→Question 83 Marks
In what time will $Rs. 3600$ amount to $Rs. 4410$ at $9\%$ per annum simple interest$?$
Answer$P = Rs. 3600$
$A = Rs. 4410$
$R = 9\%$
$S.I. = A - P$
$= Rs. 4410 - Rs. 3600 = Rs. 810$
$\Rightarrow\text{T}=\frac{\text{S.I.}\times100}{\text{P}\times\text{R}}$
$=\frac{810\times100}{3600\times9}$
$=\frac{90}{36}$
$=2\frac12\text{ years}$
View full question & answer→Question 93 Marks
Find the time when:Principal $= Rs. 9540, S.I. = Rs. 1908$ and Rate $= 8\%\ p.a.$
AnswerPrincipal $(P) = Rs. 9540$
$S.I. = Rs. 1908$
Rate $(r) = 8\%$ p.a.
$\therefore\text{Time}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1908\times100}{9540\times12}=\frac{5}{2}\text{ years}$
$=2\frac12\text{ years}$
View full question & answer→Question 103 Marks
What sum will amount to $Rs. 6600$ in $2$ years $Rs.$ at $10\%$ per annum simple interest$?$
AnswerLet the sum be $Rs. x$
$\text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$=\frac{\text{x}\times10\times2}{100}$
$=\frac{\text{x}}{5}$ Now, $A = P + S.I.$
$\text{A}=\text{x}+\frac{\text{x}}{5}=\frac{6\text{x}}{5}$
But amount is $Rs. 6600$
$\therefore\frac{6\text{x}}{5}=6600$
$\Rightarrow\text{x}=\frac{6600\times5}{6}={5500}{}$
Hence, the required sum is $Rs. 5500$
View full question & answer→Question 113 Marks
Find the time when: Principal $= Rs. 6400, S.I. = Rs. 1152$ and rate $= 6\%\ p.a.$
AnswerPrincipal $(P) = Rs. 6400$
$S.I. = Rs. 1152$
Rate $(r) = 6\% p.a.$
$\therefore\text{Times}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1152\times100}{6400\times6}$
$=\text{3 years}$
View full question & answer→Question 123 Marks
A sum of money becomes $\frac43$ of it self in $6$ years at a certain rate of simple interest. Find the rate of interest.
AnswerLet the sum be $Rs. x$ Amount $=\frac{4}{3}\text{x}$
$\text{S.I.}=\text{A}-\text{P}=\frac{4}{3}\text{x}-\text{x}$
Let the rate be $R\%\ \text{S.I.}=\frac{\text{P}\times\text{R}\times\text{T}}{100}$
$\Rightarrow\frac{\text{x}}{3}=\frac{\text{x}\times\text{R}\times\text{6}}{100}$
$\Rightarrow\text{R}=\frac{\text{x}\times100}{\text{x}\times6\times3}$
$=\frac{100}{18}=5.55$
Hence, the rate of interest is $5.55\%$
View full question & answer→Question 133 Marks
At what rate percent annum simple interest will a sum double it self in $12$ years$?$
AnswerLet the sum be $Rs. x$
Amount $= Rs. 2x$
$S.I. = (2x - x)$
Time $= 12$ years
$P = x$
$S.I. = x$
$T = 12$ years
$\therefore\text{R}=\frac{100\times\text{S.I.}}{\text{P}\times\text{T}}$
$=\frac{100\times\text{x}}{\text{x}\times12}=8.3\%$
View full question & answer→Question 143 Marks
Find the simple interest and the amount when: Principal $= Rs. 6400,$ rate $= 6\%$ p.a. and tlme $= 2$ years.
AnswerPrincipal $(P) = Rs. 6400$
Rate $(r) = 6\% p.a.$
Time $(t) = 2$ years
$\therefore\text{S.I}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{6400\times6\times2}{100}$
$=\text{Rs. }768$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }6400+\text{Rs. }768=\text{Rs. }7168$
View full question & answer→Question 153 Marks
Find the time when: A sum of money lent at simple interest amounts to $Rs. 4745$ in $3$ years and to $Rs. 5475$ in $5$ years. Find the sum and the rate per cent per annum.
AnswerAmount in $5$ years $= Rs. 5475$
Amount in $3$ years $= Rs. 4745$
Interest for $2$ years $= Rs. 5475 - 4745 = Rs. 730$ And
interest for $3$ years $=\text{Rs. }\frac{730\times3}{2}$
$\therefore\text{Principal}=\text{Rs. }4745-1095$
$=\text{Rs. }3650$ And rate $=\frac{\text{S.I.}\times100}{\text{P}\times\text{t}}=\frac{1095\times100}{3650\times3}$
$=10\%\text{ p.a.}$
View full question & answer→Question 163 Marks
Find the simple interest and the amount when:Principal $= Rs. 1500$, rate $= 12\%$ p.a. and time $= 3$ years $3$ months.
AnswerPrincipal $(P) = Rs. 1500 $
Rate $(r) = 12\%\ p.a.$
$\text{Time(t)}=3\text{ years }3\text{ month}=3\frac14=\frac{13}{4}\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{1500\times12\times13}{100\times4}$
$\text{Rs. }585$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$\text{Rs. }1500+\text{Rs. }585$
$=\text{Rs. }2085$
View full question & answer→Question 173 Marks
Find the time when: At what rate per cent per annum will $Rs. 3600$ amount to $Rs. 4734$ in $3\frac12\text{ years?}$
AnswerAmount $(A) = Rs. 4734$
Principal $(P) = Rs. 3600$
$S.I. = A - P$
$= Rs. 4734 - Rs. 3600 = Rs. 1134$
$\text{Time}=3\frac12=\frac72\text{ years}$
$\therefore\text{Rate}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1134\times100\times2}{3600\times7}$
$=9\%$
View full question & answer→Question 183 Marks
Find the time when: Hari borrowed $Rs. 12600$ from a money lender at $15\%$ per annum simple interest. After $3$ years, he paid $Rs. 7070$ and gave a goat to clear of the debt. What is the cost of the goat$?$
AnswerPrincipal $= Rs. 12600$
Rate $(A) = 15\%$ p.a.
Time $(t) = 3$ years $\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{12600\times15\times3}{100}$
$=\text{Rs. }5670$ Amount $= P + S.I. = Rs. 12600 + Rs. 5670 = Rs. 18270$
Amount paid in cash $= Rs. 7070$
Balance $= Rs. 18270 - 7070 = Rs. 11200$
Price of goat $= Rs. 11200$
View full question & answer→Question 193 Marks
Find the time when:
Shanta borrowed $Rs. 6000$ from the State Bank of India for $3$ years $8$ months at $12\%$ per annum. What amount will clear off her debt$?$
AnswerPrincipal $(P) = Rs. 6000$
Rate $(r) = 12\%\ p.a.$
$\text{Time}=3\text{ years}\ 8\text{ years}=3\frac{8}{12}$
$=3\frac23\text{ years}=\frac{11}{3}\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{6000\times12\times11}{100\times3}$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }6000+\text{Rs. }2640$
$=\text{Rs. }8640$
View full question & answer→Question 203 Marks
Find the simple interest and the amount when: Principal $= Rs. 5000,$ rate $= 9\% p.a.$ and time $= 146$ days.
AnswerPrincipal $(P) = Rs. 5000$
Rate $(r) = 9\% p.a.$
$\text{Times(t)}=146\text{ days}=\frac{146}{365}=\frac{2}{5}\text{ years}$
$\text{S.I.}=\frac{\text{P}\times\text{t}}{100}$
$=\frac{5000\times9\times2}{100\times5}$
$=\text{Rs. }180$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }5000+\text{Rs. }180$
$=\text{Rs. }5180$
View full question & answer→Question 213 Marks
Find the simple interest and the amount when: Principal $= Rs. 9600,$ rate $=7\frac12\text{%}\text{ p.a.}$ and time $= 5$ months.
AnswerPrincipal $(P) = Rs. 9600$
$\text{Rate(r)}=7\frac12\text{%}=\frac{15}{2}\%$
$\text{Time(t)}=5\text{ months}=\frac{5}{12}\text{ years}$
$\therefore\text{S.I.}=\frac{\text{P}\times\text{r}\times\text{t}}{100}$
$=\frac{9600\times15\times5}{100\times2\times12}$
$=\text{Rs. }300$
$\therefore\text{Amount}=\text{P}+\text{S.I.}$
$=\text{Rs. }9600+\text{Rs. }300$
$=\text{Rs. }9900$
View full question & answer→Question 223 Marks
Find the time when: Principal $= Rs. 5000,$ Amount $= Rs. 6450$ and rate $= 12\%$ p.a.
AnswerAmount $(A) = Rs. 6450$
Principal $(P) = Rs. 5000$
$S.I. = A - P$
$= Rs. (6450 - 5000) = Rs. 1450$
Rate $(r) = 12\%\ p.a.$
$\text{Time}=\frac{\text{S.I.}\times100}{\text{P}\times\text{r}}$
$=\frac{1450\times100}{5000\times12}$
$=\frac{29}{12}\text{ years}=2\frac{5}{12}\text{ years}$
$ = 2$ years $5$ month
View full question & answer→