Download App

Maths 3 Mark Question

Interest · Maharashtra Board STD 7 (MSBSHSE - Semi English Medium). 22 questions.

Questions

3 Mark Question

Take a timed test

22 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Find the simple interest and the amount when:
Principal = Rs. 2650, rate = 8% p.a. and time $=2\frac12\text{ years.}$
Answer
Principal (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.
Answer
Principal (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?
Answer
Principal (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 $=2\frac12\text{ years}.$
Answer
Principal (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.
Answer
Principal (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.
Answer
Principal (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?
Answer
Let 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.
Answer
Principal (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.
Answer
Let 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?
Answer
Let 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.
Answer
Principal (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.
Answer
Amount 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.
Answer
Principal (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 $3\frac12\text{ years?}$
Answer
Amount (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?
Answer
Principal = 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?
Answer
Principal (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.
Answer
Principal (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.
Answer
Principal (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.
Answer
Amount (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→
Generate Paper FreeAll Interest topics