Exercise 2(C)
Question 1
What amount of money should Mohan invest in a bank in order to get ₹ 1323 in 2 years at 5% compounded annually?
Sol :
After 2 years amount =1323
Rate=5%
Time= 2 years
Amount $P\left[1+\frac{R}{100}\right]^{n}$
$1323=P\left(1+\frac{5}{100}\right)^{2}$
$1323=P\left[1+\frac{1}{20}\right\rfloor^{2}$
$1323=P\left[\frac{21}{20}\right\rfloor^{2}$
$P=1323 \times\left(\frac{20}{21}\right)^{2}$
$P \Rightarrow 1323 \times \frac{20}{21} \times \frac{20}{21}$
P=1200
Question 2
Find the sum which amounts to ₹ 1352 in 2 years at 4% compound interest.
Sol :
$1352=P\left[1+\frac{4}{100}\right]$
$1352= P\left[1+\frac{1}{25}\right\rfloor^{2}$
$1352=P\left[\frac{26}{25}\right]^{2}$
$P=1352 \times\left(\frac{25}{26}\right)^{2}$
$P=1352 x \frac{25}{26} \times \frac{25}{26}$
P=1250
Question 3
What principal will amount to ₹ 9768 in two years. If the rates of interest for the successive years are 10% p.a. and 11% p.a. respectively.
Sol :
Amount $\Rightarrow P\left[1+\frac{r_{1}}{100}\right]\left[1+\frac{r_{2}}{100}\right]$
$\begin{aligned} 9768 &=P\left[1+\frac{10}{100}\right)\left(1+\frac{11}{100}\right) \\ 9768 &=P\left(\frac{11}{10}\right)^{10}\left[\frac{111}{100}\right\rfloor \\ P & \Rightarrow 9768 \times \frac{10}{11} \times \frac{100}{111} \end{aligned}$
P=8000
Question 4
On what sum of money does the difference between the simple interest and compound interest in 2 years at 5% per annum is Rs. 15?
Sol :
Let Principal =100
Rate=5%
Time (n)=2 years
Difference between S.I. and C.I=15
$S.I.= \frac{P R T}{100}$
$\frac{100 \times 5 \times R}{100}$
S.I.=10
Amount$P\left(1+\frac{R}{100}\right)^{n}$
$=100 \left[1+\frac{5}{100}\right)^{2}$
$=100\left[\frac{21}{20}\right]^2$
$=100\left[\frac{21}{20} \times \frac{21}{20}\right]$
$=\frac{441}{4}$
C.I.=A-P
$=\frac{441}{4}-100$
$=\frac{441-400}{4}=\frac{41}{4}$
Difference between S.I. and C.I.
$=\frac{41}{4}- 10$
$=\frac{41-40}{4}=\frac{1}{4}$
If difference is $\frac{1}{4}$ then principal is 100
If difference is 1 then principal is equal to $\frac{100 \times 4}{1} \Rightarrow 400$
According tot he question if difference is 15 then principal 400×15=6000
Question 5
The difference between simple and compound interest on the same sum of money at $6\frac{2}{3}/%$ for 3 years is ₹ 184. Determine the sum.
Sol :
Time(n) = 3 years
$S.I. = \frac{P R T}{100}$
$\frac{100 \times 20 \times 3}{100 \times 3}$
Amount$=P\left(1+\frac{R}{100}\right)^{n}$
$100\left(1+\frac{20}{3 \times 100}\right)^{3}$
$100\left(1+\frac{2}{30}\right)^{3}$
$100\left(1+\frac{1}{15}\right)^{3}$
$100 \times\left(\frac{16}{15}\right)^{3}$
$100 \times \frac{16}{15} \times \frac{16}{15} \times \frac{16}{15}$
$\frac{16384}{135} $
C.I.=A-P
$\frac{16384}{135}-100$
$\frac{16384-13500}{135}$
$C.I.=\frac{2804}{135}$
Difference between C.I. and S.I.
$\frac{2884}{135}-20$
$\frac{2884-2400}{135}=\frac{184}{135}$
If difference is $\frac{184}{135}$ then principal is 100
If difference is 1 then principal
$=100 \times \frac{135}{184}$
According to the question difference is 184 then principal
$=100 \times \frac{135}{184} \times 184$
P=13500
Question 6
On what sum of money will the difference between the simple interest and the compound interest for 2 years at 5% per annum be equal to ₹ 50.
Sol :
$=\frac{100\times 5 \times 2}{100}=10$
Amount$=P\left(1+\frac{R}{100}\right)^{n}$
$=100\left(1+\frac{5}{100}\right)^{2}$
$=\left.10211+\frac{1}{20}\right)^{2}$
$=100\left(\frac{21}{20}\right)^{2}$
$=100 \times \frac{21}{20}\times \frac{21}{20}$
$=\frac{441}{4}$
C.I=A-P
$=\frac{41}{4}-10$
$=\frac{41-40}{4}$
$=\frac{1}{4}$
If difference $\frac{1}{4}$ then principal is 100
If difference is 1 then principal
$\frac{100 \times 4}{1} \Rightarrow 400$
According to the question , difference is 50 then
P=400×50=20000
Question 7
Find the rate per cent per annum, if compounded yearly
(i) Principal = ₹ 196, Amount = ₹ 225, time = 2 years
(ii) Principal = ₹ 3136, Compound interest = ₹ 345, Time = 2 years
Sol :
$\frac{225}{196}=\left(1+\frac{R}{100}\right)^{2}$
$\left(\frac{15}{14}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$
$\frac{15}{14}=1+\frac{R}{10}$
$\frac{R}{100}=\frac{15}{14}-1$
$\frac{R}{100}=\frac{15-14}{14}$
$\frac{R}{100}=\frac{1}{14}$
$R=\frac{100}{14}=\frac{50}{7}=7\frac{1}{7}\%$
(ii) Principa=3136
C.I=345
∴Amount=P+C.I.
=3136+345
=34841
Time=2 years
We know that
$A \Rightarrow P\left[1+\frac{R}{100}\right]^{n}$
$\frac{A}{P} \Rightarrow\left[1+\frac{R}{1000}\right]^{n}$
$\frac{3481}{3136}=\left(1+\frac{R}{100}\right)^{2}$
$\left( \frac{59}{56}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$
$\frac{59}{56}=1+\frac{R}{100}$
$\frac{R}{100}=\frac{59}{56}-1$
$\frac{R}{100}=\frac{59-56}{56}$
$\frac{R}{100}=\frac{3}{56}$
$R \Rightarrow \frac{3 \times 100}{56}$
$R=\frac{75}{14}$
∴Rate$=\frac{75}{14}\%$
$=5\frac{5}{14}\%$
Question 8
Hari purchased Relief Bonds for 1000, a sum which will fetch him ₹ 2000 after 5 years. Find the rate of interest if the interest is compounded half-yearly.
(Given that $\sqrt[10]{2}$ = 1.072)
Sol :
$A \Rightarrow P\left[1+\frac{R}{100}\right]^{n}$
$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$
$\frac{2000}{1000}=\left(1+\frac{R}{100}\right)^{10}$
$\frac{2}{1}=\left( 1+\frac{R}{100}\right)^{10}$
$1+\frac{R}{100}=10 \sqrt{2}$ (∵ Given that 10√2=1.072)
$1+\frac{R}{100}=1.072$
$\frac{R}{100}=1.072-1.000$
$\frac{R}{10}=0.072$
R=7.2
Rate half yearly=7.2
Annually=7.2×2=14.4%
Question 9
₹ 8000 became ₹ 9261 in a certain interval of time at the rate of 5% per annum C.l. Find the time.
Sol :
$\frac{A}{P}=\left[1+\frac{R}{100}\right]^{n}$
$\frac{9261}{8000}=\left(1+\frac{5}{200}\right)^{n}$
$\frac{9261}{8000}=\left(1+\frac{1}{20}\right)^{n}$
$\frac{9261}{8000}=\left(\frac{21}{20}\right)^n$
$\left(\frac{21}{20}\right)^{30}=\left(\frac{21}{20}\right)^{n}$
n=3
∴Time = 3 years
Question 10
In how many years will a sum of ₹ 3000 at 20% per annum compounded semi-annually become ₹ 3993.
Sol :
$\frac{3993}{3000}=\left(1+\frac{10}{1000}\right)^{n}$
$\frac{3993}{3000}=\left(1+\frac{1}{10}\right)^{n}$
$\frac{3993}{3000}=\left(\frac{11}{10}\right)^{n}$
$\frac{3}{3}\left(\frac{1331}{1000}\right)=\left(\frac{11}{10}\right)^{n}$
$\left(\frac{11}{10}\right)^{3}=\left(\frac{11}{10}\right)^{n}$
Question 11
A sum of money put out at compound interest amounts in 2 years to ₹ 578.40 and in 3 years to ₹ 614.55. Find the rate of interest.
Sol :
$=\frac{36.15 \times 100}{572.40 \times 1}$
$=\frac{3615 \times 100}{57040}$
$=\frac{25}{4}$
Question 12
A sum compounded annually becomes $\frac{25}{16}$ times of itself in 2 years. Determine the rate of interest per annum?
Sol :
$\frac{A}{P}=\left(1+\frac{R}{100}\right)^{n}$
$\frac{25}{16 \times 1}=\left(1+\frac{R}{100}\right)^{2}$
$\frac{25}{16}=\left(1+\frac{R}{100}\right)^{2}$
$\left(\frac{5}{4}\right)^{2}=\left(1+\frac{R}{100}\right)^{2}$
$\frac{R}{100}=\frac{5}{4}-1$
$R \Rightarrow \frac{1}{4} \times 100$
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