Exercise 4.3 - Chapter 4 Life Mathematics 8th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Question 1.
Fill in the blanks:
(i) The compound interest on $₹ 5000$ at $12 \%$ p.a for 2 years, compounded annually is
Answer:
₹ 1272
Hint
Compound Interest (CI) formula is
$\mathrm{CI}=$ Amount $-$ Principal
$\begin{aligned}
\text { Amount } &=A\left(1+\frac{r}{100}\right)^{n} \\
&=5000\left(1+\frac{12}{100}\right)^{2}=5000 \times\left(\frac{112}{100}\right)^{2}=6272 \\
\therefore 6272-5000 &=₹ 1272
\end{aligned}$
(ii) The compound interest on $₹ 8000$ at $10 \%$ p.a for 1 year, compounded half yearly is
Answer:
₹ 820

(iii) The annual rate of growth in population of a town is $10 \%$. If its present population is 26620 , then the population 3 years ago was
Answer:
₹ 20,000
Hint:
Rate of growth of population $r=10 \%$
Present population $=26620$
Let population 3 years ago be $x$
$\therefore$ Applying the formula for population growth which is similar to compound interest,
$\begin{aligned}
26620 &=x\left(1+\frac{r}{100}\right)^{3} \\
\therefore 26620 &=x\left(1+\frac{10}{100}\right)^{3}=x\left(\frac{110}{100}\right)^{3} \\
\therefore x &=26620 \times\left(\frac{100}{110}\right)^{3}=₹ 20,000
\end{aligned}$
The population 3 years ago was $₹ 20,000$
(iv) If the compound interest is calculated quarterly, the amount is found using the formula
Answer:
$\mathrm{A}=P\left(1+\frac{r}{400}\right)^{4 n}$
Hint:
Quarterly means 4 times in a year.
$\therefore$ The formula for compound interest is
$\mathrm{A}=P\left(1+\frac{r}{400}\right)^{4 n}$

(v) The difference betwecn the C.I and S.I for 2 ycars for a principal of $₹ 5000$ at the rate of interest $8 \%$ p.a is
Answer:
₹ 32
Hint:
Difference between S.I \& C.I is given by the formula
$\mathrm{CI}-\mathrm{SI}=\left(\frac{r}{100}\right)^{2}$
Principal $(\mathrm{P})=5000 . \mathrm{r}=8 \% \mathrm{p} . \mathrm{a}$
$\therefore \mathrm{CI}-\mathrm{SI}=5000\left(\frac{8}{100}\right)^{2}=5000 \times\left(\frac{8}{100}\right)^{2} \times\left(\frac{8}{100}\right)^{2}=₹ 32$

Question $2 .$
Say True or False.
(i) Depreciation value is calculated by the formula, $P\left(1-\frac{r}{100}\right)^{n}$.
Answer:
True
Hint:
Depreciation formula is $P\left(1-\frac{r}{100}\right)^{n}$

(ii) If the present population ola city is $\mathrm{P}$ and it increases at the rate of $r \%$ p.a, then the population n years ago would be $P\left(1-\frac{r}{100}\right)^{n}$
Answer:
False

(iii) The present value of a machine is $₹ 16800$. It depreciates at $25 \%$ p.a. Its worth after 2 years is $₹ 9450$.
Answer:
True

(iv) The time taken for $₹ 1000$ to become $₹ 1331$ at $20 \%$ p.a, compounded annually is 3 years.
Answer:
False

(v) The compound interest on ₹ 16000 for 9 months at $20 \% \mathrm{p} .$ a, compounded quarterly is ₹ 2522 .
Answer:
True

Question $3 .$
Find the compound interest on $₹ 3200$ at $2.5 \%$ p.a for 2 years, compounded annually.
Answer:
Principal $(P)=₹ 3200$
$\mathrm{r}=2.5 \% \mathrm{p} . \mathrm{a}$
$\mathrm{n}=2$ years comp. annually
$\therefore$ Amount $(\mathrm{A})=\left(1+\frac{r}{100}\right)^{n}$
$=3200\left(1+\frac{25}{100}\right)^{2}$
$=3200 \times(1.025)^{2}=3362$
Compound interest $(\mathrm{CI})=$ Amount $-$ Principal $=3362-3200=162$
 

Question $4 .$
Find the compound interest for $2 \frac{1}{2}$ years on $₹ 4000$ at $10 \%$ p.a, if the interest is compounded yearly.
Answer:
Principal $(P)=₹ 4000$
$\mathrm{r}=10 \% \mathrm{p} . \mathrm{a}$
Compounded yearly
$\mathrm{n}=2 \frac{1}{2}$ years. Since it is of the form a $\frac{b}{c}$ years

$\begin{aligned}
&\text { Amount (A) }=\left(1+\frac{r}{100}\right)^{a}\left(1+\frac{\frac{b}{c} \times r}{100}\right)^{1} \\
&=4000\left(1+\frac{10}{100}\right)^{2}\left(1+\frac{\frac{1}{2} \times 10}{100}\right)^{1} \\
&=4000 \times\left(\frac{110}{100}\right)^{2} \times\left(\frac{105}{100}\right)^{1} \\
&=4000 \times 1.1 \times 1.1 \times 1.05=5082 \\
&\therefore \mathrm{CI}=\text { Amount }-\text { principal }=5082-4000=1082
\end{aligned}$
 

Question $5 .$
A principal becomes ₹ 2028 in 2 years at $4 \%$ p.a compound interest. Find the principal.
Answer:
$\mathrm{n}=2 \text { years }$
$r=$ rate of interest $=4 \%$ p.a
Amount $\mathrm{A}=₹ 2028$

$\begin{aligned}
\text { Amount }(\mathrm{A}) &=\mathrm{P}\left(1+\frac{r}{100}\right)^{n} \\
2028 &=\mathrm{P}\left(1+\frac{4}{100}\right)^{2} \\
2028 &=\mathrm{P}\left(\frac{104}{100}\right)^{2} \\
\therefore \mathrm{P} &=\frac{2028 \times 100 \times 100}{104 \times 104}=₹ 1875
\end{aligned}$
 

Question 6 .
In how many years will ₹ 3375 become ₹ 4096 at $13 \%$ p.a if the interest is compounded half-yearly?
Answer:
Principal $=₹ 3375$
Amount $=₹ 4096$
$\mathrm{r}=13 \frac{1}{3} \% \mathrm{p} . \mathrm{a}=\frac{40}{3} \% \mathrm{p} . \mathrm{a}$

Compounded half yearly $r=\frac{\frac{40}{3}}{2}=\frac{20}{3} \%$ half yearly
Let no. of years be $n$
for compounding half yearly, formula is
$\begin{aligned}
A &=P\left(1+\frac{r}{100}\right)^{2 n} \\
\therefore 4096 &=3375\left(1+\frac{\frac{20}{100}}{}\right)^{2 n} \\
\therefore \frac{4096}{3375} &=\left(1+\frac{20}{3 \times 100}\right)^{2 n}=\left(1+\frac{1}{15}\right)^{2 n} \\
\left(\frac{15+1}{15}\right)^{2 n} &=\left(\frac{16}{15}\right)^{2 n} \Rightarrow \therefore \frac{4096}{3375}=\left(\frac{16}{15}\right)^{2 n}
\end{aligned}$
Taking cubic root on both sides,
$\begin{aligned}
&\left(\frac{16}{15}\right)^{\frac{2 n}{3}}=\frac{\sqrt[3]{4096}}{\sqrt[3]{3375}}=\frac{16}{15} \quad\left(\frac{16}{15}\right)^{\frac{2 n}{3}}=\frac{\sqrt[3]{4096}}{\sqrt[3]{3375}}=\frac{16}{15} \\
&\therefore \frac{2 n}{3}=1 \quad \therefore n=\frac{3}{2}=1.5 \text { yrs. } \therefore \frac{2 n}{3}=1 \quad \therefore n=\frac{3}{2}=1.5 \text { yrs. }
\end{aligned}$

Question $7 .$
Find the CI on ₹ 15000 for 3 years if the rates of interest are $15 \%, 20 \%$ and $25 \%$ for the I, II and III years respectively.

Answer:
Principal $(\mathrm{P})=₹ 15000$
rate of interest 1 (a) $=15 \%$ for year I
rate of interest $2(b)=20 \%$ for year II
rate of interest $3(\mathrm{c})=25 \%$ for year III
Formula for amount when rate of interest is different for different years is
$A=\left(1+\frac{a}{100}\right)^{1}\left(1+\frac{b}{100}\right)^{1}\left(1+\frac{c}{100}\right)^{1}$
Substituting in the above formula, we get
$\begin{aligned}
A &=15000\left(1+\frac{15}{100}\right)\left(1+\frac{20}{100}\right)\left(1+\frac{25}{100}\right) \\
&=15000 \times \frac{115}{100} \times \frac{120}{100} \times \frac{125}{100}=25,875
\end{aligned}$
$\therefore$ Compound Interest(CI) $=\mathrm{A}-\mathrm{P}=25,875-15,000=₹ 10,875$
$\mathrm{CI}=₹ 10.875$

Question 8 .
Find the difference between C.I and S.I on ₹ 5000 for 1 year at $2 \%$ p.a, if the interest is compounded half yearly.
Answer:

time period $(n)=1$ yr.
Rate of interest (r) $=2 \%$ p.a
for half yearly $r=1 \%$
Difference between CI \& SI is given by the formula
time period $(\mathrm{n})=1 \mathrm{yr}$.
Rate of interest $(\mathrm{r})=2 \%$ p.a
for half yearly $\mathrm{r}=1 \%$
Difference between CI \& SI is given by the formula
$\mathrm{CI}-\mathrm{SI}=\mathrm{P}\left(\frac{r}{100}\right)^{2 n} \text { [for half yearly compounding] }$
$\therefore \mathrm{CI}-\mathrm{SI}=5000\left(\frac{1}{100}\right)^{2 \times 1}=5000 \times \frac{1}{100} \times \frac{1}{100}=₹(1.50)$
 

Question 9 .
Find the rate of interest if the difference between C.I and S.I on ₹ 8000 compounded
annually for 2 years is ₹ 20 .
Answer:
Principal $(\mathrm{P})=₹ 8000$
time period $(\mathrm{n})=2$ yrs.
$\mathrm{CI}-\mathrm{SI}=\mathrm{P}\left(\frac{r}{100}\right)^{2 n} \quad$ [for half yearly compounding]
$\therefore \mathrm{CI}-\mathrm{SI}=5000\left(\frac{1}{100}\right)^{2 \times 1}=5000 \times \frac{1}{100} \times \frac{1}{100}=₹(0.50)$
 

Question $9 .$
Find the rate of interest if the difference between C.I and S.I on $₹ 8000$ compounded annually for 2 years is $₹ 20$.
Answer:
Principal $(\mathrm{P})=₹ 8000$
time period $(\mathrm{n})=2 \mathrm{yrs}$.
rate of interest $(r)=$ ?

Difference between CI \& SI is given by the formula
$\mathrm{CI}-\mathrm{SI}=p\left(1+\frac{r}{100}\right)^{n}$
Difference between CI \& SI is given as 20
$\begin{aligned}
&\therefore 20=8000 \times\left(\frac{r}{100}\right)^{2} \\
&\therefore\left(\frac{r}{100}\right)^{2}=\frac{20}{8000}=\frac{1}{400}
\end{aligned}$
Taking square root on both sides
$\begin{aligned}
&\frac{r}{100}=\sqrt{\frac{1}{400}}=\frac{1}{20} \\
&\therefore \mathrm{r}==5 \%
\end{aligned}$

Question $10 .$
Find the principal if the difference between C.I and S.l on it at $15 \%$ p.a for 3 years is ₹ 1134 .
Answer:
Rate of interest $(\mathrm{r})=15 \% \mathrm{p}$.a
time period $(n)=3$ years
Difference between CI \& SI is given as 1134
Principal $=? \rightarrow$ required to find
Using formula for difference
Simple Interest SI $=\frac{P n r}{100}$
Compound Interest $\mathrm{CI}=\mathrm{p}(1+\mathrm{i})^{\mathrm{n}}-\mathrm{p}$
$\begin{aligned}
\mathrm{CI}-\mathrm{SI} &=\mathrm{P}\left[(1+\mathrm{i})^{n}-1-\frac{n r}{100}\right] \\
1134 &=\mathrm{P}\left[\left(1+\frac{15}{100}\right)^{3}-1-\frac{3 \times 15}{100}\right] \\
1134 &=\mathrm{P}\left[\left(\frac{115}{100}\right)^{3}-1-\frac{45}{100}\right]
\end{aligned}$

$\begin{aligned}
&\therefore \mathrm{p}=\frac{1134 \times 100}{0.07 \times 100}=\frac{113400}{7} \\
&\mathrm{P}=₹ 16200
\end{aligned}$

Objective Type Questions
Question 11.
The number of conversion periods in a year, if the interest on a principal is compounded every two months is
(A) 2
(B) 4
(C) 6
(D) 12
Answer:
(C) 6

Question $12 .$
The time taken for $₹ 4400$ to become $₹ 4851$ at $10 \%$, compounded half yearly is
(A) 6 months
(B) 1 year
(C) $1 \frac{1}{2}$ years
(D) 2 years
Answer:
(B) 1 year

Question $13 .$
The cost of a machine is ₹ 18000 and it depreciates at $16 \frac{2}{3} \%$ annually. Its value after 2 years will be
(A) ₹ 2000
(B) ₹ 12500
(C) ₹ 15000
(D) ₹ 16500
Answer:
(B) ₹ 12500

Question $14 .$
The sum which amounts to ₹ 2662 at $10 \%$ p.a in 3 years, compounded yearly is
(A) ₹ 2000
(B) ₹ 1800
(C) ₹ 1500
(D) ₹ 2500
Answer:
(A) ₹ 2000

Question $15 .$
The difference between compound and simple interest on a certain sum of money for 2 years at $2 \%$ p.ais U. The sum of money is
(A) ₹ 2000
(B) ₹ 1500
(C) ₹ 3000
(D) ₹ 2500
Answer:
(D) ₹ 2500