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## Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 2 Chapter 1 Life Mathematics Ex 1.3

Question 1.
Fill in the blanks

Question (i)
The compound interest on ₹ 5000 at 12% p.a for 2 years compounded annually is ………..
₹ 1272
Hint:
Compound Interest (Cl) formula is
Cl = Amount – Principal
Amount = A (1 + $$\frac{r}{100}$$)n = 5000 (1 + $$\frac{12}{100}$$)2
= 5000 (1 + $$\frac{112}{100}$$)2 = 6272
∴ Cl = 6272 – 5000 = ₹ 1272

Question (ii)
The compound interest on ₹ 8000 at 10% p.a for 1 year, compounded half yearly is …………
₹ 820
Hint:
Compound interest (CI) = Amount – Principal
∴ Amount = P (1 + $$\frac{r}{100}$$)2n [2n as it is compounded half yearly]
r = 10% p.a, for half yearly r = 1 + $$\frac{10}{2}$$ = 5
A = 8000 (1 + $$\frac{5}{100}$$)2×1 = 8000 x ($$\frac{105}{100}$$)2 = 8820
CI = Amount – principal
= 8820 – 8000 = ₹ 820

Question (iii)
The annual rate of growth in population of a town is 10%. If its present population is 26620, the population 3 years ago was ………..
₹ 20,000
Hint:
Rate of growth of population r = 10%; Present population = 26620
Let population 3 years ago be x
∴ Applying the formula for population growth which is similar to compound interest,
26620 = x (1 + $$\frac{r}{100}$$)3
∴ 26620 = x (1 + $$\frac{10}{100}$$)3 = x ($$\frac{110}{100}$$)3
∴ x = 26620 x ($$\frac{110}{100}$$)3
= ₹ 20,000
The population 3 years ago was ₹ 20,000

Questions (iv)
The amount if the compound interest is calculated quarterly, is found using the formula ………….
A = P (1 + $$\frac{r}{400}$$)4n
Hint:
Quarterly means 4 times in a year.
∴ The formula for compound interest is
A = P (1 + $$\frac{r}{400}$$)4n

Question (v)
The difference between the S.I and C.I for 2 years for a principal of ₹ 5000 at the rate of interest 8% p.a is …….
₹ 32
Hint:
Difference between S.I & C.I is given by the formula
CI – SI = P ($$\frac{r}{100}$$)2
Principal (P) = 5000, r = 8% p.a
∴ CI – SI = 5000 ($$\frac{8}{100}$$)2 = 5000 x $$\frac{8}{100}$$ x $$\frac{8}{100}$$ = ₹ 32

Question 2.
Say True or False

Question (i)
Depreciation value is calculated by the formula P (1 – $$\frac{r}{100}$$)n
True
Hint:
Depreciation formula is P (1 – $$\frac{r}{100}$$)n

Question (ii)
If the present population of a city is P and it increases at the rate of r % p.a, then the population n years ago would be P (1 + $$\frac{r}{100}$$)n.
False
Hint:
Let the population ‘n’ yrs ago be ‘x’
Present population (P) = x × (1 + $$\frac{r}{100}$$)n
x = $$\frac { P }{ (1+\frac { r }{ 100 } )^{ n } }$$

Question (iii)
The present value of a machine is ₹ 16800. It depreciates @25% p.a. Its worth after 2 years is ₹ 9450.
True
Hint:
Present value of machine = ₹ 16800
Depreciation rate (n) = 25%
Value after 2 years = P (1 – $$\frac{r}{100}$$)n = 16800 (1 – $$\frac{25}{100}$$)2
= 16800 x (1 – $$\frac{1}{4}$$)2 = 16800 x $$\frac{3}{4}$$ x $$\frac{3}{4}$$ = 9450

Question (iv)
The time taken for ₹ 1000 to become ₹ 1331 @20% p.a compounded annually is 3 years.
False
Principal money = 1000
rate of interest Amount = 20%
Amount = 1331, applying in formula we get
A = (1 + $$\frac{r}{100}$$)n
1331 = 1000(1 + $$\frac{r}{100}$$)n
∴ $$\frac{1331}{1000}$$ = (1 – $$\frac{1}{5}$$)n
$$\frac{1331}{1000}$$ = ($$\frac{6}{5}$$)n
∴ n ≠ 3 (False)

Question (v)
The compound interest on ₹ 16000 for 9 months @20% p.a, compounded quarterly is ₹ 2522.
True
Hint:
Principal (P) = 16000
n = 9 months = $$\frac{9}{12}$$ years
r = 20% p.a
For compounding quarterly, we have to use below formula.
Amount (A) = P x (1 + $$\frac{r}{100}$$)4n
Since quarterly we have to divide r by 4
r = $$\frac{20}{4}$$

∴ Interest = A – P = 18522 – 16000 = 2522 (True)

Question 3.
Find the compound interest on ₹ 3200 at 2.5% p.a for 2 years, compounded annually.
Solution:
Principal (P) = ₹ 3200
r = 2.5% p.a
n = 2 years comp, annually
∴ Amount (A) = (1 + $$\frac{r}{100}$$)n = (1 + $$\frac{2.5}{100}$$)2
= 3200 x (1.025)2 = 3362
Compound interest (Cl) = 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.
Solution:
Principal (P) = ₹ 4000
r = 10% p.a
Compounded yearly n = 2$$\frac{1}{2}$$ years. Since it is of the form a$$\frac{b}{c}$$ years

= 4000x 1.1 x 1.1 x 1.05 = 5082
∴ Cl = Amount – Principal = 5082 – 4000 = 1082

Question 5.
Magesh invested ₹ 5000 at 12% p.a for one year. If the interest is compounded half yearly, find the amount he gets at the end of the year.
Solution:
Principal (P) = ₹ 5000
Interest compounded half yearly
r = 12% p.a = $$\frac{12}{2}$$ = 6% for half yearly
t = 1 yr.
Since compounded half yearly, the formula to be used is
Amount A = P (1 + $$\frac{r}{100}$$)2n
A = 5000 (1 + $$\frac{6}{100}$$)2×1 = 5000 x ($$\frac{106}{100}$$)2 = ₹ 5618

Question 6.
At what time will a sum of ₹ 3000 will amount to ₹ 3993 at 10% p.a compounded annually?
Solution:
Amount A = ₹ 3993
Principal = ₹ 3000
r = 10% p.a
n = ?

Question 7.
A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the Principal.
Solution:
n = 2 years
r = rate of interest = 4% p.a
Amount A = ₹ 2028
Amount (A) = P (1 + $$\frac{r}{100}$$)n
2028 = P (1 + $$\frac{4}{100}$$)n
2028 = P ($$\frac{r}{100}$$)2
∴ P = $$\frac{2028x100x100}{104×104}$$ = ₹ 1875

Question 8.
At what rate percentage p.a will ₹ 5625 amount to ₹ 6084 in 2 years at compound interest?
Solution:
Principal (P) = ₹ 5625
Amount (A) = ₹ 6084
n = 2 years
r = ?
Amount (A) = P (1 + $$\frac{r}{100}$$)n [Applying in formula]
6084 = 5625 (1 + $$\frac{r}{100}$$)2
(1 + $$\frac{r}{100}$$)2 = $$\frac{6084}{5625}$$
Taking square root on both sides, we get
1 + $$\frac{r}{100}$$ = $$\frac{78}{75}$$
$$\frac{r}{100}$$ = $$\frac{78}{75}$$ – 1 = $$\frac{3}{75}$$ = $$\frac{1}{25}$$
∴ r = $$\frac{1}{25}$$ x 100 = 4%

Question 9.
In how many years will ₹ 3375 amount to ₹ 4096 at 13$$\frac{1}{3}$$% p.a where interest is compounded half-yearly?
Solution:
Principal = ₹ 3375
Amount = ₹ 4096
r = 13$$\frac{1}{3}$$% p.a = $$\frac{40}{3}$$% p.a
Compounded half yearly r = $$\frac { \frac { 40 }{ 3 } }{ 2 }$$ = $$\frac{2}{3}$$
Let no. of years be n
for compounding half yearly, formula is

Question 10.
Find the C.I on ₹ 15000 for 3 years if the rates of interest is 15%, 20% and 25% for I, II and III years respectively.
Solution:
Principal (P) = ₹ 15000
rate of interest 1 (a) = 15% for year I
rate of interest 2 (b) = 20% for year II
rate of interest 3 (c) = 25% for year III
Formula for amount when rate of interest is different for different years is

Compound Interest (Cl) = A – P = 25,875 – 15,000 = 10,875
Cl = ₹ 10,875

Question 11.
The present height of a tree is 847 cm. Find its height two years ago, if it increases at 10 % p.a.

Solution:
Present height of tree = 847 cm
Present height = ‘h’
n = 2 yrs
rate of growth = 10% p.a
Applying in formula, we get

∴ Original height of tree = 70 cm

Question 12.
Find the difference between the C.I and the S.I on ₹ 5000 for 1 year at 2% p.a, if the interest is compounded half yearly.
Solution:
Principal (P) = ₹ 5000
time period (n) = 1 yr.
Rate of interest (r) = 2% p.a
for half yearly r = 1%
Difference between Cl & SI is given by the formula
CI – SI = P ($$\frac{r}{100}$$)2n [for half yearly compounding]
CI – SI = P ($$\frac{1}{100}$$)2×1
= 5000 x $$\frac{1}{100}$$ x $$\frac{1}{100}$$ = ₹ 0.50

Question 13.
What is the difference in simple interest and compound interest on 115000 for 2 years at 6% p.a compounded annually.
Solution:
Principal (P) = ₹ 15,000
Time period (n) = 2 yrs.
Rate of interest (r) = 6% p.a compounded annually
Difference between CI and SI given by
CI – SI = P ($$\frac{r}{100}$$)n = 15000 ($$\frac{6}{100}$$)2
= 15000 x $$\frac{6}{100}$$ x $$\frac{6}{100}$$
= ₹ 54

Question 14.
Find the rate of interest if the difference between the C.I and S.I on ₹ 8000 compounded annually for 2 years is ₹ 20.
Solution:
Principal (P) = ₹ 8000
time period (n) = 2 yrs.
rate of interest (r) = ?
Difference between Cl & SI is given by the formula
CI – SI = P (1 + $$\frac{r}{100}$$)n
Difference between Cl & SI is given as 20
∴ 20 = 8000 x ($$\frac{r}{100}$$)2
∴ ($$\frac{r}{100}$$)2 = $$\frac{20}{8000}$$ = $$\frac{1}{400}$$
Taking square root on both sides
$$\frac{r}{100}$$ = $$\sqrt { \frac { 1 }{ 400 } }$$ = $$\frac{1}{20}$$
∴ r = $$\frac{100}{20}$$

Question 15.
Find the principal if the difference between C.I and S.I on it at 15% p.a for 3 years is ₹ 1134.
Solution:
Rate of interest (r) = 15% p.a
time period (n) = 3 years
Difference between Cl & SI is given as 1134
Principal = ? → required to find
Simple Interest SI = $$\frac{Pnr}{100}$$
Compound Interest CI = P (1 + i)n – P
Cl – SI = P [(1 + i)n – 1 – $$\frac{nr}{100}$$]

Objective Type Questions

Question 16.
The number of conversion periods, if the interest on a principal is compounded every two months is ……………
(a) 2
(b) 4
(c) 6
(d) 12
(c) 6
Hint:
Conversion period is the time period after which the interest is added to the principal. If principal is compounded every two months then in a year, there will be 6$$\frac{12}{2}$$ conversation periods.

Question 17.
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) 2years
(b) 1 year
Hint:
Principal = ₹ 4400
Amount = ₹ 4851
Rate of interest = 10% p.a
for half yearly, divide by 2,
r = $$\frac{10}{2}$$ = 5 %
Compounded half yearly, so the formula is
A = P (1 + $$\frac{r}{100}$$)2n
Substuting in the above formula, we get

Taking square root on both sides, we get
($$\frac{21}{20}$$)2n = ($$\frac{21}{20}$$)2
Equating power on both sides
∴ 2n = 2,
n = 1

Question 18.
The cost of a machine is ₹ 18000 and it depreciates at 16$$\frac{2}{3}$$% annually. Its value after 2 years will be ………..
(a) ₹ 12000
(b) ₹ 12500
(c) ₹ 15000
(d) ₹ 16500
(b) ₹ 12500
Hint:
Cost of machine = ₹ 18000
Depreciation rate = 16$$\frac{2}{3}$$% = $$\frac{50}{3}$$% p.a
time period = 2 years
∴ As per depreciation formula,
Depriciated value = Original value (1 – $$\frac{r}{100}$$)n
Substituting in above formula, we get

Question 19.
The sum which amounts to ₹ 2662 at 10% p.a in 3 years compounded yearly is ………..
(a) ₹ 2000
(b) ₹ 1800
(c) ₹ 1500
(d) ₹ 2500
(a) ₹ 2000
Hint:
Amount = ₹ 2662
rate of interest = 10% p.a
Time period = 3 yrs. Compounded yearly
Principal (P) → required to find?
Applying formula A = P (1 + $$\frac{r}{100}$$)n

Question 20.
The difference between simple and compound interest on a certain sum of money for 2 years at 2% p.a is ₹ 1 the sum of money is ……….
(a) ₹ 2000
(b) ₹ 1500
(c) ₹ 3000
(d) ₹ 2500
CI – SI = P x (1 + $$\frac{r}{100}$$)n
1 = P x ($$\frac{2}{100}$$)2
∴ P = 1 x ($$\frac{100}{2}$$)2