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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 ………..

Answer:

₹ 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 …………

Answer:

₹ 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 ………..

Answer:

₹ 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 ………….

Answer:

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 …….

Answer:

₹ 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}

Answer:

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}.

Answer:

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.

Answer:

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.

Answer:

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.

Answer:

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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(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

Answer:

(d) ₹ 2500

Difference between Cl and SI is given as Re 1

Time period (n) = 2 yrs.

Rate of interest (r) = 2% p.a

Formula for difference is

CI – SI = P x (1 + \(\frac{r}{100}\))^{n}

Substituting the values in above formula, we get

1 = P x (\(\frac{2}{100}\))^{2}

∴ P = 1 x (\(\frac{100}{2}\))^{2}

= 1 x (50)^{2} = ₹ 2500