Tamilnadu State Board New Syllabus Samacheer Kalvi 8th Maths Guide Pdf Chapter 4 Life Mathematics Ex 4.3 Text Book Back Questions and Answers, Notes.

## Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics Ex 4.3

This calculator computes any of the values in the half-life formula given the rest values.

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

CI = Amount – Principal

∴ 6272 – 5000 = ₹ 1272

(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 \(\left(1+\frac{r}{100}\right)^{2 n}\) [2n as it is compounded half yearly]

r = 10% p.a, for half yearly r = \(\frac{10}{2}\) = 5

∴ A = 8000 \(\left(1+\frac{5}{100}\right)^{2 \times 1}\) = 8000 × \(\left(\frac{105}{100}\right)^{2}\) = 8820

CI = Amount – principal = 8820 – 8000 = ₹ 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

∴ Applying the formula for population growth which is similar to compound interest,

The population 3 years ago was ₹ 20,000

(iv) If the compound interest is calculated quarterly, the amount is found using the formula ________ .

Answer:

A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)

Hint:

Quarterly means 4 times in a year.

∴ The formula for compound interest is

A = \(P\left(1+\frac{r}{400}\right)^{4 n}\)

(v) The difference between the C.I and S.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 = \(\left(\frac{r}{100}\right)^{2}\)

Principal (P) = 5000. r = 8% p.a

∴ CI – SI = 5000\(\left(\frac{8}{100}\right)^{2}\) = 5000 × \(\left(\frac{8}{100}\right)^{2}\) × \(\left(\frac{8}{100}\right)^{2}\) = ₹ 32

CBSE Class 10 Maths formulas and equations are available chapter wise.

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 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

Hint:

Let the population ‘n’ yrs ago be ‘x’

∴ Present popuLation (P) = x × \(\left(1+\frac{r}{100}\right)^{n}\)

∴ x = \(\frac{P}{\left(1+\frac{r}{100}\right)^{n}}\)

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

Answer:

True

Hint:

Present value of machine = ₹ 16800

Depreciation rate (n) = 25%

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

Answer:

False

Hint:

Pnncipal money = 1000

rate of interest = 20%

Amount = 1331, applying in formula we get

(v) The compound interest on ₹ 16000 for 9 months at 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 × \(\left(1+\frac{r}{100}\right)^{4 n}\)

Since quarterly we have to divide ‘r’ by 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.

Answer:

Principal (P) = ₹ 3200

r = 2.5% p.a

n = 2 years comp. annually

∴ Amount (A) = \(\left(1+\frac{r}{100}\right)^{n}\)

= 3200 \(\left(1+\frac{25}{100}\right)^{2}\)

= 3200 × (1.025)^{2} = 3362

Compound interest (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

r = 10 %p.a

Compounded yearly

n = 2\(\frac { 1 }{ 2 }\) years. Since it is of the form a \(\frac{b}{c}\) years

∴ CI = Amount – principal = 5082 – 4000 = 1082

Question 5.

A principal becomes ₹ 2028 in 2 years at 4% p.a compound interest. Find the principal.

Answer:

n = 2 years

r = rate of interest = 4% p.a

Amount A = ₹ 2028

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

r = 13\(\frac{1}{3}\)%p.a = \(\frac{40}{3}\)%p.a

Let no. of years be n

for compounding half yearly, formula is

Taking cubic root on both sides,

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 (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

Substituting in the above formula, we get

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

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:

Principal (P) = ₹ 5000

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

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 (P) = ₹ 8000

time period (n) = 2 yrs.

rate of interest (r) = ?

Difference between CI & SI is given by the formula

CI – SI = \(p\left(1+\frac{r}{100}\right)^{n}\)

Difference between CI & SI is given as 20

∴ 20 = 8000 × \(\left(\frac{r}{100}\right)^{2}\)

∴ \(\left(\frac{r}{100}\right)^{2}=\frac{20}{8000}=\frac{1}{400}\)

Taking square root on both sides

\(\frac{r}{100}=\sqrt{\frac{1}{400}}=\frac{1}{20}\)

∴ r = \(\) = 5 %

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 (r) = 15% p.a

time period (n) = 3 years

Difference between CI & SI is given as 1134

Principal = ? → required to find

Using formula for difference

Simple Interest SI = \(\frac{P n r}{100}\)

Compound Interest CI = p(1 + i)^{n} – p

1134 = P[(1.15)^{3} – 1 – 0.45] = P(1.52 – 1.45) = P (0.07)

∴ p = \(\frac{1134 \times 100}{0.07 \times 100}=\frac{113400}{7}\)

P = ₹ 16200

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

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\(\left(\frac{12}{2}\right)\) conversion periods.

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

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\(P\left(1+\frac{r}{100}\right)^{2 n}\)

Substituting in the above formula, we get

Taking square root on both sides, we get

\(\left(\frac{21}{20}\right)^{2 n}=\left(\frac{21}{20}\right)^{2}\)

Equating power on both sides

∴ 2n = 2, n = 1

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

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 \(\left(1-\frac{r}{100}\right)^{n}\)

Substituting in above formula, we get

Depreciated value after 2 years

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

Hint:

Amount = ₹ 2662

rate of interest = 10 % p.a

Time period = 3 yrs. Compounded yearly

Principal (P) → required to find?

Question 15.

The difference between compound and simple interest on a certain sum of money for 2 years at2%p.ais U. The sum of money is _________ .

(A) ₹ 2000

(B) ₹ 1500

(C) ₹ 3000

(D) ₹ 2500

Answer:

(D) ₹ 2500

Hint:

Difference between CI and SI is given as Re I

Time period (n) = 2 yrs.

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

Formula for difference is

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

Substituting the values in above formula, we get

1 = p × \(\left(\frac{2}{100}\right)^{2}\)

∴ p = 1 × \(\left(\frac{100}{2}\right)^{2}\) = 1 × (50)^{2} = ₹ 2500