TN 8 Maths

Samacheer Kalvi 8th Maths Guide Chapter 4 Life Mathematics Ex 4.3

Samacheer Kalvi 8th Maths Guide Chapter 4 Life Mathematics Ex 4.3

Tamilnadu Samacheer Kalvi 8th Maths Solutions Chapter 4 Life Mathematics 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
CI = Amount – Principal

(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,

(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

(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%
Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 3

 

(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

Let no. of years be n
for compounding half yearly, formula is

Taking cubic root on both sides,
Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 10Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 10

 

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
Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 11
Substituting in the above formula, we get
Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 12
∴ 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
Samacheer Kalvi 8th Maths Guide Answers Chapter 4 Life Mathematics Ex 4.3 13

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) = ?

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 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

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