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Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Additional Questions

Additional Questions and Answers

Exercise 2.1

Question 1.
72% of 25 students are good at science. How many are not good at science?
Solution:
Number of students who are good at science
= 72% of 25 = \(\frac { 72 }{ 100 } \) × 25 = 18 students
∴ Number of students who are not good at science
= 25 – 18 = 7 students

Question 2.
A flower garden has 1000 plants. 5% of the plants are roses and 1% are daisy plants. What is the total number of other plants.
Solution:
Total plants = 1000
Number of rose plants = 5 % of 1000 = \(\frac { 5 }{ 100 } \) × 1000 = 50
Number of Daisy plants = 1 % of 1000 = \(\frac { 1 }{ 100 } \) × 1000 = 10
Total of rose and daisy = 50 + 10 = 60
Number of other plants = 1000 – 60 = 940

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Additional Questions

Question 3.
Find 135 % of 80 ₹.
Solution:
135 % of 80 = \(\frac { 135 }{ 100 } \) × 80 = ₹ 108

Exercise 2.2

Question 1.
Neka bought 72.3m of cloth from a role of 100m. Express the cloth bought in terms of percentage.
Solution:
Total length of the cloth = 100 m
Length of cloth bought = 72.3 m
Percentage of cloth bought = \(\frac { 72.3 }{ 100 } \) = 72.3 %

Question 2.
Convert (i) 88 % (ii) 1.86 % into decimals.
Solution:
(i) 88 % = \(\frac { 88 }{ 100 } \) = 0.88
(ii) 1.86 % = \(\frac { 1.86 }{ 100 } \) = 0.0186

Question 3.
Convert (i) 3.35 (ii) 0.5 into percentage.
Solution:
(i) 3.35 = \(\frac { 335 }{ 100 } \) × 100 % = 335 %
(ii) 0.5 = \(\frac { 5 }{ 10 } \) × 100 % = 50 %

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Additional Questions

Exercise 2.3

Question 1.
If Gayathri had ₹ 600 left after spending 75% of her money, how much did she have in the beginning?
Solution:
Suppose Gayathri had ₹ X in the beginning.
Then money Spend = 75 % of X = \(\frac { 75 }{ 100 } \) X = \(\frac { 3X }{ 4 } \)
Money left with her = X – \(\frac { 3X }{ 4 } \) = \(\frac { 4X-3X }{ 4 } \) = \(\frac { X }{ 4 } \)
But it is given that money left = ₹ 600
i.e. \(\frac { X }{ 4 } \) = 600
X = 600 × 4 = 2400
∴ Gayathri had ₹ 2,400

Question 2.
Mohan gets 98 marks in her exams. This amounts to 56% of the total marks, What are the maximum marks?
Solution:
Let the maximum marks be X. 56 % of X = 98
\(\frac { 56 }{ 100 } \) × (X) = 98
⇒ X = 98 × \(\frac { 100 }{ 56 } \)
X = 175
∴ Maximum marks = 175

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Additional Questions

Exercise 2.4

Question 1.
On what sum of money lent out at 9% per annum for 6 years does the simple interest amount to ₹ 810?
Solution:
Given Simple Interest I = ₹ 810
Let the sum of money (Principal) be P
Rate of interest r = 9 % Per annum.
Time n = 6 years
I = \(\frac { pnr }{ 100 } \)
810 = \(\frac{P \times 6 \times 9}{100}\)
P = \(\frac{810 \times 100}{6 \times 9}\)
P = ₹ 1500
Sum of money required = ₹ 1500

Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest Additional Questions

Question 2.
Find the simple interest on ₹ 1120 for 2 \(\frac { 2 }{ 5 } \) years at the rate of 5% per annum.
Solution:
Simple Interest I = \(\frac { pnr }{ 100 } \)
Principal P = ₹ 1120
Time n = 2 \(\frac { 2 }{ 5 } \) years
= \(\frac { 12 }{ 5 } \) years
Rate of Interest r = 5 %
Samacheer Kalvi 7th Maths Solutions Term 3 Chapter 2 Percentage and Simple Interest add 1
∴ I = 1120 × \(\frac { 12 }{ 5 } \) × \(\frac { 5 }{ 100 } \)
= \(\frac { 672 }{ 5 } \)
= ₹ 134.4
Simple interest = = ₹ 134.4